NOTATION
OF
DECIMALS
.
CHAP.
I.
1.
WHAT
Arithmetick
,
and
the
Subject
thereof
,
(
viz.
Number
)
is
,
I
have
largely
defined
in
the
First
Chapter
of
my
Vulgar
Arithmetick
,
in
which
Treatise
I
have
applyed
the
species
of
Numeration
to
the
various
Rules
of
Vulgar
Arithmetick
,
both
in
Integers
,
and
Fractions
;
for
the
solution
of
various
Practical
Questions
solvable
thereby
,
by
such
plain
and
easie
Rules
as
many
years
experience
in
the
practice
thereof
had
made
me
capable
of
,
and
which
I
hope
might
render
it
Intelligible
,
and
serviceable
to
the
meanest
Capacity
.
And
in
this
I
shall
shew
you
the
use
of
Decimal
Fractions
in
all
the
Rules
of
Arithmetick
,
but
Principally
in
the
solving
Questions
of
Interest
and
Rebate
,
according
to
several
Rates
of
Interest
,
both
Simple
and
Compound
,
with
the
true
Valuation
of
Leases
and
Annuities
,
either
present
or
in
Reversion
,
and
likewise
their
use
in
the
calculating
of
Tables
for
that
purpose
,
&c.
II.
In
Decimal
Fractions
we
suppose
the
unite
or
Integer
to
be
divided
into
ten
equal
parts
,
and
each
of
those
tenth
parts
are
again
divided
into
ten
other
equal
parts
,
so
that
then
the
Unit
or
Integer
will
be
divided
into
100
equal
parts
,
and
then
again
each
of
those
hundred
parts
is
supposed
to
be
divided
into
10
other
equal
parts
,
so
that
then
the
Unit
or
Integer
will
be
divided
into
1000
equal
parts
,
&c.
And
so
by
Decimating
the
first
and
subdecimating
the
second
we
proceed
ad
infinitum
.
III.
And
hence
it
is
evident
that
a
Decimal
Fraction
is
alwayes
either
so
many
tenths
,
or
it
is
so
many
tenths
of
1/10
,
or
't
is
so
many
tenths
of
1/10
of
1/10
,
or
so
many
tenths
of
1/10
of
1/10
of
1/10
,
&c.
which
compound
Decimal
Fraction
being
Reduced
,
as
is
taught
in
the
6
Rule
of
the
19
Chapter
of
my
Vulgar
Arithmetick
,
will
give
its
equivalent
simple
Decimal
Fraction
;
As
for
Example
,
8/10
of
1/10
of
1/10
is
.008
that
is
8/1000
and
hence
it
follows
that
always
a
Decimal
Fraction
hath
for
its
Denominator
an
unite
with
a
Cypher
,
or
else
Cyphers
annexed
to
it
on
the
right
hand
,
viz.
either
10
,
or
100
,
or
1000
,
or
10000
,
or
100000
,
&c.
ad
infinitum
.
IIII.
In
Decimal
Fractions
the
Denominator
is
never
express'd
,
but
may
at
first
sight
be
understood
by
the
Number
of
places
contained
in
the
Numerator
;
the
Denominator
being
alwayes
an
unite
with
as
many
Cyphers
annexed
to
it
,
as
there
are
real
places
in
the
Numerator
;
as
8
being
a
Decimal
is
8/10
,
viz.
its
Denominator
is
an
unite
with
one
Cypher
annexed
to
it
,
and
85/100
is
thus
written
.85
and
146/1000
thus
written
.146
;
But
if
the
Numerator
of
a
Decimal
Fraction
consisteth
not
of
so
many
places
as
there
are
Cyphers
in
the
Denominator
,
then
such
defect
is
supplyed
by
prefixing
so
many
Cyphers
before
the
Numerator
,
(
viz.
)
on
the
left
hand
as
there
are
places
deficient
;
as
for
example
,
8/100
if
it
were
only
set
down
thus
,
(
.8
)
then
it
would
be
but
8/10
,
but
by
praefixing
a
Cypher
before
it
thus
(
.08
)
it
is
8/100
and
8/1000
is
thus
expressed
(
.008
)
and
25/1000
is
thus
written
(
.025
)
and
25/10000
thus
(
.0025
)
&c.
V.
A
Decimal
Fraction
being
written
without
its
Denominator
,
is
known
from
a
whole
Number
,
by
having
a
point
or
prick
prefixed
before
it
thus
;
.25
is
25/100
but
if
it
had
been
expressed
without
a
point
thus
(
25
)
it
would
have
signified
so
many
unites
:
The
same
is
to
be
observed
in
mixt
Numbers
,
for
2916/100
being
written
Decimally
,
will
stand
thus
(
29.16
)
and
4825/1000
thus
,
(
48.025
)
and
4828/10000
thus
,
(
48.0028
)
and
the
like
of
any
others
.
But
some
Authors
distinguish
Decimals
from
whole
Numbers
,
by
prefixing
a
virgula
,
or
perpendicular
line
before
the
Decimal
,
(
whether
it
be
alone
,
or
joyned
with
a
whole
Number
)
thus
,
〈◊〉
is
8/10
,
and
〈◊〉
is
25/1000
,
and
〈◊〉
is
2916/100
,
&c.
Others
express
the
same
Decimal
Fraction
and
mixt
Numbers
thus
,
(
viz.
)
〈◊〉
〈◊〉
〈◊〉
,
&c.
Others
with
a
point
over
the
place
of
Units
in
the
whole
Number
;
and
then
the
former
Fractions
and
mixt
number
will
be
thus
written
;
viz.
08
,
0025
,
2916
the
like
of
others
:
And
some
Authors
again
put
points
over
all
the
places
or
figures
in
a
Decimal
Fraction
thus
:
8
,
025
,
2916
,
48025
,
&c.
but
being
written
according
to
the
first
Direction
,
I
conceive
they
may
be
most
fit
for
Calculation
.
VI.
As
whole
Numbers
do
increase
their
value
in
a
decuple
proportion
,
by
annexing
a
Cypher
or
Figure
to
the
place
of
Units
,
so
by
prefixing
a
Cypher
or
Figure
on
the
left
hand
of
a
Decimal
,
so
as
actually
to
take
place
in
the
Decimal
,
its
value
is
decreased
in
a
subdecuple
proportion
,
so
the
Number
4
,
by
annexing
a
Figure
or
Cypher
to
it
;
it
is
increased
from
4
to
40
,
&c.
But
if
4
had
been
a
Decimal
,
viz.
4.
and
if
there
had
been
o
prefixed
before
it
on
the
left
hand
,
its
value
had
been
decreased
from
4/10
to
4/100
or
.04
,
and
by
prefixing
5
,
it
is
.54
;
and
still
by
prefixing
more
Figures
or
Cyphers
,
its
value
will
decrease
in
the
same
Ratio
ad
infinitum
.
VII
.
And
as
Cyphers
being
prefixed
before
a
whole
Number
,
(
viz.
)
on
the
left
hand
thereof
)
do
neither
increase
or
decrease
its
value
;
(
for
4
,
and
04
,
and
004
being
Integers
,
do
still
retain
one
and
the
same
value
;
)
So
a
Decimal
,
by
having
a
Cypher
,
or
Cyphers
annexed
to
the
Right
hand
thereof
,
have
not
their
value
either
Increased
,
or
Decreased
.
Whence
it
is
evident
,
that
all
Decimal
Fractions
may
be
Reduced
to
an
equal
Denomination
at
first
sight
;
for
suppose
.15
,
and
.068
,
and
.73465
were
Decimals
given
to
be
Reduced
to
one
Denomination
;
In
this
case
I
consider
that
the
Denominator
for
the
given
Decimal
consisting
of
the
most
places
is
100000
,
and
.15
and
.068
whose
Denominators
are
100
and
1000
may
be
reduced
to
Decimals
of
the
same
value
,
having
likewise
100000
for
their
Denominator
,
by
annexing
so
many
Cyphers
on
the
Right
hand
of
the
Numerators
,
as
(
according
to
the
4
Definition
foregoing
)
may
make
each
of
them
to
have
100000
for
a
Denominator
,
so
.15
will
be
.15000
,
and
.068
will
be
.06800
.
VIII
.
As
the
order
of
places
in
whole
Numbers
is
from
the
Right
Hand
to
the
left
,
so
the
order
of
places
in
a
Decimal
Fraction
is
from
the
Left
Hand
to
the
Right
;
the
first
place
being
accounted
tenth
parts
of
an
Unity
,
and
by
some
it
is
called
primes
,
the
second
place
is
so
many
hundredth
parts
of
unitie
,
or
it
is
called
seconds
,
the
third
place
is
so
many
thousandth
partes
of
unitie
,
or
it
is
called
thirds
,
&c.
which
will
more
fully
appear
by
the
following
Table
.
A
Table
of
Notation
of
Integers
and
Decimals
.
Integers
|
|
Of
Unites
|
Ninth
place
|
3
|
C
of
Mill.
|
Eighth
place
|
8
|
X
of
Mill.
|
Seventh
place
|
4
|
Millions
|
Sixth
place
|
3
|
C
of
Thous
.
|
Fifth
place
|
7
|
X
of
Thous
.
|
Fourth
place
|
5
|
Thousands
|
Third
place
|
8
|
Hundreds
|
Second
place
|
6
|
Tens
|
First
place
.
|
4
|
Unites
|
Decimals
|
|
Of
Unitie
.
|
Primes
|
8
|
Tenth
partes
|
Seconds
|
2
|
Hundr
.
partes
|
Thirds
|
3
|
Thous
.
partes
|
Fourths
|
0
|
X
Thous
.
partes
|
Fifths
|
5
|
C
Thous
.
partes
|
Sixths
|
6
|
Mill.
partes
.
|
Sevenths
|
3
|
X
Mill.
partes
|
Eighths
|
5
|
C
Mill.
partes
|
Ninths
|
4
|
Tho.
Mill.
partes
|
In
the
foregoing
Table
is
given
a
mixt
Number
of
Integers
and
Decimals
;
the
Integers
being
separated
from
the
Decimals
by
a
point
,
or
prick
,
according
to
the
fifth
definition
beforegoing
;
so
that
384375864
signifie
so
many
Integers
or
Unites
,
and
823056345
signifie
so
many
parts
of
Unity
,
the
figure
8
in
the
first
place
being
so
many
tenth
parts
of
unitie
;
and
the
next
figure
,
viz.
the
figure
2
is
so
many
hundredth
parts
of
unitie
,
&c.
So
in
the
Decimal
Fraction
.4378
,
the
figure
4
possesseth
the
first
place
,
and
is
4
primes
,
or
four
tenth
parts
of
an
unite
,
and
3
the
second
figure
is
called
3
seconds
,
or
three
hundredth
parts
of
an
unite
,
and
7
the
third
figure
is
called
seven
thirds
,
or
seven
thousandth
parts
of
an
unite
,
and
8
the
fourth
figure
is
called
eight
fourths
,
or
eight
ten
thousandth
parts
of
an
unite
,
&c.
Whence
it
appeareth
that
every
place
in
a
Decimal
Fraction
being
considered
a
part
by
it self
,
without
any
respect
to
the
rest
,
will
of
it self
make
a
particular
Decimal
Fraction
;
so
in
the
last
mentioned
Decimal
Fraction
,
viz.
4378
,
each
place
being
considered
by
it self
,
will
make
these
following
Decimal
Fractions
,
viz.
.
4.03.007
and
.0008
,
or
4/10
,
3/100
,
7/1000
,
and
8/10000
;
which
Fractions
being
added
together
,
according
to
the
Rules
of
Addition
of
Decimals
hereafter
delivered
in
the
third
Chapter
,
their
sum
will
be
.4378
,
which
is
the
given
Decimal
of
which
they
are
composed
.
IX
.
A
Decimal
Fraction
is
expressed
by
some
Authors
,
by
Primes
,
Seconds
,
Thirds
,
Fourths
,
&c.
As
if
this
Decimal
.748
were
to
be
expressed
,
they
say
it
is
seven
primes
,
four
Seconds
,
and
eight
thirds
:
Others
there
are
which
express
it
thus
,
viz.
seven
hundred
forty
eight
thirds
,
but
the
most
approved
way
to
express
or
read
a
Decimal
Fraction
,
is
according
to
the
method
of
reading
a
vulgar
Fraction
,
and
to
give
it
the
Denomination
of
the
Figure
in
the
last
place
of
the
Decimal
,
and
then
the
Decimal
.748
will
be
thus
read
,
viz.
seven
hundred
forty
eight
thousandths
,
and
.036
is
thus
read
,
thirty
six
thousandths
,
and
so
of
any
other
.
This
Chapter
being
well
understood
,
all
the
parts
of
Numeration
,
viz.
Addition
,
Subtraction
,
Multiplication
,
and
Division
of
Decimals
will
prove
very
easie
.
CHAP.
II.
Reduction
of
Decimals
.
To
Reduce
a
given
Vulgar
Fraction
to
a
Decimal
,
that
shall
be
equivalent
thereto
.
1.
WHen
in
any
Arithmetical
Operation
your
work
is
so
mingled
with
vulgar
Fractions
,
as
to
render
it
tedious
,
or
difficult
;
the
best
remedy
you
can
have
,
is
to
reduce
your
vulgar
Fraction
or
Fractions
into
a
Decimal
or
Decimals
,
which
having
done
the
work
,
will
be
as
easie
in
every
Respect
,
as
if
you
had
to
do
with
nothing
but
whole
Numbers
,
which
you
may
effect
by
the
following
Proportion
,
viz.
As
the
Denominator
of
the
given
vulgar
Fraction
,
is
to
its
Numerator
,
So
is
an
Unite
,
with
so
many
Cyphers
as
you
intend
your
Decimal
shall
have
places
,
To
the
Decimal
Required
.
So
if
the
Fraction
to
be
Reduced
were
¾
,
and
you
would
reduce
it
to
a
decimal
consisting
of
4
places
,
I
say
,
the
Proportion
is
As
4
(
the
Denominator
of
the
given
Fraction
.
)
Is
to
3
(
its
Numerator
.
)
So
is
10000
(
the
denominator
of
the
decimal
required
.
)
To
.7500
(
the
Decimal
required
.
)
So
that
I
conclude
¾
will
be
reduced
to
its
equivalent
decimal
.7500
,
or
.75
;
for
Cyphers
on
the
Right
hand
of
a
decimal
do
neither
increase
nor
diminish
its
value
,
by
the
seventh
definition
of
the
First
Chapter
.
Now
according
to
the
foresaid
proportion
,
it
is
evident
,
that
if
to
the
Numerator
of
any
Fraction
given
to
be
Reduced
to
a
decimal
,
you
an●…ex
as
many
Cyphers
as
you
intend
its
equivalent
decimal
shall
have
places
,
and
then
divide
it
by
its
denominator
,
the
Quote
will
be
the
decimal
required
.
So
let
there
(
again
)
be
given
¾
to
be
Reduced
(
as
before
)
to
a
decimal
of
4
places
,
in
order
thereunto
I
annex
4
Cyphers
to
the
Numerator
3
,
and
it
makes
30000
,
which
I
divide
by
the
denominator
4
,
and
it
Quotes
.7500
,
or
.75
,
for
the
decimal
equivalent
to
the
vulgar
Fraction
¾
.
Note
that
all
vulgar
Fractions
cannot
be
Reduced
to
decimals
,
having
exactly
the
same
value
,
although
they
may
come
infinitely
near
,
and
the
more
places
that
you
make
your
decimal
to
consist
of
,
so
much
the
nearer
doth
it
come
to
the
truth
,
but
4
or
5
places
is
exact
enough
for
most
operations
,
so
if
it
be
required
to
reduce
the
vulgar
Fraction
9/11
to
a
decimal
of
4
places
,
it
will
be
found
to
be
.
818●…
which
is
not
exact
,
but
yet
it
wanteth
not
1/10000
part
of
an
unit
of
the
truth
,
and
if
you
make
it
.8182
,
it
will
be
somewhat
more
than
the
truth
.
Again
if
you
annex
5
Cyphers
to
the
Numerator
,
and
so
make
the
decimal
consist
of
5
places
,
it
will
then
be
.81818
,
yet
it
will
want
of
the
truth
,
but
not
so
much
as
when
it
had
but
4
places
,
for
now
it
will
not
want
1/100000
part
of
an
unite
of
the
exact
truth
,
and
if
you
make
it
to
be
.81819
,
it
will
then
exceed
the
truth
.
Thus
by
increasing
the
number
of
places
in
the
Decimal
,
you
may
come
infinitely
near
the
truth
,
but
never
find
a
decimal
exactly
equivalent
in
many
cases
.
Note
also
,
that
if
after
you
have
Reduced
your
vulgar
Fraction
to
a
decimal
,
according
to
the
foregoing
Rule
,
there
be
not
as
many
places
in
the
decimal
,
as
you
annexed
Cyphers
to
the
Numerator
of
the
given
vulgar
Fraction
,
then
you
are
to
supply
such
defect
by
prefixing
so
many
Cyphers
on
the
left
hand
of
the
significant
Figures
,
as
there
are
places
wanting
,
according
to
the
fourth
Rule
of
the
First
Chapter
.
So
if
11/941
were
given
to
be
Reduced
to
a
decimal
of
any
number
of
places
,
as
suppose
6
;
in
order
to
it
,
I
annex
6
Cyphers
to
the
Numerator
11
,
and
it
makes
11000000
,
for
a
dividend
,
which
divided
by
941
,
it
quotes
11689
,
which
consisteth
but
of
5
places
,
but
it
should
have
6
places
,
wherefore
to
make
it
compleat
,
I
prefix
a
Cypher
before
it
,
and
it
makes
.011689
for
the
true
decimal
Required
;
and
if
it
had
been
required
to
consist
of
4
places
,
then
I
annex
4
Cyphers
to
the
numerator
,
yet
after
division
is
ended
,
there
will
be
but
3
places
in
the
Quotient
,
viz.
116
,
therefore
to
make
it
consist
of
4
places
,
I
prefix
a
Cypher
before
it
,
and
it
makes
.0116
for
the
decimal
●…ought
.
Again
let
there
be
given
23/5642
to
be
Reduced
to
a
decimal
of
(
suppose
)
5
places
it
will
be
found
to
be
.00407
;
and
14/64837
will
be
reduced
to
.000215
To
Reduce
the
known
parts
of
Money
,
Weight
,
Measure
,
Time
,
&c.
to
Decimal
Fractions
.
II.
Hence
it
is
evident
that
the
known
parts
of
Money
,
Weight
,
Measure
,
Time
,
and
Motion
,
&c.
may
be
Reduced
to
decimal
Fractions
of
the
same
value
,
or
infinitely
near
it
,
for
if
(
in
Money
)
a
Pound
Sterling
be
an
Integer
,
whatsoever
is
less
than
a
Pound
,
is
either
a
part
or
parts
of
the
same
;
and
when
you
know
what
part
or
parts
thereof
it
is
,
you
may
reduce
it
to
a
decimal
of
the
same
value
,
by
the
first
Rule
of
this
Chapter
;
so
if
you
would
know
what
is
the
decimal
of
a
Pound
Sterling
equal
to
7
Shillings
;
consider
that
7
s.
is
7/20
of
a
Pound
,
and
by
the
said
Rule
,
the
decimal
answering
thereto
is
.35
l.
And
if
I
would
know
the
decimal
equal
to
3
d.
I
consider
that
3
d.
is
3/12
of
1/20
of
a
Pound
,
or
3/240
of
a
Pound
,
and
the
decimal
equivalent
thereto
,
will
be
found
(
by
the
said
Rule
)
to
be
.0
125
;
likewise
if
there
were
given
7
s
.3
d
to
find
the
decimal
equal
thereunto
:
First
,
I
consider
,
that
7
s
.3
d.
is
87
pence
,
which
is
87/240
of
a
pound
,
and
the
decimal
equal
thereto
will
be
found
to
be
.3625
l.
In
like
manner
if
it
were
required
to
find
the
decimal
of
a
Pound
Troy
weight
equivalent
to
6
oz.
−
12
pw
.
I
first
find
that
6
oz.
−
12
pw
.
make
132
pwts
.
which
is
132/240
of
a
pound
Troy
weight
,
and
the
decimal
equivalent
thereunto
,
will
be
found
to
be
.55
by
the
said
first
Rule
of
this
Chapter
.
The
like
is
to
be
understood
in
the
Reducing
of
any
of
the
known
parts
of
Coyn
,
Weight
,
Measure
,
&c.
into
Decimals
.
To
find
the
value
of
a
Decimal
Fraction
,
in
the
known
parts
of
Money
,
Weight
,
Measure
,
&c.
III.
When
you
would
find
the
value
of
a
decimal
Fraction
in
the
known
parts
of
Coyn
,
Weight
,
Measure
,
Time
,
Motion
,
or
the
like
,
observe
the
following
RULE
Multiply
the
given
Decimal
by
the
Number
of
parts
in
the
next
inferiour
denomination
that
are
equal
to
an
Integer
in
the
same
denomination
with
the
given
decimal
,
and
see
how
many
places
are
in
the
product
,
more
than
were
in
the
said
given
decimal
;
and
cut
so
many
off
from
the
left
hand
with
a
dash
of
your
pen
,
and
those
Figures
so
cut
off
,
are
the
value
of
the
said
decimal
in
the
next
inferiour
denomination
to
it
,
and
the
Figures
(
if
there
be
any
)
Remaining
,
are
the
decimal
of
an
Integer
in
the
said
denomination
,
and
may
be
Reduced
as
low
as
you
please
by
the
same
Rule
;
as
in
the
following
Example
.
Let
it
be
required
to
find
the
value
of
this
decimal
of
a
pound
sterling
,
viz.
7635.
First
,
I
multiply
the
given
decimal
by
20
,
and
the
product
is
152700
which
is
of
6
places
,
and
the
given
decimal
is
but
of
4
places
,
wherefore
I
cut
off
2
Figures
at
the
left
hand
,
viz.
15.
which
is
so
many
shillings
,
now
when
the
said
15
is
Cut
off
from
the
rest
,
there
are
yet
remaining
2700
,
which
I
multiply
by
12
to
find
the
value
thereof
in
pence
,
and
the
product
is
32400
,
which
consisting
of
5
places
,
I
cut
off
one
Figure
,
(
viz.
3
)
from
the
left
hand
,
which
is
so
many
Pence
;
so
that
I
conclude
the
value
of
the
given
decimal
to
be
15
s.
−
03
d.
and
the
remaining
Figures
,
viz.
2400
are
the
decimal
parts
of
a
peny
,
which
because
they
do
not
amount
to
the
value
of
a
Farthing
,
I
do
not
Reduce
any
lower
,
see
the
work
in
the
Margent
.
So
if
.6847
l.
be
given
,
and
it
be
required
to
find
its
value
,
if
you
work
as
is
before
directed
,
you
will
find
it
to
be
13
s.
−
08
d.
−
1.312
quarters
.
And
374
l.
being
so
Reduced
,
you
will
find
it
to
make
7
s.
−
05
d.
−
3.040
quarters
.
In
like
manner
,
if
it
were
required
to
Reduce
this
decimal
of
a
pound
Troy
weight
,
viz.
.84576
l.
into
known
parts
;
First
,
I
multiply
it
by
12
,
and
it
produceth
1014912
,
from
which
I
cut
off
the
two
first
Figures
to
the
left
hand
,
(
viz.
10.
)
for
Ounces
,
and
the
Remaining
Figures
,
which
are
14912
do
I
multiply
by
●…0
,
and
the
product
is
298240
,
from
which
I
cut
off
the
first
Figure
,
viz.
2
,
which
is
two
peny
weight
,
and
98240
remaineth
,
which
I
multiply
by
24
,
and
the
product
is
2357760
,
which
is
23.57760
grains
;
so
that
I
conclude
the
value
of
the
given
Decimal
.84576
pound
Troy
weight
to
be
10
oz.
−
02
pw
.
−
23
gr.
.57760
;
the
same
is
to
be
observed
in
finding
the
value
of
any
other
decimal
whatsoever
,
whether
of
Coin
,
Weight
,
Measure
,
Time
,
or
Motion
.
I
might
here
have
added
Tables
of
Reduction
,
shewing
the
decimal
Fractions
of
any
of
the
parts
of
Money
,
Weight
,
&c.
as
divers
Authors
have
already
done
;
but
(
because
they
are
(
though
useful
)
seldom
made
use
of
,
and
partly
by
reason
of
the
ease
in
finding
the
equivalent
decimal
of
any
Fraction
whatsoever
,
according
to
the
Rules
herein
delivered
)
I
shall
forbear
it
.
IV.
There
is
a
briefer
way
of
discovering
the
value
of
a
decimal
of
a
Pound
sterling
,
viz.
The
Figure
which
standeth
in
the
first
place
of
the
decimal
,
(
viz.
in
the
place
of
primes
)
being
doubled
,
gives
you
the
number
of
shillings
;
then
let
the
Figure
possessing
the
second
place
of
the
decimal
,
(
viz.
the
place
of
seconds
)
be
esteemed
so
many
tens
,
and
the
Figure
in
the
third
place
account
so
many
units
,
which
said
te●…s
and
units
being
accounted
one
intire
number
,
and
made
less
by
one
,
will
be
so
many
farthings
,
which
said
shillings
and
farthings
are
the
value
of
the
given
decimal
;
but
if
the
Figure
in
the
second
place
be
5
,
or
else
exceed
5
,
then
reckon
one
shilling
for
that
,
and
for
the
excess
above
5
,
esteem
every
unit
10
,
as
before
.
Example
1.
What
is
the
value
of
.7365
l
?
The
Figure
7
(
standing
in
the
place
of
primes
)
being
doubled
,
gives
14
,
which
is
so
many
shillings
,
and
the
Figure
in
the
second
place
,
(
which
is
3
)
being
accounted
so
many
tens
is
30
,
and
the
Figure
in
the
third
place
(
viz.
6
)
being
esteemed
units
,
and
annexed
to
the
tens
beforesaid
,
makes
36
,
which
being
lessened
by
1
,
makes
35
farthings
,
which
is
8d
.
¾
,
so
is
14s
.
−
08d
.
¾
the
value
of
the
given
Decimal
.7365
l.
Example
2.
What
is
the
value
of
.8896
l
?
The
first
Figure
(8)
being
doubled
,
makes
16
;
and
because
the
next
Figure
is
above
5
,
I
add
1
to
16
,
which
makes
17
shillings
:
Then
the
excess
of
the
second
Figure
above
5
being
3
,
I
esteem
it
so
many
Tens
,
and
the
Figure
(9)
in
the
third
place
being
units
,
makes
39
;
which
lessened
by
1
,
makes
38
Farthings
,
which
is
9d
.
½
,
so
is
17s
.
−
9d
.
½
,
the
value
of
the
given
decimal
.8896
.
And
after
the
same
manner
may
the
value
of
any
decimal
of
a
pound
sterling
,
be
discovered
at
first
sight
without
loss
of
a
farthing
.
CHAP.
III.
Addition
of
Decimals
.
I.
THE
work
of
Addition
of
Decimal
Fractions
is
in
every
respect
the
very
same
with
that
of
whole
Numbers
of
one
Denomination
in
common
Arithmetick
,
respect
being
had
to
the
right
ordering
or
placing
of
the
decimals
required
to
be
added
,
which
that
you
may
understand
,
observe
this
General
Rule
,
II.
When
two
or
more
decimals
are
given
to
be
added
together
,
you
are
so
to
dispose
of
them
one
under
the
other
,
as
that
all
the
Figures
on
the
left
hand
may
stand
in
order
one
under
the
other
,
that
is
to
say
,
primes
under
primes
,
or
tenth
under
tenths
,
(
whether
they
be
Cyphers
or
significant
Figures
)
and
seconds
or
hundredths
,
under
seconds
or
hundredths
,
&c.
observing
the
same
order
if
they
consist
some
of
them
of
never
so
many
places
,
and
others
of
never
so
few
.
Example
.
Let
there
be
given
these
following
Decimals
to
be
added
together
,
viz.
.00746
,
and
.0832
,
and
.62
and
.8
:
First
,
I
dispose
of
them
in
order
to
the
work
,
as
you
see
in
the
Margent
,
where
you
see
the
lowermost
Figure
8
,
which
is
primes
,
is
placed
under
6
,
0
and
0
,
which
are
likewise
primes
,
and
the
Figure
2
in
62
being
in
in
the
place
of
seconds
is
placed
under
8
and
0
which
are
likewise
seconds
,
or
hundredths
,
and
the
Figure
3
in
the
place
of
thirds
,
or
thousandths
is
placed
under
7
,
which
is
also
so
many
thirds
,
&c.
The
same
order
is
to
be
observed
in
placing
of
the
decimals
of
mixt
Numbers
to
be
added
,
as
suppose
there
were
given
these
following
mixt
Numbers
to
be
added
together
,
viz.
168.3572
,
and
36.864
,
and
7.42
.
and
.6
:
Now
in
order
to
the
finding
out
their
sum
,
I
dispose
of
them
in
order
one
under
the
other
as
followeth
.
Where
you
may
observe
that
the
whole
Numbers
themselves
,
or
Integral
parts
of
the
given
mixt
Numbers
are
placed
one
under
the
other
,
as
is
directed
in
Addition
of
whole
Numbers
,
without
any
respect
at
all
had
to
the
decimals
annexed
to
them
,
and
the
decimals
are
placed
under
each
other
,
according
to
the
Directions
given
in
the
last
Rule
,
without
any
respect
had
to
the
Integers
,
properly
belonging
to
them
.
III.
Having
placed
your
given
Decimals
in
order
,
according
to
this
Rule
,
draw
a
line
under
them
,
as
in
Addition
of
whole
Numbers
,
under
which
line
you
are
to
place
their
sum
;
Then
proceed
in
your
work
in
every
respect
,
as
in
Addition
of
Integers
,
beginning
at
the
right
hand
,
and
so
proceeding
through
the
Decimals
without
any
regard
to
them
as
Decimals
,
but
as
if
they
were
all
whole
Numbers
:
As
for
example
,
let
us
take
the
Decimals
given
in
the
first
example
of
the
last
Rule
foregoing
;
And
first
I
put
down
6
under
the
line
,
because
there
is
no
other
figure
or
number
to
add
to
it
,
then
I
proceed
to
the
next
,
saying
2
and
4
make
6
,
which
I
also
set
down
in
order
under
the
line
,
then
I
say
3
and
7
make
10
,
so
I
set
down
0
,
and
carry
1
to
the
next
,
saying
1
that
carry
,
and
2
,
and
8
,
make
11
,
for
which
I
set
down
1
,
and
carry
1
to
the
next
,
saying
,
1
that
I
carry
and
8
,
and
6
,
make
15
,
which
I
put
in
its
place
under
the
line
,
because
it
is
the
last
;
and
because
the
figure
5
standeth
under
the
place
of
primes
,
I
put
a
point
before
it
,
that
is
to
say
between
1
and
5
,
and
the
work
is
finished
;
the
number
1
being
an
integer
,
and
the
rest
a
decimal
,
whereby
I
find
the
sum
to
be
1.51066
,
that
is
1
integer
,
and
.51066
parts
of
an
integer
.
After
the
same
manner
if
the
mixt
numbers
in
the
second
example
of
the
foregoing
Rule
were
given
to
be
added
,
their
sum
will
be
found
to
be
213.2412
,
that
is
213
integers
and
.2412
decimal
parts
of
an
integer
,
as
you
may
see
by
the
following
work
.
Other
Examples
for
the
Learners
practice
may
〈◊〉
such
as
follow
.
CHAP.
IV.
Subtraction
of
Decimal
Fractions
.
1.
WHen
two
Decimal
Fractions
are
given
,
and
their
difference
or
excess
is
required
,
you
must
place
them
(
in
order
to
the
work
)
as
you
were
taught
in
the
foregoing
Chapter
of
Addition
,
and
the
operation
is
the
very
same
in
every
respect
as
in
Subtraction
of
whole
Numbers
of
one
denomination
,
beginning
at
the
Right
hand
,
as
in
the
following
Example
.
Let
it
be
required
to
Subtract
the
Decimal
.634
from
the
Decimal
.728
;
In
order
to
the
work
I
put
them
one
under
the
other
,
viz.
the
biggest
uppermost
,
and
take
each
figure
in
the
lowermost
out
of
its
Correspondent
figure
in
the
uppermost
,
putting
their
respective
differences
in
order
below
the
line
,
and
I
find
,
(
that
when
I
have
finished
the
operation
)
the
Remainder
or
difference
to
be
.094
as
by
the
work
appeareth
.
In
like
manner
if
the
mixt
Number
were
given
to
be
subtracted
from
the
mixt
number
76.123
.
I
place
them
in
the
same
order
as
is
directed
in
Addition
before-going
,
only
with
this
Caution
,
be
sure
to
place
the
biggest
uppermost
,
then
proceed
to
take
each
figure
in
the
lowermost
out
of
its
correspondent
figure
in
the
uppermost
,
as
if
they
were
whole
numbers
,
and
having
finished
the
work
,
the
Remainder
,
or
difference
will
be
found
to
be
33.776
as
you
see
it
done
in
the
Margent
.
When
the
Decimal
given
to
be
Subtracted
do
not
consist
of
an
equal
number
of
places
,
such
defect
must
be
supplyed
by
annexing
Cyphers
,
or
supposing
as
many
Cyphers
to
be
annexed
(
as
are
wanting
)
on
the
Right
hand
,
and
then
the
work
will
be
as
in
the
former
Examples
.
Example
.
Let
it
be
Required
to
Subtract
.037486
from
from
.84
;
Now
because
.84
hath
in
it
but
2
places
,
and
the
other
hath
6
,
I
supply
that
defect
by
annexing
4
Cyphers
thereto
as
in
the
Margent
,
and
the
work
being
finished
,
I
find
the
Remainder
,
or
difference
to
be
.802514
.
The
same
is
to
be
observed
when
a
decimal
Fraction
or
mixt
number
is
given
to
be
Subtracted
from
a
whole
number
,
as
suppose
15.486
were
given
to
be
subtracted
from
64
,
because
there
is
no
decimal
annexed
to
64
,
you
are
to
supply
the
Decimal
places
with
Cyphers
,
and
then
proceed
in
the
work
as
before
is
directed
,
and
having
finished
the
work
of
Subtraction
,
the
Remainder
will
be
found
to
be
48.514
as
by
the
work
in
the
margent
appeareth
.
Other
Examples
for
Practice
may
be
these
following
.
CHAP.
V.
Multiplication
of
Decimal
Fractions
.
I.
IN
Multiplication
of
Decimals
,
whether
both
the
Factors
are
decimal
Fractions
,
or
whether
they
be
mixt
Numbers
,
or
if
the
one
be
a
decimal
Fraction
,
and
the
other
a
whole
or
mixt
Number
the
Multiplier
is
to
be
placed
under
the
Multiplicand
in
the
very
same
manner
as
in
multiplication
of
whole
Numbers
,
and
when
they
are
so
placed
,
the
operation
is
the
same
in
every
Respect
,
as
in
Multiplication
of
whole
Numbers
,
and
when
you
have
added
the
several
particular
products
together
,
as
is
usual
in
whole
Numbers
,
the
value
of
the
product
is
to
be
found
out
by
this
General
Rule
,
Look
how
many
Decimal
places
are
in
both
the
Factors
,
(
viz.
the
Multiplicand
and
Multiplier
)
so
many
decimal
places
must
be
in
the
product
.
Wherefore
cut
off
so
many
figures
from
the
right
hand
of
the
product
for
decimals
,
and
the
figure
or
figures
Remaining
on
the
left
hand
(
if
there
be
any
)
are
Integers
,
as
in
the
following
Example
.
Let
it
be
Required
to
Multiply
34.82
by
7.26
,
it
matters
not
which
you
make
the
Multiplicand
,
or
the
Multiplyar
,
but
I
take
7.26
for
the
Multiplyar
,
because
it
hath
fewest
places
,
and
put
it
in
order
under
34.82
,
as
if
they
were
both
whole
numbers
,
and
having
finished
the
work
of
Multiplication
I
find
the
product
to
be
2527932
as
you
may
see
by
the
following
work
.
Then
to
find
the
value
of
the
product
,
I
look
how
many
Decimal
places
are
in
(
both
)
the
Multiplicand
and
Multiplyar
,
and
I
find
4
,
wherefore
I
mark
the
4
first
places
to
the
Right
hand
for
Decimals
,
by
putting
a
point
between
them
and
the
other
figures
on
the
left
hand
,
and
then
the
Product
will
appear
to
be
really
252.7932
that
is
252
integers
,
and
.7932
Decimal
parts
of
an
integer
.
A
second
Example
may
be
of
a
mixt
Number
given
to
be
multiplyed
by
a
Decimal
Fraction
;
as
thus
,
let
it
be
required
to
Multiply
38.5746
,
by
.00463
;
I
prepare
the
given
Numbers
for
operation
as
is
before
directed
,
and
having
finished
the
work
I
find
the
product
to
amount
to
178600398.
Then
to
find
the
true
value
of
the
product
I
consider
the
number
of
decimal
places
,
in
both
the
Factors
,
which
I
find
to
b●…
9
,
viz.
4
in
the
Multiplicand
and
5
in
the
Multiplyar
,
therefore
I
mark
out
9
places
towards
the
Right
hand
of
the
product
for
a
Decimal
Fraction
,
which
indeed
is
the
whole
product
,
and
therefore
I
conclude
the
true
value
of
the
product
to
be
.178600398
,
as
by
the
following
operation
appeareth
,
viz.
A
Third
Example
shall
be
of
2
Decimal
Fractions
,
the
one
being
given
to
be
multiplyed
by
the
other
,
as
,
let
there
be
given
.63478
to
be
Multiplyed
by
.8264
,
having
disposed
of
the
given
numbers
according
to
order
,
and
finished
the
work
of
Multiplication
as
is
before
directed
,
I
find
the
product
to
Amount
to
524582192
,
which
being
done
,
to
find
the
true
value
thereof
,
I
consider
that
there
are
9
decimal
places
in
both
the
Factors
,
viz.
5
in
the
Multiplicand
and
4
in
the
Multiplyar
;
wherefore
I
note
out
9
places
in
the
product
for
a
decimal
Fraction
,
and
so
I
find
the
true
value
of
the
product
to
be
.524582192
,
as
by
the
following
operation
appeareth
.
The
like
is
to
be
understood
in
any
of
the
like
Cases
whatsoever
.
II.
If
it
so
happen
(
as
oftentimes
it
may
)
that
after
your
Multiplication
is
finished
,
the
figures
in
the
product
do
not
consist
of
so
many
places
as
there
are
decimal
figures
in
the
Multiplicand
and
Multiplyar
,
such
defect
must
be
supplyed
by
prefixing
as
many
Cyphers
before
it
towards
the
left
hand
,
as
it
wanteth
places
,
and
then
mark
such
product
with
the
said
prefixed
Cyphers
,
for
a
decimal
Fraction
and
the
true
product
Required
;
as
in
the
following
Example
.
Let
it
be
Required
to
Multiply
.0476
by
.0642
,
after
the
Multiplication
is
finished
,
I
find
the
product
to
be
305692
,
consisting
but
of
6
places
,
but
the
Number
of
Decimal
places
in
the
Multiplicand
and
Multiplyar
is
8
,
wherefore
to
make
the
product
to
consist
of
8
places
,
I
prefix
2
Cyphers
before
it
,
and
then
the
true
product
will
be
.00305592
;
the
work
followeth
,
In
like
manner
if
.376523
were
given
to
be
Multiplyed
by
.1346
,
you
will
find
the
product
to
be
506799958
consisting
of
9
places
,
but
there
are
10
Decimal
places
in
both
the
given
Factors
;
wherefore
the
product
must
be
increased
to
10
places
by
prefixing
a
Cypher
which
will
make
it
.0506799958
,
as
by
the
following
work
.
By
this
time
I
doubt
not
but
the
diligent
Learner
is
well
acquainted
with
Multiplication
of
Decimal
Fractions
,
the
work
being
as
plain
and
easie
as
in
whole
Numbers
;
The
next
we
come
to
is
Division
.
CHAP.
VI.
Division
of
Decimal
Fractions
.
HAving
gone
through
Addition
,
Subtraction
,
and
Multiplication
,
(
The
operation
being
(
as
you
see
)
in
every
Respect
the
very
same
as
in
whole
Numbers
)
we
come
now
to
Division
;
and
although
in
Decimals
,
(
as
well
as
in
whole
Numbers
)
Division
may
seem
somewhat
difficult
to
the
young
Practitioner
,
yet
we
shall
endeavour
to
render
it
as
plain
and
easie
as
possible
may
be
.
I.
The
operation
in
Division
of
Decimals
is
in
every
respect
the
same
with
that
of
whole
numbers
,
therefore
the
difficulty
in
Division
of
Decimals
lyeth
not
in
the
operation
,
but
in
finding
out
the
value
of
the
Quotient
after
the
work
of
Division
is
ended
,
a
general
Rule
for
finding
of
which
shall
be
given
by
and
by
.
II.
It
is
necessary
many
times
to
annex
a
cypher
or
cyphers
to
the
Dividend
,
whether
it
be
a
whole
Number
,
or
a
mixt
Number
,
or
a
Decimal
Fraction
,
for
many
times
the
Divisor
consisteth
of
more
places
then
the
Dividend
,
and
in
that
case
there
must
be
a
competent
Number
of
Cyphers
annexed
to
the
Dividend
,
as
,
suppose
it
were
required
to
divide
73.564
by
46.24897
,
here
you
cannot
conveniently
proceed
in
the
work
till
you
have
annexed
Cyphers
to
the
Dividend
,
to
increase
the
number
of
places
in
the
decimal
part
thereof
,
and
you
may
annex
as
many
as
you
please
,
for
by
the
7
Rule
of
the
first
Chapter
,
Cyphers
annexed
to
a
Decimal
Fraction
do
neither
Augment
,
nor
diminish
its
value
.
III.
When
a
question
to
be
wrought
by
Division
of
decimals
is
proposed
,
consider
whether
there
are
as
many
decimal
figures
in
the
dividend
,
as
there
are
in
the
divisor
;
if
there
be
any
wanting
,
make
them
full
as
many
,
or
rather
more
by
annexing
Cyphers
thereto
According
to
the
Rule
foregoing
,
but
in
some
Cases
there
must
of
necessity
be
more
,
for
when
there
is
an
equal
number
of
decimal
places
in
the
dividend
,
and
in
the
divisor
,
and
a
division
can
be
made
,
then
the
Quotient
will
infallibly
be
a
whole
Number
without
any
Fraction
,
except
what
is
in
the
Remainder
.
IV.
In
Multiplication
of
Decimal
Fractions
,
the
product
containeth
as
many
Decimal
figures
as
there
are
decimal
places
in
the
Multiplicand
and
Multiplier
,
and
in
Division
if
you
multiply
the
Quotient
by
the
divisor
the
product
will
be
equal
to
the
dividend
,
upon
which
consideration
the
true
value
of
the
Quotient
of
any
division
may
infallibly
be
known
by
this
General
Rule
.
After
the
work
of
Division
is
ended
,
consider
how
many
decimal
places
are
in
the
dividend
more
then
there
are
in
the
divisor
,
and
how
many
soever
the
excess
is
,
let
so
many
in
the
Quotient
be
separated
from
the
Rest
for
a
Decimal
.
But
if
there
are
not
so
many
figures
in
the
Quotient
,
as
the
said
excess
is
,
such
defect
must
be
supplyed
,
by
prefixing
as
many
Cyphers
on
the
left
hand
,
putting
a
point
before
them
,
as
hath
been
Taught
already
;
then
shall
such
Decimal
as
aforesaid
,
be
the
true
value
of
the
Quotient
sought
.
I
shall
explain
this
Rule
by
Examples
of
the
several
Cases
that
may
happen
in
the
division
of
Decimals
,
which
are
9
,
as
followeth
.
1
|
a
whole
Number
|
being
given
to
be
divided
by
|
a
whole
Number
|
2
|
a
mixt
Number
|
3
|
a
Decimal
Fraction
|
4
|
a
mixt
Number
|
a
whole
Number
|
5
|
a
mixt
Number
|
6
|
a
Decimal
Fraction
|
7
|
a
Decimal
Fraction
|
a
whole
Number
|
8
|
a
mixt
Number
|
9
|
a
Decimal
Fraction
|
Case
1.
A
whole
number
given
to
be
divided
by
a
whole
number
.
V.
When
you
are
to
divide
one
whole
Number
by
another
and
they
are
not
commensurable
,
though
there
are
no
decimals
in
either
the
dividend
or
the
divisor
,
yet
if
you
annex
a
Competent
number
of
Cyphers
to
the
dividend
,
there
will
be
a
decimal
in
the
Quotient
consisting
of
as
many
places
as
you
annexed
Cyphers
to
the
dividend
.
Example
1.
Let
there
be
given
5729
to
be
divided
by
438
;
According
to
the
foregoing
Rule
,
I
annex
a
Number
of
Cyphers
,
(
suppose
4
)
to
the
given
dividend
which
will
supply
4
decimal
places
,
and
it
will
be
5729.0000
,
and
after
the
work
of
division
is
finished
I
find
the
Quotient
to
be
130799
Now
to
find
out
the
value
of
the
Quotient
by
the
General
Rule
before-going
,
I
consider
that
there
are
no
decimals
in
the
divisor
,
but
there
are
4
in
the
dividend
,
and
consequently
by
the
said
Rule
there
must
be
4
decimal
places
noted
out
in
the
Quotient
by
setting
a
point
before
them
,
and
then
the
true
value
of
the
Quotient
will
be
found
to
be
13.0799
.
Example
2.
Let
there
be
given
48
to
be
divided
by
43796
,
you
cannot
here
make
any
work
till
you
have
annexed
Cyphers
to
the
Dividend
,
because
the
divisor
is
bigger
than
the
dividend
and
therefore
annex
as
many
as
you
think
convenient
,
suppose
6
,
and
having
finished
the
work
of
Division
you
will
find
the
Quotient
to
be
1095
,
now
to
find
out
its
true
value
,
consider
that
there
are
no
decimal
places
in
the
divisor
,
but
there
are
6
,
in
the
dividend
,
therefore
there
must
be
6
decimal
places
in
the
Quotient
,
but
the
Quotient
as
yet
possesseth
but
4
places
,
therefore
to
make
them
up
6
according
to
the
said
general
Rule
,
I
prefix
two
Cyphers
before
the
other
figures
,
on
the
left
hand
of
the
same
,
so
as
they
may
take
place
in
the
decimal
by
putting
a
point
before
them
,
so
will
the
true
Quotient
be
.001095
,
&c.
This
First
Case
may
very
well
serve
for
a
further
illustration
of
the
first
Rule
of
the
second
Chapter
of
this
book
.
Case
2.
Example
.
3.
A
whole
number
given
to
be
divided
by
a
mixt
number
.
Let
the
whole
number
586
be
given
to
be
divided
by
the
mixt
Number
36.4865
;
here
you
may
observe
that
although
the
dividend
be
greater
then
the
divisor
,
yet
there
can
be
no
operation
until
the
dividend
is
prepared
by
annexing
a
competent
number
of
Cyphers
to
it
,
and
according
to
the
third
Rule
of
this
Chapter
,
I
must
annex
at
least
4
,
but
here
I
shall
take
6
(
or
more
at
pleasure
)
and
then
the
dividend
will
be
586.000000
,
and
the
work
being
finished
as
in
Division
of
whole
numbers
,
the
Quotient
will
be
found
to
be
1606
,
&c.
Now
to
discover
the
value
of
this
Quotient
,
according
to
the
general
Rule
aforegoing
,
I
consider
that
there
are
4
Decimal
figures
in
the
divisor
,
and
6
decimal
places
in
the
dividend
,
the
excess
being
2
,
and
consequently
there
must
be
two
decimal
places
noted
in
the
Quotient
by
putting
a
point
before
them
,
and
then
the
true
Quotient
will
be
16.06
as
you
may
prove
at
your
leisure
.
Example
4.
Another
Example
of
the
second
Case
may
be
this
,
Let
there
be
given
the
number
2
,
to
be
divided
by
the
mixt
number
28.74
,
having
prepared
the
dividend
,
by
annexing
6
Cyphers
to
it
,
(
or
more
at
pleasure
)
and
finished
the
work
of
division
as
in
whole
numbers
,
I
find
the
Quotient
to
be
695
,
&c.
Now
to
find
out
the
true
value
of
this
Quotient
I
consider
according
to
the
General
Rule
,
that
there
are
but
two
Decimal
places
in
the
Divisor
,
and
6
in
the
dividend
,
therefore
(
the
excess
being
4
)
there
must
be
4
decimal
places
in
the
Quotient
,
but
there
are
but
three
places
,
wherefore
I
make
them
up
4
,
by
prefixing
a
Cypher
before
them
,
according
to
the
latter
part
of
the
said
General
Rule
.
Case
3.
Exàmple
5.
A
whole
numb
.
given
to
be
divided
by
a
decimal
Fract
.
Let
there
be
given
the
whole
Number
48
to
be
divided
by
the
decimal
.0675
,
after
the
dividend
is
prepared
by
annexing
a
competent
number
of
Cyphers
,
as
suppose
7
,
after
the
work
of
Division
is
ended
,
I
find
the
Quotient
to
amount
to
.711111
as
followeth
.
Now
to
find
out
the
value
of
the
said
Quotient
,
by
the
foregoing
general
Rule
,
I
consider
that
there
are
4
decimal
places
in
the
Divisor
,
and
7
in
the
Dividend
,
the
excess
being
3
;
wherefore
I
conclude
that
according
to
the
said
Rule
,
there
must
be
3
decimal
figures
in
the
Quote
,
out
off
or
separated
from
the
rest
by
a
point
,
and
then
the
true
value
of
the
Quotient
will
be
711.111
that
is
711
integers
,
and
111
decimal
parts
of
an
integer
or
very
near
.
Case
4.
Example
6.
A
mixt
number
given
to
be
divided
by
a
whole
number
.
Let
there
be
given
the
mixt
Number
743.574
,
to
be
divided
by
the
whole
Number
75.
After
the
Dividend
is
prepared
by
annexing
Cyphers
at
pleasure
,
and
the
operation
(
according
to
division
of
whole
numbers
finished
)
you
will
find
this
Quotient
,
viz.
991432.
Now
to
find
out
the
true
value
of
the
said
Quotient
,
I
consider
that
after
there
are
2
Cyphers
annexed
to
the
dividend
,
that
the
decimal
part
thereof
will
possess
5
places
;
and
because
there
are
none
in
the
divisor
,
therefore
the
excess
is
5
,
and
consequently
(
according
to
the
said
General
Rule
)
I
note
5
places
in
the
Quotient
for
the
decimal
part
,
which
being
done
,
I
find
the
true
value
of
it
to
be
9.91432
.
Example
7.
Again
,
Let
the
dividend
in
the
last
Example
,
viz.
743.574
be
given
to
be
divided
by
the
whole
Number
43576
,
and
the
Quotient
will
be
found
to
be
17063
,
if
there
be
3
Cyphers
annexed
to
the
dividend
,
and
there
will
be
6
decimal
places
in
it
,
and
not
one
in
the
divisor
,
wherefore
there
must
be
6
decimal
places
in
the
Quotient
,
but
there
are
but
5
,
therefore
to
make
them
6
,
according
to
the
said
General
Rule
,
I
prefix
a
Cypher
,
and
then
the
true
value
of
the
Quotient
will
be
.017063
as
upon
proof
you
will
easily
find
.
Case
5.
Example
8.
A
mixt
number
given
to
be
divided
by
a
mixt
nmuber
.
Let
the
following
mixt
Number
viz.
3.748
be
given
to
be
divided
by
the
mixt
Number
46.375
.
Here
according
to
former
directions
I
annex
Cyphers
(
at
pleasure
)
to
the
dividend
,
suppose
5
,
then
will
the
dividend
be
3.74800000
and
having
finished
the
work
of
division
,
as
if
they
were
whole
numbers
,
I
find
the
Quote
to
be
8084
,
&c.
but
the
true
value
of
this
Quotient
thus
found
I
as
yet
know
not
,
therefore
to
make
a
discovery
of
its
value
I
consider
that
in
the
dividend
there
are
8
decimal
places
,
and
in
the
divisor
there
are
but
three
such
places
,
therefore
the
number
of
decimal
places
in
the
dividend
exceeds
the
number
of
places
in
the
divisor
by
5
,
so
that
by
the
foregoing
general
Rule
I
know
that
there
must
be
5
decimal
places
in
the
Quotient
,
but
there
are
only
4
figures
,
viz.
8084
,
but
to
make
them
5
according
to
the
General
Rule
,
I
prefix
a
Cypher
before
the
other
figures
and
it
makes
.08084
,
which
is
the
true
Quotient
sought
.
Case
6.
A
mixt
number
given
to
be
divided
by
a
Dec.
Fraction
.
Example
9.
Let
there
be
given
the
mixt
number
54.379
to
be
divided
by
the
Decimal
Fraction
.34687
,
having
annexed
a
competent
number
of
Cyphers
,
so
that
there
may
be
3
,
or
4
,
or
5
Decimal
places
in
the
dividend
more
then
there
are
in
the
divisor
,
wherefore
I
annex
6
Cyphers
,
and
then
the
dividend
will
be
54.379000000
,
and
when
the
work
of
division
is
ended
the
Quotient
will
be
found
to
be
1567705.
Which
being
done
the
next
thing
in
order
to
the
compleating
of
the
work
,
is
to
find
out
the
true
value
of
the
said
Quotient
,
which
is
easily
done
by
the
said
general
rule
,
for
I
consider
that
in
the
divisor
there
are
5
decimal
places
but
in
the
dividend
there
are
9
(
viz.
3
given
significant
figures
and
6
Cyphers
annexed
)
so
that
the
excess
is
4
,
therefore
I
conclude
that
there
must
be
4
decimal
places
in
the
Quotient
,
and
the
rest
are
of
the
Integral
part
,
so
that
I
find
the
true
Quotient
is
156.7705
,
that
is
156
Integers
,
and
.7705
or
7705/10000
partes
of
an
Integer
,
which
you
may
easily
prove
at
your
Leisure
.
Example
10.
If
there
were
given
the
mixt
Number
45.384
to
be
divided
by
.000247
;
here
are
not
so
many
decimal
places
in
the
dividend
as
there
are
in
the
divisor
,
therefore
do
I
increase
their
Number
by
annexing
5
Cyphers
thereto
,
and
then
the
dividend
will
be
45.38400000
,
then
do
I
proceed
to
the
operation
,
taking
no
notice
at
all
of
the
Cyphers
which
are
before
the
divison
,
but
work
as
if
there
were
none
at
all
,
and
when
the
work
of
division
is
finished
,
I
find
the
Quotient
to
be
183740.89
,
&c.
Now
the
Quotient
being
found
,
I
come
next
to
find
out
its
value
,
which
to
do
I
consider
that
there
are
6
decimal
places
in
the
divisor
,
and
8
in
the
dividend
,
so
that
the
excess
is
2
places
,
therefore
I
conclude
according
to
the
said
general
Rule
,
that
there
must
be
two
decimal
places
noted
in
the
Quotient
,
so
that
then
its
true
value
will
be
found
to
be
183740.89
,
&c.
Case
7.
A
Decimal
Fraction
given
to
be
divided
by
a
whole
number
.
Example
11.
Let
it
be
Required
to
divide
the
decimal
Fraction
.07864
by
the
whole
Number
25.
here
in
this
Example
,
there
is
no
need
of
annexing
any
Cyphers
to
the
dividend
to
prepare
it
for
operation
,
but
yet
you
may
at
your
pleasure
,
only
because
there
is
no
necessity
I
shall
forbear
it
,
and
proceed
to
the
work
according
to
the
Rule
of
Division
in
whole
Numbers
,
and
the
work
being
finished
,
I
find
the
Quotient
to
be
314
,
then
I
proceed
to
find
out
the
value
of
this
Quotient
by
the
General
Rule
foregoing
,
and
because
there
is
no
decimal
in
the
divisor
,
and
5
in
the
dividend
,
therefore
there
must
be
5
decimal
places
in
the
Quotient
,
and
there
are
but
3
places
as
yet
,
therefore
do
I
prefix
two
Cyphers
before
the
Quotient
thus
found
,
and
note
them
for
the
true
Quotient
sought
,
which
is
.00314
,
as
by
the
operation
appeareth
.
Case
8.
A
Decimal
Fraction
given
to
be
divided
by
a
mixt
Number
.
Example
12.
Let
there
be
given
the
Decimal
Fraction
846
,
to
be
divided
by
the
mixt
Number
3.476
,
here
l
1.
prepare
the
dividend
for
the
work
by
annexing
4
Cyphers
thereto
,
and
having
finished
the
work
of
division
I
find
the
Quotient
to
be
2433
,
and
to
discover
its
value
according
to
the
general
Rule
,
I
consider
that
there
are
in
the
dividend
(
after
the
4
Cyphers
are
thereto
annexed
)
7
decimal
places
,
&
in
the
Divisor
there
are
but
3
,
so
that
the
excess
is
4
,
therefore
I
conclude
that
there
must
be
four
decimal
places
in
the
Quotient
,
so
that
the
true
Quotient
is
the
Decimal
Number
.2433
,
&c.
as
followeth
.
But
if
to
the
said
dividend
.846
there
had
been
annexed
5
Cyphers
then
the
true
Quotient
would
have
been
.24338
,
&c.
and
if
there
had
been
6
Cyphers
annexed
thereto
then
had
the
Quotient
been
.243383
,
&c.
Example
13.
Let
it
be
Required
to
divide
this
Decimal
Fraction
,
viz.
.846
by
the
mixt
Number
34.76
,
after
I
have
annexed
Cyphers
to
the
Dividend
to
prepare
it
for
the
work
;
and
the
work
of
Division
being
finished
,
I
find
(
as
before
)
the
Quotient
to
be
2433
,
but
the
value
of
it
being
found
out
,
by
the
general
Rule
,
will
be
different
from
the
former
Quote
,
for
having
taken
the
number
of
decimal
places
,
in
the
dividend
and
the
divisor
,
I
find
the
excess
to
be
5
in
the
dividend
,
so
that
there
should
be
5
decimal
places
in
the
Quotient
,
but
there
are
now
but
4
places
,
wherefore
to
supply
that
defect
I
prefix
a
Cypher
before
the
said
Quote
,
and
put
a
point
before
it
so
as
it
may
take
place
in
the
Decimal
,
and
then
the
true
value
of
the
Quotient
will
be
.02433
,
&c.
as
followeth
And
if
the
Divisor
had
been
347.6
then
the
Quote
would
have
been
.002433
,
&c.
Case
9.
A
Decimal
Fraction
given
to
be
divided
by
a
Decimal
Fraction
.
Example
14.
Let
there
be
given
the
Decimal
Fraction
.835796
to
be
divided
by
.243
,
here
I
may
annex
Cyphers
at
pleasure
to
the
Dividend
to
prepare
it
for
operation
,
but
because
there
is
no
necessity
for
it
I
shall
forbear
,
and
proceed
to
the
work
as
in
Division
of
whole
numbers
,
which
being
finished
I
find
in
the
Quotient
the
number
3439
,
and
now
I
have
nothing
to
do
but
to
find
out
the
true
value
of
this
Quotient
,
and
in
order
thereunto
I
consider
that
in
the
dividend
are
6
decimal
places
,
and
in
the
divisor
but
3
,
wherefore
the
excess
is
3
,
which
is
the
number
of
decimal
places
in
the
Quotient
,
which
being
separated
from
the
rest
by
a
point
according
to
former
directions
,
the
true
value
of
the
Quotient
will
be
found
to
be
3.439
,
&c.
But
if
the
dividend
had
had
a
Cypher
annexed
to
it
,
then
the
Quotient
would
have
been
3.4394
&c.
and
if
two
Cyphers
had
been
annexed
to
it
,
then
the
Quotient
would
have
been
3.43948
,
&c.
But
if
the
dividend
had
been
.0835796
,
and
the
divisor
the
same
as
before
,
the
operation
would
have
been
still
the
same
,
and
the
same
figures
would
be
in
the
Quotient
but
not
of
the
same
value
,
for
they
would
have
been
all
Decimals
,
viz.
.3439
,
&c.
But
if
the
dividend
had
been
(
as
before
)
.835796
,
and
the
divisor
had
been
(
.0243
,
)
the
same
as
before
with
a
Cypher
prefixed
before
it
to
depress
its
value
,
though
the
operation
be
the
very
same
,
yet
the
value
of
the
Quotient
would
have
been
34.39
,
&c.
And
if
the
Divisor
had
had
two
Cyphers
prefixed
before
it
thus
.00243
then
the
Quotient
would
have
been
343.9
,
&c.
And
if
the
divisor
had
been
(
.000243
)
the
same
as
before
with
3
Cyphers
prefixed
before
it
then
the
Quotient
would
have
been
3439.
consisting
intirely
of
Integers
,
except
you
had
annexed
Cyphers
to
the
Dividend
.
Thus
have
I
largely
gone
over
all
the
Cases
that
can
happen
in
division
of
Decimals
,
and
have
given
one
or
more
examples
in
every
Case
,
so
that
I
hope
by
this
time
the
diligent
Reader
is
made
capable
of
performing
any
operation
,
either
in
Addition
,
Subtraction
,
Multiplication
,
or
Division
of
Decimals
,
and
if
he
be
so
perfected
,
perhaps
he
may
be
desirous
to
know
something
of
their
use
and
applicatio●…
in
the
practical
parts
of
Arithmetick
,
before
he
comes
to
the
more
difficult
part
of
the
extraction
of
Roots
,
and
because
I
would
not
dull
the
edge
of
his
Appetite
,
I
shall
give
him
a
taste
of
their
excellent
use
in
the
Rule
of
Proportion
,
and
in
the
mensuration
of
some
Supersicies
and
Solids
,
and
then
come
to
shew
their
use
in
the
extracting
the
Cube
and
Square
Roots
,
and
the
calculating
of
Interest
,
&c.
CHAP.
VII
.
The
Rule
of
3
in
Decimals
.
I
Shall
not
here
meddle
with
the
Rule
of
3
in
its
distinct
kinds
,
viz.
Single
,
Double
,
Direct
,
or
Inverse
,
supposing
the
Learner
to
be
acquainted
with
that
already
in
the
Practice
of
vulgar
Arithmetick
.
I.
In
the
Rule
of
3
in
Decimals
,
the
operation
is
in
every
Respect
the
same
as
in
whole
Numbers
,
so
is
it
in
all
the
parts
,
or
Rules
of
Arithmetick
,
only
when
you
work
in
Decimals
,
you
must
have
Respect
to
the
Decimal
Rules
before
taught
,
for
in
Decimals
you
must
Add
,
Subtract
,
Multiply
,
and
Divide
,
when
,
and
after
the
same
manner
as
you
do
in
whole
Numbers
,
a
few
Examples
will
make
you
perfect
in
the
knowledge
thereof
.
Example
.
1.
If
1¾
l.
of
Tobacco
cost
3
s.
6
d.
how
much
will
326¼
l.
cost
at
that
Rate
?
When
the
Fractional
parts
of
the
Numbers
in
this
Question
are
turned
into
Decimals
,
then
it
will
be
read
thus
,
viz.
If
1.75
l.
of
Tobacco
cost
3.5
s.
how
much
will
326.25
l.
of
the
same
cost
at
that
Rate
?
The
numbers
being
orderly
placed
as
is
directed
in
the
6
Rule
of
the
10
Chapter
of
my
Vulgar
Arithmetick
will
stand
as
followeth
,
viz.
And
if
you
multiply
the
third
number
by
the
second
,
or
the
second
by
the
third
,
which
is
all
one
,
and
divide
the
product
thereof
by
the
first
,
as
is
directed
in
the
10
Rule
of
the
7
Chap.
of
my
Vulgar
Arithmetick
;
only
in
Multiplying
and
Dividing
,
you
must
have
regard
to
Multiplication
and
Division
of
Decimals
delivered
in
the
two
Chapters
foregoing
,
and
when
the
work
is
finished
,
the
answer
will
be
found
to
be
652.5
shillings
,
or
32
l.
12
s.
6
d.
see
the
following
work
.
Example
2.
If
9
C.
of
Tobacco
cost
25
l.
7.
s.
what
will
be
the
Price
of
17
C.
weight
of
the
same
at
that
Rate
?
The
given
numbers
being
Rightly
stated
,
together
with
the
whole
operation
take
as
followeth
.
Here
you
see
that
the
Answer
in
Decimals
is
47883
l.
now
the
value
of
this
decimal
Fraction
,
may
be
discovered
at
first
sight
(
by
that
brief
way
of
finding
the
value
of
a
decimal
part
of
a
pound
sterling
,
delivered
in
the
4
Rule
of
the
2
Chapter
foregoing
)
to
be
17
s.
8.
d.
fere
,
for
8
primes
is
16
shillings
,
and
8
seconds
is
1
shilling
more
,
and
3
seconds
over
,
which
with
the
3
in
the
place
of
thirds
make
33
,
from
which
abating
1
,
because
it
is
above
25
,
there
remains
32
farthings
or
8
pence
.
Example
3.
If
an
ounce
of
Gold
be
worth
2
l.
19
s.
4
d.
I
demand
the
price
of
19
oz.
3
pw.
5
gr.
at
that
Rate
?
By
the
4
Rule
of
the
2
Chapter
3
pw.
5
gr
.
may
be
Reduced
to
this
decimal
of
an
ounce
viz.
160416
,
so
that
19.160416
is
a
mixt
number
equal
in
value
to
19
oz.
3
pw.
5
gr.
And
by
the
same
Rule
.9666
is
found
to
be
the
Decimal
part
of
a
pound
sterling
,
equal
in
value
to
19
s.
4
d.
so
that
the
Decimals
being
found
out
,
and
the
numbers
given
in
the
Question
being
stated
in
order
will
be
as
followeth
,
viz.
So
that
I
find
the
Answer
to
the
Question
to
be
56.84129
,
&c.
or
56
l.
16
s.
10
d.
very
near
as
it
may
be
discovered
by
the
brief
way
of
finding
the
value
of
the
decimal
of
a
pound
sterling
delivered
before
in
the
13
page
.
CHAP.
VIII
.
The
further
use
of
Decimals
in
the
Mensuration
of
Superficies
and
Solids
.
PROP.
I.
To
Measure
a
long
Square
.
Multiply
the
length
of
it
in
Feet
,
or
Inches
,
by
the
breadth
of
it
in
Feet
,
or
in
Inches
,
and
the
product
will
give
you
the
true
Superficial
Content
of
it
in
Feet
or
Inches
.
Example
.
There
is
a
Table
whose
length
is
18.75
Feet
and
its
Breadth
3.5
Feet
,
I
demand
its
content
in
Feet
?
To
Answer
this
Question
,
I
take
18.75
feet
(
the
length
of
it
)
and
multiply
it
by
3.5
feet
(
the
breadth
of
it
)
and
the
product
is
65.625
feet
,
which
is
the
Content
of
the
Table
,
as
was
Required
.
See
the
work
Here
by
the
way
take
notice
that
although
amongst
Artificers
,
the
two
foot
Rule
is
generally
divided
each
foot
into
12
Inches
,
&c.
Yet
for
him
that
is
at
any
time
employed
in
the
practice
of
Measuring
,
it
would
be
most
necessary
for
him
to
have
his
two
foot
Rule
,
each
foot
divided
into
10
equal
parts
,
and
each
of
those
parts
divided
again
into
10
other
equal
parts
,
so
would
the
whole
foot
be
divided
into
100
equal
parts
,
and
thereby
would
it
be
made
fit
to
take
the
dimensions
of
any
thing
whatsoever
,
in
feet
,
and
decimal
parts
of
a
foot
,
and
thereby
the
Content
of
any
thing
may
be
found
as
exactly
if
not
more
exactly
and
near
,
then
if
the
foot
were
divided
into
Inches
,
quarters
and
half
quarters
,
and
thereby
many
times
would
there
by
much
labour
and
paines
avoided
,
which
the
Artist
is
Content
to
undergo
through
the
want
of
such
Decimal
division
of
his
Rule
,
as
we
will
shew
in
the
solving
of
the
former
proposition
,
after
the
vulgar
way
.
The
question
is
There
is
a
Table
whose
length
is
18
foot
9
inches
,
and
its
breadth
3
foot
6
inches
,
now
I
demand
its
content
in
feet
?
Now
,
before
I
can
find
its
content
,
I
must
find
its
length
and
breadth
in
inches
,
and
then
multiply
the
inches
of
the
length
by
the
inches
of
the
breadth
,
and
then
the
product
will
be
9450
which
is
its
Content
in
square
inches
,
and
to
find
its
Content
in
feet
I
must
divide
the
inches
by
144
(
the
number
of
square
inches
in
a
foot
)
and
the
Quotient
is
the
Content
in
feet
:
See
the
work
following
.
So
that
you
see
according
to
this
way
the
answer
is
65
square
feet
and
90
inches
,
or
〈◊〉
of
a
foot
which
is
the
very
same
with
that
answer
in
Decimals
,
and
if
the
Division
by
144
had
been
continued
by
annexing
Cyphers
to
the
Dividend
9450
,
there
would
have
come
out
in
the
Quotient
the
Decimal
.624
as
before
.
But
how
tedious
a
work
it
is
to
answer
it
after
the
Vulgar
way
,
compared
with
the
decimal
way
I
leave
the
Judicious
Reader
to
judge
,
and
much
more
tedious
would
it
have
been
if
there
had
been
either
halfs
or
quarters
of
inches
either
in
the
length
or
breadth
,
or
both
,
but
the
work
would
still
have
been
the
same
in
the
Decimal
way
,
that
is
,
in
every
Respect
as
easie
.
After
the
same
manner
is
found
out
the
Content
of
the
true
Geometrical
Square
,
which
is
a
figure
fitly
Represented
by
an
exact
square
Trencher
,
that
is
,
having
its
length
and
breadth
both
equal
.
PROP.
II.
To
find
the
Content
of
a
right
angled
Triangle
.
A
Right
angled
Triangle
is
a
plain
figure
having
3
sides
and
3
Angles
as
the
figure
B
,
C
,
D
,
in
the
margent
,
two
of
which
sides
viz.
BC
,
and
CD
,
are
perpendicular
to
each
other
,
now
if
from
the
top
of
the
perpendicular
at
D
,
there
be
a
line
drawn
parallel
to
the
base
B
,
C
,
as
is
the
Prickt
line
A
,
D
,
and
from
the
end
of
the
Base
at
B
,
there
be
drawn
the
prickt
line
BA
parallel
to
the
perpendicular
till
it
meet
the
line
AD
in
A
,
then
will
there
be
made
the
parallelogram
,
or
long
square
A
,
B
,
C
,
D
,
of
which
the
Triangle
B
,
C
,
D
,
is
half
,
the
Diagonal
B
,
D
,
dividing
the
whole
paralelogram
into
2
equal
parts
.
Now
it
is
plain
from
the
first
proposition
,
that
if
you
multiply
the
side
B
,
C
,
by
the
side
C
,
D
,
then
the
product
will
be
the
Content
of
the
whole
parallelogram
A
,
B
,
C
,
D
,
and
then
the
half
of
that
Content
will
be
the
Content
of
the
given
Triangle
B
,
C
,
D.
Or
if
you
take
half
CD
,
which
is
C
,
F
,
and
half
of
BA
,
which
is
B
,
E
,
and
draw
the
line
E
,
F
,
then
will
E
,
F
,
divide
the
parallelogram
A
,
B
,
C
,
D
,
into
two
equal
parallelograms
,
and
either
of
them
is
equal
to
the
given
Triangle
B
,
C
,
D
,
now
if
by
the
first
proposition
I
can
find
the
Content
of
the
paralelogram
E
,
B
,
C
,
F
,
I
find
also
the
Content
of
the
Triangle
B
,
C
,
D
,
because
they
are
equal
,
whence
it
comes
to
pass
that
if
you
multiply
the
base
by
half
the
perpendicular
,
or
the
perpendicular
by
half
the
base
of
a
Rectangular
Triangle
(
which
is
all
one
)
the
product
will
be
the
true
Content
thereof
.
Example
.
In
the
former
Triangle
the
base
B
,
C
,
is
18.28
Feet
,
and
the
perpendicular
C
,
D
,
is
12.26
Feet
,
I
demand
its
Content
in
Feet
?
Here
I
first
find
the
Content
of
the
whole
Parallelogram
,
by
multiplying
the
sides
together
,
and
the
product
is
224.1128
Feet
,
and
the
half
of
that
product
,
is
the
Content
of
the
Triangle
B
,
C
,
D
,
which
is
112.0564
Feet
.
See
the
following
work
:
The
Content
would
have
been
the
same
,
if
I
had
multiplyed
the
one
side
by
half
the
other
,
which
is
indeed
the
shortest
way
,
and
most
practical
;
See
the
work
:
The
Answer
would
have
been
the
same
,
if
I
had
taken
the
whole
side
C
,
D
,
and
multiplyed
it
by
half
the
side
B
,
C.
PROP.
III.
To
find
the
Content
of
any
plain
Triangle
,
not
Rectangular
.
THE
best
and
easiest
way
is
to
let
fall
a
perpendicular
,
upon
the
longest
side
from
the
angle
that
is
opposite
to
it
,
which
will
divide
it
into
two
Right-angled
plain
Triangles
,
As
,
suppose
there
were
given
the
plain
Triangle
ABC
,
as
followeth
,
Here
the
side
AB
,
being
the
longest
side
,
I
let
fall
a
perpendicular
from
its
opposite
angle
at
C
,
which
falleth
upon
D
,
in
the
line
AB
,
so
is
the
line
CD
the
nearest
distance
between
the
angular
point
C
,
and
the
line
AB
,
and
divideth
the
given
Triangle
ABC
,
into
two
Right-angled
Triangles
,
(
viz.
)
ADC
and
CDB
;
and
if
you
find
the
Content
of
these
two
Right-angled
Triangles
(
according
to
the
directions
in
the
second
proposition
)
and
add
them
together
,
their
sum
will
be
the
Content
of
the
given
Triangle
ABC
.
But
it
may
be
more
Artificially
found
out
thus
,
Multiply
half
the
line
CD
,
into
the
whole
line
AB
,
the
product
will
give
the
Content
of
the
Triangle
which
was
sought
,
or
if
you
multiply
the
whole
line
CD
,
into
half
the
line
AB
,
the
product
will
be
the
Content
of
the
given
Triangle
,
which
is
very
plain
from
a
due
consideration
of
the
method
used
in
solving
the
se●…ond
proposition
.
Example
Let
the
base
or
longest
side
AB
be
48.5
Feet
long
,
and
let
the
length
of
the
perpendicular
CD
be
21.6
Feet
,
I
desire
to
know
how
many
square
,
or
superficial
feet
are
contained
in
the
said
Triangle
?
To
resolve
this
Question
according
to
the
foregoing
Rule
,
I
first
Bi-part
the
base
AB
48.5
which
is
24.25
which
I
multiply
by
the
length
of
the
perpendicular
CD
21.6
and
the
product
is
523.8
:
See
the
following
work
,
So
that
you
see
the
content
is
the
same
which
of
the
foresaid
ways
soever
you
work
;
observe
the
same
method
in
finding
the
content
of
an●…
oblique
Triangle
given
.
PROP.
IV.
To
find
the
Content
of
a
Trapezium
.
A
Trapezium
is
a
plane
figure
having
fo●…
unequal
sides
,
and
as
many
unequal
Angle
it
matters
not
how
unequal
they
are
,
and
to
find
out
its
Content
observe
the
following
directions
,
viz.
Divide
it
into
two
oblique
Triangles
,
by
drawing
a
line
from
any
one
of
the
angles
,
to
the
angle
that
is
opposite
thereto
,
which
line
shall
be
a
common
base
to
both
the
Triangles
.
Then
if
you
find
out
the
content
of
both
these
Triangles
,
according
to
the
method
prescribed
in
the
third
proposition
,
the
sum
of
their
contents
,
is
the
content
of
the
given
Trapezium
.
Or
it
may
be
more
Artificially
found
out
thus
,
viz.
Multiply
the
length
of
the
common
base
by
half
the
sum
of
the
perpendiculars
let
fall
from
the
Angles
opposite
to
the
said
common
base
,
and
the
product
will
be
the
content
of
the
whole
Trapezium
:
or
else
,
Multiply
the
sum
of
the
said
Perpendiculars
by
half
the
said
common
base
,
and
it
will
produce
the
same
effect
.
Example
.
In
the
following
Trapezium
ABCD
,
draw
the
base
AC
,
which
suppose
to
be
9.5
Feet
,
then
let
fall
the
perpendicular
at
D
,
which
let
be
3.45
Feet
,
and
that
at
B
4.25
Feet
,
the
sum
of
the
said
perpendiculars
is
7.7
half
of
which
is
3.85
,
by
which
if
the
common
base
AC
be
multiplyed
,
the
product
(
which
is
)
36.575
is
the
content
of
the
Trapezium
required
Or
if
you
multiply
7.7
the
sum
of
the
perpendiculars
by
half
of
the
common
base
9.5
,
which
is
4.75
the
product
will
be
the
same
.
See
the
following
work
.
PROP.
V.
To
find
the
Content
of
any
regular
Polygon
.
A
Regular
Polygon
is
a
plane
figure
consisting
of
equal
sides
and
equal
Angles
,
viz.
a
Pentagon
,
consisting
of
5
equal
sides
,
and
5
equal
Angles
;
an
Hexagon
,
consisting
of
6
equal
sides
,
and
6
equal
Angles
,
an
Heptagon
of
seven
equal
sides
,
an
Octogon
of
eight
,
&c.
and
to
measure
any
1
of
these
Regular
planes
,
do
thus
draw
a
line
from
the
Centre
of
the
figure
to
the
middle
of
any
one
of
the
sides
,
and
multiply
that
line
into
half
the
sum
of
the
sides
,
and
the
product
thence
arising
is
the
content
of
the
given
plane
.
I
shall
give
you
an
Example
of
this
in
the
Mensuration
of
an
Hexagon
,
or
plane
of
6
equal
sides
.
The
Reason
of
this
manner
of
working
is
very
plain
,
if
from
the
Centre
you
draw
the
lines
GA
,
and
GB
,
thereby
making
the
Triangle
GAB
,
whose
content
(
by
the
third
proposition
)
is
found
by
multiplying
the
perpendicular
GH
into
half
the
side
AB
,
viz.
into
HA
,
or
HB
,
but
there
are
6
such
Triangles
in
the
given
Polygon
;
therefore
GH
,
multiplied
into
6
times
HB
,
produceth
the
content
required
.
PROP.
VI.
To
find
the
Content
of
any
Irregular
Polygon
.
LET
it
be
Required
to
measure
the
following
Figure
ABCDEFGHI
.
First
take
Care
that
the
whole
plane
be
divided
in
Trapeziums
and
Triangles
according
to
your
own
Fancy
,
and
as
the
nature
of
the
plane
will
bear
,
and
then
measure
those
Trapeziums
and
Triangles
,
as
is
directed
in
the
third
and
fourth
Propositions
before
going
,
and
add
the
several
contents
together
,
so
will
the
sum
give
you
the
whole
content
of
that
Irregular
Polygon
.
PROP.
VII
.
To
find
the
circumference
of
a
Circle
having
the
Diameter
given
.
A
Circle
is
a
Geometrical
figure
exactly
Round
,
so
that
if
from
a
point
in
the
middle
of
it
called
the
Centre
,
there
be
never
so
many
lines
drawn
to
the
Circumference
,
they
will
all
be
of
equal
length
.
But
between
the
diameter
and
circumference
of
a
circle
there
cannot
be
found
a
true
and
exact
proportion
.
Archimedes
hath
demonstrated
the
proportion
to
be
near
as
7
is
to
22
;
but
that
of
Van
Ceulen
is
the
most
exact
,
who
makes
it
to
be
as
1
,
is
to
3.14159265358979323846
&c.
but
for
practice
this
following
proportion
is
sufficiently
exact
,
viz.
As
1.
is
to
3.1416
So
is
the
Diameter
of
any
Circle
,
To
its
Circumference
.
To
Answer
which
I
say
by
the
proportion
foregoing
;
As
1
is
to
3.1416
,
so
is
28
the
diameter
to
87.9648
,
which
is
the
circumference
ABCD.
The
work
followeth
.
PROP.
VIII
.
To
find
the
content
of
a
Circle
having
the
Diameter
given
.
FIrst
find
out
the
circumference
,
by
the
last
Proposition
,
then
multiply
half
the
circumference
by
half
the
diameter
,
and
that
product
is
the
Content
.
Example
There
is
a
Circle
whose
Diameter
is
14
Inches
,
I
demand
how
many
square
Inches
are
the
content
of
that
Circle
?
By
the
foregoing
Proposition
,
I
find
the
circumference
to
be
43.9824
Inches
,
the
half
of
which
is
21.9912
which
being
multiplyed
by
7
,
(
half
the
given
Diameter
)
the
product
is
153.9384
which
is
the
content
Required
.
See
the
work
.
PROP.
IX
.
To
find
the
folid
content
of
a
square
piece
of
Timber
,
Stone
,
&c.
whose
bases
are
equal
,
that
is
,
whose
ends
are
of
the
same
bigness
.
SUch
a
solid
piece
by
Geometricians
is
called
a
Parallelepipedon
,
and
it●…●…ontent
is
thus
found
out
,
viz.
First
find
out
the
●…uperficial
content
of
the
base
or
end
,
(
by
the
first
proposition
)
then
multiply
that
content
by
the
whole
length
,
and
that
product
is
the
Solid
content
of
the
whole
piece
.
Example
There
is
a
square
piece
of
Timber
,
the
two
contiguous-sides
at
the
end
of
which
are
2.5
Feet
,
and
1.8
Feet
,
and
its
length
is
22
Feet
,
I
demand
how
many
solid
Feet
are
in
that
piece
of
Timber
.
First
I
multiply
2.5
by
1.8
the
sides
of
the
base
,
and
that
produceth
4.5
for
the
content
of
the
base
or
end
,
and
that
product
I
multiply
by
22
the
length
,
and
that
produceth
99
Feet
,
and
so
many
is
there
contained
in
that
piece
of
Timber
.
As
you
may
see
in
the
following
work
.
Here
note
,
if
the
sides
of
the
end
,
or
base
,
be
given
in
Inches
,
and
its
length
in
Feet
,
then
Reduce
the
sides
of
the
base
into
the
Decimal
parts
of
a
Foot
,
and
proceed
as
before
,
or
you
may
find
out
the
content
of
the
base
in
Inches
,
and
multiply
that
content
by
the
length
in
Feet
,
and
that
product
divided
by
144
,
will
give
you
the
content
in
feet
,
or
else
reduce
the
length
into
Inches
,
and
multiply
the
content
of
the
base
thereby
,
and
Divide
that
product
by
1728
(
for
there
are
so
many
Cubical
Inches
in
a
Foot
)
and
the
Quotient
will
give
you
the
solid
content
in
Feet
.
But
the
Decimal
way
is
preferred
.
PROP.
X.
To
find
the
solid
content
of
a
Cylinder
,
having
the
Diameter
of
its
Base
given
.
A
Cylinder
is
a
Solid
whose
bases
are
Circular
,
equal
,
and
parallel
,
and
may
fitly
be
represented
by
a
round
pillar
,
or
a
Roling-stone
of
a
Garden
,
and
to
find
the
solid
content
of
such
a
body
this
is
The
Rule
.
First
find
the
plane
of
the
base
,
by
the
7
and
8
propositions
foregoing
,
and
then
multiply
that
by
the
length
thereof
,
which
product
will
give
you
the
solid
content
of
the
given
Cylinder
.
Example
.
There
is
a
Cylinder
,
(
suppose
a
Roling-stone
)
whose
length
is
8.75
Feet
,
and
the
Diameter
of
its
base
2.8
Feet
,
I
demand
the
solid
content
thereof
?
As
1
Is
to
3.1416
So
is
2.8
the
given
Diameter
To
8.79648
the
Circumference
of
the
base
,
half
of
which
(
viz.
4.39824
)
being
multiplyed
by
1.4
the
semi-diameter
will
produce
●…
.
157536
for
the
content
of
the
Base
,
which
being
multiplyed
by
8.75
(
the
length
)
it
produceth
53.87844
for
the
Solid
Content
Required
.
If
there
had
been
given
the
circumference
of
the
Cylinder
,
then
the
Diameter
of
the
base
must
have
been
found
out
by
the
converse
of
the
seventh
Proposition
,
as
suppose
there
had
been
given
8.75
the
length
of
the
Cylinder
,
and
8.79648
its
circumference
to
find
the
Solidity
thereof
.
First
I
find
out
the
Diameter
by
the
following
proportion
,
viz.
As
3.1416
Is
to
1
So
is
8.79648
the
given
Circumference
To
2.8
the
Required
Diameter
.
And
then
the
Rest
of
the
work
is
the
same
with
that
before
.
PROP.
XI
.
To
find
the
Solid
Content
of
a
Cone
.
A
Cone
is
a
Solid
Body
,
having
a
Circle
for
its
base
,
and
its
superficies
Circular
,
decreasing
its
equidistant
Diameters
from
the
base
proportionably
,
till
it
terminateth
in
a
point
over
the
Centre
of
its
base
,
and
may
fitly
be
Represented
by
a
Sugar-loaf
,
or
a
round
Spire
Steeple
;
and
to
find
its
solid
Content
this
is
The
Rule
.
By
the
7
and
8
Propositions
foregoing
find
out
the
plane
of
its
base
,
and
multiply
that
by
1/3
of
its
height
,
and
that
product
is
the
Solid
content
of
the
Cone
Required
.
Example
.
There
is
a
Cone
the
circumference
of
whose
base
is
22.5
and
its
height
is
16
,
I
demand
the
solid
content
of
such
a
Cone
?
As
3.1416
Is
to
1.
So
is
22.5
the
circumference
of
the
base
To
7.162
the
Diameter
of
the
base
.
Then
I
multiply
half
22.5
which
is
11.25
by
half
7.162
which
is
3.581
,
and
it
produceth
40.28625
which
is
the
superficial
content
of
the
base
;
then
I
take
1/3
of
the
height
of
the
Cone
(
16
)
which
is
5.333
very
near
,
by
which
I
multiply
40.28625
(
the
superficial
content
of
the
base
)
and
it
produceth
214.84657125
.
See
the
work
as
followeth
;
The
same
is
to
be
observed
in
the
mensuration
of
any
other
Cone
.
But
here
you
are
to
observe
that
the
slanting
side
of
the
Cone
,
(
viz.
)
the
length
from
the
vertex
to
the
extremity
of
its
Base
)
is
not
to
be
taken
for
its
true
height
,
but
a
perpendicular
let
fall
from
its
vertex
to
the
Centre
of
its
base
is
its
true
height
;
and
how
you
may
find
out
that
perpendiculars
length
shall
be
shown
you
in
the
work
of
the
fourth
Proposition
of
the
eleventh
Chapter
.
PROP.
XII
.
To
find
the
solid
Content
of
a
Pyramid
.
BEtween
the
Cone
and
Pyramid
,
this
is
the
Difference
,
As
the
Cone
hath
a
circular
base
and
superficies
,
the
Pyramid
hath
a
Polygon
for
its
base
,
so
that
its
base
and
superficies
are
Angular
,
its
vertex
terminating
in
a
point
just
over
the
centre
of
its
base
,
and
to
find
out
its
solidity
,
here
followeth
The
Rule
.
Find
out
the
superficial
content
of
the
base
,
by
the
fifth
Proposition
foregoing
,
and
multiply
that
by
〈◊〉
of
its
height
,
and
it
produceth
the
solid
content
of
the
Pyramid
.
Example
.
There
is
a
Pyramid
whose
base
is
an
Hexagon
the
side
of
which
is
30
,
and
its
perpendicular
height
is
54
;
I
demand
the
solid
content
of
such
a
Pyramid
?
Here
by
the
fifth
Proposition
I
find
the
superficial
content
of
the
base
to
be
2340
;
then
do
I
take
,
of
the
perpendicular
height
of
the
Pyramid
,
which
is
18
,
and
thereby
do
I
multiply
2340
,
(
the
plane
of
the
base
)
and
the
product
is
42120
,
which
is
the
solid
content
of
the
given
Pyramid
.
Here
note
(
by
the
way
)
that
a
line
drawn
from
the
point
at
the
top
of
the
Pyramid
,
to
the
extremity
of
any
part
of
the
basis
,
is
not
the
true
height
of
any
Pyramid
,
but
a
perpendicular
let
fall
from
the
Cuspis
(
or
top
)
to
the
centre
of
the
base
is
the
true
height
,
and
how
to
find
out
such
perpendicular
heights
shall
be
shewn
in
the
fourth
Proposition
of
the
11
Chapter
.
PROP.
XIII
.
To
measure
the
Frustum
of
a
Pyramid
or
Cone
.
It
is
evident
that
if
I
find
the
solidity
of
the
whole
Pyramid
AGD
,
&
also
the
solidity
of
the
lesser
Pyramid
E
F
D
,
and
then
substract
the
content
of
E
F
D
,
from
the
content
of
A
G
D
,
that
there
will
remain
the
solidity
of
the
Frustum
A
G
E
F
;
and
certainly
this
way
of
measuring
the
Frustum
of
a
Pyramid
or
Cone
,
is
the
most
exact
of
any
;
And
it
may
be
easily
measured
thus
,
first
of
all
find
out
the
height
of
the
whole
Pyramid
C
D
,
which
you
may
do
by
the
following
proportion
,
viz.
As
the
Semidifference
of
the
sides
of
the
Bases
,
Is
to
the
height
of
the
Frustum
,
So
is
the
half
side
of
the
greater
Base
,
To
the
height
of
the
whole
Pyramid
.
And
this
proportion
will
hold
good
if
you
work
by
the
Semi-difference
of
the
Diameters
of
the
bases
,
as
well
as
by
the
Semi-difference
of
the
sides
of
the
Bases
.
As
in
the
foregoing
figure
,
let
A
G
be
the
Diameter
of
the
greater
base
,
and
E
F
the
Diameter
of
the
lesser
base
,
from
E
and
F
let
fall
the
perpendiculars
E
B
and
F
O
,
then
shall
B
O
be
equal
to
E
F
,
and
the
sum
of
A
B
and
O
G
are
the
difference
of
the
Diameters
of
the
bases
,
E
F
and
A
G
;
and
consequently
A
B
is
the
Semi-difference
,
and
B
E
is
the
height
of
the
Frustum
,
and
A
C
is
half
the
side
of
the
greater
base
,
and
C
D
is
the
height
of
the
whole
Pyramid
.
Then
by
Eucl.
6.4
.
As
A
B
(
the
Semi-difference
of
Diameters
)
Is
to
B
E
(
the
height
of
the
Frustum
)
So
is
A
C
(
half
the
greater
Diameter
,
)
To
C
D
(
the
height
of
the
whole
Pyramid
.
)
So
the
height
of
the
whole
Pyramid
A
G
D
,
will
be
fo●…nd
to
be
30
foot
;
for
the
greater
Diaameter
A
G
,
is
24
Inches
,
the
lesser
8
,
the
difference
16
,
the
Semi-difference
8
,
therefore
shall
C
D
be
30
foot
;
for
Now
having
found
the
height
of
the
whole
Pyramid
to
be
30
foot
,
I
thereby
(
according
to
the
12th
.
Proposition
foregoing
)
find
the
content
of
the
whole
Pyramid
to
be
40
foot
,
then
in
the
lesser
pyramid
E
F
D
there
is
given
the
side
of
its
base
E
F
,
8
Inches
,
and
its
height
I
D
10
Inches
for
C
D
30
−
C
I
20
=
I
D
10
,
and
by
the
said
12th
.
proposition
I
find
the
solid
content
of
it
to
be
1.48
Feet
,
which
being
Subtracted
from
40
(
the
content
of
the
greater
Pyramid
)
there
will
remain
38.52
feet
for
the
true
solid
content
of
the
given
Frustum
A
G
E
F.
After
the
same
manner
is
found
the
solidity
of
the
Frustum
of
a
Cone
,
the
height
of
the
whole
Cone
being
found
out
by
the
difference
of
the
Diameters
of
its
bases
;
and
by
the
11th
.
proposition
find
the
solidity
of
the
whole
Cone
,
and
also
the
solidity
of
the
lesser
Cone
,
that
is
cut
off
from
the
Frustum
,
then
Subtract
the
content
of
the
lesser
from
the
content
of
the
greater
,
and
the
remainder
will
be
the
solid
content
of
the
Frustum
.
This
last
proposition
is
useful
in
the
measuring
of
tapering
Timber
,
Round
or
Squared
,
and
for
finding
the
liquid
capacity
of
Brewers
Conical
,
or
Pyramidal
Tuns
.
Thus
have
I
shewed
the
Use
of
Decimals
in
the
Mensuration
of
the
most
usual
Planes
and
Solids
,
I
might
proceed
farther
to
shew
you
their
Application
in
the
particular
Mensuration
of
Board
,
Glass
,
Pavement
,
Plaistering
,
Painting
,
Wainscot
,
Tiling
,
Flooring
,
Tapistry
,
Brickwork
,
Timber
and
Stone
;
But
it
requireth
(
rather
)
a
particular
Treatise
,
then
the
narow
bounds
here
allowed
for
such
a
work
.
CHAP.
IX
.
The
Extraction
of
the
Square
Root
.
IN
the
Solution
of
any
Question
,
or
in
the
working
of
any
sum
whatsoever
belonging
to
any
of
the
Rules
of
Vulgar
or
Decimal
Arithmetick
,
there
have
been
(
at
least
)
two
things
or
numbers
given
,
whereby
the
answer
might
be
found
;
but
in
the
extraction
of
the
Square
,
Cube
,
and
all
other
Roots
,
there
is
but
one
number
given
to
find
out
the
Number
sought
,
viz.
there
is
a
square
Number
given
to
find
its
Root
,
a
Cube
Number
to
find
its
Root
,
&c.
and
I.
A
square
Number
is
that
which
is
produced
by
multiplying
any
number
by
(
or
into
)
it self
,
which
Number
given
to
be
so
multiplyed
is
called
the
Root
;
As
if
the
Number
8
were
given
to
be
multiplyed
by
it self
,
it
produceth
64
,
then
is
8
called
the
Root
,
and
64
is
its
square
,
so
the
Root
12
,
hath
for
its
square
144.
II.
When
a
square
Number
is
given
,
and
its
Root
is
Required
,
the
Operation
it self
is
called
the
Extraction
of
the
Square
Root
.
III.
Square
Numbers
are
of
two
kinds
,
viz.
either
Single
or
Compound
.
IV.
A
single
square
Number
is
that
which
is
produced
by
the
Multiplication
of
a
Digit
,
or
single
Number
into
it self
,
and
consequently
such
a
square
Number
must
be
under
100
,
which
is
the
square
of
10
,
so
25
being
given
for
a
square
number
,
it
is
a
single
square
having
for
its
Root
5.
And
81
is
a
single
square
Number
,
having
for
its
Root
the
Digit
9.
All
the
single
square
numbers
with
their
Roots
,
are
contained
in
the
following
Tablet
.
Roots
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
Squares
|
1
|
4
|
9
|
16
|
25
|
36
|
49
|
64
|
81
|
V.
When
the
Root
of
any
square
Number
is
Required
it
being
lesser
than
100
,
and
yet
is
not
exactly
a
single
square
,
expressed
in
the
Tablet
above
,
then
you
are
to
take
the
Root
of
that
single
square
Number
expressed
in
the
said
Tablet
,
which
(
being
less
)
is
nearest
to
the
given
square
;
As
if
it
were
74
whose
Root
is
required
I
find
that
81
(
the
square
of
9
)
is
too
much
,
and
64
(
the
square
of
8
)
is
too
little
,
but
yet
it
is
the
nearest
square
Number
that
is
lesser
than
74
,
and
therefore
I
take
8
to
be
the
square
Root
of
74
,
but
yet
it
is
plain
that
8
is
too
little
for
the
Root
of
74.
And
to
find
out
the
Fractional
part
of
this
Root
,
you
shall
be
plainly
taught
by
and
by
.
VI.
A
compound
square
number
is
that
which
hath
above
9
for
its
Root
.
VII
.
The
Root
of
a
single
square
Number
may
be
discovered
at
the
first
sight
,
but
the
extraction
of
a
compound
square
Number
is
more
tedious
and
difficult
,
its
Root
consisting
of
two
places
,
at
the
least
,
and
the
square
it self
,
of
100
at
the
least
.
VIII
.
When
a
compound
square
number
is
given
,
and
it
is
required
to
have
its
square
Root
extracted
,
before
you
can
proceed
to
the
Operation
,
your
square
number
must
be
prepared
,
by
pointing
it
at
every
second
figure
,
beginning
at
the
place
of
Units
.
As
,
Suppose
you
were
to
extract
the
square
Root
of
2304
,
first
I
put
a
point
over
4
(
it
standing
in
the
place
of
Units
)
and
then
passing
over
the
second
place
(
or
place
of
Tens
)
which
is
〈◊〉
,
I
put
a
point
over
the
figure
standing
in
the
third
place
,
(
or
place
of
Hundreds
)
which
is
3
,
and
the
preparative
work
is
done
,
as
you
may
see
in
the
Margent
.
Now
if
there
had
been
more
places
in
the
given
number
,
then
I
must
have
put
a
point
over
the
Figure
standing
in
the
fifth
place
,
and
another
over
that
in
the
seventh
,
&c.
And
here
note
,
that
as
many
points
as
you
put
over
the
given
square
Number
,
so
many
Figures
there
will
be
in
the
Root
,
that
is
,
the
Root
will
consist
of
so
many
places
.
So
if
there
were
given
the
number
33016516
,
to
have
its
square
Root
extracted
,
after
I
have
pointed
it
according
to
the
Direction
before
given
,
it
will
stand
as
in
the
Margent
,
and
because
the
points
that
are
put
over
it
are
in
number
4
,
I
conclude
the
Root
it self
will
consist
of
4
places
,
or
figures
.
IX
.
When
you
have
thus
prepared
your
number
,
then
draw
a
crooked-line
on
the
Right-hand
of
your
number
,
behind
which
to
place
your
Root
,
as
you
do
for
a
Quotient
in
Division
.
Note
that
when
your
number
is
prepared
for
Operation
,
as
in
the
8
Rule
,
the
numbers
contained
between
point
and
point
,
may
not
unfitly
be
termed
Squares
,
and
in
the
ensuing
work
,
we
shall
so
call
them
,
as
in
the
foresaid
number
33016516
,
being
pointed
as
before
,
I
call
33
the
first
square
,
●…01
,
the
second
,
65
,
the
third
,
and
16
the
fourth
,
and
last
square
;
every
square
(
except
sometimes
the
first
consisting
of
two
figures
,
or
places
,
the
last
of
which
toward
the
Right-hand
hath
always
a
point
over
it
,
and
if
it
so
happen
(
as
it
often
doth
)
that
the
last
figure
(
in
any
given
square
number
)
towards
the
left-hand
hath
a
point
over
it
then
that
number
alone
shall
be
accounted
the
first
square
.
As
if
the
number
676
,
were
given
,
when
it
is
pointed
for
the
work
according
to
Direction
,
and
as
you
see
in
the
Margent
,
I
account
6
for
the
first
square
,
and
76
,
for
the
second
.
These
things
being
understood
,
we
shall
lay
down
those
general
Rules
requisite
for
the
management
of
the
work
it self
.
X.
When
your
number
is
prepared
,
find
out
the
square
Root
of
the
first
square
,
according
to
the
5
Rule
foregoing
,
and
place
that
Root
behind
the
said
crooked
line
:
as
Let
it
be
Required
to
extract
the
square
Root
of
the
said
number
2304
,
here
the
first
square
number
is
23
,
and
(
according
to
the
said
5
Rule
)
its
Root
is
4
,
which
I
place
behind
the
crooked
line
as
you
see
in
the
Margent
.
XI
.
Then
square
the
said
Root
,
and
place
its
square
which
is
16
under
the
said
first
square
23
,
and
having
drawn
a
line
underneath
,
subtract
the
said
square
16
,
from
23
,
and
place
the
Remainder
,
which
is
7
underneath
the
said
line
as
you
may
perceive
by
the
work
in
the
Margent
:
XII
.
Then
to
the
said
Remainder
bring
down
the
Figures
of
the
next
square
and
annex
them
thereto
on
the
Right-hand
,
so
that
they
may
make
one
intire
number
,
which
(
for
distinctions
sake
)
we
shall
call
the
Resolvend
.
As
in
this
Example
to
the
Remainder
7
,
I
bring
down
the
next
square
04
,
and
annex
it
thereto
,
and
it
maketh
704
for
a
Resolvend
,
as
you
may
see
in
the
Margent
.
XIII
.
Always
let
the
whole
Resolvend
(
except
the
last
figure
on
the
Right
hand
)
be
esteemed
a
Dividend
,
on
the
left
hand
of
which
draw
a
Crooked
line
before
which
to
place
a
Divisor
,
as
in
Division
.
So
in
this
example
,
the
Resolvend
704
is
to
be
made
a
Dividend
,
all
but
the
last
place
which
is
4
,
so
that
the
Dividend
is
70
,
before
which
I
draw
a
crooked
line
,
as
you
see
in
the
Margent
.
XIV
.
Let
the
Quotient
expressing
the
Root
(
or
part
of
the
Root
sought
)
be
doubled
,
or
multiplyed
by
2
,
and
that
double
or
product
shall
be
a
Divisor
,
and
must
be
placed
on
the
left
hand
of
the
Resolvend
,
before
the
said
crooked
line
.
So
in
our
Example
,
the
number
4
which
was
put
for
part
of
the
Root
being
doubled
makes
8
,
which
I
put
before
the
Resolvend
for
a
Divisor
,
as
it
appears
in
the
Margent
.
XV.
Then
(
according
to
the
Rule
of
Division
in
whole
numbers
)
seek
how
often
the
said
Divisor
is
contained
in
the
said
Dividend
,
and
put
the
answer
down
in
the
Quotient
,
and
also
on
the
Right
hand
of
the
Divisor
.
As
in
our
Example
I
seek
how
often
the
Divisor
8
is
contained
in
the
Dividend
70
,
which
I
find
to
be
8
times
,
therefore
I
put
8
in
the
Quotient
for
part
of
the
Root
,
and
also
on
the
Right
hand
of
the
Divisor
.
See
the
work
in
the
Margent
.
XVI
.
Then
by
the
figure
last
put
for
part
of
the
Root
,
multiply
the
said
Divisor
,
together
with
the
figure
that
you
annexed
to
it
(
accounting
them
both
as
one
intire
number
)
and
place
the
product
underneath
the
said
Resolvend
,
drawing
a
line
under
it
,
and
then
subtract
it
out
of
the
said
Resolvend
,
placing
the
Remainder
beneath
the
line
.
As
in
our
Example
,
having
placed
8
in
the
Quotient
,
and
also
on
the
Right-hand
of
the
Divisor
,
then
in
the
place
of
the
Divisor
,
there
stands
88
,
which
I
multiply
by
8
,
the
number
last
put
in
the
Quotient
,
and
the
product
is
704
,
which
I
place
in
order
under
the
Resolvend
704
,
and
having
drawn
a
line
underneath
,
I
subtract
the
said
product
704
,
from
the
Resolvend
704
,
and
there
remaineth
〈◊〉
,
so
is
the
work
finished
,
and
I
find
the
square
Root
of
2304
to
be
48.
See
the
work
in
the
Margent
.
Here
note
that
if
at
any
time
when
you
have
multiplyed
the
number
standing
in
the
place
of
the
Divisor
,
by
the
figure
last
placed
in
the
Quotient
,
or
Root
(
as
is
directed
1.
Note
.
in
the
last
Rule
)
if
the
product
be
greater
than
the
Resolvend
,
then
conclude
the
worke
to
be
erroneous
,
to
correct
which
put
a
lesser
figure
in
the
Root
,
and
proceed
as
is
before
directed
.
Note
also
that
the
work
of
the
12
,
13
,
14
,
15
,
and
16
,
Rules
must
be
repeated
as
often
as
there
are
points
over
the
figures
,
except
for
the
first
square
,
which
is
to
2.
Note
.
be
wrought
according
to
the
Directions
given
in
the
10
and
11
Rules
foregoing
,
and
the
work
of
those
two
Rules
is
to
be
observed
,
but
once
in
the
extraction
of
a
square
Root
,
tho
▪
it
consist
of
never
so
many
squares
or
points
.
These
things
will
appear
plain
and
easie
in
the
working
of
one
or
two
more
Examples
.
Example
2.
Let
it
be
Required
to
extract
the
square
Root
of
33016516.
Here
in
order
to
the
work
,
I
first
prepare
my
number
by
distinguishing
it
into
squares
,
by
pointing
it
according
to
the
8
Rule
foregoing
and
thereby
I
find
that
33
,
is
the
first
square
,
and
(
according
to
the
10
Rule
)
I
take
the
square
Root
of
33
,
which
is
5
,
and
place
it
for
the
first
figure
of
the
Root
,
then
(
according
to
the
eleventh
Rule
)
I
square
the
Root
(5)
and
it
makes
25
,
which
I
place
under
the
said
first
square
number
33
,
and
subtract
it
therefrom
,
and
the
remainder
(8)
I
place
below
the
line
,
as
in
the
following
work
.
Then
(
according
to
the
twelfth
Rule
)
I
annex
to
the
said
remainder
(8)
the
next
square
(
01
)
and
it
makes
801
for
a
Resolvend
,
then
must
80
(
according
to
the
thirteenth
Rule
)
be
my
dividend
,
and
(
according
to
the
fourteenth
Rule
)
I
double
the
number
(5)
in
the
Root
,
and
it
makes
10
for
a
Divisor
,
and
thereby
I
divide
the
said
dividend
(
80
)
and
I
find
that
it
Quotes
7
,
which
(
according
to
the
fifteenth
Rule
)
I
put
in
the
place
of
the
Root
after
5
,
and
likewise
before
the
Divisor
(
10
)
so
that
in
the
place
of
the
Divisor
instead
of
10
,
there
is
now
107.
Then
(
according
to
the
sixteenth
Rule
)
I
multiply
the
said
107
,
by
7
,
(
the
figure
last
placed
in
the
Root
)
and
the
product
is
749
,
which
I
place
orderly
under
the
said
Resolvend
,
and
subtract
it
therefrom
,
and
the
remainder
is
52
which
I
put
below
the
line
,
as
in
the
following
work
.
Then
I
repeat
the
same
work
over
again
,
in
finding
the
next
figure
of
the
Root
,
as
I
did
in
finding
the
last
,
viz.
to
the
remainder
(
52
)
(
according
to
the
twelfth
Rule
)
I
bring
down
,
and
thereto
annex
the
next
(
third
)
square
(
65
)
and
it
makes
5265
for
a
new
resolvend
,
then
(
according
to
the
thirteenth
Rule
)
is
526
a
new
dividend
,
and
(
according
to
the
fourteenth
Rule
)
I
take
the
Root
(
57
)
and
double
it
for
,
a
new
Divisor
,
and
it
makes
114
,
which
I
place
before
the
resolvend
(
5265.
)
Then
(
according
to
the
fifteenth
Rule
)
I
seek
how
often
the
divisor
(
114
)
is
contained
in
the
dividend
,
(
526
)
and
I
find
it
will
bear
4
,
which
I
place
in
the
Root
orderly
,
and
also
on
the
right
hand
of
the
divisor
,
(
114
)
and
then
there
will
be
in
the
place
of
the
divisor
,
the
number
1144
,
which
(
according
to
the
sixteenth
Rule
)
I
multiply
by
the
figure
(4)
last
put
in
the
Root
,
and
the
product
is
4576
,
which
I
place
orderly
under
the
resolvend
(
5265
)
and
subtract
it
therefrom
,
and
the
remainder
is
689
which
I
place
under
the
line
,
as
is
before
directed
.
See
the
whole
work
as
followeth
.
Then
I
again
repeat
the
work
of
the
12
,
13
,
14
,
15
,
and
16
Rules
of
this
Chapter
for
finding
the
next
figure
of
the
Root
,
viz.
first
I
bring
down
(
16
)
the
next
square
number
,
and
annex
it
to
the
remainder
689
,
(
according
to
the
twelfth
Rule
)
and
it
makes
68916
,
for
a
new
resolvend
,
of
which
(
by
the
thirteenth
Rule
)
6891
is
a
new
Dividend
then
(
according
to
the
fourteenth
Rule
)
I
double
the
Root
,
and
it
makes
1148
for
a
divisor
,
which
I
place
on
the
left
side
the
resolvend
,
and
then
seek
how
often
it
is
contained
in
the
said
dividend
(
6891
)
and
the
answer
is
6
,
which
I
place
for
part
of
the
Root
(
in
order
)
and
also
on
the
right
hand
of
the
said
Divisor
,
so
that
in
the
place
of
the
divisor
1148
,
will
then
stand
the
number
11486
,
which
(
by
the
sixteenth
Rule
)
I
multiply
by
6
,
(
the
figure
last
placed
in
the
Root
,
and
the
product
is
11486
,
which
I
place
in
order
under
the
resolvend
,
and
subtract
it
therefrom
,
and
the
remainder
is
o
,
and
so
the
work
is
finished
,
whereby
I
find
the
square
Root
of
33016516
,
to
be
5746
,
as
by
the
whole
operation
appeareth
.
And
if
the
Root
had
consisted
of
never
so
many
places
,
yet
for
every
figure
put
therein
(
except
the
first
,
for
which
you
are
to
observe
the
tenth
and
eleventh
Rules
)
the
work
of
the
12
,
13
,
14
,
15
,
and
16
Rules
must
be
repeated
according
to
the
second
note
after
the
sixteenth
Rule
foregoing
.
Example
.
3.
A
third
Example
may
be
this
,
let
it
be
required
to
extract
the
square
Root
of
8328996
,
In
the
working
of
this
Example
you
will
see
the
use
of
the
first
note
upon
the
sixteenth
Rule
,
for
only
the
number
8
is
the
first
square
,
as
you
may
see
by
the
pointing
of
the
given
number
,
and
after
the
whole
work
of
Extraction
is
finished
,
you
will
find
the
square
Root
of
the
given
number
,
to
be
2886
,
as
in
the
following
operation
.
XVII
.
When
there
is
given
a
number
that
is
●…ot
a
square
number
,
that
is
,
whose
root
cannot
●…e
exactly
found
,
and
you
are
desirous
to
find
●…he
Fractional
part
of
the
root
as
near
as
may
be
,
you
are
to
observe
the
eighth
rule
in
preparing
your
number
for
extraction
,
and
then
to
annex
thereto
an
even
number
of
Cyphers
at
pleasure
,
and
note
,
that
as
many
pairs
of
Cyphers
as
you
annex
thereto
,
so
many
Decimals
will
there
be
in
the
root
expressed
,
(
which
though
it
come
not
to
be
the
exact
root
,
yet
will
it
come
so
near
the
truth
,
that
if
the
last
Decimal
figure
placed
in
the
root
,
be
increased
by
an
unite
,
it
will
be
too
much
)
and
as
many
points
as
there
are
over
the
given
Integral
square
number
,
so
many
places
will
there
always
be
in
the
integral
part
of
the
Root
,
as
in
the
following
Example
,
where
it
is
required
to
extract
the
square
root
of
129596.
First
I
proceed
to
the
work
of
extraction
according
to
the
former
rules
as
if
it
were
an
exact
square
number
,
and
find
the
integral
root
to
be
359
,
as
followeth
.
But
because
(
when
the
work
is
finished
)
there
is
a
remainder
of
715
,
I
annex
a
competent
eve●…
number
of
Cyphers
,
to
the
given
number
,
as
〈◊〉
4
,
or
6
,
or
8
,
and
point
them
out
in
the
sa●…
manner
as
if
they
were
significant
figures
in
an
integer
,
then
bring
two
of
them
down
to
the
said
remainder
(
715
)
and
annex
them
thereto
,
so
have
you
71500
for
a
new
resolvend
;
Then
find
out
a
new
divisor
by
doubling
the
root
,
as
is
before
directed
,
and
proceed
as
if
the
annexed
Cyphers
were
significant
figures
,
or
whole
numbers
,
as
far
as
you
please
,
as
in
this
example
,
where
the
work
is
carried
on
till
there
are
3
decimal
figures
in
the
root
;
and
the
work
being
finished
,
I
find
the
root
to
be
359.994
,
and
there
is
a
remainder
of
319964.
See
the
work
.
But
if
you
proceed
to
put
another
decimal
in
the
root
you
will
find
it
to
be
359.9944
,
and
the
remainder
will
be
3196864.
Now
you
may
perceive
that
the
said
root
is
too
little
,
because
there
is
a
remainder
,
but
yet
it
is
so
near
the
truth
that
if
the
last
figure
thereof
were
increased
by
an
unite
,
and
so
made
359.9945
it
would
then
be
too
much
,
as
you
may
prove
at
your
leisure
.
XVIII
.
The
Square
root
of
a
vulgar
Fraction
that
is
commensurable
to
its
root
,
is
thus
found
,
viz.
extract
the
square
root
of
the
Numerator
,
for
a
new
Numerator
,
and
likewise
the
square
root
of
the
Denominator
,
for
a
new
Denominator
;
so
shall
that
new
Fraction
be
the
square
root
of
the
given
Fraction
;
as
for
Example
Let
it
be
required
to
extract
the
square
root
of
25/36
,
first
I
take
the
square
root
of
25
,
which
is
5
,
and
place
it
for
a
new
Numerator
,
then
I
take
the
square
root
of
the
denominator
36
,
which
is
6
,
and
place
it
for
a
new
Denominator
,
so
is
5/6
the
square
root
of
25/36
,
which
was
required
;
In
like
manner
if
4/9
were
given
to
have
its
square
root
extracted
,
its
root
would
be
found
to
be
2/3
,
and
¾
is
the
square
root
of
9/16
,
the
like
is
to
be
observed
for
any
other
.
But
here
note
diligently
;
before
you
proceed
to
extract
the
Square
root
of
any
Note
Fraction
,
that
you
reduce
it
to
its
lowest
Termes
,
for
it
may
happen
that
in
its
given
Termes
,
it
may
be
incommensurable
to
its
root
,
but
being
reduced
to
its
lowest
Termes
it
may
be
commensurable
,
and
its
root
exactly
found
out
,
so
〈◊〉
,
is
incommensurable
to
its
root
,
but
being
reduced
to
25/36
,
its
square
root
will
be
found
to
be
5/6
as
before
.
XIX
.
The
square
root
of
a
mixt
number
that
is
commensurable
to
its
root
is
thus
found
out
,
viz.
reduce
the
mixt
number
to
an
improper
Fraction
,
and
then
extract
the
square
root
of
the
Numerator
,
and
the
square
root
of
the
Denominator
,
for
a
new
Numerator
,
and
a
new
Denominator
,
as
in
the
last
Rule
.
So
if
it
were
required
to
extract
the
square
root
of
1
11/25
,
first
I
reduce
it
to
an
improper
Fraction
,
and
it
is
36/25
,
whose
square
root
is
〈◊〉
=
1
1/5
,
so
if
it
were
required
to
extract
the
square
root
of
3
13/81
,
first
I
reduce
the
given
,
mixt
number
,
to
the
improper
Fraction
256/81
,
and
then
extract
the
square
root
of
the
Numerator
256
,
and
it
I
find
to
be
16
,
for
a
new
Numerator
,
and
●…ikewise
the
square
root
of
81
,
the
denominator
,
which
I
find
to
be
9
,
for
a
new
denominator
,
so
●…s
16/9
=
1
7/9
the
square
root
of
the
given
mixt
number
3
13/81
,
which
was
required
.
XX.
When
you
are
to
extract
the
square
root
of
a
Fraction
that
is
incommensurable
to
its
●…oot
,
prefix
before
the
given
fraction
,
this
Cha●…acter
√
,
or
√
q
signifying
the
square
root
of
●…hat
before
which
it
is
prefixed
,
so
the
square
●…oot
of
24/29
is
thus
expressed
,
or
,
the
●…ke
of
any
other
.
But
if
you
would
know
as
●…ear
as
may
be
the
square
root
of
any
such
fracti●…n
,
reduce
it
to
a
decimal
of
the
same
value
by
●…he
first
Rule
of
the
second
Chapter
,
but
let
the
●…ecimal
confist
of
an
even
number
of
places
,
viz.
●…ther
of
two
,
four
,
six
,
or
eight
,
&c.
places
;
and
the
more
places
it
consisteth
of
,
so
much
the
nearer
the
truth
will
the
root
be
;
Then
extract
the
square
root
of
that
decimal
(
according
to
the
Rules
before
delivered
,
)
in
every
respect
as
if
it
were
a
whole
number
,
so
shall
this
root
so
found
be
very
near
the
true
root
;
and
so
near
,
that
if
it
consist
of
3
places
it
shall
not
want
1/1000
part
of
an
unit
of
the
true
root
;
and
if
of
4
places
,
it
shall
not
want
1/10000
part
of
an
unit
of
the
truth
.
So
if
I
would
extract
the
square
root
of
〈◊〉
first
I
reduce
it
to
a
decimal
,
which
I
find
to
be
●…75
and
because
I
would
have
the
root
to
consist
of
4
places
,
I
annex
6
Cyphers
thereto
and
it
makes
.75000000
,
then
extracting
the
square
root
thereof
as
if
it
were
a
whole
number
,
I
find
it
to
be
.8660
,
and
there
is
a
remainder
of
4400
,
but
if
I
would
have
it
consist
of
5
places
,
then
I
annex
2
more
Cyphers
to
the
said
remainder
,
and
make
it
440000
,
and
proceed
,
and
then
I
find
the
root
to
be
.86602
,
and
the
remainder
to
be
93596.
XXI
.
In
like
manner
if
it
were
required
to
extract
the
square
root
of
a
mixt
number
incommensurable
to
its
root
,
as
near
as
may
be
,
first
reduce
the
Fractional
part
to
a
decimal
,
but
let
it
consist
of
an
even
number
of
places
,
viz.
of
2
,
4
,
6
,
or
8
&c.
places
,
then
proceed
to
extract
its
square
root
,
according
to
the
Rules
formerly
delivered
in
this
Chapter
,
in
every
respect
,
〈◊〉
if
it
were
a
whole
number
,
so
shall
the
root
so
found
,
be
very
near
the
truth
,
and
the
more
places
it
consisteth
of
,
so
much
the
nearer
will
〈◊〉
be
to
the
true
roo●…
.
And
note
that
in
the
root
there
will
be
so
many
decimal
places
,
as
yo●…
placed
points
over
the
Decimal
part
of
the
square
number
.
So
if
it
were
required
to
extract
the
square
root
of
28
6/13
,
first
I
reduce
the
fractional
part
6/13
to
a
decimal
,
and
it
makes
.461538
,
so
then
the
mixt
number
whose
square
root
I
am
to
extract
is
28.461538
,
which
being
pointed
and
the
work
of
extraction
finished
,
according
to
the
former
Rules
,
I
find
its
square
root
to
be
very
near
5.334
and
there
is
a
remainder
of
9982
,
But
if
I
had
proceeded
yet
farther
,
and
made
the
decimal
part
to
have
consisted
of
7
places
,
it
would
have
had
for
its
square
root
5.3349
,
which
doth
not
want
1/10000
part
of
an
unite
of
the
true
root
.
But
if
you
would
not
extract
the
square
root
of
such
a
mixt
number
,
then
prefix
before
it
this
character
,
√
or
√
q
so
if
the
said
mixt
number
28
〈◊〉
were
given
I
would
express
its
square
root
thus
,
viz.
or
the
like
is
to
be
understood
of
any
other
.
XXII
.
When
you
are
to
extract
the
square
root
of
a
decimal
Fraction
,
which
hath
2
or
3
Cyphers
possessing
the
two
or
three
first
places
on
the
left
hand
of
the
given
decimal
,
then
cut
off
2
of
them
with
a
dash
of
the
pen
,
and
put
a
Cypher
to
possess
the
first
place
of
the
root
,
and
proceed
to
extract
the
square
root
of
the
remaining
figures
,
according
to
the
former
Rules
as
if
there
had
been
no
such
Cyphers
before
the
given
Decimal
;
and
if
the
given
decimal
have
4
Cyphers
before
it
,
cut
them
off
with
a
dash
of
the
pen
,
and
put
2
Cyphers
in
the
root
,
and
then
proceed
as
before
.
So
if
it
were
required
to
extract
the
square
root
of
5/846
,
first
I
reduce
it
to
a
decimal
Fraction
and
it
makes
.005910
,
Then
I
cut
off
the
two
first
Cyphers
,
and
place
one
Cypher
in
the
root
,
then
I
proceed
to
extract
the
square
root
of
the
remaining
figures
viz.
5910
,
as
if
there
had
been
no
such
Cyphers
before
them
,
and
I
find
the
root
to
be
very
near
.077
as
you
may
try
at
your
leisure
.
XXIII
.
The
operation
in
the
extraction
of
the
square
root
is
thus
proved
;
viz.
multiply
the
root
into
it self
,
and
(
if
there
be
no
remainder
after
the
work
The
proof
of
the
extraction
of
the
Square
Root
.
of
extraction
is
finished
)
the
product
(
if
the
work
be
truly
done
)
will
be
equal
to
the
number
first
given
.
As
in
the
first
Example
,
where
it
is
required
to
extract
the
square
root
of
2304
,
which
is
there
found
to
be
48
,
Now
if
I
multiply
48
by
it self
,
it
produceth
2304
,
which
is
the
given
number
,
and
therefore
I
conclude
the
operation
to
be
true
.
But
if
after
the
work
of
extraction
is
finished
,
there
is
any
Remainder
,
then
,
when
you
have
multiplied
the
root
by
it self
,
to
the
product
add
the
said
Remainder
,
and
if
the
sum
be
equal
to
the
given
number
,
the
operation
is
right
,
otherwise
not
.
As
in
the
Example
of
the
seventeenth
Rule
,
where
it
is
required
to
extract
the
square
root
of
129596
,
and
is
there
found
to
be
●…59
.
994
,
and
the
remainder
is
319964
;
Now
to
prove
the
work
,
I
multiply
the
root
(
359.994
)
by
it self
,
and
it
produceth
129595.680036
,
which
should
be
129596
,
therefore
to
the
said
product
I
add
the
said
Remainder
(
319964
,
)
and
the
sum
is
129596
,
and
therefore
I
conclude
the
work
to
be
truly
wrought
.
CHAP.
X.
The
Extraction
of
the
Cube
Root
.
I.
A
Cube
Number
is
that
which
is
produced
by
multiplying
any
number
into
it self
and
again
into
that
product
,
which
said
given
number
is
called
the
Cube
Root
.
As
,
Suppose
5
were
given
to
find
its
Cube
,
first
I
multiply
5
into
it self
,
and
it
produceth
25
,
which
is
called
the
Square
of
5
,
then
I
again
multiply
25
,
(
the
said
square
)
by
5
,
and
it
produceth
125
,
which
is
called
the
Cube
of
5.
And
here
note
that
as
125
is
called
the
Cube
of
5
,
so
is
5
called
the
Cube
Root
of
125.
II.
The
extraction
of
the
Cube
Root
is
nothing
else
then
when
by
having
a
Cube
number
given
,
we
find
out
its
Cube
root
,
which
said
Cube
number
given
is
alwayes
supposed
to
be
a
certain
number
of
little
Cubes
,
comprehended
within
one
intire
great
Cube
,
which
said
Cube
may
very
well
be
represented
by
a
dye
,
or
any
other
solid
body
,
having
its
length
,
breadth
and
depth
equal
;
This
being
supposed
,
let
there
be
laid
9
Dyes
constituting
a
square
,
whose
side
shall
be
3
,
and
upon
them
let
there
be
laid
9
more
Dyes
,
and
upon
them
let
there
be
laid
9
more
,
then
will
there
be
in
all
27
Dyes
,
which
will
constitute
one
greater
Cube
,
whose
length
,
breadth
and
depth
will
be
3
Dyes
,
and
this
greater
Cube
comprehendeth
27
lesser
Cubes
,
Now
the
extraction
of
the
Cube
root
is
by
having
the
number
of
little
Cubes
(
27
)
comprehended
in
the
greater
given
Cube
,
to
find
out
how
many
of
the
lesser
Cubes
make
up
the
side
of
the
greater
.
III.
A
Cube
number
is
either
Simple
or
Compound
.
IV.
A
Simple
Cube
number
is
that
which
hath
for
its
root
or
side
,
one
of
the
9
Digits
,
and
it
is
therefore
always
lesser
than
1000
;
so
shall
you
find
that
343
is
a
Simple
Cube
number
,
whose
side
or
root
is
7
,
for
7×7×7
=
343
,
All
which
said
simple
Cubes
,
and
Squares
,
as
also
their
Roots
are
expressed
in
the
Tablet
following
Roots
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
Squares
|
1
|
4
|
9
|
16
|
25
|
36
|
49
|
64
|
8
|
Cubes
|
1
|
8
|
27
|
64
|
125
|
216
|
343
|
512
|
7●…9
|
V.
A
compound
Cube
number
is
that
which
is
produced
by
the
multiplication
of
a
number
consisting
of
two
places
(
at
the
least
)
3
times
into
it self
continually
,
and
is
therefore
never
less
then
1000
,
so
1728
is
a
compound
Cube
number
,
produced
by
the
multiplication
of
12
into
its
self
3
times
,
for
12×12×12
=
1728.
VI.
When
a
compound
Cube
Number
is
given
to
have
its
Cube
root
extracted
,
before
you
can
go
about
it
,
you
must
prepare
it
for
the
work
by
pointing
it
;
which
is
thus
done
,
viz.
put
a
point
over
the
first
figure
towards
the
right
hand
,
viz.
over
the
place
of
Units
,
then
(
passing
the
two
next
places
)
put
a
point
over
the
fourth
figure
,
or
place
of
Thousands
,
and
so
proceed
by
putting
a
point
over
every
third
figure
,
as
you
did
over
every
second
figure
in
the
extraction
of
the
Square
root
,
till
you
have
finished
your
pointing
,
That
being
done
,
on
the
right
hand
of
the
said
Cube
number
draw
a
crooked
line
,
behind
which
to
place
its
Cube
root
,
as
you
do
to
place
the
Quotient
in
Division
,
as
in
the
following
Example
.
Example
.
1.
Let
it
be
Required
to
extract
the
Cube
Root
of
262144.
In
order
to
prepare
this
Cube
Number
,
for
the
Extraction
of
its
Cube
Root
,
I
first
put
a
point
over
the
first
figure
(4)
towards
the
Right
hand
,
and
then
overpassing
the
two
next
figures
(
14
)
I
put
another
point
over
the
fourth
figure
(2)
and
then
is
the
given
number
distributed
into
several
parts
not
unfitly
called
Cubes
,
viz.
262
(
as
far
as
the
first
point
goeth
)
is
the
first
Cube
,
and
144
(
from
thence
to
the
second
point
)
is
the
second
Cube
,
and
then
I
draw
a
crooked
line
behind
it
as
you
see
in
the
Margent
.
VII
.
Having
proceeded
thus
far
,
find
out
the
Cube
Root
of
the
first
Cube
(
262
)
but
because
it
is
not
an
exact
Cube
number
,
take
the
Cube
root
of
that
number
in
the
foregoing
Tablet
,
which
being
lesser
than
it
is
,
yet
is
nearest
to
it
,
(
which
I
here
find
to
be
6
,
)
and
place
it
behind
the
crooked
line
for
the
first
figure
in
the
Root
,
as
you
see
in
the
following
work
VIII
.
This
being
done
,
Cube
the
said
number
which
is
placed
in
the
Root
,
and
subscribe
its
Cube
under
the
first
cube
of
the
given
number
.
So
in
this
Example
216
being
the
cube
of
6
,
I
place
it
under
262
the
first
cube
of
the
given
number
262144
,
as
followeth
IX
.
Draw
a
line
under
the
Cube
thus
subscribed
,
and
subtract
it
from
the
first
cube
of
the
given
number
,
placing
the
remainder
orderly
underneath
the
said
line
.
So
216
(
the
cube
of
6
)
being
subtracted
from
262
,
the
remainder
is
46
,
which
I
place
underneath
the
line
as
followeth
X.
Bring
down
the
next
cube
number
and
annex
it
to
the
said
remainder
on
the
right
hand
thereof
.
So
144
being
the
next
cube
,
I
bring
it
down
and
annex
it
to
the
remainder
46
,
and
it
makes
46144
,
which
by
Artists
is
usually
called
the
Resolvend
.
XI
.
Draw
a
line
underneath
the
Resolvend
,
Then
Triple
the
Root
,
that
is
,
multiply
it
by
3
,
and
place
its
Triple
under
the
Resolvend
in
such
order
,
that
the
place
of
units
in
the
said
triple
may
stand
under
the
place
of
tens
in
the
Resolvend
.
So
the
triple
of
6
,
is
18
,
which
I
place
under
the
Resolvend
so
,
that
8
(
the
place
of
unites
in
the
said
triple
)
may
stand
under
4
in
the
place
of
tens
of
the
Resolvend
,
as
you
see
following
.
XII
.
Square
the
said
root
,
and
then
triple
the
said
square
of
the
Root
,
and
place
the
said
triple
square
under
the
said
triple
Root
in
such
order
that
the
place
of
unites
in
the
triple
square
of
the
Root
,
may
stand
underneath
the
place
of
Tens
in
the
triple
Root
,
so
in
this
Example
,
the
square
of
the
Root
6
,
is
36
,
and
the
triple
thereof
is
108
,
which
I
place
under
18
,
the
triple
Root
so
,
that
8
the
place
of
unites
in
the
said
triple
square
of
the
Root
,
may
stand
under
1
,
the
place
of
Tens
in
18
,
the
said
triple
Root
,
as
followeth
XIII
.
Draw
a
line
underneath
the
said
triple
Root
,
and
triple
square
of
the
Root
,
as
they
are
placed
,
and
add
them
together
in
the
same
order
as
they
stand
,
so
shall
their
sum
be
a
Divisor
.
So
in
our
Example
,
a
line
being
drawn
under
18
and
108
,
and
they
added
together
in
the
same
order
as
they
stand
,
their
sum
is
1098
for
a
Divisor
,
as
in
the
following
work
.
XIV
.
Draw
a
crooked
line
on
the
left
hand
of
the
Resolvend
,
before
which
to
place
the
said
Divisor
,
and
let
the
whole
Resolvend
(
except
the
place
of
unites
therein
)
be
esteemed
a
Dividend
,
then
seek
how
often
the
said
Divisor
is
contained
in
the
Dividend
,
and
put
the
answer
for
the
next
figure
in
the
Root
.
So
in
our
Example
,
seek
how
often
1098
the
divisor
is
contained
in
4614
the
Dividend
(
observing
here
the
usual
Rules
of
Division
)
and
the
answer
I
find
to
be
4
which
I
place
for
the
next
figure
in
the
Root
,
as
in
the
Example
.
XV.
Draw
a
line
underneath
the
whole
work
,
and
then
Cube
the
figure
last
placed
in
the
Root
,
and
place
its
cube
underneath
the
Resolvend
in
such
sort
that
the
place
of
unites
of
the
one
may
stand
under
the
place
of
units
in
the
other
;
so
in
our
example
64
being
the
cube
of
4
(
the
figure
last
placed
in
the
Root
)
I
place
it
under
the
Resolvend
in
such
manner
that
the
figure
4
in
the
place
of
unites
of
the
cube
64
,
may
stand
under
4
,
the
place
of
unites
in
the
Resolvend
,
and
then
the
work
will
stand
as
followeth
XVI
.
Square
the
figure
last
placed
in
the
Root
,
and
multiply
its
square
by
the
triple
Root
subscribed
underneath
the
Resolvend
,
(
as
is
directed
in
the
eleventh
Rule
of
this
Chapter
)
and
subscribe
the
product
under
the
Cube
last
put
down
,
in
such
order
,
that
the
place
of
Units
in
the
said
product
,
may
stand
under
the
place
of
Tens
,
in
the
said
Cube
.
So
in
our
Example
,
the
figure
last
placed
in
the
Root
is
4
,
which
squared
is
16
,
and
16
multiplyed
by
18
(
the
triple
Root
before
set
down
)
the
product
is
288
,
which
I
place
under
64
(
the
cube
of
4
)
in
such
sort
that
8
(
in
the
place
of
Units
of
the
said
product
)
may
stand
under
6
(
the
place
of
Tens
)
in
the
said
cube
of
4
;
view
the
work
.
XVII
.
Multiply
the
triple
square
of
the
Root
,
(
subscribed
as
is
before
directed
in
the
twelfth
Rule
of
this
Chapter
)
by
the
figure
last
placed
in
the
Root
,
and
place
the
product
under
the
number
last
subscribed
,
(
which
is
the
product
of
the
square
of
the
figure
last
placed
in
the
Root
multiplied
by
the
said
Triple
Root
)
in
such
manner
that
the
place
of
Units
of
this
,
may
stand
under
the
place
of
Tens
in
that
;
As
in
this
Example
,
The
Triple
square
of
the
Root
is
108
,
which
multiplyed
by
4
(
the
figure
last
placed
in
the
Root
)
the
product
is
432
,
which
I
place
under
288
(
the
number
last
subscribed
)
in
such
order
that
the
figure
2
(
in
the
place
of
Units
of
the
said
last
product
)
may
stand
under
8
,
which
is
in
the
place
of
Tens
,
of
the
said
number
last
subscribed
;
and
then
the
work
will
stand
as
followeth
.
XVIII
.
Draw
another
line
under
the
work
,
and
●…dd
the
3
numbers
together
,
that
were
last
placed
●…nder
the
Divisor
,
in
the
same
order
as
they
there
●…tand
,
and
let
their
sum
be
called
the
Subtra●…end
,
which
let
be
subtracted
out
of
the
Resol●…end
,
noting
the
Remainder
;
So
in
this
Exam●…le
I
add
64
,
288
,
432
together
in
the
same
or●…er
as
they
stand
,
and
their
sum
is
46144
,
for
a
●…ubtrahend
,
which
I
subtract
out
of
46144
the
●…esolvend
,
and
there
is
nothing
remaineth
,
so
●…e
whole
work
is
finished
;
and
I
find
the
Cube
●…oot
of
262144
to
be
64
,
without
any
Remain●…er
;
See
the
whole
work
as
followeth
,
Now
the
Learner
is
to
observe
three
things
in
general
from
the
Rules
before
delivered
,
concerning
the
extraction
of
the
Cube
Root
.
Observe
1.
That
the
work
contained
in
the
7
,
8
and
9
Rules
for
finding
out
the
first
figure
of
the
Root
,
is
not
again
to
be
repeated
,
throughout
the
whole
work
of
Extraction
,
although
the
Root
consist
of
never
so
many
places
,
but
the
work
of
all
the
Rules
following
is
to
be
repeated
as
often
as
a
new
figure
is
put
in
the
root
.
Observe
2.
For
every
particular
Cube
in
the
number
given
,
distinguished
by
the
points
,
(
except
the
first
)
there
is
to
be
found
out
a
new
resolvend
,
by
annexing
the
next
cube
to
the
remainder
(
according
to
the
10th
.
Rule
)
and
as
often
as
there
is
a
resolvend
,
so
often
must
there
be
found
a
new
Divisor
(
by
the
11
,
12
and
13
Rules
)
and
as
often
as
there
is
found
a
new
Divisor
so
often
must
there
be
found
a
new
Subtrahend
(
according
to
the
15
,
16
,
17
,
and
18
Rule
before-going
.
)
Observe
3.
When
the
Subtrahend
chanceth
to
be
greater
than
the
resolvend
,
then
you
may
conclude
there
is
an
error
in
your
work
,
which
must
be
corrected
by
putting
a
lesser
figure
in
the
Root
.
Example
.
2.
Let
it
be
required
to
extract
the
Cube
Root
of
48627125.
Having
prepared
the
given
number
for
the
work
of
extraction
,
according
to
the
6th
.
Rule
of
this
Chapter
,
I
find
it
to
be
distributed
into
3
Cubes
,
viz.
48
,
the
first
,
627
,
the
second
,
and
125
the
third
.
Then
I
proceed
to
the
work
;
and
first
I
find
the
Cube
root
of
48
,
(
the
first
Cube
)
which
is
3
,
then
do
I
cube
3
,
and
place
its
cube
which
is
27
,
under
(
48
)
the
first
cube
,
and
subtract
it
therefrom
and
the
remainder
is
21
,
according
to
the
7
,
8
,
and
9
Rules
of
this
Chapter
,
and
then
will
the
work
stand
as
you
see
in
the
Margent
.
Then
,
to
the
said
remainder
21
,
do
I
bring
down
,
and
thereto
annex
the
next
cube
,
which
is
627
,
and
it
makes
21627
for
a
Resolvend
,
according
to
the
10th
.
Rule
foregoing
.
Then
do
I
find
out
a
Divisor
according
to
the
11
,
12
and
13
Rules
of
this
Chapter
,
and
first
I
triple
the
Root
(3)
and
it
makes
9
,
which
I
place
under
(2)
the
place
of
Tens
in
the
Resolvend
;
Then
do
I
square
the
said
Root
(3)
and
that
makes
9
,
then
do
I
triple
its
square
(9)
and
that
makes
27
,
which
I
place
under
the
said
Triple
in
such
order
as
is
directed
in
the
12th
.
Rule
,
then
drawing
a
line
underneath
the
work
,
I
add
the
two
said
numbers
together
,
(
viz.
the
Triple
Root
,
and
the
Triple
Square
of
the
Root
)
in
such
order
as
they
are
there
placed
,
and
their
sum
is
279
for
a
Divisor
;
as
per
Margent
.
Then
according
to
the
14th
.
Rule
I
seek
how
often
the
said
Divisor
279
is
contained
in
2162
the
Dividend
,
and
I
find
the
answer
to
be
6
,
which
I
place
for
the
second
figure
in
the
Root
,
then
do
I
in
the
next
place
go
about
to
find
out
a
subtrahend
,
and
in
order
thereunto
first
(
according
to
the
fifteenth
Rule
of
this
Chapter
)
I
cube
the
figure
(6)
last
placed
in
the
root
,
and
it
maketh
216
,
which
I
place
under
the
Resolvend
in
such
order
(
as
is
directed
in
the
said
fifteenth
Rule
)
that
the
place
of
Units
of
the
one
may
stand
under
the
place
of
Units
of
the
other
;
Then
(
according
to
the
16th
.
Rule
of
this
Chapter
)
I
square
the
figure
(
viz.
6
)
last
placed
in
the
root
which
makes
36
,
and
then
multiply
it
by
(9)
the
said
triple
root
,
and
the
product
is
324
which
I
place
under
(
216
)
the
said
cube
of
6
,
in
such
order
as
is
directed
in
the
said
16th
.
Rule
.
Then
do
I
multiply
the
said
Triple
square
(
viz.
27
)
by
the
figure
(6)
last
placed
in
the
root
,
and
it
produceth
162
,
which
I
place
under
the
last
product
(
324
)
in
such
manner
as
is
directed
in
the
17th
.
Rule
.
Then
I
add
these
3
several
numbers
together
in
the
same
order
as
they
stand
,
and
their
Sum
is
19656
for
a
Subtrahend
,
which
subtracted
out
of
(
21627
)
the
Resolvend
,
the
remainder
is
1971
,
as
you
may
see
by
the
following
work
.
Then
to
the
said
remainder
(
1971
)
do
I
annex
the
next
cube
number
(
125
)
according
to
the
10th
.
Rule
,
and
it
makes
1971125
for
a
resolvend
;
Then
I
proceed
according
to
the
11
,
12
and
13
Rules
to
find
a
Divisor
,
and
therefore
I
first
triple
the
whole
Quotient
(
36
)
and
it
is
108
,
Then
do
I
square
the
whole
Quote
(
36
)
and
it
makes
1296
,
which
being
tripled
is
3888
,
which
being
orderly
placed
under
the
triple
Quote
,
and
added
thereto
in
that
order
,
the
sum
is
38988
for
a
new
Divisor
:
See
the
work
in
the
Margent
.
Then
do
I
seek
how
often
the
said
Divisor
is
contained
in
the
Dividend
(
197112
)
and
I
find
it
to
be
5
times
contained
therein
,
and
accordingly
I
place
5
in
the
root
,
and
proceed
according
to
the
15
,
16
,
17
and
18
Rules
to
find
out
a
Subtrahend
,
and
therefore
first
,
I
cube
the
number
(5)
last
placed
in
the
root
,
and
it
makes
125
,
then
do
I
square
the
said
5
,
and
it
makes
25
,
by
which
I
multiply
(
108
)
the
said
triple
root
,
and
place
the
product
(
2700
)
under
the
said
cube
,
as
is
before
directed
,
then
do
I
by
the
said
5
,
multiply
(
3888
)
the
triple
square
of
the
root
,
and
the
product
(
19440
)
do
I
place
under
the
former
product
(
2700
)
according
to
former
directions
,
and
add
the
3
Numbers
together
,
in
the
same
order
as
they
stand
,
and
their
sum
is
1971125
,
(
as
appears
per
Margent
)
for
a
Subtrahend
,
which
taken
out
of
the
said
Resolvend
there
remaineth
(0)
and
so
the
work
is
finished
,
and
I
find
the
cube
root
of
the
given
number
48627125
to
be
365
;
view
the
whole
work
laid
down
as
followeth
.
XIX
.
When
it
is
required
to
extract
the
cube
Root
of
a
Number
that
is
incommensurable
to
its
root
,
and
you
are
desirous
to
know
the
Fractional
part
of
the
root
as
near
as
may
be
,
you
are
to
annex
to
the
given
number
,
a
competent
number
of
Cyphers
,
which
number
of
Cyphers
must
be
always
a
multiple
of
3
,
viz.
either
3
,
6
,
9
,
12
,
&c.
Cyphers
,
that
is
000
,
000000
,
or
000000000
,
&c.
And
having
observed
the
6
Rule
for
the
punctation
of
the
given
number
,
likewise
poynt
the
annexed
Cyphers
,
in
the
same
manner
as
if
they
were
significant
figures
,
or
integers
;
and
observe
,
that
as
many
points
as
you
put
over
the
integral
part
,
so
many
places
will
the
integral
part
of
the
root
consist
of
,
and
so
many
points
as
are
put
over
the
Cyphers
,
or
Decimals
,
so
many
decimal
places
will
there
be
in
the
root
,
this
being
observed
the
work
it self
in
the
extracting
the
Cube
root
of
a
Decimal
Fraction
,
or
of
a
mixt
number
of
integers
and
decimals
,
is
the
same
in
every
respect
as
if
the
number
given
were
an
integral
Cube
number
,
according
to
the
Rules
before
delivered
in
this
Chapter
.
As
in
the
following
Example
.
Let
it
be
required
to
extract
the
Cube
root
of
13798
which
is
a
number
incommensurable
to
its
Cube
root
,
and
to
find
out
its
root
as
near
as
may
be
I
annex
to
it
9
Cyphers
(
so
by
that
means
I
shall
have
3
Decimals
in
the
root
)
and
prepare
it
for
Extraction
by
pointing
it
as
is
before
directed
,
and
as
you
see
following
.
And
having
performed
the
work
of
extraction
according
to
the
former
rules
,
I
find
its
Cube
root
to
be
23.984
,
which
(
as
by
the
remainder
you
may
perceive
)
is
somewhat
too
little
,
but
yet
so
near
the
truth
,
that
,
if
the
Decimal
part
were
increased
by
an
unite
and
so
made
23.985
it
would
then
be
too
much
,
and
so
consequently
it
cannot
want
(
as
it
is
)
1/1000
part
of
an
unite
of
the
truth
,
and
if
the
root
had
had
another
figure
placed
in
it
,
it
would
then
have
come
so
near
the
truth
that
it
would
not
have
wanted
1/10000
part
of
an
unite
;
for
your
further
satisfaction
see
the
whole
work
performed
as
followeth
.
XX.
If
at
any
time
it
is
required
to
extract
the
cube
root
of
a
vulgar
Fraction
,
let
such
Fraction
be
first
To
extract
the
Cube
Root
of
a
vulgar
Fraction
.
reduced
to
its
lowest
Terms
;
because
it
may
not
be
commensurable
to
its
root
in
the
given
Terms
,
but
being
reduced
to
its
lowest
Terms
it
may
,
and
having
so
done
to
perform
the
work
,
this
is
The
Rule
.
Extract
the
Cube
root
of
the
Numerator
,
(
by
the
former
Rules
)
and
place
that
for
a
new
Numerator
,
then
extract
the
Cube
root
of
the
Denominator
,
and
place
that
root
for
a
new
Denominator
,
so
shall
this
new
Fraction
be
the
cube
root
of
the
given
Fraction
.
As
for
Example
,
Let
it
be
required
to
extract
the
Cube
root
of
27/〈◊〉
,
first
I
take
the
cube
root
of
27
(
the
Numerator
)
which
is
3
and
place
it
for
a
new
Numerator
,
Then
I
take
the
cube
root
of
64
,
(
the
Denominator
)
which
is
4
,
and
place
it
for
a
new
Denominator
,
so
shall
this
new
Fraction
¾
be
the
cube
root
of
the
given
Fraction
27/64.
In
like
manner
if
there
were
given
108/256
to
have
its
cube
root
extracted
,
I
can
easily
discover
that
there
cannot
be
found
any
cube
root
exactly
either
for
the
Numerator
or
Denominator
,
in
the
Termes
they
are
given
in
,
but
being
reduced
to
their
lowest
Termes
,
they
are
27/64
,
whose
cube
root
is
¾
as
before
.
In
like
manner
the
cube
root
of
●…/1024
will
be
found
to
be
¼
=
&c.
〈◊〉
.
XXI
.
But
when
there
is
given
a
vulgar
Fraction
to
have
its
cube
root
extracted
,
it
being
incommensurable
To
extract
the
Cube
Root
of
a
vulgar
Fraction
that
is
incommensurable
to
its
Root
.
to
its
root
,
you
may
find
its
cube
root
very
near
▪
if
you
reduce
the
given
vulgar
Fraction
to
a
decimal
and
then
extract
the
cube
root
of
that
decimal
(
by
the
Rules
before
delivered
)
in
every
respect
as
if
it
were
a
whole
Number
,
and
then
shall
that
be
a
decimal
cube
root
,
less
than
the
truth
,
yet
so
near
the
truth
that
if
you
add
an
unite
to
the
last
decimal
figure
it
will
then
be
greater
than
the
truth
.
Here
take
notice
by
the
way
that
your
vulgar
fraction
being
reduced
to
a
decimal
in
order
to
have
its
cube
root
extracted
,
Note
its
equivalent
decimal
must
consist
of
such
a
number
of
places
as
may
be
a
multiple
of
3
,
that
is
,
it
must
consist
of
3
,
6
,
9
,
12
,
15
,
&c.
places
,
and
the
more
places
there
is
in
the
decimal
,
the
nearer
the
truth
will
the
root
be
.
Example
Let
it
be
required
to
extract
the
cube
root
of
●…/8
:
In
order
whereunto
I
reduce
it
to
this
Decimal
,
viz.
.625
,
which
because
it
consisteth
but
of
3
places
,
(
and
so
consequently
can
have
but
1
figure
in
its
Root
)
I
increase
to
9
places
by
annexing
6
Cyphers
thereto
thus
.625000000
and
then
the
root
will
consist
of
3
places
,
then
do
I
proceed
to
extract
its
cube
root
,
(
according
to
the
former
Rules
)
and
find
it
to
be
.854
,
&c.
and
there
will
be
a
remainder
of
2164136
as
you
may
prove
at
your
leisure
.
XXII
.
When
your
given
vulgar
Fraction
is
reduced
to
a
Decimal
of
the
same
value
,
and
the
3
,
or
4
first
places
towards
the
left
hand
are
possessed
by
Cyphers
,
then
in
this
case
you
are
to
cut
off
3
of
them
with
a
dash
of
the
pen
,
and
for
them
place
a
Cypher
to
possess
the
first
place
in
the
root
,
and
then
proceed
to
extract
the
cube
root
of
the
remaining
figures
,
according
to
the
former
Rules
,
as
if
there
had
been
no
such
Cyphers
at
all
.
As
for
Example
.
Let
there
be
given
4/8237
to
have
its
cube
root
extracted
;
First
reduce
it
to
a
Decimal
Fraction
by
the
first
Rule
of
the
second
Chapter
of
this
Book
,
and
it
makes
.000485613
,
&c.
now
to
extract
the
cube
root
of
this
Fraction
,
I
first
prepare
it
,
by
pointing
it
in
every
respect
as
if
it
were
a
whole
number
,
then
with
a
dash
of
my
Pen
,
I
cut
off
the
3
first
Cyphers
,
and
put
a
(0)
to
possess
the
first
place
in
the
root
,
then
I
proceed
to
extract
the
cube
root
of
the
remaining
figures
(
485613
)
as
if
there
had
been
no
cyphers
at
all
before
them
;
and
having
finished
the
work
I
find
its
cube
root
to
be
.078
as
by
the
following
work
.
In
like
manner
if
the
decimal
which
is
given
to
have
its
cube
root
extracted
,
have
6
Cyphers
placed
before
the
significant
figures
on
the
left
hand
,
then
cut
off
those
6
Cyphers
with
a
dash
of
the
Pen
,
and
for
them
put
two
Cyphers
to
possess
the
two
first
places
in
the
root
,
Then
proceed
to
extract
the
cube
root
of
the
remaining
figures
as
if
there
had
been
no
such
Cyphers
,
&c.
XXIII
.
When
it
is
required
to
extract
the
Cube
Root
of
To
extract
the
Cube
Root
of
a
mixt
number
.
a
mixt
number
,
reduce
it
to
an
improper
Fraction
,
and
if
it
hath
a
perfect
cube
root
,
then
extract
the
cube
root
of
the
Numerator
,
and
place
it
for
a
new
Numerator
,
and
also
extract
the
Cube
Root
of
the
Denominator
,
and
place
it
for
a
new
Denominator
,
so
shall
this
new
Fraction
be
the
Cube
Root
of
the
given
mixt
Number
.
Example
.
Let
it
be
required
to
extract
the
Cube
Root
of
513/343
,
having
reduced
it
to
an
improper
Fraction
,
I
find
it
to
be
〈◊〉
,
and
having
extracted
the
Cube
Root
of
the
Numerator
(
1728
)
I
find
its
Root
to
be
12
,
for
a
Numerator
,
and
the
Cube
Root
of
343
the
Denominator
is
7
,
for
a
Denominator
,
so
that
I
conclude
〈◊〉
or
15/7
to
be
the
Cube
Root
of
the
given
mixt
Number
513/343
,
as
you
may
prove
at
your
leisure
.
XXIV
.
But
if
the
given
mixt
Number
,
whose
Cube
Root
is
required
,
have
not
a
perfect
Root
,
then
you
are
to
reduce
the
fractional
part
into
a
Decimal
of
the
same
value
,
(
but
let
the
number
of
decimal
places
be
alwaies
a
multiple
of
3
)
and
then
proceed
to
extract
the
Cube
Root
of
that
mixt
number
,
as
if
it
were
a
whole
Number
,
alwaies
reserving
so
many
decimal
places
in
the
Root
,
as
there
are
points
over
the
decimal
part
of
the
mixt
number
.
Example
.
Let
it
be
required
to
extract
the
Cube
Root
of
28¾
.
First
,
Reduce
¾
into
its
equivalent
decimal
,
which
is
.75
,
but
to
make
it
consist
of
six
places
,
I
annex
thereto
four
Cyphers
,
and
then
the
said
mixt
number
will
be
28.750000
,
which
being
done
,
I
proceed
to
the
work
as
followeth
.
So
that
I
find
by
the
work
,
the
Cube
Root
of
28.750000
to
be
3.06
&c.
XXV
.
It
is
usual
amongst
Artists
to
express
the
Cube
Root
of
a
whole
Number
,
mixt
number
,
or
Fraction
,
either
Vulgar
,
or
Decimal
,
that
is
incommensurable
to
its
Root
,
by
prefixing
this
Character
,
(
viz.
)
before
the
incommensurable
number
or
quantity
,
so
the
Cube
Root
of
328
may
be
thus
expressed
,
and
the
Cube
Root
of
24¾
,
thus
,
or
in
a
decimal
mixt
number
thus
and
of
the
fraction
¾
thus
¾
&c.
XXVI
.
The
operation
in
the
extraction
of
the
Cube
Root
is
proved
thus
,
viz.
Cube
the
Root
found
out
,
that
The
proof
of
the
extraction
of
the
Cube
Root
.
is
,
Multiply
it
three
times
into
it self
,
and
if
any
thing
remain
after
the
work
is
done
,
add
it
to
the
last
product
,
and
if
that
sum
be
equal
to
the
given
number
,
then
the
work
is
truly
performed
,
otherwise
not
.
As
in
our
first
Example
,
where
it
is
required
to
extract
the
Cube
Root
of
110592
,
and
which
is
found
to
be
48
;
and
to
prove
the
work
,
multiply
48
by
it self
,
whose
product
is
2304
,
which
being
again
multiplyed
by
48
,
it
produceth
110592
,
which
is
equal
to
the
given
number
,
and
therefore
I
conclude
the
work
to
be
right
.
Likewise
to
prove
the
Example
of
the
Nineteenth
Rule
,
where
it
is
required
to
extract
the
Cube
Root
of
13798
which
is
found
to
be
the
mixt
number
23.984
.
Now
to
prove
the
work
,
I
Cube
the
Root
,
as
is
before
directed
,
and
find
it
to
be
13796.370427904
to
which
I
add
the
remainder
1629572096
and
their
sum
maketh
the
given
number
13798
which
proves
the
work
to
be
right
.
CHAP.
XI
.
The
Use
of
the
Square
and
Cube
Roots
in
solving
some
Questions
Arithmetical
and
Geometrical
.
PROP.
I.
To
find
a
mean
proportional
between
two
given
Numbers
.
MUltiply
the
given
Numbers
the
one
by
the
other
,
and
extract
the
square
Root
of
the
product
,
so
shall
that
square
Root
be
the
mean
proportional
sought
.
Example
.
Let
the
given
Numbers
be
12
and
48
,
and
let
it
be
required
to
find
a
mean
proportional
between
them
;
first
multiply
the
given
numbers
12
and
48
the
one
into
the
other
,
and
their
product
is
576
,
the
Square
Root
of
which
is
24
,
so
that
I
conclude
24
to
be
a
mean
proportional
between
12
and
48
,
for
,
the
square
of
the
mean
being
equal
to
the
product
of
the
extreams
.
This
proposition
is
useful
in
finding
the
side
of
a
square
that
shall
be
equal
to
any
given
paralellogram
;
for
,
(
according
to
the
first
Proposition
of
the
Eighth
Chapter
of
this
Book
,
)
if
you
multiply
the
contiguous
sides
of
a
Rectangular
paralelogram
the
one
by
the
other
,
that
product
will
be
its
content
,
and
if
you
extract
the
square
root
of
that
content
,
it
will
give
you
the
side
of
a
square
,
(
in
the
same
measure
your
paralelogram
was
)
which
will
be
equal
to
the
given
Paralelogram
.
PROP.
II.
To
find
the
side
of
a
Square
that
shall
be
equal
to
the
Content
of
any
given
superficies
.
FInd
out
the
Content
of
the
given
superficies
by
the
Rules
laid
down
in
the
Eighth
Chapter
,
and
then
extract
the
Square
Root
of
the
Content
,
so
will
that
Root
be
the
side
of
a
square
equal
to
the
given
superficies
.
Example
.
There
is
a
Rectangled
Triangle
whose
base
and
perpendicular
are
16
and
18
,
I
demand
the
side
of
a
Square
that
will
be
equal
to
the
given
Triangle
.
According
to
the
second
Proposition
of
the
Eighth
Chapter
,
I
find
the
Content
of
this
Triangle
tobe
144
,
the
square
root
of
which
is
12
,
and
is
the
side
of
a
square
equal
to
the
said
Triangle
.
In
like
manner
,
if
you
extract
the
square
root
of
the
Content
of
a
Circle
,
Pentagon
,
Hexagon
,
&c.
or
of
any
other
Figure
regular
,
or
irregular
,
it
will
give
the
side
of
a
square
equal
to
that
superficies
.
PROP.
III.
Having
any
two
of
the
sides
of
a
Right-angled
plain
Triangle
,
given
to
find
the
third
side
.
THis
most
excellent
and
useful
proposition
is
generally
called
Pythagoras
his
Theoreme
,
and
in
the
47
Prop.
of
E●…clides
Elements
of
Geom.
it
is
demonstrated
,
and
proved
that
the
Square
made
of
the
Hypothenuse
,
or
slant
side
of
a
Right
angled
plain
Triangle
is
equal
to
the
sum
of
the
squares
made
of
the
base
and
perpendicular
.
As
for
Example
.
In
the
Triangle
ABC
,
the
Base
AB
is
48
,
and
the
perpendicular
BC
is
36
,
now
I
demand
the
length
of
the
Hypothenuse
AC
.
PROP.
IV.
THere
is
a
Tower
about
which
there
is
a
Moat
that
is
48
foot
wide
,
and
a
scaleing
Ladder
that
is
60
Foot
long
,
will
reach
from
the
outside
of
the
Moat
,
to
the
top
of
a
Wall
,
that
is
within
the
said
Moat
,
now
I
demand
the
height
of
the
said
Wall
above
the
Water
?
Let
the
Base
AB
in
the
foregoing
Triangle
be
the
breadth
of
the
Moat
,
and
let
the
Hypothenuse
AC
be
the
scaling
.
Ladder
,
then
is
the
perpendicular
BC
the
height
of
the
Wall
above
the
Water
.
Now
it
is
plain
that
(
because
the
Square
of
AC
is
equal
to
the
sum
of
the
squares
of
AB
and
BC
)
if
from
the
square
of
AC
which
is
3600
you
subtract
the
square
of
AB
which
is
2304
,
there
will
remain
1296
,
which
is
the
square
of
CB
,
therefore
I
extract
the
square
Root
of
1296
,
and
find
it
to
be
36
,
which
is
the
height
of
the
said
Wall
above
the
Water
as
was
required
.
By
the
help
of
this
Proposition
may
be
found
the
true
perpendicular
height
of
a
Cone
,
or
of
a
Pyramid
;
for
,
in
a
Cone
,
if
you
square
the
slant
height
,
(
which
is
the
length
of
a
line
drawn
from
its
vertical
point
,
to
the
Circumference
of
its
base
)
and
from
the
square
of
that
,
subtract
the
square
of
the
semidiameter
of
its
base
,
there
will
remain
the
square
of
the
perpendicular
height
of
that
Cone
.
Also
,
In
a
Pyramid
,
if
from
the
square
of
the
slant
height
of
it
,
you
subtract
the
square
of
that
line
which
being
drawn
from
the
Centre
of
its
base
,
shall
touch
the
end
of
the
said
slant
line
,
(
whether
they
meet
at
an
Angle
or
not
)
the
remainder
will
be
the
square
of
the
perpendicular
height
of
that
Pyramid
,
and
its
square
Root
will
give
the
height
it self
.
PROP.
V.
By
the
Content
of
a
Circle
to
find
its
Diameter
.
The
proportion
is
AS
22.
Is
to
28.
So
is
the
given
Content
to
the
square
of
the
Diameter
.
Example
.
There
is
a
Circle
whose
superficial
Content
is
153.9385
,
I
demand
its
Diameter
?
The
Square
Root
of
which
is
13.99
(
very
near
14
)
for
the
Diameter
required
.
PROP.
VI.
By
the
Content
of
a
Circle
to
find
its
Circumference
.
The
proportion
is
AS
7
Is
to
88
So
is
the
given
Content
to
the
square
of
the
Circumference
.
The
Square
Root
of
which
is
the
Circumference
required
.
Example
.
There
is
a
Circle
whose
superficial
content
is
153.9385
,
I
demand
the
Circumference
of
that
Circle
?
The
square
Root
of
which
is
44
fere
which
is
the
Circumference
required
.
II.
The
Cube
Root
is
that
by
help
of
which
we
resolve
all
Questions
Mathematical
that
concern
solidity
,
and
by
which
we
increase
solid
bodies
according
to
any
given
proportion
.
By
it
we
discover
the
solidity
of
a
body
that
is
capable
of
length
,
breadth
,
and
depth
,
(
or
thickness
,
)
and
by
having
the
solidity
given
,
we
discover
the
side
or
Diameter
of
such
a
body
.
Some
Questions
pertinent
thereto
may
be
such
as
follow
.
PROP.
VII
.
THere
is
a
Cube
whose
side
is
4
,
I
demand
what
shall
be
the
side
of
a
Cube
whose
solidity
is
double
to
the
solidity
of
that
Cube
?
To
answer
this
proposition
,
find
out
the
Cube
of
4
(
the
side
of
the
given
Cube
)
which
is
64
,
and
double
it
,
which
is
128
,
then
extract
the
Cube
Root
of
128
,
and
it
makes
5.0397
fere
,
and
that
is
the
side
of
the
Cube
which
is
double
to
the
Cube
whose
side
is
4.
PROP.
VIII
.
THere
is
a
Cube
whose
solidity
is
128
foot
,
I
demand
the
side
of
a
Cube
whose
solidity
is
half
as
much
?
Take
½
of
128
=
64
the
Cube
Root
of
which
(
viz
,
4.
)
answers
the
Question
.
PROP.
IX
.
HAving
the
solid
Content
of
a
Globe
to
find
the
side
of
a
Cube
whose
solidity
shall
be
equal
to
the
given
Globe
?
Extract
the
Cube
Root
of
the
given
solid
Content
of
the
Globe
,
and
it
will
give
you
the
side
of
the
Cube
Required
.
Example
.
There
is
a
Globe
whose
solid
Content
is
1728
Inches
,
I
demand
the
side
of
a
Cube
equal
thereto
?
Having
extracted
the
Cube
Root
of
1728
,
I
find
it
to
be
12
,
which
is
the
side
of
the
Cube
required
.
PROP.
X.
HAving
the
Diameter
and
Weight
of
a
Bullet
,
to
sind
the
Weight
of
another
Bullet
,
whose
Diameter
is
given
.
As
the
Cube
of
the
given
Bullets
Diameter
,
Is
to
its
weight
,
or
solidity
.
So
is
the
Cube
of
the
Diameter
of
any
other
Bullet
,
To
its
weight
,
or
solidity
.
Examp'e
.
There
is
a
Bullet
whose
Diameter
is
4
Inches
,
and
its
weight
is
9
Pound
,
I
demand
the
weight
of
another
Bullet
,
whose
Diameter
is
6●…
or
6.25
Inches
?
The
Cube
of
4
is
64.
The
Cube
of
6.25
is
244.140625
Then
I
say
So
that
the
weight
required
is
34.33227
pounds
,
and
if
you
Reduce
the
Decimal
to
the
known
parts
of
Averdupois
weight
,
you
will
find
the
answer
to
be
34
lb.
−
05
oz.
−
05
dr
.
This
kind
of
proportion
is
by
Artists
Termed
triplicate
proportion
.
In
like
manner
,
the
Diameters
of
two
Bullets
,
or
Globes
being
given
,
and
the
solidity
of
one
of
them
to
find
out
the
solidity
of
the
other
,
it
may
be
done
by
the
same
proportion
,
only
changing
the
middlemost
Term.
PROP.
XI
.
TO
find
the
side
of
a
Cube
equal
to
a
given
paralelepipedon
.
Find
out
the
solidity
of
the
given
Paralelepipedon
by
the
Eighth
Prop.
of
the
Eighth
Chapter
,
then
is
the
Cube
Root
thereof
,
the
required
side
.
Example
.
There
is
a
paralelepipedon
having
the
sides
of
its
base
10
Foot
4
Inches
,
and
5
Foot
2
Inches
,
and
its
length
is
20
Foot
8
Inches
,
I
desire
to
know
what
is
the
side
of
a
Cube
whose
content
shall
be
equal
to
the
given
paralelepipedon
?
The
superficial
Content
of
the
base
is
7688
inches
,
which
drawn
into
248
the
length
in
inches
,
the
product
is
1906624
inches
for
its
solid
Content
,
the
Cube
Root
of
which
is
124
inches
for
the
side
of
a
Cube
equal
to
the
given
paralelepipedon
.
In
like
manner
if
you
would
find
at
any
time
the
side
of
a
Cube
equal
to
any
solid
Body
whether
Regular
,
or
irregular
:
First
,
Find
the
solid
Content
of
that
Body
,
and
then
extracting
the
Cube
Root
of
its
solid
Content
you
have
your
desire
.
PROP.
XII
.
BEtween
two
given
Numbers
to
find
two
mean
proportionals
.
Divide
the
greater
extream
by
the
lesser
,
and
extract
the
Cube
Root
of
the
Quotient
,
and
by
the
said
Cube
Root
multiply
the
lesser
extream
,
then
will
the
product
give
you
the
lesser
mean
proportional
,
then
multiply
the
said
lesser
mean
by
the
said
Cubique
Root
,
and
that
product
will
give
you
the
greater
mean
proportional
.
Example
.
Let
the
two
given
extreams
be
6
and
48
between
which
it
is
required
to
find
2
mean
proportionals
.
First
,
I
divide
48
(
the
Greater
Extream
)
by
6
(
the
Lesser
Extream
)
and
the
Quotient
is
8
,
the
Cube
Root
of
which
is
2
,
then
by
(
the
Cube
Root
)
2
do
I
multiply
6
(
the
lesser
extream
,
and
the
product
is
12
for
the
lesser
mean
proportional
,
and
12
being
multiplyed
by
2
(
the
Cube
Root
)
the
product
is
24
,
for
the
greater
mean
proportional
sought
.
Thus
have
I
found
12
and
24
to
be
two
mean
proportionals
between
6
and
48
,
for
In
like
manner
between
3
and
81
will
be
found
9
and
27
,
for
two
mean
proportionals
.
PROP.
XIII
.
THE
,
Concave
Diameter
of
two
Guns
being
known
,
and
the
quantity
of
Gun-powder
that
will
charge
one
of
them
,
to
find
out
how
much
will
be
sufficient
to
charge
the
other
.
The
Capacities
are
one
to
another
,
as
are
the
Cubes
of
their
Diameters
,
and
also
the
proportion
is
direct
.
Example
.
If
.25
pound
of
Gun-powder
be
sufficient
to
charge
a
Gun
,
whose
Concave
Diameter
is
1½
Inches
,
or
1.5
Inch.
how
much
powder
will
be
sufficient
to
charge
a
Gun
,
whose
Concave
Diameter
is
7
inches
?
Answer
,
25.47
.
The
Cube
of
1.5
is
3.375
and
the
Cube
of
7
is
343.
wherefore
the
proportion
is
as
followeth
.
Or
thus
,
PROP.
XIV
.
THE
Concave
Diameters
of
two
Guns
being
given
,
and
the
quantity
of
a
weaker
sort
of
Gun-powder
sufficient
to
charge
one
of
them
,
to
find
out
how
much
Gun-powder
of
a
stronger
sort
(
the
proportion
of
the
strength
and
weakness
of
the
Gun-powder
being
also
given
)
will
be
sufficient
to
charge
the
other
Gun
.
This
is
solved
by
two
operations
in
the
Rule
of
proportion
,
first
to
find
out
how
much
of
the
stronger
sort
of
Gun-powder
will
be
of
equivalent
strength
with
the
given
quantity
of
the
weaker
sort
,
and
this
proportion
is
Reciprocal
;
The
second
is
the
same
with
that
in
the
fore-going
Proposition
.
Example
.
There
is
a
Gun
whose
Concave
Diameter
is
1½
inches
,
and
it
requireth
25
pound
of
powder
to
Charge
it
,
now
there
is
another
sort
of
Gun-powder
which
is
much
stronger
than
the
former
,
and
the
proportion
between
their
strength
is
as
5
to
2
,
now
I
demand
how
much
of
the
strongest
powder
is
sufficient
to
charge
a
Gun
,
whose
concave
Diameter
is
7
inches
?
To
answer
this
,
First
,
I
find
out
how
much
of
the
strongest
powder
will
charge
that
Gun
▪
which
is
1½
inch
in
its
Concave
Diameter
,
which
is
done
by
the
following
proportion
,
viz.
Thus
have
I
found
that
1/10
of
a
pound
of
the
strongest
powder
will
charge
a
Gun
whose
Concave
diameter
is
〈◊〉
inch
.
And
according
to
the
last
proposition
,
I
find
by
a
direct
Proportion
that
10.16
pounds
of
the
same
will
be
sufficient
to
charge
a
Gun
whose
concave
diameter
is
7
inches
,
viz.
CHAP.
XII
.
Concerning
Simple
Interest
.
I.
WHen
Money
pertaining
,
or
belonging
to
one
person
is
in
the
hands
,
possession
,
or
keeping
,
or
is
lent
to
another
,
and
the
Debtor
payeth
or
alloweth
to
the
Creditor
,
a
certain
sum
in
Consideration
of
forbearance
for
a
certain
time
,
such
consideration
for
forbearance
is
called
Interest
,
loane
,
or
use
money
;
and
the
money
so
lent
,
and
forborne
is
called
the
principal
.
II.
Interest
is
either
Simple
,
or
Compound
.
III.
When
for
a
sum
of
money
lent
there
is
loane
,
or
interest
allowed
,
and
the
same
is
not
paid
when
it
becomes
due
,
and
if
such
Interest
doth
not
then
become
a
part
of
the
Principal
,
it
is
called
Simple
Interest
.
IV
,
In
the
taking
of
Interest
for
the
continuance
or
forbearance
of
Money
,
respect
must
be
had
to
the
rate
limited
by
Act
of
Parliament
,
which
Act
now
in
force
,
forbiddeth
,
or
restraineth
all
persons
whatsoever
,
from
taking
more
than
6
l.
for
the
Interest
of
an
100
l.
for
a
year
,
and
according
to
the
same
proportion
for
a
greater
or
a
lesser
sum
,
not
confining
the
lender
or
borrower
to
the
space
of
one
year
,
no
more
than
it
confineth
him
or
them
to
the
limitation
of
the
sum
to
be
lent
,
or
borrowed
,
but
that
the
sum
may
be
either
more
or
less
than
100
l
,
and
may
continue
in
the
hands
of
the
Debtor
,
either
a
longer
,
or
a
shorter
time
than
one
year
,
according
as
the
Lender
and
Borrower
do
agree
,
and
oblige
each
other
;
Now
for
any
time
greater
than
one
year
,
the
rate
or
proportion
of
Interest
is
by
Act
of
Parliament
limited
,
but
the
Act
doth
not
say
what
part
of
6
l.
shall
be
the
interest
of
an
100
l.
for
half
a
year
,
a
quarter
of
a
year
,
a
month
,
a
day
,
or
for
any
time
lesser
than
one
year
,
and
in
this
case
several
Artists
do
differ
in
their
opinions
,
some
would
have
the
true
proportional
interest
for
any
time
less
than
a
year
to
be
discovered
by
continual
mean
proportionals
;
as
suppose
it
were
required
to
know
the
interest
of
100
l.
for
half
a
year
at
6
per
Cent.
per
Annum
,
they
would
have
the
Interest
to
be
reckoned
after
the
Rule
of
Compound
interest
,
and
so
3l
.
is
not
the
interest
of
a
100
l.
for
half
a
year
,
but
is
too
much
:
But
say
they
,
to
find
out
the
true
interest
thereof
,
you
are
to
find
a
mean
proportional
between
100
,
and
106
,
and
that
made
less
by
100
,
will
give
you
the
interest
of
100
l.
for
half
a
year
,
and
so
by
extracting
of
Roots
they
find
out
the
interest
for
any
time
less
than
one
year
,
but
this
is
sufficiently
laborious
and
painful
if
it
be
done
without
the
help
of
Logarithms
;
but
to
perform
this
work
to
the
12
power
for
a
Moneth
,
or
to
the
52
for
a
Weak
,
is
very
tedious
,
and
to
the
365
power
for
one
Day
is
scarcely
possible
to
be
effected
by
natural
Numbers
;
but
custom
and
dayly
practice
tell
us
that
the
interest
of
Money
for
any
time
less
than
one
year
ought
to
be
computed
according
to
the
Rules
of
Simple
Interest
,
and
so
3
l.
is
the
undoubted
interest
of
100
l.
for
6
moneths
,
and
30
shillings
is
the
interest
of
100
l.
for
a
quarter
of
a
year
;
but
here
note
by
the
way
that
by
6
moneths
is
not
meant
6
times
4
weeks
,
or
6
times
28
dayes
,
but
by
six
moneths
,
or
half
a
year
is
to
be
understood
the
half
of
365
dayes
,
and
a
quarter
of
a
year
is
¼
of
3●…5
dayes
,
and
by
1
moneth
is
understood
1/12
of
365
dayes
,
so
that
a
moneth
consisteth
of
305/12
dayes
.
Upon
the
foresaid
custom
of
computing
the
interest
of
money
for
time
less
than
one
year
,
this
following
Analogie
seems
to
be
assumed
for
a
safe
exposition
Vide
Sect.
6
of
the
5
chap
of
Mr.
Kersies
Appendix
to
Wing
.
Arith.
of
the
statute
(
and
which
is
indeed
the
ground
,
and
reason
it self
of
Simple
Interest
)
viz.
That
such
proportion
as
365
dayes
(
or
one
year
)
hath
to
the
interest
of
any
sum
for
a
year
,
such
proportion
hath
any
part
of
one
year
,
or
any
number
of
dayes
propounded
,
to
the
interest
of
the
same
sum
,
for
that
time
propounded
.
And
this
(
as
was
said
before
)
is
the
whole
ground
work
,
and
very
foundation
of
the
manner
of
computing
of
Simple
Interest
.
V.
Rebate
,
or
Discount
,
is
,
when
there
is
an
allowance
of
so
much
per
Cent.
for
money
paid
before
it
be
due
,
and
Of
Rebate
,
what
it
is
.
as
the
increase
of
money
at
interest
is
found
out
by
continual
proportionals
Arithmetical
or
Geometrical
increasing
,
so
is
the
Rebate
or
discount
of
Money
found
out
by
continual
proportionals
decreasing
Arithmetically
or
Geometrically
,
that
is
according
as
the
allowance
is
,
either
after
Simple
or
Compound
Interest
;
Now
the
nature
of
Rebate
or
discount
is
thus
;
When
there
is
a
sum
of
Money
,
(
suppose
100
l.
)
to
become
due
at
the
end
of
a
certain
time
to
come
,
(
viz.
at
the
end
of
12
Moneths
;
)
and
it
is
agreed
upon
by
the
Debtor
and
Creditor
that
there
shall
be
made
present
payment
of
the
whole
Debt
,
and
it
is
likewise
agreed
that
in
consideration
of
this
present
payment
,
that
the
Creditor
shall
allow
the
Debtor
after
the
rate
of
6
per
cent
.
per
annum
:
Now
upon
this
agreement
the
Creditor
ought
to
receive
so
much
money
as
being
put
out
at
interest
for
the
same
time
it
was
paid
before
't
was
due
,
and
at
the
same
rate
of
interest
,
that
the
discount
was
reckoned
at
,
then
would
it
amount
or
be
increased
to
the
sum
that
was
first
due
.
The
manner
of
working
Questions
in
Rebate
at
Simple
Interest
shall
be
shewn
in
the
ninth
Rule
of
this
Chapter
,
and
of
working
Questions
in
Rebate
at
Compound
Interest
shall
be
shewen
in
the
Fourth
Rule
of
the
next
Chapter
.
VI.
When
the
interest
of
a
100
l.
for
a
year
is
known
,
the
interest
of
any
other
sum
,
for
the
same
time
,
is
also
found
out
,
by
one
single
rule
of
direct
proportion
,
viz.
The
Interest
of
a
100
l.
for
a
year
by
the
statute
is
6l
,
I
demand
what
is
the
interest
of
75
l.
for
the
same
time
,
and
at
the
same
Rate
of
Interest
?
The
proportion
is
as
followeth
.
Or
if
you
would
have
the
Answer
to
produce
both
principal
and
interest
,
then
make
the
second
number
to
be
the
sum
of
the
given
principal
and
interest
,
and
the
fourth
proportional
will
answer
your
desire
.
Thus
,
VII
.
When
the
Interest
of
100
for
a
year
is
given
,
and
the
interest
of
any
other
sum
of
pounds
,
shillings
,
and
pence
is
required
for
a
year
,
the
answer
may
be
easily
found
after
the
practical
method
delivered
in
the
following
Example
.
Let
it
be
required
to
find
the
interest
of
148l
.
−
13s.
04.
for
one
year
after
the
rate
of
6
per
cent
.
per
annum
,
simple
interest
?
First
,
I
place
the
given
Numbers
according
to
the
Direction
given
for
the
Rule
of
3
,
which
will
then
stand
thus
,
viz.
Now
it
is
evident
that
if
I
multiply
148l
.
−
13s
−
04d
.
(
which
is
the
third
number
)
by
6
(
which
is
the
second
number
)
and
divide
the
product
by
100
(
which
is
the
first
number
)
the
Quotient
will
be
the
answer
;
Therefore
I
proceed
thus
,
viz
first
I
multiply
the
pence
by
6
,
which
makes
24
pence
,
or
two
shillings
,
therefore
I
set
down
o
under
the
pence
,
and
carry
2
to
the
next
,
then
I
go
to
the
13s
.
saying
6
times
13
is
78
,
and
2
that
I
carryed
is
80
s.
which
is
4
l.
therefore
I
set
down
o
under
the
shillings
,
and
carry
4
to
the
pounds
,
then
I
proceed
,
saying
6
times
8
is
48
,
and
4
that
I
carry
is
52
,
then
I
set
down
2
,
and
carry
5
,
&c.
proceeding
thus
till
the
work
be
finished
,
and
then
will
the
product
be
890l
.
−
00s
.
−
00d
,
which
product
should
be
divided
by
100
(
the
first
number
)
but
it
being
an
Unite
with
two
Cyphers
,
I
cut
off
two
figures
from
the
right
hand
of
the
pounds
,
with
a
dash
of
the
pen
,
and
the
figures
on
the
left
hand
of
the
said
dash
,
are
so
many
pounds
,
and
those
on
the
right
hand
of
it
,
are
the
Decimal
parts
of
a
pound
,
whose
value
may
be
found
out
by
the
3
R●…le
of
the
2
Chap.
But
remember
,
that
if
there
be
any
shillings
or
pence
,
in
the
product
you
are
to
add
them
to
their
respective
products
in
your
Reduction
.
The
work
of
the
foregoing
Example
is
as
followeth
.
So
that
by
the
work
I
find
the
interest
of
148
l.
−
13s
.
−
4d
.
for
one
year
after
the
rate
of
6
per
Cent.
per
An.
to
be
8l
.
−
18s
.
−
04d
−
3
.
2qu
.
Another
Example
may
be
this
,
viz.
I
demand
the
Interest
of
368l
.
15s
.
−
3d.
for
one
year
,
at
6
per
Cent.
per
An.
Answer
22l
.
−
02s
.
−
6d
.
as
by
the
work
following
.
VIII
.
The
Interest
of
100l
.
being
known
for
a
year
,
or
365
dayes
,
the
interest
of
any
other
sum
may
be
known
for
any
other
time
,
or
number
of
dayes
,
more
or
less
than
a
year
,
by
two
single
Rules
of
3
Direct
,
viz.
First
,
find
out
what
is
the
interest
of
the
given
sum
,
for
one
year
,
or
365
dayes
,
according
to
the
last
Rule
,
then
having
found
out
that
,
you
may
(
by
another
single
Rule
of
3
Direct
)
find
out
its
interest
for
any
other
time
more
or
less
.
Example
.
What
is
the
interest
of
322l
.
for
6
years
after
the
rate
of
6
per
Cent.
per
Ann●…m
●…imple
Interest
?
First
,
I
find
what
is
the
Interest
of
322l
.
for
a
year
by
the
following
proportion
,
Thus
having
found
the
Interest
of
322l
.
for
a
year
to
be
19.32l
.
at
6
per
Cent.
by
the
following
proportion
,
I
find
out
its
interest
for
6
years
,
to
be
115l
.
−
18s
.
−
04¾d
.
and
that
added
to
the
principal
,
makes
437l
.
−
18s
.
−
04¾d
.
for
the
sum
due
to
the
Creditor
at
the
end
of
the
said
time
.
And
here
take
notice
that
the
second
number
in
this
last
proportion
,
must
always
be
only
the
interest
of
the
sum
proposed
,
and
not
the
sum
of
the
principal
and
interest
,
as
in
the
second
proportion
under
the
sixth
Rule
.
After
the
same
manner
is
the
interest
of
1l
.
(
at
the
rate
of
6
per
Cent.
per
Annum
,
or
any
other
rate
of
interest
,
)
discovered
for
a
day
,
by
the
help
of
which
the
interest
of
any
sum
whatsoever
may
be
discovered
for
any
number
of
dayes
as
shall
be
shown
by
and
by
.
So
that
by
the
foregoing
proportions
I
have
found
that
the
interest
of
1l
.
at
6
per
Cent.
per
Annum
for
a
day
is
.0001643835l
.
Now
if
you
would
know
the
interest
of
any
other
sum
for
any
number
of
dayes
more
or
less
than
365
,
you
may
do
it
by
help
of
the
said
number
after
this
manner
,
viz.
Multiply
the
sum
whose
interest
is
required
by
the
said
number
,
and
that
product
will
give
you
the
interest
of
the
said
sum
for
one
day
,
then
multiply
that
product
by
the
number
of
dayes
given
,
and
the
last
product
will
give
you
the
interest
of
the
said
sum
for
the
number
of
dayes
in
the
Question
.
Take
the
following
Question
for
an
example
,
viz.
What
is
the
interest
of
568l
.
for
213
dayes
after
the
Rate
of
6
per
Cent.
per
Annum
.
Having
finished
the
work
as
you
see
,
I
find
the
answer
to
be
19.8877
&c.
which
upon
sight
I
discover
to
be
19l
.
−
17s
−
09d
.
by
the
brief
way
of
valluing
a
Decimal
Fraction
of
Coyne
laid
down
in
the
4
Rule
of
the
2
Chapter
before-going
.
But
when
the
interest
of
any
sum
of
Money
is
required
for
any
number
of
dayes
as
aforesaid
,
at
any
other
rate
of
interest
then
at
6
per
Cent.
per
Annum
,
the
foresaid
number
will
not
then
serve
for
the
work
,
but
you
are
to
find
out
particular
multiplyars
for
the
several
Rates
of
Interest
as
is
before
directed
.
All
which
I
have
expressed
from
4
to
10
per
Cent.
in
the
following
Table
.
When
you
would
find
the
Interest
of
any
sum
for
any
number
of
dayes
at
the
rate
of
|
4
|
per.
Cent
per
An.
the
Multiplyar
is
|
.0001095890
|
5
|
.0001369863
|
6
|
.0001643835
|
7
|
.0001917808
|
8
|
.0002191780
|
9
|
.0002465753
|
10
|
.0002739726
|
So
that
when
you
would
find
out
the
interest
of
any
sum
of
Money
for
any
number
of
dayes
according
to
the
direction
before
given
,
at
any
Rate
from
4
to
10
per
Cent.
per
Annum
,
Simple
Interest
,
you
may
performe
the
work
by
the
multiplyar
in
the
foregoing
Table
which
is
placed
against
each
respective
Rate
of
interest
.
IX
.
When
the
present
worth
of
a
sum
of
money
due
at
the
end
of
any
time
to
come
is
required
,
Rebate
being
allowed
at
any
rate
of
Simple
Interest
,
it
may
be
found
out
by
the
following
method
;
viz.
First
,
Find
out
the
Interest
of
100l
.
for
the
time
that
the
Rebate
is
to
be
allowed
for
,
and
at
the
same
rate
of
interest
propounded
,
then
make
the
sum
of
an
100
pound
,
and
its
interest
for
the
proposed
time
,
to
be
the
first
number
in
the
Rule
of
3
,
and
100l
.
the
second
number
,
and
the
given
sum
whose
present
worth
is
required
,
let
be
the
third
number
,
and
the
fourth
number
in
a
direct
proportion
shall
answer
the
question
,
as
in
the
following
Example
,
viz.
What
present
Money
will
satisfie
a
debt
of
100l
.
that
is
due
at
the
end
of
a
year
yet
to
come
,
discount
,
or
Rebate
being
allowed
at
the
Rate
of
6
per
Cent.
per
Annum
.
According
to
the
foregoing
Directions
,
I
state
the
numbers
as
followeth
,
and
the
fourth
proportional
number
or
answer
to
the
question
is
94.33962
l.
=
94
l.
−
06s
.
−
09
½
d
fere
.
The
reason
of
the
said
Analogy
will
appear
if
you
consider
,
that
there
ought
to
be
so
much
ready
money
paid
,
that
if
it
were
put
out
to
interest
at
the
same
rate
of
Int.
that
Rebate
was
allowed
for
,
and
for
the
same
time
,
the
same
would
then
be
augmented
to
the
sum
that
was
at
first
due
,
as
in
the
last
question
,
there
is
given
100
l.
which
is
due
at
the
end
of
12
moneths
,
now
I
say
,
that
there
ought
to
be
so
much
money
paid
down
to
satisfie
this
debt
,
as
being
put
out
to
interest
at
6
perCent
.
for
12
moneths
,
would
then
be
increased
to
100
l.
which
is
the
sum
first
due
,
and
again
it
is
as
evident
that
if
there
were
106
l.
due
at
the
end
of
12
moneths
,
or
a
year
,
and
present
payment
is
agreed
upon
,
allowing
Rebate
at
6
per
Cent.
per
Annum
,
that
then
there
ought
to
be
paid
the
sum
of
100
l
,
in
full
discharge
of
the
said
debt
of
106
l.
for
if
when
I
have
received
the
said
sum
of
100
l
,
I
put
it
out
to
interest
for
one
year
at
the
rate
of
6
per.
Cent.
it
will
then
be
increased
to
●…06
l.
Therefore
to
solve
the
said
question
,
the
proportion
here
used
is
no
more
than
if
I
should
say
,
If
106
l.
be
decreased
to
100
l.
what
will
100
l.
be
decreased
to
?
The
answer
is
,
to
94l
.
−
6s
.
−
9d
.
●…
,
and
for
proof
,
if
yon
will
seek
what
that
sum
will
be
increased
to
at
the
end
of
12
moneths
,
at
the
rate
of
6
per
Cent.
you
will
find
it
to
be
100l
.
Example
2.
How
much
present
money
will
satisfie
a
debt
of
82l
.
−
15s
.
due
at
the
end
of
126
dayes
,
yet
to
come
allowing
Rebate
after
the
rate
of
6
per
Cent.
per
Annum
?
First
,
I
find
the
interest
of
100l
.
at
the
same
rate
of
interest
for
126
dayes
,
by
the
following
proportion
.
Then
do
I
add
2.0712l
.
(
the
interest
of
100l
.
)
to
100
l.
and
the
sum
is
102.0712
which
I
make
the
first
number
in
the
Rule
of
3
,
and
100l
.
the
second
,
and
82.75l
.
(
the
sum
given
to
be
Rebated
)
the
third
number
,
and
the
fourth
number
in
a
direct
proportion
is
the
answer
to
the
question
,
see
the
work
as
followeth
.
So
that
by
the
work
it
appears
that
82l
.
−
15s
.
due
at
the
end
of
126
dayes
yet
to
come
,
will
be
satisfied
with
the
present
payment
of
81l
.
−
01s
.
−
04¾d
.
Rebate
be
allowed
after
the
rate
of
6
per
Cent.
per
An.
The
proof
of
the
Rule
.
Find
out
(
by
the
eighth
Rule
foregoing
)
how
much
the
present
money
that
is
paid
upon
Rebate
,
will
amount
to
being
put
out
to
interest
for
the
same
time
,
and
at
the
same
Rate
of
interest
that
Rebate
was
allowed
for
,
and
if
the
amount
be
equal
to
the
sum
that
was
due
at
the
end
of
that
time
,
then
you
may
conclude
the
work
to
be
rightly
performed
,
otherwise
not
.
As
for
Example
.
In
the
foregoing
Question
it
was
found
that
81.0708
l.
being
paid
presently
would
satisfie
a
debt
of
82.75
due
at
the
end
of
126
dayes
to
come
,
and
to
prove
it
,
let
us
see
whether
81.0708
being
put
out
to
interest
for
126
dayes
at
the
rate
of
6
per
Cent.
per
Annum
,
will
be
increased
to
82.75
l.
(
the
sum
which
was
said
to
be
due
at
the
end
of
126
dayes
to
come
)
which
I
do
by
these
two
proportions
following
according
to
the
Eighth
Rule
.
So
you
see
that
I
have
found
the
interest
of
81.0708
for
126
dayes
to
be
1.6791
&c.
which
added
to
the
principal
81.0708
the
sum
is
82.7499
which
by
the
brief
way
of
valuing
the
Decimal
of
a
pound
sterling
is
82
l.
−
15s
.
and
indeed
it
doth
not
want
1/10
part
a
farthing
of
the
exact
sum
,
which
is
occasioned
by
the
defective
Decimal
wherefore
I
conclude
the
work
to
be
rightly
performed
.
Upon
the
foregoing
ninth
Rule
is
grounded
the
manner
of
calculating
the
ensuing
Table
of
Multiplyars
,
which
sheweth
in
Decimal
parts
of
a
pound
,
the
present
worth
of
a
pound
sterling
due
at
the
end
of
any
number
of
years
to
come
,
not
exceeding
30
,
Simple
interest
being
Computed
at
6
per
Cent.
per
annum
.
The
first
number
in
the
Table
being
found
out
by
this
following
proportion
,
viz.
As
106l
.
is
to
100l
,
so
is
1
l.
to
.943396
,
and
the
second
number
in
the
Table
being
the
present
worth
of
1
l
due
at
the
end
of
two
years
to
come
,
is
thus
found
out
,
viz.
First
I
consider
that
12
l.
is
the
simple
interest
of
100l
,
for
2
years
,
which
added
to
100
l.
makes
112
l.
wherefore
I
say
,
as
112
l.
is
to
100
l.
so
is
1l
.
to
.892857
l.
which
is
the
present
worth
of
1l
.
due
at
the
end
of
2
years
to
come
.
The
several
proportions
and
operations
for
the
whole
Calculation
being
as
followeth
,
viz.
And
after
the
same
manner
are
all
the
numbers
in
the
following
Table
Calculated
;
which
being
well
understood
,
the
way
of
calculating
most
of
the
ensuing
Tables
will
easily
be
obtained
;
and
its
use
you
will
find
immediately
after
the
Table
it self
.
TABLE
I.
Which
sheweth
in
Decimal
parts
of
a
pound
the
present
worth
of
1l
.
due
at
the
end
of
any
number
of
years
to
come
under
31
,
at
the
rate
of
6
per
Cent.
per
Ann.
Simple
Interest
.
years
|
|
1
|
.943396
|
2
|
.892857
|
3
|
.847457
|
4
|
.806451
|
5
|
.769230
|
6
|
.735294
|
7
|
.704225
|
8
|
.675675
|
9
|
.649350
|
10
|
.625000
|
11
|
.602409
|
12
|
.581395
|
13
|
.561797
|
14
|
.543478
|
15
|
.526315
|
16
|
.510204
|
17
|
.495049
|
18
|
.480769
|
19
|
.467289
|
20
|
.454545
|
21
|
.
4●…2477
|
22
|
.431034
|
23
|
.420168
|
24
|
.409836
|
25
|
.400000
|
26
|
.390625
|
27
|
.381679
|
28
|
.373134
|
29
|
.364963
|
30
|
.357143
|
After
the
same
method
might
this
Table
be
continued
to
any
number
of
years
at
pleasure
,
I
might
also
have
calculated
for
other
rates
of
interest
,
as
those
are
in
the
next
Chapter
concerning
Compound
Interest
,
but
Simple
Interest
being
not
so
generally
in
practice
,
I
shall
therefore
forbear
.
The
use
of
the
preceding
TABLE
.
It
is
evident
(
by
the
ninth
Rule
foregoing
)
that
if
any
sum
be
paid
with
an
allowance
of
Rebate
,
you
are
to
make
100
l.
with
its
interest
(
for
the
same
time
you
Rebate
for
)
both
in
one
sum
,
to
be
the
first
number
in
the
Rule
of
3
,
100
the
second
,
and
the
sum
to
be
rebated
the
third
,
then
will
the
fourth
proportional
be
the
answer
;
and
the
same
may
be
wrought
by
any
other
number
and
its
interest
,
as
well
as
by
100
l.
and
its
interest
mutatis
mutandis
;
Now
in
the
Table
before-going
there
is
expressed
in
Decimal
parts
of
a
pound
,
the
present
worth
of
1
l.
due
at
the
end
of
any
number
of
years
to
come
under
31
,
&c.
that
is
to
say
,
if
you
take
the
money
signified
by
those
Decimals
,
and
put
it
out
to
interest
at
6
per
Cent.
per
Annum
,
Simple
Interest
for
so
many
years
as
are
expressed
in
the
Collum
of
years
against
the
said
Decimal
,
then
will
that
sum
at
the
end
of
the
said
Term
be
augmented
to
1
l
,
wherefore
if
you
have
any
sum
whatsoever
to
be
Rebated
for
any
number
of
years
within
the
limits
of
the
Table
,
make
1
l.
the
first
number
in
the
Rule
of
3
,
and
the
Decimal
in
the
Table
against
the
number
of
years
to
be
Rebated
for
,
make
that
the
second
,
and
the
sum
whose
present
worth
is
required
the
third
number
,
so
will
the
fourth
proportional
be
the
Answer
.
But
(
because
the
first
number
(
being
Unity
)
neither
multiplyeth
nor
divideth
)
if
you
take
the
number
in
the
Table
,
correspondent
to
the
number
of
years
for
which
you
would
reckon
Rebate
,
and
thereby
multiply
the
sum
whose
present
worth
is
required
,
the
product
will
give
you
the
Answer
.
Example
.
There
is
a
sum
of
money
,
viz.
560
l.
due
at
the
end
of
8
years
to
come
,
but
the
Debtor
and
Creditor
agree
that
present
payment
shall
be
made
,
and
the
Debtor
to
be
allowed
Rebate
after
the
rate
of
6
per
Cent.
per
Annum
,
Simple
Interest
.
Now
I
demand
how
much
present
money
will
satisfie
the
said
Debt
?
Answer
,
378.378l
.
=
378l
.
−
07s
.
−
06¾d
.
see
the
following
work
.
First
(
the
Rebate
being
to
be
reckoned
for
8
years
)
I
look
for
8
in
the
Collum
of
years
,
and
just
against
it
on
the
Right
hand
,
I
fiud
.675675
which
I
multiply
by
560
(
the
sum
whose
present
worth
is
required
,
)
and
the
product
is
378.378
,
which
(
by
the
brief
way
of
valuing
the
fraction
of
a
pound
sterling
)
I
find
at
first
sight
to
be
378
l.
=
07s
.
−
06●…d
.
This
Question
,
if
it
had
been
wrought
by
the
foregoing
ninth
Rule
,
would
have
produced
the
same
answer
,
for
,
The
Int.
of
100l
.
for
8
Mon.
is
48l
.
and
100
+
48
=
148
wherefore
by
the
Rule
of
3
I
say
X.
When
an
Annuity
or
yearly
income
is
in
arrears
for
any
number
of
years
,
and
you
would
know
the
increase
,
or
amount
of
it
,
allowing
Simple
Interest
at
a
certain
rate
per
Cent.
per
annum
,
for
each
yearly
payment
from
the
time
it
first
became
due
,
the
operation
will
be
somewhat
more
tedious
than
to
find
the
amount
of
one
single
sum
,
according
to
the
eighth
Rule
of
this
Chap.
which
will
clearly
appear
by
solving
the
following
question
,
viz.
There
is
an
Annuity
,
or
an
income
of
100
l.
per
annum
forborne
to
the
end
of
6
years
,
I
demand
how
much
is
due
at
the
end
of
the
said
Term
,
allowing
interest
at
the
rate
of
6
per
Cent.
per
Annum
Simple
Interest
?
Answer
690
l.
In
order
to
the
solution
of
this
Question
,
I
consider
,
First
,
that
It
is
evident
that
for
the
last
year
,
viz.
the
sixth
years
payment
,
there
must
be
no
interest
at
all
Reckoned
,
because
it
becomes
not
due
till
the
end
of
the
sixth
year
;
Secondly
,
there
must
be
reckoned
the
interest
of
100
l.
for
one
year
,
viz
that
which
is
due
at
the
end
of
the
fifth
year
;
Thirdly
,
there
must
be
reckoned
the
interest
of
100
l.
for
two
years
,
viz.
that
which
is
due
at
the
end
of
the
fourth
year
.
Fourthly
,
There
must
be
reckoned
the
interest
of
100
l.
for
three
years
,
viz.
that
which
is
due
at
the
end
of
the
third
year
,
Fifthly
,
the
interest
of
a
100
l.
for
4
years
,
viz.
that
which
is
due
at
the
end
of
the
second
year
:
And
Sixthly
,
The
interest
of
100
l.
for
5
years
,
viz.
that
which
is
due
at
the
end
of
the
first
year
,
and
is
forborne
the
second
,
third
,
fourth
,
fifth
,
and
sixth
years
;
all
which
interests
being
added
together
,
and
their
sum
added
to
the
sum
of
each
years
income
,
the
sum
will
exhibit
the
total
sum
,
due
at
the
end
of
the
said
six
years
,
which
you
may
perceive
by
the
following
work
to
be
690
l.
which
is
the
answer
to
the
foregoing
Question
.
The
Construction
of
Table
II.
Upon
the
foregoing
reason
is
grounded
the
Calculation
of
the
following
Table
,
which
sheweth
the
amount
of
1l
.
annuity
,
being
forborne
to
the
end
of
any
number
of
years
under
31
,
Interest
being
allowed
for
each
yearly
payment
after
the
rate
of
6
per
Cent.
per
Annum
,
Simple
Interest
.
The
first
number
in
the
Table
being
1
l.
which
is
that
due
at
the
end
of
the
first
year
,
no
interest
being
due
for
that
;
the
second
number
in
the
Table
is
2.06
,
which
is
the
first
and
second
years
payment
,
and
the
interest
of
1l
.
for
one
year
,
being
that
which
was
due
at
the
end
of
the
first
year
;
The
third
number
in
the
Table
is
3.18
l.
being
the
increase
of
1
l.
for
2
years
added
to
the
second
number
in
that
Table
which
is
2.06
,
for
the
amount
of
1
l.
at
the
end
of
3
years
is
1.12
which
added
to
2.06
the
second
number
it
makes
3.18
for
the
third
number
;
The
fourth
number
is
the
amount
of
1
l.
for
3
years
which
is
1.18
added
to
the
number
before
it
,
viz.
the
third
number
,
proceeding
in
the
same
method
,
till
you
have
composed
the
Table
at
your
pleasure
,
each
number
in
the
Table
being
1
l.
and
the
amount
of
1
l.
(
for
so
many
years
as
it
standeth
against
in
the
Table
made
less
by
one
.
)
added
to
the
number
immediately
preceding
it
.
TABLE
II.
Which
sheweth
in
pounds
and
Decimal
parts
of
a
pound
the
amount
of
1
l
,
annuity
being
forborne
to
the
end
of
any
number
of
years
under
31
,
Simple
Interest
being
computed
after
the
Rate
of
6
per
Cent.
per
Annum
.
Years
|
|
1
|
1.00
|
2
|
2.06
|
3
|
3.18
|
4
|
4.36
|
5
|
5.60
|
6
|
6.90
|
7
|
8.26
|
8
|
9.68
|
9
|
11.16
|
10
|
1●…
.
70
|
11
|
14.30
|
12
|
15.96
|
13
|
17.68
|
14
|
19.46
|
15
|
21.30
|
16
|
23.20
|
17
|
25.16
|
18
|
●…7
.
18
|
19
|
29.26
|
20
|
31.40
|
21
|
33.60
|
22
|
35.86
|
23
|
38.18
|
24
|
40.56
|
25
|
43.00
|
26
|
45.50
|
27
|
48.06
|
28
|
50.68
|
29
|
53.36
|
30
|
56.10
|
The
Use
of
Table
II.
In
the
preceding
Table
in
the
Collum
under
the
word
Years
,
are
set
down
every
year
successively
from
1
to
30
,
and
the
number
in
the
Table
placed
against
each
year
,
is
the
amount
of
1
l
annuity
,
in
pounds
and
Decimal
parts
of
a
pound
,
being
forborne
so
many
years
as
it
is
placed
against
.
The
use
of
it
will
plainly
appear
by
the
solving
of
one
,
or
two
Questions
,
viz.
There
is
an
Annuity
of
134
l.
−
10s
.
−
6d
.
all
forborne
to
the
end
of
4
years
;
I
demand
how
much
is
due
to
the
Creditor
at
the
end
of
the
said
Term
,
Simple
Interest
being
allowed
after
the
rate
of
6
per
Cent.
per
Annum
?
Facit
586l
.
−
10s
.
−
07d
.
To
answer
this
Question
,
First
,
I
look
for
4
years
,
in
the
Collum
of
years
,
and
the
number
against
it
is
4.36
which
is
the
amount
of
1
l.
Annuity
for
4
years
;
therefore
having
turned
the
1
s.
−
6d
.
(
in
the
given
annuity
)
into
a
Decimal
(
which
is
.525
)
I
say
by
the
Rule
of
3
thus
.
Thus
by
the
work
I
find
the
answer
to
be
58.529
l.
the
value
of
which
Decimal
by
the
brief
way
of
valuing
a
Decimal
laid
down
in
the
4th
Rule
of
the
2d
Chapter
,
I
find
to
be
580l
.
10s.
7d
.
And
it
is
plain
that
in
solving
Questions
by
this
Table
,
that
(
the
first
number
in
the
Rule
of
3
being
unite
)
if
you
multiply
the
given
Annuity
by
the
proper
Tabular
Number
,
that
then
the
product
will
be
the
answer
.
Example
2.
What
is
the
amount
of
an
Annuity
of
150
l.
10
s.
being
forborne
to
the
end
of
7
years
,
allowing
Simple
Interest
after
the
Rate
of
6
per
Cent
,
per
Annum
?
Answer
,
1243
l.
−
02
s.
−
07
¼
d.
fere
.
XI
.
When
an
Annuity
or
yearly
Income
,
for
a
certain
number
of
years
to
come
,
is
to
be
sold
for
ready
Money
,
and
the
seller
is
to
allow
the
Buyer
Rebate
at
The
Rebate
of
Annuities
at
Simple
Interest
Simple
Interest
for
his
present
payment
,
then
in
this
case
the
buyer
ought
to
pay
so
much
present
money
for
each
yearly
payment
,
as
being
put
out
at
Simple
Interest
for
so
many
years
as
it
is
Rebated
for
,
it
would
then
amount
to
one
yearly
payment
,
and
the
sum
of
all
those
present
worths
will
be
the
present
worth
of
the
Annuity
required
,
the
Rule
will
appear
very
plain
by
the
following
Example
.
There
is
an
Annuity
or
Lease
of
100
l.
per
Annum
to
continue
6
years
yet
to
come
to
be
sold
for
ready
Money
,
the
Seller
being
to
allow
the
Buyer
Rebate
at
6
per
Cent.
per
Annum
,
Simple
Interest
,
now
I
desire
to
know
how
much
present
Money
will
buy
out
the
said
Lease
?
Facit
499
l.
−
09
s.
−
04
¼
d.
fere
.
It
is
evident
that
if
we
find
out
the
present
worth
of
100
l.
due
at
the
end
of
the
first
year
,
and
also
the
present
worth
of
a
100
l.
due
at
the
end
of
the
second
year
,
and
the
present
worth
of
100
l.
due
at
the
end
of
the
third
year
,
and
likewise
the
present
worth
of
100
l.
due
at
the
end
of
the
fourth
,
fifth
,
and
sixth
years
,
and
add
all
these
present
worths
together
,
their
sum
will
be
the
present
worth
of
the
given
Annuity
;
which
several
present
worths
are
found
out
according
to
the
ninth
Rule
,
by
the
several
proportions
following
,
viz.
So
that
you
see
by
the
foregoing
proportions
,
the
present
worth
of
100
l.
per
Annum
to
continue
six
years
,
allowing
Rebate
at
6
per
Cent.
per
Annum
,
Simple
Interest
,
is
499.468754
l.
=
499
l.
9
s.
04
¼d
.
Upon
the
foregoing
eleventh
Rule
is
grounded
the
construction
and
calculation
of
the
following
Table
which
The
construction
of
the
3
Table
.
sheweth
the
present
worth
of
1
pound
annuity
to
continue
any
number
of
years
under
31
Simple
Interest
being
computed
after
the
rate
of
6
per
Cent.
per
Annum
;
The
first
number
in
the
Table
is
.943396
which
is
the
present
worth
of
1
pound
due
at
the
end
of
a
year
to
come
.
The
second
number
in
the
Table
is
1.836253
,
which
is
the
sum
of
the
present
worths
of
1
l.
due
at
the
end
of
two
years
to
come
,
and
of
1
l.
due
at
the
end
of
one
year
to
come
added
together
;
And
the
third
number
in
the
Table
is
2.683710
which
is
the
sum
of
the
present
worths
of
1
l.
due
at
the
end
of
3
,
2
,
and
1
years
to
come
.
And
after
the
same
method
is
the
whole
Table
calculated
.
But
the
numbers
in
the
said
Table
may
more
easily
be
found
out
thus
,
viz.
Look
in
the
first
Table
,
and
let
the
first
number
of
that
be
the
first
number
of
this
third
Table
,
and
let
the
sum
of
the
first
number
in
this
,
and
the
second
number
in
that
be
the
second
number
in
this
Table
,
and
for
the
third
number
in
this
Table
take
the
sum
of
the
second
in
this
,
and
the
third
in
that
Table
,
and
in
this
manner
you
may
proceed
till
you
have
composed
the
whole
Table
.
TABLE
III.
Which
sheweth
the
present
worth
of
1
l.
annuity
to
continue
any
number
of
years
under
31
,
Simple
Interest
being
computed
at
6
per
Cent.
per
An.
Years
|
|
1
|
.943396
|
2
|
1.836253
|
3
|
2.683710
|
4
|
3.490161
|
5
|
4.259391
|
6
|
4.994685
|
7
|
5.698900
|
8
|
6.374575
|
9
|
7.023925
|
10
|
7.648925
|
11
|
8.251334
|
12
|
8.832729
|
13
|
9.394526
|
14
|
9.938004
|
15
|
10.464319
|
16
|
10.974523
|
17
|
11.469572
|
18
|
11.950341
|
19
|
12.437630
|
20
|
12.892175
|
21
|
13.334652
|
22
|
13.765686
|
23
|
14.175524
|
24
|
14.585360
|
25
|
14.985360
|
26
|
15.375985
|
27
|
15.757664
|
28
|
16.120798
|
29
|
16.485761
|
30
|
16.842904
|
The
Use
of
the
foregoing
Table
.
III.
In
the
foregoing
third
Table
,
in
the
left
hand
Collum
under
the
Title
of
years
,
are
expressed
all
the
integral
numbers
,
from
1
to
30
,
which
signifie
so
many
years
,
and
the
numbers
in
the
Right
hand
Collum
which
are
placed
against
the
number
of
years
are
pounds
,
and
decimal
parts
of
a
pound
sterling
,
and
every
one
of
them
are
the
present
worth
of
1
pound
Annuity
to
continue
so
many
years
to
come
as
are
placed
against
them
in
the
Collum
of
years
,
Rebate
being
allowed
at
Simple
Interest
6
per
Cent.
per
An.
As
,
suppose
there
were
a
Lease
of
20
shillings
per
annum
to
continue
6
years
,
to
be
sold
for
present
money
,
allowing
the
buyer
Rebate
at
6
per
Cent.
per
annum
Simple
Interest
.
I
desire
to
know
how
much
is
its
present
worth
?
To
answer
this
,
I
look
in
the
Collume
of
years
for
6
,
and
in
the
next
Collume
on
the
right
hand
just
against
6
you
have
4.99468
5
l.
=
4
l.
−
19
s.
−
10
¾
d.
which
is
the
answer
to
the
Question
.
And
by
the
help
of
this
Table
may
the
present
worth
of
any
Annuity
to
continue
any
number
of
years
under
31
be
found
out
,
allowing
Rebate
at
6
per
Cent.
per
Annum
,
Simple
Interest
,
by
one
single
Rule
of
3
Direct
,
according
to
the
manner
of
solving
the
following
question
,
viz.
Quest.
1.
There
is
a
Lease
of
18
years
yet
to
come
,
of
the
yearly
value
of
130
l.
to
be
sold
for
ready
Money
,
and
the
purchaser
is
to
be
allowed
Rebate
after
the
rate
of
6
per
Cent.
per
Annum
,
Simple
Interest
,
now
I
demand
how
much
is
the
present
worth
of
this
Lease
?
Facit
1553
l.
−
10
s.
−
10
¾
d.
First
,
I
look
in
the
Table
for
18
years
,
and
over
against
it
on
the
right
hand
I
find
●…1
.
950341
which
is
the
present
worth
of
1
pound
annuity
to
continue
18
years
,
&c.
Therefore
by
the
Rule
of
3
Direct
,
I
say
So
that
by
the
work
you
find
the
answer
to
be
1553.544
l.
&c.
or
1553
l.
−
10
s.
−
10
¾
d.
very
near
,
which
said
answer
is
nothing
else
but
the
product
of
the
Tabular
number
,
(
11.950341l
.
)
multiplyed
by
the
given
annuity
(
130
l.
)
For
it
is
evident
,
that
if
the
present
worth
of
1
pound
annuity
to
continue
18
years
be
11.950341
l.
then
the
present
worth
of
130
l.
per
annum
,
to
continue
the
same
number
of
years
(
and
Rebate
being
allowed
at
the
same
Rate
per
Cent.
per
an
.
for
the
one
as
for
the
other
)
must
be
130
times
as
much
.
But
when
Rebate
is
to
be
allowed
after
any
other
rate
then
6
per
Cent.
per
annum
,
then
the
foregoing
Table
will
not
at
all
be
useful
,
but
you
must
have
recourse
to
a
Table
calculated
for
the
same
rate
of
interest
,
which
you
may
easily
perform
at
leisure
by
the
foregoing
Rules
.
Quest.
2.
What
Annuity
to
continue
18
years
will
1553.5443
30
purchase
,
allowing
the
Buyer
Simple
Interest
at
6
per
Cent.
per
annum
?
Facit
130
l.
This
Question
is
but
the
converse
of
the
former
,
and
may
be
thus
Resolved
,
viz.
Táke
the
Tabular
number
corresponding
to
18
years
,
which
is
11.950341
by
which
divide
the
given
purchase
Money
,
and
the
Quotient
will
give
you
the
annuity
that
it
will
purchase
,
viz.
So
that
by
the
work
I
find
it
will
purchase
an
Annuity
of
130l
.
to
continue
18
years
.
The
reason
of
the
work
is
plain
,
for
if
the
Tabular
number
correspondent
to
18
years
be
the
present
worth
of
1
l.
Annuity
to
continue
18
years
to
come
,
then
it
is
certain
that
so
much
money
as
is
expressed
by
that
Tabular
number
,
will
purchase
an
Annuity
of
1
l.
to
continue
18
years
:
And
consequently
we
may
find
by
help
of
the
said
Table
what
annuity
any
other
sum
of
money
will
purchase
,
to
continue
any
number
of
years
not
exceeding
30
,
by
a
single
Rule
of
3
Direct
,
as
in
the
last
Question
,
the
proportion
is
as
followeth
,
viz.
And
it
is
no
more
in
effect
than
a
sum
in
Division
,
for
the
second
number
(
being
1
)
neither
multiplyeth
,
nor
divideth
,
&c.
By
what
hath
been
said
concerning
the
use
of
the
foregoing
Table
,
you
may
perceive
that
the
p●…esent
worth
of
an
Annuity
is
found
out
by
multiplication
,
and
to
know
what
annuity
any
sum
will
purchase
is
performed
by
Division
.
I
might
have
made
Tables
for
other
Rates
of
Interest
,
but
Simple
Interest
being
seldom
allowed
in
the
purchasing
or
valuing
of
Leases
and
Annuities
,
(
they
being
generally
purchased
at
Compound
Interest
,
or
Interest
upon
Interest
)
makes
me
forbear
,
and
indeed
at
Simple
Interest
a
Lease
is
over-valued
.
CHAP.
XIII
.
Of
Compound
Interest
.
I.
WHat
hath
been
said
in
the
last
Chapter
,
I
judge
sufficient
for
the
understanding
of
the
Nature
and
use
of
Simple
Interest
,
and
that
being
well
understood
,
the
nature
of
Compound
Interest
will
not
seem
difficult
to
the
studious
Learner
,
and
the
better
he
is
acquainted
with
the
nature
of
Simple
Interest
,
so
much
the
easier
will
he
come
to
the
knowledge
of
the
nature
,
and
use
of
Compound
Interest
.
II.
Compound
Interest
is
,
when
a
sum
of
money
is
put
out
to
interest
,
and
the
interest
thereof
becoming
due
is
still
continued
in
the
hands
of
the
Debtor
,
so
as
to
become
part
of
the
principal
,
Interest
being
reckoned
for
it
from
the
time
it
becometh
due
,
for
which
reason
it
is
called
Interest
upon
Interest
:
And
as
Simple
Interest
increaseth
by
a
series
of
Arithmetical
proportionals
continued
;
so
doth
Compound
interest
increase
by
a
rank
or
series
of
continual
Geometrical
proportionals
.
For
when
a
sum
of
money
is
put
out
to
interest
at
any
Rate
per
Cent.
per
Annum
,
(
as
suppose
100
l.
to
be
put
out
to
receive
at
the
end
of
one
year
6
l.
for
its
interest
)
it
is
evident
that
if
the
interest
(
being
6
l.
)
be
continued
in
the
hands
of
the
Debtor
,
there
will
be
at
the
end
of
the
second
year
the
increase
of
106
l.
which
is
112.36
l.
and
at
the
third
years
end
there
will
be
the
increase
of
112.36
l.
so
that
every
number
proceedeth
from
that
going
before
it
,
after
the
same
Rate
or
Reason
as
100
proceedeth
from
100
,
as
you
see
following
.
So
that
by
the
Augmentation
of
100
l.
in
4
years
,
you
have
this
rank
of
Geometrical
proportionals
continued
viz.
100
,
106
,
112.36
,
119.1016
.
and
126.247696
which
is
in
number
5
,
viz.
more
by
one
than
is
the
number
of
years
,
the
last
of
which
is
the
amount
of
100
l.
at
6
per
Cent.
for
4
years
reckoning
Compound
Interest
,
or
Interest
upon
Interest
,
and
each
of
these
proportionals
proceedeth
from
that
going
before
it
as
106
proceedeth
from
100
,
that
is
to
say
,
every
of
the
said
proportionals
,
is
in
such
proportion
to
that
which
goeth
before
it
as
106
is
to
100
,
or
as
100
is
to
106
,
so
is
any
one
of
them
,
to
that
which
followeth
it
,
or
if
you
take
any
3
of
them
which
are
placed
together
,
there
is
this
proportion
between
them
,
viz
▪
As
the
first
of
those
three
is
to
the
second
,
so
is
the
second
to
the
third
,
and
the
third
to
the
fourth
,
and
the
fourth
to
the
fifth
,
and
the
fifth
to
the
sixth
,
&c.
whence
it
is
evident
that
they
have
amongst
themselves
this
following
Qualification
,
viz.
that
the
Square
of
any
●…ne
of
them
is
equal
to
the
Rectangle
,
or
Product
,
made
by
that
which
is
placed
immedi●…tely
before
it
,
and
that
immediately
after
it
,
and
the
same
would
it
be
if
there
were
never
so
many
Terms
,
and
is
a
peculiar
property
of
all
numbers
that
are
Geometrical
proportionals
continued
.
III.
The
Interest
of
100
l.
for
a
year
being
known
,
the
Compound
Interest
of
any
other
sum
for
any
number
of
years
may
be
likewise
found
out
by
so
many
single
Rules
of
3
,
as
there
are
given
years
,
for
,
As
100
l
,
is
to
its
increase
for
one
year
,
so
is
any
other
sum
to
its
increase
for
the
same
time
,
and
so
is
the
first
years
increase
to
the
second
,
and
the
second
years
increase
to
the
third
,
and
so
is
the
third
years
increase
to
the
fourth
,
&c.
Example
.
Let
it
be
required
to
find
how
much
350
l.
will
be
increased
to
being
put
out
to
Interest
at
6
per
Cent.
per
Annum
Compound
Interest
for
5
years
?
Answer
,
468l
.
−
7s
.
−
4
,
d.
fere
.
Sec
the
following
work
.
Whereby
you
see
that
350
l.
being
put
out
to
Interest
after
the
rate
of
6
per
Cent
will
at
the
first
years
end
be
increased
to
371
l.
And
371
l.
being
put
out
for
the
second
year
,
will
be
increased
to
393.26
l.
and
393.26
l.
being
made
a
principal
,
and
put
out
at
the
same
Rate
for
the
third
year
,
will
at
the
end
thereof
be
increased
to
416
8556
l.
and
at
the
end
of
5
years
it
will
be
increased
to
468
37895216
l.
And
upon
the
aforesaid
grounds
is
Calculated
the
following
Table
1
,
whose
Construction
and
use
immediately
followeth
the
same
.
TABLE
I.
Which
sheweth
what
one
Pound
will
amount
to
,
being
forborne
to
the
end
of
any
number
of
years
to
come
,
not
exceeding
30
,
Compound
Int.
being
computed
at
5
,
6
,
7
,
8
,
9
,
or
10
per
Cent.
per
annum
.
Years
|
5.
|
6.
|
7.
|
8.
|
9.
|
10
|
1
|
1.05000
|
1.06000
|
1.07000
|
1.08000
|
1.09000
|
1.10000
|
2
|
1.10250
|
1.12360
|
1.14490
|
1.16640
|
1.18810
|
1.21000
|
3
|
1.15762
|
1.19101
|
1.22504
|
1.25971
|
1.29502
|
1.33100
|
4
|
1.21550
|
1.26247
|
1.31079
|
1.36048
|
1.41158
|
1.46410
|
5
|
1.27628
|
1.33822
|
1.40255
|
1.46932
|
1.53862
|
1.61051
|
6
|
1.34009
|
1
.
4●…851
|
1.50073
|
1.58687
|
1.67710
|
1.77156
|
7
|
1.40710
|
1.50363
|
1.60578
|
1.71382
|
1.82803
|
1.94871
|
8
|
1.47745
|
1.59384
|
1.71818
|
1.85093
|
1.99256
|
2.14358
|
9
|
1
.
55●…32
|
1.68947
|
1.83845
|
1.99900
|
2.17189
|
2.35794
|
10
|
1.62889
|
1.79084
|
1.96715
|
2.15892
|
2.36736
|
2.59374
|
11
|
1.71033
|
1.89829
|
2.10485
|
2
33163
|
2.58042
|
2.85311
|
12
|
1.79585
|
2.01219
|
2.25219
|
2.51817
|
2.81266
|
3.13842
|
13
|
1.88564
|
2.13292
|
2.40984
|
2
.
7196●…
|
3.06580
|
3
.
452●…7
|
14
|
1.97993
|
2.26090
|
2.57853
|
2.93719
|
3.34172
|
3.79749
|
15
|
2.07892
|
2.39655
|
2.75903
|
3.17216
|
3
64248
|
4.17724
|
16
|
2.18287
|
2.54035
|
2.95216
|
3.42594
|
3.97030
|
4.59497
|
17
|
2
.
29●…01
|
2.69277
|
3.15881
|
3.70001
|
4.32763
|
5
05447
|
18
|
2.40661
|
2.85433
|
3.37993
|
3.99601
|
4
71712
|
5.55991
|
19
|
2.52695
|
3.02559
|
3.61652
|
4.31570
|
5.14166
|
6
.
1159●…
|
20
|
2.65329
|
3.20713
|
3.86968
|
4.66095
|
5.60441
|
6.72749
|
21
|
2.78596
|
3.39956
|
4.14056
|
5.03383
|
6.10880
|
7.40024
|
22
|
2.92526
|
3.60353
|
4
43040
|
5.43654
|
6.65860
|
8.14027
|
23
|
3.07152
|
3.81975
|
4.74053
|
5.87146
|
7.25787
|
8.95430
|
24
|
3.22509
|
4.04893
|
5.07236
|
6
34118
|
7.91108
|
9.84973
|
25
|
3.38635
|
4.29187
|
5.42743
|
6.84847
|
8.62308
|
10.83470
|
26
|
3.55567
|
4.54938
|
5.80735
|
7.39635
|
9.39915
|
11.91817
|
27
|
3.73345
|
4.82234
|
6.21386
|
7.98806
|
10.24508
|
13.10999
|
28
|
3.92012
|
5.11178
|
6.64883
|
8.62710
|
11.16713
|
14.42099
|
29
|
4.11613
|
5.41838
|
7.11425
|
9.31727
|
12.17218
|
15.86309
|
30
|
4.32194
|
5.74349
|
7.61225
|
10.06265
|
13.26767
|
17.44940
|
The
Construction
of
the
foregoing
TABLE
I.
By
the
third
Rule
foregoing
it
is
evident
that
the
Interest
of
100
l.
for
a
year
being
known
,
the
Compound
Interest
for
any
other
sum
may
be
found
out
for
any
number
of
years
;
According
to
which
Rule
all
the
numbers
in
the
said
Table
are
found
out
,
being
the
amount
of
1
l.
at
Compound
Interest
for
any
number
of
years
,
not
exceeding
30
,
being
put
out
at
any
of
these
Rates
,
viz.
5
,
6
,
7
,
8
,
9
,
or
10
per
Cent.
per
Annum
,
which
numbers
are
found
out
by
the
Rule
of
Proportion
thus
,
By
which
means
the
four
first
numbers
in
the
econd
Collum
of
the
Table
(
being
placed
under
the
number
5
are
found
,
and
by
a
Continuation
of
the
same
operation
are
all
the
rest
of
the
numbers
in
that
Collum
found
out
;
which
is
indeed
nothing
else
but
a
continual
multiplication
of
the
first
number
,
(
viz.
1.05
)
into
it self
29
times
,
and
so
the
last
number
in
that
Collum
is
the
thirtieth
power
of
1.05
,
and
the
same
Collum
may
be
continued
to
any
other
number
of
years
at
pleasure
above
30
;
the
numbers
in
this
Collum
being
the
interest
of
1
l.
at
5
per
Cent.
per
Annum
Compound
Interest
for
30
years
.
The
numbers
in
the
third
Collum
under
the
Figure
6
,
are
the
increase
of
1
l.
at
●…
per
Cent.
per
Annum
.
Comp.
Int.
for
30
years
,
and
are
found
out
by
multiplying
1
.
0●…
into
it self
29
times
,
according
to
the
Rule
of
Continual
Multiplication
.
The
like
is
to
be
understood
of
all
the
rest
.
The
use
of
the
foregoing
TABLE
.
In
the
first
Collum
of
the
Table
under
the
Title
years
,
are
expressed
the
number
of
years
from
1
to
30
,
and
in
the
second
Collum
under
the
figure
5
,
and
against
every
respective
year
are
expressed
the
increase
of
1
l.
being
put
out
at
5
per
Cent.
per
Annum
,
Compound
Interest
.
In
the
third
Collume
,
under
the
number
6
is
expressed
the
yearly
increase
of
1
l.
being
put
out
at
6
per
Cent.
per
Annum
,
Compound
Int.
And
so
in
the
4
,
5
,
6
,
and
7
Collumes
,
are
the
yearly
amounts
of
1
l.
at
7
,
8
,
9
,
and
10
per
Cent.
per
Annum
,
Compound
Interest
.
All
which
numbers
in
the
said
Table
are
multiplyars
,
for
the
producing
of
the
amount
or
Increase
of
any
other
Sum
being
put
out
at
Compound
interest
,
at
any
rate
of
Interest
,
and
for
any
number
of
years
therein
expressed
,
as
will
appear
by
the
following
Examples
.
Example
.
1.
I
demand
the
full
amount
of
365
l.
being
put
to
Interest
for
9
years
,
Interest
being
computed
after
the
rate
of
6
per
Cent.
per
Annum
,
Compound
Interest
?
Facit
616
l.
−
13
s.
01
d.
Here
because
the
sum
proposed
is
put
out
at
6
per
Cent.
and
for
9
years
,
I
look
in
the
Collume
of
6
per
Cent.
which
is
the
third
Collum
of
the
Table
under
the
figure
6
,
and
just
against
9
in
the
Collum
of
years
,
I
find
1.68947
which
is
the
increase
of
1
l.
being
forborne
the
same
time
,
and
at
the
same
Rate
of
Interest
,
wherefore
,
by
the
Rule
of
3
I
say
So
that
by
the
work
I
find
that
if
the
sum
of
365
l.
be
all
forborne
to
the
end
of
9
years
,
and
interest
be
computed
for
the
same
at
6
per
Cent.
per
Annum
Compound
Interest
,
it
will
then
be
increased
to
616.65655
which
is
616
l.
13
s.
1
½
d.
Example
2.
What
will
128
l.
−
16
s.
−
08
d.
be
increased
to
?
The
utmost
improvement
thereof
being
made
for
15
years
at
7
per
Cent.
per
annum
,
Compound
Interest
?
Facit
355
l.
−
09
s.
−
01
d.
First
,
turn
the
16
s.
−
8
d.
into
the
Decimal
of
a
pound
by
the
2d
Rule
of
the
2d
Chapter
foregoing
,
and
you
will
find
it
to
be
.8333
,
so
that
the
given
sum
is
128.8333
,
&c.
Now
to
answer
the
Question
,
I
look
into
the
foregoing
Table
,
and
in
the
Collum
of
7
per
Cent
:
and
just
against
15
years
I
find
2.75903
which
is
the
uttermost
increase
of
1
l.
for
15
years
at
7
per
Cent.
Compound
Interest
,
by
which
if
you
multiply
the
given
sum
,
the
Product
will
be
the
answer
to
the
Question
,
as
by
the
following
work
will
plainly
appear
.
By
the
foregoing
work
the
answer
is
found
to
be
355.4549
&c.
=
355
l.
−
09
s.
−
01
d.
But
if
any
sum
be
put
out
at
Compound
Int.
for
moneths
,
or
dayes
over
and
above
the
given
number
of
years
,
then
the
work
will
be
somewhat
different
from
the
former
;
for
first
you
must
find
out
the
amount
of
the
given
sum
,
for
the
given
number
of
years
,
and
then
by
the
8th
Rule
of
the
foregoing
Chapter
find
out
the
Interest
of
that
amount
for
the
odd
time
,
being
either
moneths
or
dayes
under
a
year
,
and
that
Int.
being
added
to
the
foresaid
amount
,
that
sum
will
be
the
answer
to
the
Question
;
This
is
so
obvious
that
it
needeth
no
Example
.
IV.
When
a
sum
of
Money
due
at
the
end
of
any
number
of
years
to
come
is
to
be
satisfied
with
present
money
,
Of
Rebate
or
discount
at
Compound
Interest
.
allowing
Rebate
at
Compound
Interest
,
there
must
be
found
a
Rank
or
series
of
Continual
proportionals
,
more
in
number
by
one
than
the
number
of
years
for
which
the
discount
is
proposed
,
of
which
rank
or
series
of
proportionals
,
the
sum
to
be
satisfied
by
present
payment
must
be
the
first
,
and
the
second
must
decrease
from
that
after
the
same
rate
or
proportion
as
100
decreaseth
from
the
sum
of
100
added
to
its
interest
for
one
year
,
after
the
Rate
of
interest
propounded
;
that
is
to
say
,
as
100
proceedeth
from
106
,
or
108
if
the
interest
be
6
or
8
per
Cent.
and
after
the
same
Rate
,
or
Reason
must
the
third
decrease
from
the
second
,
and
the
fourth
from
third
,
&c.
When
a
Question
is
stated
for
the
Rebate
of
Money
at
Compound
Interest
,
it
is
solvable
by
as
many
single
Rules
of
3
,
as
the
number
of
years
for
which
the
sum
proposed
is
to
be
Rebated
,
and
it
is
nothing
else
but
the
inverse
of
the
third
Rule
of
this
Chapter
,
as
may
be
proved
by
the
working
of
the
following
Question
,
taken
out
of
the
said
Rule
,
where
it
is
proved
that
350
l.
being
forborne
in
the
Debtors
hands
for
5
years
at
6
per
Cent.
Compound
Interest
,
it
will
then
be
increased
to
468
.
3800121●…
;
now
let
the
said
Question
be
inverted
thus
,
viz.
There
is
a
sum
of
Money
,
viz.
468.38001216
due
at
the
end
of
5
years
to
come
,
now
I
demand
how
much
present
Money
will
satisfie
the
said
Debt
,
Rebate
being
allowed
after
the
Rate
of
6
per
Cent.
per
annum
,
Compound
Interest
?
First
,
I
say
,
as
106
is
to
100
,
so
is
the
sum
due
at
the
end
of
5
years
,
viz.
468.38001216
,
to
441.867936
,
which
is
the
sum
due
at
the
fourth
years
end
,
and
so
is
the
sum
due
at
the
fourth
years
end
,
to
the
sum
due
at
the
third
years
end
,
&c.
as
by
the
work
appeareth
.
So
that
by
the
foregoing
work
you
see
that
if
468.38001216
l.
be
due
at
the
end
of
5
years
to
come
,
and
is
to
be
satisfied
by
the
payment
of
present
money
,
Rebate
being
allowed
at
6
per
Cent.
per
annum
,
Compound
Interest
,
350
l.
is
the
sum
required
.
And
upon
this
Rule
is
grounded
the
Calculation
of
the
following
Table
,
which
sheweth
what
〈◊〉
l.
due
at
the
end
of
any
number
of
years
to
come
,
not
exceeding
30
in
worth
in
present
Money
,
Rebate
being
reckoned
at
any
of
these
Rates
,
viz.
5
,
6
,
7
,
8
,
9
,
or
10
per
Cent.
per
an
.
Compound
Interest
.
TABLE
II.
Which
sheweth
the
present
worth
of
one
pound
payable
at
the
end
of
any
number
of
years
to
come
,
not
exceeding
30
,
Rebate
being
yearly
allowed
at
5
,
6
,
7
,
8
,
9
,
or
10
per
Cent.
per
Annum
,
Compound
Interest
.
Years
|
5.
|
6.
|
7.
|
8.
|
9.
|
10.
|
1
|
.952381
|
.943396
|
.934579
|
.925925
|
.917431
|
.909090
|
2
|
.907029
|
.889996
|
.873438
|
.857338
|
.841680
|
.826446
|
3
|
.363837
|
.839619
|
.816297
|
.793832
|
.772183
|
.751314
|
4
|
.822702
|
.792093
|
.762895
|
.735029
|
.708425
|
.683013
|
5
|
.783526
|
.747258
|
.712986
|
.680583
|
.649331
|
.620921
|
6
|
.746215
|
.704960
|
.666342
|
.630169
|
.596267
|
.564474
|
7
|
.710681
|
.665057
|
.622749
|
.583490
|
.
547●…34
|
.513158
|
8
|
.676839
|
.627412
|
.582009
|
.540268
|
.501866
|
.466507
|
9
|
.644608
|
.591898
|
.543933
|
.500248
|
.460427
|
.424097
|
10
|
613913
|
.558391
|
.508349
|
.463193
|
.422410
|
.385543
|
11
|
.584679
|
.526787
|
.475092
|
.428882
|
.387532
|
.350494
|
12
|
.556837
|
.496989
|
.444012
|
.397113
|
.355534
|
.318630
|
13
|
.530321
|
.468839
|
.414964
|
.367697
|
.326178
|
.289664
|
14
|
.505067
|
.442300
|
.387817
|
.340461
|
.299246
|
.263331
|
15
|
.481017
|
.417265
|
.362446
|
.315241
|
.274538
|
.239392
|
16
|
.458111
|
.393646
|
.338734
|
.291890
|
.251869
|
.217629
|
17
|
.436296
|
.371364
|
.316574
|
.270269
|
.231073
|
.197844
|
18
|
.415520
|
.350343
|
.295864
|
.250249
|
.211993
|
.179858
|
19
|
.395733
|
.330512
|
.276508
|
.231712
|
.194489
|
.163508
|
20
|
.376889
|
.311804
|
.258419
|
.214548
|
.178430
|
.148643
|
21
|
.358942
|
.294155
|
.241513
|
.198655
|
.163698
|
.135130
|
22
|
.341849
|
.277505
|
.225713
|
.183940
|
.150181
|
.122846
|
23
|
.325571
|
.261797
|
.210947
|
.170315
|
.137781
|
.111678
|
24
|
.310067
|
.246978
|
.197146
|
.157699
|
.126405
|
.101525
|
25
|
.295302
|
.232998
|
.184249
|
.146018
|
.115967
|
.092296
|
26
|
281240
|
.219810
|
.172195
|
.135201
|
.106392
|
.083905
|
27
|
.267848
|
.207367
|
.160930
|
.125186
|
.097607
|
.076277
|
28
|
.255093
|
.195630
|
.150402
|
.115913
|
.089548
|
.069343
|
29
|
.242946
|
.184556
|
.140562
|
.107327
|
.082154
|
.063039
|
30
|
.231377
|
.174110
|
.131367
|
.099377
|
.075371
|
.057308
|
The
Construction
of
the
foregoing
TABLE
.
By
the
Fourth
Rule
of
this
Chapter
is
plainly
shewn
the
manner
of
finding
the
present
worth
of
any
sum
of
money
due
at
the
end
of
any
number
of
years
to
come
,
Rebate
being
computed
at
Compound
Interest
,
and
after
the
same
manner
are
all
the
numbers
in
the
foregoing
Table
found
as
you
may
see
in
the
following
Example
,
where
the
four
first
numbers
in
the
third
Collum
are
methodically
found
out
by
the
Rule
,
that
being
the
Collum
of
Rebate
at
6
per
Cent.
per
annum
.
So
that
by
the
foregoing
proportions
,
I
say
first
,
if
106
l.
be
decreased
to
100
l.
what
will
1
l.
be
decreased
to
?
Answer
,
to
.94339
l.
&c
=
18
s.
10
〈◊〉
.
The
five
first
Figures
thereof
being
the
first
number
in
the
third
Collum
of
the
foregoing
Table
and
it
sheweth
that
the
present
worth
of
〈◊〉
l.
due
at
the
end
of
one
year
to
come
,
Rebate
being
allowed
at
6
per
Cent.
is
.94339
l.
=
18
s
−
10
d.
Secondly
,
I
say
by
the
Rule
of
3
,
If
106
l.
be
decreased
to
100
l.
what
will
.94339
&c.
be
decreased
to
?
The
Answer
is
.88999
l.
&c.
=
17
s.
●…9
〈◊〉
.
fere
.
And
this
is
the
second
number
in
the
third
Collum
of
the
said
Table
,
and
is
placed
against
2
years
in
the
first
Collum
,
and
sheweth
the
present
worth
of
1
l.
due
at
the
end
of
two
years
to
come
,
Rebate
being
allowed
after
the
Rate
of
6
per
Cent.
per
Annum
,
Compound
Interest
.
And
after
the
same
manner
are
all
the
rest
of
the
numbers
in
the
said
Collum
of
6
per
Cent.
found
out
,
and
also
all
the
other
Decimal
Fractions
in
the
second
,
fourth
,
fifth
,
sixth
,
&c.
Collums
,
(
shewing
the
Rebate
of
one
pound
for
any
number
of
years
not
exceeding
30
,
at
5
,
7
,
8
,
9
,
and
10
per
Cent.
(
mutatis
mutandis
.
)
The
Use
of
the
foregoing
TABLE
II.
The
first
Collum
is
the
number
of
years
for
the
Rebate
of
1
l.
and
the
numbers
in
the
rest
of
the
Collums
,
are
Decimal
Fractions
,
shewing
the
present
worth
of
1
l.
due
at
the
end
of
so
many
years
to
come
,
as
they
are
placed
against
in
the
Collum
of
years
;
Rebate
being
allowed
at
the
same
rate
of
Interest
under
which
they
are
placed
,
the
figures
5
,
6
,
7
,
8.
9
,
and
10
placed
at
the
top
,
denoting
the
same
.
An
Example
or
two
will
make
its
use
more
plain
.
Example
.
1.
I
demand
how
much
present
money
will
satisfie
a
Debt
of
684
l.
due
at
the
end
of
6
years
to
come
,
Allowing
Rebate
after
the
rate
of
8
per
Cent.
per
annum
,
Compound
Interest
?
To
answer
this
Question
,
look
in
the
Collum
of
8
per
Cent.
and
against
6
years
I
find
this
number
,
viz.
.63017
which
sheweth
that
if
1
l.
be
due
at
the
end
of
6
years
to
come
,
its
present
worth
is
.63017l
.
Rebate
being
allowed
after
the
Rate
of
8
per
Cent.
per
annum
,
Compound
Interest
.
Therefore
I
say
by
the
Rule
of
3
,
If
1l
.
be
decreased
to
.63017l
.
what
will
684l
.
be
decreased
to
at
that
rate
?
Facit
431.03628l
.
=
431l
.
0s.
8½
d.
as
by
the
work
appeareth
?
So
that
you
see
the
sum
proposed
being
multiplyed
by
the
proper
Tabular
number
,
produceth
the
Answer
to
the
Question
,
for
the
number
1
,
which
is
here
the
first
number
in
the
Rule
of
3
doth
not
either
multiply
or
divide
,
and
therefore
the
answer
is
found
out
by
multiplication
only
.
Observe
the
work
of
the
next
Example
.
Example
.
What
is
the
present
worth
of
164l
.
−
15s
.
due
at
the
end
of
9
years
to
come
,
allowing
Rebate
after
the
Rate
of
6
per
Cent.
per
annum
,
Compound
Interest
?
Facit
97l
.
−
10s
.
−
03.
〈◊〉
.
Look
in
the
Table
aforesaid
in
the
Collum
of
6
per
Cent.
and
against
9
in
the
Collum
of
years
you
will
find
this
number
,
viz.
.59189
which
is
the
present
worth
of
1l
.
due
at
the
end
of
9
years
to
come
,
and
is
the
proper
multiplyar
for
finding
the
answer
to
this
Question
,
as
by
the
work
.
The
answer
found
by
the
foregoing
operation
is
97.5138775
=
97l
.
−
10s
.
−
03¼
.
But
if
the
given
time
for
the
Rebate
of
any
sum
consisteth
of
odd
moneths
or
dayes
,
besides
years
,
then
in
such
case
,
the
Rebate
(
at
the
given
Rate
of
Interest
)
for
the
odd
time
must
be
found
(
by
the
9th
Rule
of
the
12
Chap.
foregoing
)
for
the
given
sum
,
and
then
the
present
worth
of
the
given
sum
thus
decreased
,
must
be
found
for
the
number
of
years
as
in
the
two
last
Examples
.
Example
3.
There
is
640l
.
−
10s
.
due
at
the
end
of
six
year
,
and
3
moneths
to
come
,
what
is
its
present
worth
,
Rebate
being
allowed
at
the
Rate
of
7
per
Cent.
per
annum
,
Compound
Interest
?
First
,
I
find
the
decrease
of
640.5l
.
for
3
moneths
thus
,
viz.
So
that
I
find
the
Int.
of
100l
.
for
3
moneths
at
7
per
Cent.
to
be
1.75l
.
which
added
to
100l
.
makes
101.75
,
then
to
find
the
decrease
of
640.5l
.
for
3
Moneths
:
I
say
,
So
that
I
find
by
the
last
proportion
,
that
if
at
the
end
of
6
years
,
3
moneths
:
There
was
due
640.5l
yet
at
the
end
of
6
years
there
will
be
due
but
629.484l
.
whose
present
worth
by
the
foregoing
directions
will
be
found
to
be
419l
.
−
9s
.
for
,
Read
the
9th
Rule
of
the
Twelfth
Chapter
foregoing
,
and
you
will
easily
understand
the
method
here
used
for
solving
Questions
of
this
nature
.
V.
Questions
in
Rebate
at
Compound
Interest
may
be
Resolved
by
the
First
Table
of
this
Chapter
which
sheweth
the
increase
of
1l
.
at
Compound
Interest
,
&c.
But
as
in
the
second
Table
you
make
the
Tabular
numbers
multiplyars
,
to
find
out
the
present
worth
of
a
sum
;
so
if
you
will
find
out
the
present
worth
of
a
sum
by
the
First
Table
,
you
must
then
make
those
Tabular
Numbers
Divisors
;
The
Reason
whereof
is
plain
,
for
the
First
Table
sheweth
the
increase
of
1l
.
for
30
years
,
&c.
But
they
may
likewise
serve
to
shew
what
sum
of
money
due
at
the
end
of
any
number
of
years
to
come
under
31
(
allowing
Rebate
according
to
the
rates
of
Interest
therein
mentioned
)
1l
.
present
money
will
satisfie
.
Now
to
Resolve
Questions
in
Rebate
by
this
Table
,
look
in
the
Collum
of
the
proposed
Interest
,
or
Rebate
,
and
against
the
proposed
number
of
years
is
the
Tabular
number
for
your
work
,
which
must
be
according
to
the
following
proportion
,
viz.
As
the
Tabular
Number
so
found
,
Is
to
1
,
So
is
the
sum
proposed
to
be
Rebated
,
To
its
present
worth
.
To
make
this
a
little
more
plain
,
I
shall
Answer
the
first
Question
in
the
Use
of
the
second
Table
,
by
the
help
of
the
first
Table
only
,
which
is
as
followeth
,
viz.
I
demand
how
much
present
Money
will
satisfie
a
Debt
of
684l
.
due
at
the
end
of
6
years
to
come
,
allowing
Rebate
after
the
rate
of
8
per
Cent.
per
annum
,
Compound
Interest
?
Look
in
Table
1.
in
the
Collum
of
8
per
Cent.
and
against
6
years
you
will
find
this
number
▪
viz.
1.58687
,
Therefore
the
proportion
is
as
followeth
.
So
that
by
this
proportion
the
Answer
is
431.03719l
.
=
431
l.
−
00
−
08
¾
d.
very
near
to
the
answer
before
found
by
the
second
Table
of
Rebate
.
VI.
When
an
Annuity
is
in
arrear
,
and
it
is
required
to
know
its
utmost
Improvement
,
accounting
Interest
The
manner
of
valuing
Annuities
that
are
in
arrear
.
upon
Interest
for
each
particular
sum
from
the
time
it
becomes
due
,
to
the
end
of
the
given
Term
of
years
.
The
manner
how
to
work
such
Questions
will
be
apparent
by
the
working
of
the
following
Question
,
viz.
There
is
an
Annuity
of
150
l.
to
continue
to
the
end
of
5
years
,
and
the
utmost
improvement
thereof
to
be
made
after
the
rate
of
6
per
Cent.
per
annum
,
Compound
Interest
;
now
I
demand
how
much
will
then
be
due
to
the
Creditor
?
It
is
evident
that
there
must
be
found
out
,
first
the
amount
of
150
l.
for
one
year
,
viz.
that
which
is
due
at
the
end
of
the
fourth
year
,
it
lying
in
the
Debtors
hands
all
the
fifth
year
.
Secondly
,
There
must
be
accounted
the
improvement
of
150
l.
for
2
●…ears
,
viz
that
which
is
〈◊〉
〈◊〉
the
end
of
the
third
year
▪
it
lying
in
the
Debtors
hands
the
years
and
fifth
〈◊〉
Thirdly
,
There
must
be
accounted
the
improvement
of
150
l.
for
3
years
,
viz.
that
which
is
due
at
the
end
of
the
second
year
,
itlying
in
the
hands
of
the
Debtor
the
third
,
fourth
,
and
fifth
years
.
And
in
the
fourth
place
there
must
be
accounted
the
utmost
improvement
of
150
l.
for
4
years
,
viz.
that
which
is
due
at
the
end
of
the
first
year
,
it
lying
in
the
Debtors
hands
thesecond
,
third
,
fourth
,
and
fifth
years
.
And
besides
there
must
be
acconnted
150
l.
due
at
the
end
of
the
fifth
year
,
no
Interest
being
reckoned
for
that
,
because
it
becometh
not
due
till
the
expiration
of
the
last
year
,
and
then
the
sum
of
all
these
is
the
utmost
amount
of
that
annuity
.
The
solving
of
Questions
concerning
Annuities
at
Compound
Interest
,
will
not
be
any
thing
different
in
their
operation
,
from
the
manner
of
solving
a
Question
concerning
a
single
sum
of
money
put
out
for
years
at
Compound
Interest
,
by
the
third
Rule
before-going
.
As
suppose
that
in
stead
of
an
Annuity
of
150
l.
there
was
a
single
sum
of
150
l.
put
out
for
4
years
at
Compound
Interest
,
at
6
per
Cent.
what
would
be
its
utmost
improvement
at
the
end
of
the
said
Term
?
Here
you
will
easily
perceive
that
in
solving
the
one
,
the
other
is
also
solved
.
See
the
work
according
to
the
foregoing
third
Rule
.
Now
if
the
foregoing
proportions
be
well
considered
,
you
will
find
that
The
Sum
due
at
the
end
of
the
fifth
year
,
being
that
years
Rent
is
—
|
l.
150
|
And
150
l.
due
at
the
end
of
the
fourth
year
will
,
at
the
fifth
years
end
be
encreased
to
—
|
l.
159
|
And
150
l.
due
at
the
end
of
the
third
year
,
will
at
the
end
of
the
fifth
year
be
increased
to
—
|
l.
168.54
|
And
150
l.
due
at
the
second
years
end
,
will
at
the
fifth
years
end
be
increased
to
—
|
l.
178.6524
|
And
150
l.
due
at
the
first
years
end
,
will
at
the
fifth
years
end
be
increased
to
—
|
l.
189.371544
|
The
sum
of
all
these
being
due
at
the
five
years
end
is
—
|
l.
845.553944
|
So
that
if
an
Annuity
of
150
l.
be
all
forborne
to
the
end
of
five
years
,
and
it
be
improved
to
the
utmost
after
the
rate
of
6
per
Cent.
per
annum
,
Compound
Interest
,
it
will
then
be
increased
to
the
sum
of
845.553944
=
845
l.
11
s.
o
¾
d.
Now
if
the
particular
numbers
in
finding
out
the
augmentation
of
the
said
Annuity
according
to
the
manner
before
prescribed
,
be
well
viewed
,
and
the
method
in
finding
them
out
be
well
considered
,
it
will
appear
,
that
if
an
Annuity
,
payable
by
yearly
payments
,
be
all
forborne
to
the
end
of
any
number
of
years
,
and
the
utmost
improvement
thereof
be
made
at
Compound
Interest
,
the
total
then
due
at
the
end
of
the
said
Time
,
or
Term
of
years
,
will
be
the
sum
of
a
series
,
or
Rank
of
continual
proportionals
as
many
in
number
as
the
years
of
the
Annuities
forbearance
,
the
first
being
the
Annuity
,
or
yearly
payment
it self
,
and
the
second
proceeding
from
the
first
after
the
same
Rate
or
proportion
as
100
l.
and
its
Interest
for
a
year
added
together
,
proceedeth
from
100
l.
and
after
the
same
Rate
doth
the
third
proceed
from
the
second
,
and
the
fourth
from
the
third
,
&c.
The
manner
of
Calculating
the
following
Third
TABLE
.
And
upon
this
Rule
is
grounded
the
Calculation
of
the
following
Table
,
which
sheweth
what
1
l.
annuity
(
being
forborne
to
the
end
of
any
number
of
years
to
come
,
not
exceeding
30
)
will
be
increased
to
Compound
Interest
,
being
computed
after
any
of
the
Rates
mentioned
at
the
head
of
the
Table
.
But
considering
that
as
an
Annuity
increaseth
yearly
at
Compound
Interest
,
the
sum
due
at
each
years
end
,
is
the
sum
of
a
series
of
continual
proportionals
equal
in
number
to
the
yearly
payments
,
and
that
the
first
number
is
the
annual
payment
its
self
,
therefore
may
a
Table
to
shew
the
Annual
increase
of
1
l.
Annuity
with
great
ease
be
made
from
the
First
Table
,
shewing
the
yearly
increase
of
1
l.
at
Compound
Interest
,
as
will
plainly
appear
by
what
followeth
.
Let
us
pitch
upon
making
the
Collum
of
6
per
Cent.
per
Annum
,
in
the
third
Table
?
Look
in
the
first
Table
,
and
you
will
find
the
Collum
of
6
per
Cent.
to
have
for
its
first
number
1.06
,
and
the
second
number
1.12360
&c.
And
to
make
the
Collum
of
6
per
Cent.
in
the
third
Table
proceed
thus
,
for
the
first
number
in
the
said
third
Table
put
1
,
or
1.00000
,
and
for
the
second
number
in
the
third
Table
,
take
the
sum
of
the
first
number
in
the
Third
Table
(
which
is
1.00000
,
)
and
the
first
number
in
the
first
Table
(
which
is
1.06
)
and
that
makes
2.06
for
the
said
second
number
;
then
add
the
second
number
in
the
Third
Table
,
to
the
second
in
the
first
,
and
their
sum
is
the
third
number
in
the
third
Table
;
then
add
the
said
third
number
to
the
third
number
in
the
first
Table
,
and
their
sum
is
the
fourth
number
in
the
third
Table
,
&c.
And
after
this
manner
proceed
till
you
have
made
all
the
numbers
in
the
said
Collum
of
6
per
Cent.
And
after
the
same
method
are
the
rest
of
the
Collums
made
,
(
the
first
number
in
each
being
1.
or
1.00000
)
mutatis
mutandis
.
But
here
note
,
that
the
numbers
in
the
said
first
Table
ought
to
be
continued
to
more
places
than
are
there
expressed
,
to
prevent
the
errors
that
else
may
be
found
in
the
third
Table
,
by
adding
of
defective
Decimals
.
The
use
of
the
said
Table
is
shewn
immediately
after
the
same
.
TABLE
III.
Which
sheweth
in
Pounds
and
Decimal
parts
of
a
〈◊〉
the
In●…
or
Amount
of
1
l.
Annuity
,
to
continue
any
number
of
Years
not
exceeding
●…0
,
Comp●…d
Interest
being
computed
at
5
,
6
,
7
,
8
,
9
,
and
10
per
Cent.
per
Annum
.
Years
|
5.
|
6.
|
7.
|
8.
|
9.
|
10.
|
1
|
1.00000
|
1.00000
|
1.00000
|
1.00000
|
1.00000
|
1.00000
|
2
|
2
05000
|
2
06000
|
2
07000
|
2.08000
|
2.09000
|
2.10000
|
3
|
3.15250
|
3.18360
|
3.21490
|
3.24640
|
3.27810
|
3.31000
|
4
|
4
31012
|
4
37461
|
4
43994
|
4.50611
|
4.57312
|
4.64100
|
5
|
5.52563
|
5.63709
|
5.75073
|
5.86660
|
5.08471
|
6.10510
|
6
|
6.80191
|
6.97531
|
7.15329
|
7.33592
|
7.52333
|
7.71561
|
7
|
8.14200
|
8.39383
|
8.65402
|
8.92280
|
9.20043
|
9.48717
|
8
|
9
64910
|
9.89746
|
10.25980
|
10.63662
|
11.02847
|
11.43588
|
9
|
11.02656
|
11.49131
|
11.97798
|
12.48755
|
13.02103
|
13.57947
|
10
|
12.57789
|
13.18079
|
13.81644
|
14.48656
|
15.19292
|
15
93742
|
11
|
14.20678
|
14.97164
|
15.78359
|
16.64548
|
17.56029
|
18.53116
|
12
|
15.91712
|
16.86994
|
17.88845
|
18.97712
|
20.14071
|
21.38428
|
13
|
17.71298
|
18.88213
|
20
14064
|
21.49529
|
22.95338
|
24.52271
|
14
|
19.59863
|
21.01506
|
22.55048
|
24
21492
|
26.01918
|
27.97498
|
15
|
21.57856
|
23.27596
|
25.12902
|
27.15211
|
29
36091
|
31.77248
|
16
|
23.65749
|
25.67252
|
27.88805
|
30.32428
|
33.00339
|
35.94972
|
17
|
25.84036
|
28.21287
|
30.84021
|
33.75022
|
36.97370
|
40.54470
|
18
|
28.13238
|
30.90565
|
33.99903
|
37.45024
|
41.30133
|
45.59917
|
19
|
30.53900
|
33.75999
|
37.37896
|
41.44626
|
46.01845
|
51.15909
|
20
|
33.06595
|
36.78559
|
40.99549
|
45.76196
|
51.16011
|
57.27499
|
21
|
35.71925
|
39.99272
|
44.86517
|
50.42292
|
56.76453
|
64.00249
|
22
|
38.50521
|
43.39228
|
49.00573
|
55.45675
|
62.87333
|
71.40274
|
23
|
41.43047
|
46.99582
|
53.43614
|
60.89329
|
69.53193
|
79.54302
|
24
|
44.50199
|
50.81557
|
58.17667
|
66.76475
|
76.78981
|
88.49732
|
25
|
47.72709
|
54.86451
|
63
24903
|
73.10593
|
84.70089
|
98.34705
|
26
|
51.11345
|
59.15638
|
68.67646
|
79.95441
|
93.32397
|
109.18176
|
27
|
54.66912
|
63.70576
|
74.48382
|
87.35076
|
102.72313
|
121.09994
|
28
|
58.40258
|
68.52810
|
80.69769
|
95.33882
|
112.96821
|
134.20993
|
29
|
62.32271
|
73.63979
|
87.34652
|
103.96593
|
124.13535
|
148.63092
|
30
|
62.43884
|
79.05818
|
94.46078
|
113.28321
|
136.30753
|
164.49402
|
The
Use
of
the
third
TABLE
.
The
numbers
5
,
6
,
7
,
8
,
9
,
10
at
the
head
of
the
Table
are
the
several
Rates
of
Interest
,
of
100l
.
for
a
year
,
and
the
numbers
placed
in
the
several
Collums
under
those
numbers
,
shew
the
yearly
increase
of
1
pound
Annuity
,
at
the
same
Rate
of
Interest
as
it
is
placed
under
,
and
for
so
many
years
as
it
is
placed
against
in
the
Collum
of
years
on
the
left
hand
of
the
Table
;
and
the
use
of
these
numbers
will
be
manifest
by
the
method
used
in
solving
the
following
Question
,
viz.
There
is
an
Annuity
of
34l
.
−
8s
.
payable
by
yearly
payments
,
forborne
unto
the
end
of
Twelve
years
;
Now
,
I
demand
how
much
is
due
at
the
end
of
the
said
Term
,
Compound
Interest
being
allowed
at
6
per
Cent.
per
Annum
?
Facit
580l
.
−
6s
.
−
6d
.
and
somewhat
more
as
will
appear
by
the
following
operation
.
The
increase
of
the
said
Annuity
being
proposed
at
6
per
Cent.
I
look
in
the
Collum
which
hath
the
number
6
placed
at
the
head
of
it
,
and
against
the
number
12
in
the
Collum
of
years
I
find
the
number
16.86994
which
sheweth
that
if
1l
.
Annuity
be
forborne
to
the
end
of
12
years
,
and
there
be
allowed
Compound
Interest
at
6
per
Cent.
it
will
then
be
increased
to
16.86994
=
16l
.
−
17s
.
−
04½d
.
therefore
I
say
by
the
Rule
of
Proportion
.
Whereby
it
is
apparent
that
those
tabular
numbers
are
only
Multiplyars
for
the
producing
of
the
amount
of
any
given
Annuity
for
any
number
of
years
not
exceeding
30
,
any
Rate
of
Compound
Interest
,
being
allowed
from
5
to
10
per
Cent.
Inclusive
,
&c.
VII
.
Questions
concerning
the
increase
of
Annuities
at
Compound
Interest
may
be
likewise
solved
by
the
first
Table
in
this
Chapter
,
according
to
the
following
method
,
viz.
When
an
Annuity
is
in
arrear
,
and
it
is
required
to
know
to
what
sum
it
is
augmented
,
Compound
Interest
being
computed
,
&c.
Find
out
what
principal
will
in
one
year
gain
the
Annual
Rent
proposed
,
allowing
the
proposed
Rate
of
Interest
.
Then
(
as
is
Taught
in
the
use
of
the
said
first
Table
)
find
the
increase
of
the
said
principal
for
the
number
of
years
,
and
at
the
Rate
of
Interest
proposed
,
and
from
the
amount
thereof
subtract
the
said
principal
,
then
will
that
remainder
be
the
amount
of
the
given
Annuity
for
the
given
time
,
as
will
appear
by
solving
the
first
Question
of
the
sixth
Rule
before-going
,
which
is
this
,
viz.
there
is
an
Annuity
of
150l
.
forborne
to
the
end
of
5
years
,
what
is
its
amount
at
6
per
Cent
,
per
annum
,
Compound
Interest
?
Now
to
answer
this
,
I
find
out
a
principal
that
at
6
per
Cent.
will
gain
150l
.
in
one
year
,
which
I
do
by
the
following
proportion
,
viz.
So
that
I
find
2500l
.
to
be
the
answer
,
then
supposing
the
said
principal
2500l
.
to
be
put
out
to
Interest
at
6
per
Cent.
Compound
Interest
for
5
years
,
look
in
the
first
Table
in
the
Collum
of
6
per
Cent.
and
against
5
years
you
will
find
1.3382255
,
&c.
which
being
multiplyed
by
2500
,
produceth
3345.563944
from
which
if
you
subtract
the
said
principal
2500l
.
there
will
remain
845.563944
for
the
answer
which
is
the
same
with
that
found
before
.
VIII
.
When
an
Annuity
to
continue
any
number
of
years
is
to
be
bought
with
ready
money
,
there
ought
to
be
paid
so
much
money
,
as
being
put
out
at
Compound
Interest
,
at
any
Rate
,
and
for
the
time
of
the
Leases
continuance
,
its
total
amount
may
be
equal
to
the
utmost
improvement
of
the
said
Annuity
,
being
all
forborne
to
the
time
of
the
Leases
expiration
,
Compound
Interest
being
Computed
at
the
same
Rate
.
And
The
manner
of
finding
the
present
worth
of
annuities
,
Rebate
being
all●…wed
at
Comp.
Int.
the
manner
of
finding
out
such
a
present
worth
,
is
as
in
the
following
Example
,
viz.
There
is
an
Annuity
of
468.38001216l
.
to
continue
5
years
what
is
its
present
worth
,
allowing
Rebate
after
the
Rate
of
6
per
Cent.
per
annum
,
Compound
Interest
?
Here
it
is
plain
that
there
must
first
be
computed
the
present
worth
of
the
said
Annuity
,
due
at
the
end
of
the
first
,
second
,
third
,
fourth
,
and
fifth
years
,
and
the
sum
of
all
these
present
worths
,
will
be
the
present
worth
of
the
said
annuity
,
as
will
appear
by
the
following
work
which
is
wrought
by
the
fourth
Rule
of
this
Chapter
.
The
present
worth
of
468.38001216l
.
due
at
the
end
of
the
first
year
is
—
—
|
l.
441.867936
|
The
same
sum
due
at
the
end
of
two
years
,
is
in
ready
money
worth
—
—
|
416.8556
|
The
same
sum
due
at
the
end
of
three
years
is
worth
—
|
●…93
.
26
|
The
same
sum
due
at
the
end
of
four
years
is
worth
—
|
371
|
The
same
due
at
the
end
of
five
years
is
worth
in
ready
money
.
—
—
—
|
350
|
The
sum
of
all
the
said
present
worths
is
—
—
|
1972.983536
|
Which
is
the
present
worth
of
an
Annuity
of
4
8.3001216
to
continue
5
year
,
Rebate
,
being
allowed
at
the
Rate
of
6
per
Cent.
per
Annum
,
Compound
Interest
.
The
construction
of
the
following
TABLE
IV.
And
upon
the
same
grounds
with
the
solution
of
the
last
Question
is
calculated
,
the
following
fourth
Table
,
which
sheweth
the
present
worth
of
1
l.
Annuity
to
continue
any
number
of
years
,
not
exceeding
30
,
and
payable
by
yearly
Payments
,
Rebate
being
allowed
after
the
Rate
of
5
,
6
,
7
,
8
,
9
,
and
10
per
Cent.
per
An.
Compound
Interest
.
But
the
nature
of
the
following
Table
being
rightly
considered
,
you
will
find
the
making
of
it
to
be
easily
performed
by
help
of
the
numbers
in
the
second
Table
of
this
Chapter
.
As
for
Example
.
Let
us
pitch
upon
the
making
of
the
Collum
of
6
per
Cent.
First
,
I
turn
to
the
second
Table
,
and
by
the
numbers
in
the
Collum
of
6
per
Cent.
I
do
the
work
;
The
first
number
in
the
second
Table
,
I
make
to
be
the
first
number
in
the
fourth
and
so
that
fourth
I
ad●…
the
second
number
in
the
second
Table
,
and
their
sum
is
the
second
num●…er
in
the
fourt●…
Table
;
then
to
this
second
number
●…o
I
add
the
third
number
i●…
the
second
Table
,
and
●…eir
sum
is
the
third
number
in
the
fourth
Table
,
and
after
the
same
manner
are
all
the
rest
of
the
numbers
in
that
Collum
made
,
and
also
those
in
the
rest
of
the
Collums
,
mutatis
mutandis
.
But
remember
when
ever
you
caleulate
one
Table
by
the
help
of
another
,
to
continue
the
Table
you
make
use
of
,
to
more
places
than
you
intend
the
numbers
in
your
Table
to
consist
of
for
fear
of
errors
through
the
Addition
of
Defective
Decimals
.
TABLE
IV.
Which
sheweth
the
present
worth
of
one
Pound
Annuity
payable
by
yearly
payments
,
and
to
continue
any
number
of
years
not
exceeding
30
,
Reb●…e
being
allowed
at
5
,
6
,
7
,
8
,
9
,
or
10
pre
Cent.
per
annum
Compound
Interest
.
Years
|
5.
|
6.
|
7.
|
8.
|
9.
|
〈◊〉
|
1
|
.
95238
|
.
9433
|
.
93457
|
.
92952
|
.
91743
|
.
90909
|
2
|
1.85941
|
1.83339
|
1.80801
|
1.80801
|
1.75911
|
1.73553
|
3
|
2.72324
|
2.67301
|
2.62431
|
2.62431
|
2.53129
|
2.48685
|
4
|
3.54595
|
3.46510
|
3.38721
|
3.38721
|
3.23971
|
3.16986
|
5
|
4.32947
|
4.21236
|
4.10019
|
3.10019
|
3.88965
|
3.79078
|
6
|
5.07569
|
4.91732
|
4.76653
|
4.76653
|
4.48591
|
4.35526
|
7
|
5.78637
|
5.58233
|
5.38928
|
5.38928
|
5.03295
|
4.86841
|
8
|
6.46321
|
6.20979
|
5.97129
|
5.97129
|
5.53481
|
5.33492
|
9
|
7.10782
|
6.80169
|
6.51523
|
6.51523
|
5.99524
|
5.75901
|
10
|
7.72173
|
7.36008
|
7.02358
|
6.02358
|
6.41765
|
6.14456
|
11
|
8.30641
|
7.88687
|
7.49867
|
7.49867
|
6.80519
|
6.49506
|
12
|
8.86325
|
8.38384
|
7.94268
|
7.94268
|
7.16072
|
6.81369
|
13
|
9.39357
|
8.85268
|
8.35765
|
7.35765
|
7.48690
|
7.10335
|
14
|
9.89864
|
9.29498
|
8.74546
|
8.74546
|
7.78614
|
7.36668
|
15
|
10.37965
|
9.71224
|
9.10791
|
8.10791
|
8.06068
|
7.60608
|
16
|
10.83776
|
10.10589
|
9.44664
|
8.85136
|
8.31255
|
7.82371
|
17
|
11.27406
|
10.47725
|
9.76322
|
9.12163
|
8.54363
|
8.02155
|
18
|
11.68958
|
10.82760
|
10.05908
|
9.37188
|
8.75562
|
8.20141
|
19
|
12.08531
|
11
15811
|
10.33559
|
9.60359
|
8.95011
|
8.36492
|
20
|
12.46220
|
11.46992
|
10.59401
|
9.81814
|
9.12854
|
8.51356
|
21
|
12.82115
|
11.76407
|
10.83557
|
10.01681
|
9.29224
|
8.64869
|
22
|
13.16300
|
12.04158
|
11.06124
|
10.20074
|
9.44242
|
8.77154
|
23
|
13.48857
|
12.30337
|
11.27218
|
10.37105
|
9.58020
|
8.88322
|
24
|
13.79864
|
12.55035
|
11.46933
|
10.52875
|
9.70661
|
8
.
98●…74
|
25
|
14.09394
|
12.78335
|
11.65358
|
10.67477
|
9.82257
|
9.07704
|
26
|
14
.
375●…8
|
13.00316
|
11.82577
|
10.80997
|
9.92897
|
9.16094
|
27
|
14
64303
|
13.21053
|
11.98671
|
10.93516
|
10.02657
|
9.23722
|
28
|
14.89812
|
13.40616
|
12.13711
|
11.05107
|
10.11612
|
9.30656
|
29
|
15.14071
|
13.59071
|
12.27767
|
11
.
158●…0
|
10.19828
|
9.36960
|
30
|
15.39244
|
13.76482
|
12.40994
|
11.
●…5778
|
10.27365
|
9
,
42691
|
The
Use
of
the
foregoing
TABLE
.
The
first
Collum
is
the
number
of
years
from
1
to
30
;
and
the
number
5
,
6
,
7
,
8
,
9
,
10
,
at
the
head
of
the
Table
are
the
Rates
of
Interest
of
100
l.
for
a
year
,
and
the
numbers
in
each
of
these
Collums
under
the
said
Rates
of
Interest
,
are
the
present
worths
of
1
l.
Annuity
to
continue
for
the
number
of
years
which
is
placed
against
them
,
allowing
Rebate
after
the
rate
of
Interest
at
the
head
of
each
Collum
,
and
are
Multiplyars
serving
to
find
the
present
worth
of
any
other
Annuity
,
as
will
appear
by
the
following
Example
.
There
is
an
Annuity
of
48
l.
to
continue
12
years
,
and
payable
by
yearly
payments
,
to
be
sold
for
present
money
,
I
demand
what
it
is
worth
,
allowing
Rebate
at
6
per
Cent.
per
Annum
,
Compound
Interest
.
?
Facit
402.424
=
40
l.
8
s.
05
¾
d.
which
is
thus
found
out
by
the
foregoing
Table
,
viz.
look
in
the
said
Table
,
in
the
Collum
of
6
per
Cent.
and
against
12
in
the
Collum
of
years
,
you
have
this
number
,
viz.
8.38384
which
is
the
present
worth
of
1
l.
Annuity
to
continue
twelve
years
,
Rebate
being
allowed
,
&c.
therefore
by
the
Rule
of
proportion
,
I
say
,
So
that
I
find
the
answer
to
be
402.42432
,
which
is
found
by
multiplying
the
said
Tabular
number
by
48
,
as
you
see
by
the
work
.
Otherwise
find
a
principal
which
may
bear
such
proportion
to
the
given
Annuity
(
that
is
to
be
Rebated
)
as
100
beareth
to
the
Rate
of
Interest
allowed
in
the
Rebate
.
Then
find
the
present
worth
of
this
principal
so
found
,
by
the
Directions
given
in
the
use
of
the
second
Table
of
this
Chapter
,
then
subtract
the
said
present
worth
from
the
principal
found
as
before
,
and
the
remainder
will
be
the
present
worth
of
the
given
Annuity
,
Rebate
being
allowed
as
proposed
.
Example
.
What
is
the
present
worth
of
an
Annuity
of
50
l.
to
continue
3
years
,
allowing
Rebate
at
8
per
Cent.
per
Annum
,
Compound
Interest
?
First
,
I
find
a
principal
that
shall
be
to
the
given
number
50
as
100
is
to
8
which
I
find
to
be
625
l.
by
the
following
proportion
,
viz.
Then
by
the
second
Table
I
find
the
present
worth
of
625
l.
which
is
496.145
l.
which
I
subtract
from
the
said
principal
625
l.
and
there
remaineth
128.855
l.
=
128
l.
−
17
s.
−
1¼d
.
fere
.
which
is
the
present
worth
of
50
l.
per
annum
to
continue
3
years
,
Rebate
being
allowed
at
8
per
Cent.
per
annum
,
Compound
Interest
.
Moreover
by
the
numbers
in
the
foregoing
fourth
Table
,
you
may
at
first
sight
discover
how
many
years
purchase
any
Lease
to
continue
any
number
of
years
,
not
exceeding
30
,
is
worth
in
ready
money
,
Compound
Int.
being
Computed
on
both
sides
at
any
of
the
Rates
mentioned
at
the
head
of
the
Table
.
Example
.
Suppose
there
were
a
Lease
issuing
out
of
Lands
to
continue
16
years
to
be
sold
for
ready
money
,
allowing
Rebate
at
8
per
Cent.
per
Annum
,
Compound
Interest
,
I
demand
how
many
years
purchase
the
said
Lease
is
worth
?
Look
in
the
Table
4
,
in
the
Collum
of
8
per
Cent.
and
against
16
years
you
will
find
8.85136
which
sheweth
that
it
is
worth
8.85136
years
purchase
which
is
somewhat
above
8
years
,
and
3
Quarters
;
But
if
the
said
Lease
had
been
of
Houses
,
and
10
per
Cent.
were
thought
a
convenient
allowance
for
the
same
,
then
you
will
find
it
to
be
worth
7.82371
years
purchase
which
is
7
years
,
and
above
3
quarters
purchase
.
IX
.
When
there
is
a
sum
of
money
propounded
,
and
itis
required
to
know
what
annuity
to
continue
any
given
number
of
years
,
it
will
purchase
according
to
any
given
Rate
of
Interest
,
you
may
suppose
any
annuity
at
pleasure
,
then
by
the
directions
Of
the
purchase
of
Annuities
at
Comp.
Interest
.
given
in
the
use
of
the
fourth
Table
;
or
else
by
the
eighth
Rule
of
this
Chapter
,
find
the
present
worth
of
the
supposed
annuity
for
the
number
of
years
,
and
at
the
rate
of
Interest
propounded
,
which
being
done
,
you
may
find
what
Annuity
to
continue
the
same
number
of
years
,
the
sum
propounded
will
purchase
by
the
following
proportion
,
viz.
As
the
present
worth
of
the
supposed
Annuity
,
Is
to
the
said
Annuity
,
So
is
the
sum
propounded
,
To
the
Annuity
required
.
As
for
Example
.
Let
it
be
required
to
find
out
what
Annuity
to
continue
4
years
,
800
l.
present
money
will
purchase
,
Compound
Interest
being
computed
at
6
per
Cent.
per
Annum
?
Facit
230.873
l.
First
,
Suppose
an
Annuity
at
pleasure
,
to
continue
for
4
years
,
as
suppose
150
l.
then
do
I
find
by
the
eighth
Rule
of
this
Chapter
the
present
worth
of
the
said
Annuity
to
be
519.76584
l.
therefore
by
the
Rule
of
proportion
I
say
The
Construction
of
the
following
TABLE
V.
Upon
the
Reason
of
the
foregoing
Rule
is
grounded
the
Calculation
of
the
following
Table
for
the
purchasing
of
Annuities
;
and
it
may
somewhat
more
readily
be
calculated
thus
,
viz.
It
is
evident
by
the
construction
of
the
first
Table
of
this
Chapter
,
that
1
l.
present
is
equivalent
to
1.06
l.
due
at
the
end
of
a
year
to
come
;
therefore
is
1.06
the
first
number
in
the
Collum
of
6
per
Cent.
of
the
following
Table
;
because
1
l.
will
purchase
1.06
l.
Then
it
is
also
evident
by
the
fourth
Table
that
the
present
worth
of
1
l.
Annuity
to
continue
two
years
at
the
same
rate
is
1.83339
,
&c.
that
is
,
1.83339
,
&c.
will
purchase
a
Lease
of
1
l.
per
Annum
to
continue
two
years
,
Compound
Interest
,
being
allowed
at
6
per
Cent.
therefore
by
the
Rule
of
3
Direct
,
I
say
,
By
which
I
find
that
1
l.
ready
money
will
buy
a
Lease
of
.54543
l.
per
Annum
to
continue
2
years
,
therefore
it
is
the
second
number
in
the
following
Table
.
Likewise
by
the
fourth
Table
I
find
that
2.67301
is
the
present
worth
of
1
l.
Annuity
to
continue
3
years
at
the
same
rate
of
Interest
,
wherefore
by
the
Rule
of
proportion
I
say
,
Whereby
I
find
that
1
l.
will
purchase
an
Annuity
of
.37411
,
to
continue
3
years
.
Compound
Interest
being
allowed
at
6
per
Cent.
wherefore
.37411
is
the
third
number
in
the
said
Table
;
Whereby
it
is
evident
that
if
you
divide
1
,
or
Unite
by
the
several
numbers
in
the
said
Collum
of
6
per
Cent.
in
the
fourth
Table
,
successively
,
the
several
Quotients
will
give
you
the
numbers
successively
,
for
the
Collum
of
6
per
Cent.
in
the
fifth
Table
;
And
after
the
same
manner
are
all
the
numbers
in
the
other
Collums
of
the
said
fifth
Table
found
out
(
except
the
first
number
of
each
Collum
,
which
must
be
the
same
with
the
first
numbers
in
each
Collum
of
the
first
Table
)
mutatis
mutandis
.
But
it
is
absolutely
necessary
that
the
numbers
in
the
said
fourth
Table
,
be
continued
to
more
places
than
are
there
expressed
,
to
prevent
the
errors
that
otherwise
will
arise
,
by
dividing
by
defective
Decimals
.
TABLE
V.
Which
sheweth
what
Annuity
payable
by
yearly
payments
to
continue
any
number
of
years
not
exceeding
30
,
one
pound
will
purchase
,
Compound
Interest
,
being
computed
at
5
,
6
,
7
,
8
,
9
,
or
10
per
Cent.
Years
|
5.
|
6.
|
7.
|
8.
|
9.
|
10.
|
1
|
1.05000
|
1.06000
|
1
07000
|
1.08000
|
1.09000
|
1.10000
|
2
|
.
53780
|
.
54543
|
.
55309
|
.
56076
|
.
56846
|
.
57619
|
3
|
.
36720
|
.
37411
|
.
38105
|
.
38803
|
.
39505
|
.
40211
|
4
|
.
28209
|
.
28859
|
.
29519
|
.
30192
|
.
30866
|
.
31947
|
5
|
.
23097
|
.
23739
|
.
24389
|
.
25045
|
.
25709
|
.
26379
|
6
|
.
19701
|
.
20336
|
.
20979
|
.
21631
|
.
22291
|
.
22960
|
7
|
.
17281
|
.
17913
|
.
18555
|
.
19207
|
.
19869
|
.
20545
|
8
|
.
15472
|
.
16103
|
.
16746
|
.
17401
|
.
18067
|
.
18744
|
9
|
.
14069
|
.
14702
|
.
15548
|
.
16007
|
.
16679
|
.
17364
|
10
|
.
12950
|
.
13586
|
.
14237
|
.
14902
|
.
15582
|
.
16274
|
11
|
.
12038
|
.
12679
|
.
13335
|
.
14007
|
.
14694
|
.
15396
|
12
|
.
11284
|
.
11927
|
.
12590
|
.
13269
|
.
13965
|
.
14676
|
13
|
.
10645
|
.
10296
|
.
11965
|
.
12652
|
.
13356
|
.
14077
|
14
|
.
10101
|
.
10758
|
.
11434
|
.
12129
|
.
12843
|
.
13574
|
15
|
.
09634
|
.
10296
|
.
10979
|
.
11682
|
.
12405
|
.
13147
|
16
|
.
09226
|
.
09895
|
.
10585
|
.
11298
|
.
12029
|
.
12781
|
17
|
08869
|
.
09544
|
.
10242
|
.
10962
|
.
11704
|
.
12466
|
18
|
.
0855●…
|
.
09235
|
.
09941
|
.
10670
|
.
11421
|
.
12192
|
19
|
.
08274
|
.
08962
|
.
09675
|
.
10412
|
.
11173
|
.
11954
|
20
|
.
08024
|
.
08718
|
.
09439
|
.
10184
|
.
10954
|
.
11745
|
21
|
.
07799
|
.
08500
|
.
09228
|
.
09983
|
.
10761
|
.
11562
|
22
|
.
07597
|
.
08304
|
.
09040
|
.
09803
|
.
10590
|
.
11400
|
23
|
.
07413
|
.
08127
|
.
08871
|
.
09642
|
.
10438
|
.
11257
|
24
|
.
07247
|
.
07967
|
.
08718
|
.
09497
|
.
10302
|
.
11129
|
25
|
.
07095
|
.
07822
|
.
08581
|
.
09367
|
.
10180
|
.
11016
|
26
|
.
06956
|
.
07690
|
.
08456
|
.
09250
|
.
10071
|
.
10915
|
27
|
.
06829
|
.
07569
|
.
08342
|
.
09144
|
.
09973
|
.
10825
|
28
|
.
06712
|
.
07459
|
.
08239
|
.
09048
|
.
09885
|
.
10745
|
29
|
.
06604
|
.
07357
|
.
08144
|
.
08961
|
.
09805
|
.
10672
|
30
|
.
06496
|
.
07264
|
.
08058
|
.
08882
|
.
09733
|
.
10607
|
The
Use
of
the
foregoing
Table
V.
The
use
of
the
foregoing
Table
will
appear
in
the
solution
of
the
following
Question
,
viz.
A
Merchant
hath
1500
l.
by
him
,
which
he
is
willing
to
lay
out
upon
an
Annuity
,
issuing
out
of
Lands
to
continue
20
years
,
beginning
presently
,
Compound
Interest
being
Computed
on
both
sides
at
6
per
Cent.
per
Annum
.
Now
I
demand
what
Annuity
the
said
sum
will
buy
?
Facit
130.77
l.
=
130
l.
−
15
s.
−
●…5
d.
very
near
.
To
answer
this
Question
,
I
look
in
the
Collum
of
6
per
Cent.
of
the
foregoing
Fifth
Table
,
and
against
20
in
the
Collum
of
years
I
find
.08718
,
which
is
the
annuity
that
1
l.
present
money
will
purchase
to
continue
20
years
,
wherefore
by
the
Rule
of
3
Direct
,
I
say
X.
Questions
concerning
the
purchasing
of
Leases
and
Annuities
may
be
solved
very
well
by
the
numbers
in
the
fourth
Table
,
if
you
make
them
Divisors
instead
of
Multiplyars
.
Let
the
last
Question
be
proposed
,
and
solved
by
the
fourth
Table
viz.
What
Annuity
to
continue
20
years
will
1500
l.
ready
Money
purchase
,
Compound
Interest
being
allowed
at
6
per
Cent
?
To
answer
this
,
I
look
in
the
fourth
Table
,
in
the
Collum
of
6
per
Cent.
against
20
years
;
and
there
,
I
find
this
number
,
viz.
11
46992
,
which
is
the
present
worth
of
1
l.
Annuity
to
continue
20
years
,
Compound
Interest
being
allowed
at
6
per
Cent.
And
if
it
be
the
present
worth
of
1
l.
Annuity
,
I
conclude
it
will
purchase
1
l.
Annuity
to
continue
the
same
number
of
years
,
wherefore
I
say
by
the
Rule
of
3
Direct
,
So
that
the
Answer
is
the
same
with
the
former
,
which
was
found
by
help
of
the
Fifth
Table
.
All
the
foregoing
Tables
might
have
been
continued
to
any
greater
number
of
years
at
pleasure
;
But
although
these
Tables
are
calculated
but
for
30
years
;
yet
may
they
be
made
serviceable
for
years
above
30
,
as
shall
be
shewed
by
and
by
.
Arithmetical
Questions
to
exercise
the
Learner
in
the
Precedent
Tables
.
Quest.
1.
There
is
a
Lease
of
20
years
to
begin
presently
,
which
in
ready
Money
is
worth
1200
l.
But
suppose
the
said
Lease
were
not
to
begin
till
the
expiration
of
8
years
,
I
demand
what
would
then
be
the
present
worth
of
the
said
Lease
Rebate
,
being
allowed
at
8
per
Cent.
per
Annum
,
Compound
Interest
?
The
main
intent
of
this
Question
is
to
shew
the
use
of
the
second
Table
,
for
if
you
find
the
present
worth
of
1200
l.
due
at
the
end
of
8
years
,
at
8
per
Cent.
the
Question
is
answered
,
which
according
to
the
Directions
given
after
the
said
Table
,
will
be
found
to
be
648.3216
l.
=
648
l.
−
06
s.
−
05
d.
Quest.
2.
A
oweth
to
B
600
l.
to
be
paid
in
6
years
,
viz.
100
l.
every
year
,
but
being
weakned
in
his
Estate
,
is
not
able
to
performe
;
but
an
Estate
being
to
come
into
his
hands
at
the
end
of
10
years
;
B
is
willing
to
forbear
it
all
till
then
,
and
to
be
allowed
Compound
Interest
at
8
per
Cent.
for
his
forbearance
;
I
demand
how
much
will
be
due
to
B
at
the
10
years
end
?
This
Question
is
solved
by
help
of
the
third
and
first
Tables
;
for
first
100
l.
is
to
be
paid
in
the
nature
of
Annuity
for
6
years
;
therefore
by
the
third
Table
I
find
the
amount
of
an
Annuity
of
100
l.
to
continue
6
years
at
8
per
Cent.
which
is
733.592
l.
and
will
be
due
at
the
expiration
of
6
years
,
and
then
is
that
sum
to
be
forborne
to
the
end
of
10
years
,
which
is
4
years
after
the
6
years
;
which
being
a
single
sum
,
its
amount
is
found
by
the
first
Table
to
be
998.037
l.
&c.
which
is
the
Answer
to
the
Question
.
Quest.
3.
There
is
a
Lease
to
continue
21
years
to
be
sold
for
1000
l.
but
the
Lessee
desireth
rather
to
pay
an
Annual
Rent
:
Now
the
Question
is
what
that
Annual
Rent
ought
to
be
,
Compound
Interest
being
computed
at
10
l.
per
Cent.
per
Annum
?
The
intent
of
this
Question
is
to
find
what
annuity
to
continue
21
years
1000
will
purchase
,
at
10
per
Cent.
which
is
to
be
done
by
the
Fifth
Table
,
thus
,
Because
the
time
is
for
21
years
,
look
in
the
Collum
of
years
for
21
,
and
just
against
it
in
the
Collum
of
10
per
Cent.
you
will
find
.11562
,
by
which
multiply
1000
,
and
the
product
is
115.62
l.
and
so
much
will
1000
l.
purchase
for
21
years
at
10
per
Cent.
Compound
Interest
.
Quest.
4.
A
and
B
have
each
of
them
a
Lease
to
continue
20
years
;
A
hath
80
l.
per
Annum
,
and
B
120
l.
per
Annum
,
and
they
agree
to
make
an
exchange
,
upon
this
condition
,
that
A
shall
pay
in
ready
money
the
excess
of
his
Estate
,
allowing
him
Compound
Interest
at
8
per
Cent.
Now
I
demand
how
much
ready
money
A
ought
to
give
B
upon
this
exchange
,
according
to
that
condition
?
Subtract
80
l.
from
120
l.
and
the
remainder
is
40
l.
and
so
much
per
annum
is
the
Lease
of
B
worth
more
than
that
of
A
,
therefore
A
must
pay
B
so
much
money
as
will
purchase
40
l.
per
Annum
to
continue
20
years
at
8
per
Cent.
which
by
the
third
Table
will
be
found
to
be
392.7256
l.
Quest.
5.
There
is
a
House
to
be
let
by
Lease
for
21
years
,
for
which
the
Lessor
will
have
50
l.
fine
,
and
70
l.
per
Annum
,
but
the
Lessee
is
willing
to
pay
the
greater
fine
,
that
he
may
have
the
Rent
but
40
l.
per
annum
,
now
I
demand
what
fine
he
ought
to
pay
upon
that
condition
,
Compound
Interest
being
allowed
at
8
per
Cent.
per
annum
?
Take
the
Difference
between
40
and
70
,
which
is
30
,
for
the
abatement
in
the
yearly
Rent
for
21
years
;
Then
by
the
fourth
Table
find
the
present
worth
of
30
l.
per
annum
for
21
years
at
8
per
cent
.
which
is
300.5043
l.
=
300
l.
−
10
s
,
−
01
d.
which
added
to
the
said
50
l.
fine
makes
350
l.
−
10
s.
−
01
d.
for
the
fine
to
be
paid
upon
the
said
Condition
.
Quest.
6.
There
is
a
Lease
to
belet
of
20
l.
per
annum
,
and
250
l.
fine
for
24
years
,
and
the
Lessee
is
willing
to
pay
the
greater
Rent
,
that
he
may
pay
but
50
l.
fine
,
now
I
demand
what
Rent
he
ought
to
pay
upon
that
condition
,
Compound
Interest
being
computed
at
7
per
cent
.
per
annum
?
It
is
manifest
,
that
if
the
lessor
taketh
50
l.
fine
,
he
abateth
200
l
,
therefore
find
by
the
fifth
Table
what
Annuity
to
continue
24
years
,
200
l.
will
purchase
at
7
per
cent
.
The
Tabular
number
is
.08718
,
which
multiplyed
by
200
produceth
17.436
=
17
l.
−
8
s.
−
9
d.
and
so
much
must
the
Lessee
raise
his
Rent
if
he
will
have
200
l.
abated
of
his
fine
,
to
which
if
you
add
20
l.
the
proposed
Rent
,
the
sum
is
37
l.
−
08
s.
−
09
d.
for
the
yearly
Rent
to
be
paid
to
satisfie
the
said
condition
.
Quest.
7.
What
annuity
to
continue
20
years
,
may
I
grant
presently
,
for
900
l.
to
be
paid
6
years
hence
,
accompting
6
per
cent
.
per
annum
,
Compound
Interest
?
First
,
find
by
the
Second
Table
the
present
worth
of
900
l.
due
six
years
hence
,
at
6
per
cent
.
which
is
634.464
l.
=
634
l.
−
09
s
−
03½
d.
Then
by
the
Fifth
Table
find
what
Annuity
to
continue
20
years
634.464
will
purchase
at
6
per
cent
.
And
you
will
find
the
Answer
to
be
55.31257152
l.
=
55
l.
−
06
s.
−
03
d.
and
so
much
I
ought
to
grant
yearly
for
20
years
for
900
l.
to
be
paid
me
at
the
end
of
6
years
.
Quest.
8
I
have
6
years
of
an
old
Lease
,
yet
to
come
,
and
would
take
a
new
Lease
in
reversion
for
21
years
,
after
the
expiration
of
the
old
Lease
,
the
annual
Rent
whereof
is
40
l.
But
I
would
pay
such
a
sum
of
money
present
as
a
fine
,
that
for
my
Lease
in
Reversion
for
the
said
21
years
,
I
may
pay
but
15
l.
per
Annum
,
Now
I
demand
how
much
present
money
I
ought
to
pay
the
Lessor
,
to
satisfie
these
conditions
,
Compound
Interest
being
computed
at
8
per
cent
.
The
difference
between
40
and
15
is
25
,
and
so
much
the
Lessee
desireth
to
have
abated
in
his
Rent
,
wherefore
by
the
fourth
Table
find
the
present
worth
of
25
per
annum
for
21
years
at
8
per
Cent.
which
is
250.42025
l.
=
250
l.
−
08
s.
−
05
d.
Then
by
the
second
Table
find
the
present
worth
of
250.42025
l.
due
at
the
end
of
6
years
to
come
,
at
8
per
Cent.
which
is
157.807
l.
&c.
=
157
l.
−
16
s.
−
01
〈◊〉
.
And
so
much
ought
I
to
give
to
satisfie
the
said
conditions
.
Quest.
9.
There
is
a
Lease
to
be
let
for
12
years
,
for
20
l
per
Annum
,
and
200
l.
fine
,
but
the
Lessee
desireth
to
take
a
Lease
of
the
same
for
21
years
,
and
to
pay
the
same
Rent
;
the
Question
is
,
what
fine
ought
to
be
paid
for
the
Lease
of
21
years
,
accompting
Compound
Interest
at
6
per
Cent
?
Facit
280
l.
−
12
s.
−
05
d.
By
the
Fifth
Table
seek
what
Annuity
to
continue
12
years
,
200
l.
will
purchase
at
6
per
Cent.
which
you
will
find
to
be
23.854
l.
Then
by
the
Fourth
Table
find
the
present
worth
of
23.854
l.
Annuity
to
continue
21
years
at
6
per
Cent.
which
is
280.620
l.
&c.
=
280
l.
−
12
s.
−
05
d.
and
so
much
ought
the
Lessee
to
pay
for
a
fine
,
to
have
his
Lease
for
21
years
.
Quest.
10.
A
Gentleman
hath
1000
l.
which
he
would
lay
out
to
purchase
an
Annuity
of
100
l.
to
be
paid
by
yearly
payments
;
Now
the
Question
is
,
how
many
years
must
the
said
Annuity
continue
,
Compound
Interest
being
allowed
on
both
sides
at
8
per
Cent.
per
Annum
?
First
,
Divide
1000
by
100
,
and
the
Quotient
will
be
10
,
which
sheweth
that
the
Buyer
giveth
10
years
purchase
for
the
said
Annuity
.
Then
in
the
Fourth
Table
,
and
in
the
Collum
of
8
per
Cent.
look
for
the
number
10
,
which
cannot
be
exactly
found
,
but
the
nearest
to
it
and
less
than
it
,
is
9.81814
which
is
placed
against
20
years
,
and
the
nearest
to
it
greater
than
it
is
,
is
10.01681
,
therefore
I
conclude
that
the
Annuity
must
continue
above
20
years
,
but
not
21
years
,
and
to
find
out
how
much
it
must
continue
more
than
20
years
,
I
work
thus
,
viz.
First
,
I
find
the
Difference
between
the
said
Tabular
numbers
10.01681
and
9.81814
,
which
is
.19867
.
Then
I
find
the
difference
between
the
lesser
of
the
said
Tabular
Numbers
,
viz.
9.81814
and
10
,
the
Number
that
I
would
find
in
the
Table
,
which
is
.18186
,
then
by
the
Rule
of
proportion
I
say
Which
is
as
much
as
to
say
,
as
the
greater
Difference
.19867
is
to
one
year
,
so
is
the
lesser
Difference
to
.9153
parts
of
a
year
,
which
is
47
weeks
,
and
5
dayes
,
therefore
the
number
of
years
sought
in
the
Question
is
20
years
,
47
weeks
,
and
5
dayes
.
Quest.
11.
A
Gentleman
bought
a
Lease
of
100
l.
per
annum
to
continue
18
years
,
for
960
l.
now
I
demand
what
Rate
of
Compound
Interest
was
there
implyed
in
such
a
bargain
?
To
Answer
this
,
First
,
I
divide
960
by
100
,
and
the
Quotient
is
9.6
which
sheweth
how
many
years
purchase
it
was
worth
;
Then
because
the
Lease
was
to
continue
18
years
,
I
look
in
the
fourth
Table
in
the
Collum
of
years
for
18
,
and
carry
my
eye
exactly
in
the
line
against
it
,
looking
for
the
said
Quotient
9.6
which
I
cannot
find
exactly
,
but
the
next
(
lesser
)
number
to
it
,
is
,
9.37188
in
the
Collum
of
8
per
Cent.
and
the
next
(
bigger
)
number
to
it
is
10.05908
,
in
the
Collum
of
7
per
Cent.
wherefore
I
conclude
that
the
Rate
of
Interest
implyed
is
between
7
and
8
per
Cent.
and
to
know
how
much
it
is
more
than
7
,
I
do
thus
,
Take
the
Difference
between
the
two
said
Tabular
numbers
which
you
will
find
to
be
68720
also
subtract
(
9.6
)
the
said
Quotient
,
from
10.05908
(
the
greater
Tabular
number
,
and
the
remainder
is
.45908
,
Then
by
the
Rule
of
Proportion
,
I
say
,
That
is
to
say
,
As
the
Difference
between
the
two
Tabular
numbers
is
to
the
lesser
Remainder
,
so
is
1
l.
the
Difference
between
7
and
8
per
Cent.
to
.668
the
proportional
part
to
be
added
to
7
l.
which
is
13
s.
−
04
d.
so
that
7
l.
13
s.
04
d.
is
very
near
the
Interest
required
.
How
to
find
out
Tabular
Numbers
for
years
exceeding
30.
It
may
many
times
fall
out
,
that
the
number
of
years
proposed
in
a
Question
,
may
exceed
the
number
of
years
limited
in
the
foregoing
first
,
second
,
third
,
fourth
and
fifth
Tables
,
and
in
such
cases
that
defec●…
may
be
supplyed
by
the
method
used
in
the
solution
of
the
following
Questions
.
Quest.
12.
Suppose
80
l.
were
put
out
to
Interest
at
5
l.
per
Cent.
Compound
Interest
for
40
years
,
I
demand
how
much
it
will
then
be
amounted
to
?
This
Question
is
to
be
solved
by
the
first
Table
,
thus
viz.
Take
any
two
Numbers
in
the
Collum
of
years
,
which
together
will
make
up
40
,
and
then
take
the
Tabular
numbers
in
the
Collum
of
5
per
Cent.
which
stand
against
those
two
numbers
,
and
multiply
them
together
,
and
then
multiply
that
product
by
80
l.
the
given
sum
,
and
the
last
product
will
be
the
Answer
.
As
suppose
you
take
30
and
10
,
or
21
and
19
,
or
31
and
9
,
or
25
and
15
&c.
But
we
will
pitch
upon
30
and
10
,
and
the
Tabular
number
against
30
in
the
Collum
of
5
per
Cent.
is
4.32194
,
and
against
10
is
1.62889
which
two
numbers
being
multiplyed
,
produce
7.03996
,
&c.
which
is
the
amount
of
1
l.
for
40
years
at
5
per
Cent.
then
I
multiply
7.03996
,
&c.
by
80
l.
and
the
product
is
563.197
l.
&c.
=
563
l.
−
03
s.
−
11½
d.
fere
.
The
Answer
would
have
been
the
same
,
if
we
had
pitched
upon
any
other
two
numbers
to
have
made
up
40.
And
for
Tryal
hereof
,
let
us
pitch
upon
25
and
15
,
the
Tabular
number
against
25
is
3.38635
,
and
the
tabular
number
against
15
years
is
2.07892
,
and
the
product
of
these
two
tabular
numbers
is
7.0399
,
&c.
which
multiplyed
by
80
,
produceth
563.197
l.
as
before
,
and
so
much
will
80
l.
be
increased
to
in
40
years
at
5
per
Cent.
per
an
.
Compound
Interest
.
The
like
is
to
be
understood
for
any
other
number
of
years
.
Quest.
13.
Suppose
420
l.
to
be
payable
at
the
end
of
50
years
to
come
,
What
is
its
present
worth
,
Rebate
being
allowed
at
5
per
Cent.
per
Annum
,
Compound
Interest
?
This
Question
is
of
the
same
nature
with
those
belonging
to
the
second
Table
,
and
is
answered
thereby
,
according
to
the
method
used
in
solving
the
last
Question
by
the
first
Table
,
viz.
the
given
time
being
50
years
;
I
pitch
upon
30
and
20
,
and
the
tabular
number
against
30
is
.231377
and
that
against
20
is
.376889
and
the
product
of
these
two
is
.
〈◊〉
,
&c.
which
is
the
present
worth
of
〈◊〉
l.
due
50
years
hence
at
5
perCent
.
per
an
.
wherefore
I
multiply
.
〈◊〉
,
&c.
by
420
,
and
the
product
is
36.62544
,
&c.
=
36
l.
12
s.
6
d.
and
so
much
is
the
present
worth
of
4●…0l
.
due
50
years
hence
at
5
per
Cent.
per
an
.
Compound
Interest
.
Quest.
14
,
An
Heir
being
beyond
the
Sea
,
did
not
return
till
36
years
after
an
Estate
of
30
l.
per
Annum
was
fallen
to
him
by
the
Death
of
the
proprietor
;
the
Question
is
,
what
was
then
due
to
him
,
Compound
Interest
being
computed
at
6
per
Cent.
per
Annum
?
This
Question
is
of
the
nature
of
those
belonging
to
the
third
Table
,
and
the
manner
of
solving
it
is
thus
,
viz.
Find
out
(
by
the
seventh
Rule
of
this
Chapter
)
what
principal
will
in
one
year
gain
30
l.
at
6
per
Cent.
by
the
following
proportion
.
Having
found
500
l.
to
be
the
principal
,
seek
(
after
the
manner
of
the
12
Question
)
by
the
first
Table
the
Amount
or
Increase
of
500
l.
for
36
years
at
6
per
Cent.
which
you
will
find
to
be
4073.5998
l.
&c.
from
which
if
you
subtract
the
said
principal
500
l.
the
remainder
is
3573.599l
&c.
=
3573
l.
−
12
s
,
−
00
d.
fere
.
And
so
much
was
due
to
the
heir
at
his
return
.
Quest.
15.
There
is
an
Annuity
of
30
l.
to
continue
37
years
,
the
Question
is
,
what
it
is
worth
in
ready
money
,
Compound
Interest
being
computed
at
6
per
Cent.
per
Annum
?
By
the
second
way
of
solving
Questions
under
the
fourth
Table
,
find
a
principal
which
will
gain
30
l.
in
one
year
,
at
6
per
Cent.
which
is
here
500l
.
then
according
to
the
method
used
in
solving
the
thirteenth
Question
foregoing
,
find
the
present
worth
of
500
l.
for
37
years
at
6
per
Cent.
which
will
be
found
to
be
57.896537
which
subtracted
from
500
l.
leaves
442.1034
&c.
=
442
l.
2s
.
−
0¾d
.
And
so
much
is
the
present
worth
of
the
foresaid
Annuity
.
Quest.
16.
What
Annuity
to
continue
40
years
will
500
l.
purchase
Compound
Interest
,
being
computed
at
6
per
Cent.
per
Annum
?
It
is
evident
by
the
tenth
Rule
of
this
Chapter
,
that
if
you
find
out
the
present
worth
of
1
l.
Annuity
for
any
number
of
years
,
and
at
any
rate
of
Interest
,
it
may
easily
be
found
what
Annuity
to
continue
the
same
number
of
years
any
other
sum
will
purchase
at
the
same
Rate
of
interest
by
one
single
Rule
of
3
Direct
:
Therefore
,
Find
out
the
present
worth
of
1
l.
Annuity
to
continue
40
years
at
6
per
cent
.
by
the
method
used
in
solving
the
last
Question
,
which
will
be
found
to
be
15.04632
l.
=
15
l.
−
00
s.
−
11
d.
which
sum
of
Money
will
purchase
an
Annuity
of
1
l.
to
continue
40
years
at
6
per
cent
.
therefore
to
know
what
Annuity
500
l.
will
purchase
for
the
same
Time
,
say
by
the
Rule
of
Proportion
.
which
will
be
found
to
be
33.230
&c.
=
33
l.
−
04
s.
−
07¼d
.
fere
,
and
such
an
Annuity
to
Continue
40
years
will
500
l
:
purchase
Interest
being
Allowed
at
6
per
cent
.
FINIS
,
Cockers
ARTIFICIAL
ARITHMETICK
,
SHEWING
The
Genesis
or
Fabrick
of
the
Logarithmes
and
their
use
in
the
extraction
of
Roots
,
solving
of
Questions
in
Anatocisme
,
or
Compound
Interest
,
and
in
the
other
Rules
of
Arithmetick
,
in
a
Method
not
usually
practised
.
Composed
by
EDWARD
COCKER
,
late
Practitioner
in
the
Arts
of
Writing
,
Arithmetick
,
and
Engraving
.
Perused
,
Corrected
,
and
Published
By
IOHN
HAWKINS
,
School-Master
at
St.
George's
Church
in
Southwark
.
Nil
tam
difficile
est
quod
non
solertia
vincat
.
LONDON
,
Printed
by
Iohn
Richardson
in
the
Year
1684.
The
meaning
of
such
Characters
as
are
used
in
the
Ensuing
Treatise
.
+
IS
the
sign
of
Addition
,
and
is
as
much
as
to
say
plus
,
signifying
that
the
Numbers
or
Quantities
between
which
it
is
placed
,
are
to
be
added
together
as
4+7
signifyeth
that
4
and
7
are
to
be
added
together
.
−
Is
the
sign
of
subtraction
,
and
as
much
as
to
say
minus
,
signifying
that
the
Number
which
followeth
it
is
to
be
Subtracted
out
of
the
Number
which
precedeth
it
,
as
8
−
5
signifieth
that
5
is
to
be
Subtracted
from
8.
×
Is
the
sign
of
Multiplication
,
and
signifyeth
that
the
Numbers
between
which
it
is
placed
,
are
to
be
Multiplyed
together
,
as
6×8
signifieth
that
6
and
8
are
to
be
Multiplyed
together
.
=
Is
a
sign
of
Equality
,
and
signifieth
that
the
Numbers
or
Magnitudes
between
which
it
is
placed
,
are
equal
as
3+6
=
7+2
signifyeth
that
3
and
6
are
equal
to
7
and
2
:
Likewise
18
−
6
=
4+8
=
12
and
4×7
=
28
&c.
If
this
be
not
a
sufficient
Explanation
,
read
the
13
,
14
,
15
,
and
19
Sections
of
the
First
Chapter
of
my
Algebraical
Arithmetick
.
THE
SECOND
BOOK
OF
ARTIFICIAL
ARITHMETICK
.
I.
ARtificial
Arithmetick
is
performed
by
Artificial
Numbers
,
very
fitly
called
Logarithmes
.
II.
Logarithmes
are
borrowed
Numbers
which
differ
among
themselves
by
Arithmetical
proportion
,
as
the
Numbers
which
they
signified
differ
by
Geometrical
proportion
III.
Logarithmetical
Arithmetick
is
an
Artificial
use
of
numbers
,
invented
for
ease
in
Calculation
,
wherein
each
natural
Number
is
so
fitted
with
an
Artificial
,
that
what
is
usually
produced
by
Multiplication
of
Natural
Numbers
,
is
here
effected
by
the
Addition
of
their
Artificial
Numbers
:
And
what
Natural
Numbers
performe
by
Division
,
is
here
effected
by
the
Subtraction
of
their
Artificial
Numbers
,
and
what
Natural
Numbers
do
perform
by
long
and
tedious
operations
in
the
extraction
of
Square
,
Cube
,
Biquadrate
,
&c.
Roots
,
is
here
easily
effected
.
by
Bipartition
,
Tripartition
,
Quadrupartition
,
&c.
of
their
Artificial
Numbers
,
and
so
the
hardest
parts
of
Calculation
is
avoidede
by
an
easie
prosthaphaeresis
,
as
our
Trigonometrical
Calculators
of
late
have
sufficiently
experienced
,
by
avoiding
very
tedious
Multiplications
and
Divisions
in
the
use
of
the
Tables
of
Natural
Sines
,
Tangents
,
Secants
to
the
Everlasting
Credit
of
the
honourable
[
](A33564-e171820-b)*
*
The
Lord
Nepair
,
Baron
of
Merchiston
in
Scotland
.
Author
of
this
late
and
incomparable
invention
.
IV.
The
parts
of
Artificial
Arithmetick
are
the
same
with
Natural
Arithmetick
,
but
we
shall
treat
of
them
in
this
order
,
viz.
First
of
the
Nature
of
Logarithmes
;
Secondly
of
their
Genesis
,
or
the
Invention
of
the
Table
of
Logarithmes
And
Thirdly
,
of
the
use
of
the
logarithmes
in
Multiplication
,
Division
,
the
Extraction
of
Roots
,
&c.
CHAP.
II.
Of
the
nature
of
Logarithmes
.
I.
LOgarithmes
are
numbers
so
fitted
to
proportional
numbers
,
that
themselves
Retain
equal
differences
.
Let
there
be
assigned
a
series
,
or
Rank
of
numbers
in
Geometrical
proportion
,
as
those
in
the
Collume
A
viz.
1
,
2
,
4
,
8
,
16
,
32
,
A
|
B
|
C
|
D
|
E
|
1
|
0
|
2
|
5
|
0
|
2
|
1
|
4
|
8
|
3
|
4
|
2
|
6
|
11
|
6
|
8
|
3
|
8
|
14
|
9
|
16
|
4
|
10
|
17
|
12
|
32
|
5
|
12
|
20
|
15
|
64
|
6
|
14
|
23
|
18
|
128
|
7
|
16
|
26
|
21
|
256
|
8
|
18
|
29
|
24
|
512
|
9
|
20
|
32
|
27
|
1024
|
10
|
22
|
35
|
30
|
2048
|
11
|
24
|
38
|
33
|
4096
|
12
|
26
|
41
|
36
|
&c.
And
let
there
be
as
many
other
numbers
placed
over
against
them
in
Arethmetical
progression
,
that
is
having
equal
Differences
as
those
in
the
Collums
B.
C.
D.
E.
or
any
other
numbers
whatsoever
of
the
like
nature
.
Then
,
Forasmuch
as
these
numbers
in
the
Collums
B.
C.
D.
E.
are
of
equal
difference
among
themselves
,
therefore
shall
they
be
the
Log●…rithmes
of
the
numbers
in
the
Collume
A
,
each
of
the
Respective
number
against
which
it
is
placed
.
So
in
the
Collume
B.
the
number
4
,
is
the
Logarithme
of
16
in
the
Collume
A
,
and
in
the
Collume
C
the
number
10
is
the
Logarithme
of
16
in
the
Collume
A
,
and
in
the
Collume
D
,
29
is
the
Logarithme
of
256
in
the
Collum
A
,
&c.
And
as
the
Numbers
in
the
said
Collums
B
,
C
,
D
,
E
,
are
Logarithmes
of
the
Respective
Numbers
in
the
Collum
A
,
so
they
may
be
Logarithmes
of
any
other
Rank
,
or
series
of
Numbers
in
Geometrical
proportion
.
II.
If
four
numbers
are
Arithmetical
proportionals
,
either
Continued
,
or
Discontinued
,
the
sum
of
the
meanes
,
is
equal
to
the
sum
of
the
Extreames
.
Let
us
choose
8
,
10
,
12
,
14
,
in
the
Collume
C.
I
say
that
the
sum
of
the
Extreames
,
8
and
14
,
are
equal
to
the
sum
of
the
two
means
,
10
,
and
12.
For
,
8
+
14
=
10
+
12
=
22.
Or
if
they
are
discontinued
as
,
10
,
12
,
22
,
24
,
in
the
Collum
C
;
for
10
+
24
=
12
+
22
=
34.
The
like
of
any
other
,
this
being
a
peculiar
property
of
all
Numbers
that
are
Arithmetically
proportional
.
III.
If
four
Numbers
are
in
Geometrical
proportion
,
either
continued
,
or
discontinued
,
the
product
arising
from
the
Multiplication
of
the
two
extreams
,
is
equal
to
the
product
of
the
two
means
.
So
4
,
8
,
16
,
32
,
in
the
Collum
of
A
are
Geometrical
proportionals
continued
,
and
the
product
of
the
Extreams
4
and
32
,
is
equal
to
the
product
of
the
means
,
8
,
and
16
,
for
,
4
×
32
=
8
×
16
=
128.
Also
,
4
,
8
,
64
,
128
are
Geometrical
proportionals
discontinued
,
and
the
product
of
4
and
128
,
the
extreams
,
is
equal
to
the
product
of
8
and
64
the
two
means
,
for
4×128
=
8
×
64
=
512.
Hence
it
follows
,
that
what
Geometrical
proportionals
performe
by
Multiplication
,
the
same
will
the
Logarithmes
(
being
Arithmetical
proportionals
)
performe
by
Addition
.
Let
there
be
given
four
Geometrical
proportionals
in
the
Collum
A
,
viz.
8
,
16
,
128
,
and
256
,
and
let
their
Logarithmes
be
8
,
10
,
16
,
and
18
in
the
Collum
C
;
I
say
,
that
as
8
×
256
,
the
product
of
the
extreams
is
equal
to
16
×
128
the
product
of
the
means
,
so
is
8
+
18
the
sum
of
the
Logarithmes
of
the
extreams
is
equal
to
10
+
16
the
sum
of
the
Logarithmes
of
the
means
.
Therefore
,
If
3
Numbers
are
given
to
find
a
fourth
proportional
,
it
may
be
found
by
Addition
and
Subtraction
of
their
Logarithmes
,
(
for
,
as
in
Natural
Numbers
if
you
multiply
the
second
and
third
together
,
and
divide
their
product
by
the
first
,
the
Quote
will
be
the
fourth
proportional
number
so
)
if
you
add
the
Logarithmes
of
the
second
and
third
together
,
and
from
their
sum
subtract
the
Logarithms
of
the
first
,
the
remainder
will
be
the
logarithme
of
the
fourth
proportional
number
.
Example
.
Let
there
be
given
2
,
16.
and
64
,
and
let
it
be
required
to
find
a
fourth
proportional
number
thereto
,
which
is
512.
The
Logarithmes
of
the
given
numbers
are
3
,
12
,
and
18
,
Now
if
you
add
12
and
18
together
(
which
are
the
Logarithmes
of
the
second
and
third
)
their
sum
is
30
,
(
which
is
the
Logarithme
of
1024
,
the
product
of
the
second
and
third
)
and
if
from
30
the
said
sum
of
the
Logarithmes
,
you
substract
,
3
,
(
the
logarithme
of
the
first
)
there
will
remain
27
,
which
is
the
logarithme
of
the
512
the
fourth
proportional
number
sought
for
.
And
IV.
By
what
hath
been
said
,
you
may
perceive
that
to
natural
numbers
there
may
be
fitted
divers
kinds
of
Logarithmes
,
but
we
shall
pitch
only
upon
that
kind
which
were
framed
by
Mr.
Briggs
at
the
request
of
the
Baron
of
Merchiston
,
who
hath
chosen
these
Geometrical
proportionals
,
vi
.
1.10
.
100.1000
.
10000.100000
,
&c.
To
which
Numbers
he
hath
assumed
the
logarithmes
following
,
viz.
for
the
number
1
,
the
logarithme
0.000000
,
for
10
the
logar
.
1.000000
,
for
100
,
the
logar
.
2.000000
for
1000
the
log
.
3.000000
,
for
10000
,
the
log
.
4.000000
,
&c.
as
in
the
following
Table
.
A
|
B
|
1
|
0.000000
|
10
|
1.000000
|
100
|
2.000000
|
1000
|
3.000000
|
10000
|
4.000000
|
100000
|
5.000000
|
1000000
|
6.000000
|
10000000
|
7.000000
|
100000000
|
8.000000
|
1000000000
|
9.000000
|
10000000000
|
10.000000
|
The
Numbers
in
the
Collum
A
are
a
series
of
Geometrical
proportionals
,
and
the
Numbers
in
the
Collum
B
,
are
the
Respective
logarithmes
of
each
of
those
Geometrical
proportionals
,
themselves
being
Arithmetical
proportionals
,
where
note
that
the
Figures
1
,
2
,
3
,
4
,
&c.
which
are
separated
from
the
rest
by
a
point
or
prick
,
are
called
the
Indices
,
or
Characteristicks
of
the
logarithme
,
because
they
declare
how
many
places
the
numbers
by
them
signified
do
consist
of
;
the
Characteristick
of
any
logarithme
being
always
an
unite
less
than
the
number
of
places
,
which
the
number
by
it
signified
doth
consist
of
:
As
in
the
foregoing
Table
you
may
perceive
that
the
logarithme
of
1
,
is
0.000000
,
and
the
logarithme
of
10
is
1.000000
,
and
the
logarithme
of
100
is
2.000000
,
&c.
so
that
the
Index
,
or
Characteristick
of
1
,
and
of
all
numbersfrom
1
to
10
is
0.
&
the
characteristick
of
10
,
and
of
all
numbers
from
10
to
100
is
1
:
And
the
charasterick
of
100
,
and
of
all
Numbers
from
100
,
to
1000
is
2.
and
the
Characteristick
of
each
number
being
an
unite
less
than
the
number
of
places
of
which
the
number
by
it
signified
doth
consist
,
as
was
said
before
.
The
logarithmes
of
this
kind
ought
all
to
consist
of
an
equal
number
of
places
,
that
is
to
say
they
ought
not
to
be
,
one
log
.
of
10
places
,
another
of
8
,
&c.
but
all
of
them
to
be
of
6
,
of
7
,
of
8
&c.
places
.
CHAP.
III.
Of
the
Genesis
or
Fabrick
of
the
Logarithmes
.
I.
THE
Logarithme
of
1
being
assumed
to
be
0.000000
,
and
the
logarithme
of
10
to
be
1.000000
,
the
logarithme
of
100
to
be
2.000000
,
&c.
In
the
next
place
it
will
be
requisite
to
shew
the
way
and
manner
of
Calculating
the
logarithmes
of
the
intermediate
numbers
,
viz.
of
the
numbers
between
1
and
10
,
which
are
2
,
3
,
4
,
5
,
&c.
and
between
10
and
100
,
which
are
11
,
12
,
13
,
14
,
15
,
16
,
&c.
and
between
100
,
and
1000
,
which
are
101
,
102
,
103
,
104
,
&c.
which
to
do
,
observe
the
following
Rules
.
II.
Find
so
many
continual
means
between
1
and
10
,
till
that
continual
mean
which
cometh
nearest
1
,
may
be
a
mixt
number
less
than
2
,
and
so
near
1
,
that
it
may
have
as
many
Cyphers
placed
before
the
significant
Figures
of
the
Numerator
,
as
you
intend
your
logarithmes
to
consist
of
places
;
But
our
Directions
here
shall
be
for
the
making
a
Table
of
logarithmes
to
consist
of
7
places
;
wherefore
find
so
many
continual
means
between
1
and
10
,
till
the
last
may
have
7
Cyphers
placed
before
the
significant
Figures
of
its
Numerator
,
in
order
whereunto
,
annex
to
the
number
10
a
competent
number
of
Cyphers
,
(
viz.
28
,
because
the
work
may
be
the
more
exact
)
and
extract
the
Square
Root
of
that
number
so
enlarged
,
which
being
done
,
you
will
find
its
Square
Root
to
be
3.16227
766016837
,
This
being
done
,
annex
to
the
said
Root
14
Cyphers
more
,
and
extract
the
Square
Root
thereof
,
which
you
will
find
to
be
1.77827941003892
.
Again
annex
to
the
Root
last
found
14
Cyphers
more
,
and
extract
the
Square
Root
thereof
,
which
you
will
find
to
be
1.33352143216332
,
and
thus
proceeding
successively
by
annexing
of
Cyphers
,
and
a
continual
extraction
of
the
Square
Root
,
until
you
have
found
a
Square
Root
,
or
Continual
mean
,
having
7
Cyphers
placed
before
the
significant
Figures
of
its
Numerator
,
which
will
be
found
after
27
several
Extractions
to
be
1.00000001715559
.
So
the
3
last
continual
means
between
10
and
1
will
be
found
to
be
-
1.00000006862238
-
1.00000003431119
-
1.00000001715559
All
which
3
continual
means
are
less
than
2
,
and
so
near
1
,
that
there
are
7
Cyphers
placed
before
the
significant
Figures
of
each
of
their
Numerators
.
Having
found
27
several
means
between
10
,
and
1
,
place
them
successively
one
under
the
other
,
as
in
the
Collum
A
,
of
the
following
Table
;
Then
make
another
Collum
(
B
)
to
contain
the
Respective
logarithmes
of
those
continual
means
.
And
because
biparting
the
logarithme
of
any
number
produceth
the
logarithme
of
the
Square
Root
of
that
number
,
therefore
take
the
logarithme
of
10
,
which
is
1.000000
,
and
place
it
in
the
Collum
B
over
against
10
,
then
bipart
it
,
(
that
is
,
divide
it
by
2
)
and
you
will
have
0.500000
which
is
the
logarithme
of
3
.
1622776●…
016837
the
Square
Root
of
10
,
then
take
half
of
that
logarithme
,
viz
0.500000
which
is
0.250000
,
and
place
it
for
the
logarithme
of
1.778279410
,
&c.
the
second
mean
proportional
,
(
or
Square
Root
of
3
.
1622776●…0
,
&c.
)
And
so
by
continu●…
bipartition
,
you
will
at
length
find
that
0.000000007450580
,
will
be
the
logarithme
of
the
last
continual
mean
,
viz.
the
logarithme
of
1.000000017
559
,
as
in
the
following
Table
.
A
Continual
means
.
|
B
their
Logarithmes
.
|
10.00000000000000
|
1.00000000000000
|
3.16227766016837
|
0.50000000000000
|
1.77827941003892
|
0.25000000000000
|
1.33352143216332
|
0.12500000000000
|
&c.
|
&c.
|
1.00000006862238
|
0.0000000029802322
|
1.00000003431119
|
0.0000000014901161
|
1.00000001715559
|
0.0000000007405805
|
III.
Any
number
whatsoever
being
given
,
how
to
make
the
Logarithme
thereof
.
When
it
is
required
to
make
the
Logarithme
of
any
number
,
extract
so
many
continual
means
between
the
given
number
and
1
,
until
the
mean
which
cometh
nearest
1
,
may
be
a
mixt
number
less
than
2
,
and
so
near
1
,
that
it
may
have
7
Cyphers
placed
before
the
significant
Figures
of
its
Numerator
,
which
being
done
,
you
may
easily
find
out
the
Logarithme
of
that
continual
mean
,
by
help
of
the
foregoing
Table
;
and
then
by
doubling
,
and
redoubling
the
logarithme
of
the
said
continual
mean
,
as
many
times
as
you
found
continual
means
by
extraction
,
so
shall
you
at
last
have
the
logarithme
of
the
given
number
.
You
may
make
the
Logarithme
of
any
number
whatsoever
by
this
and
the
last
Rule
.
As
for
Example
.
Let
us
pitch
upon
the
number
2
,
and
make
its
Logarithme
.
To
do
which
,
annex
to
the
number
2
a
competent
number
of
Cyphers
,
viz.
28
,
and
extract
the
Square
Root
thereof
,
which
you
will
find
to
be
1
.
41421356237●…09
for
the
first
continual
mean
,
to
which
said
mean
annex
14
Cyphers
more
,
and
extract
the
Square
Root
thereof
,
and
so
proceed
,
by
annexing
of
Cyphers
and
extracting
of
Roots
,
till
the
nearest
mean
proportional
number
to
1
,
may
have
seven
Cyphers
placed
before
the
significant
Figures
of
its
Numerator
,
which
after
23
several
Extractions
you
will
find
to
be
1.00000008262958
.
Then
to
find
out
the
logarithme
of
this
continual
mean
,
say
by
the
Rule
of
3
Direct
.
As
the
significant
Figures
of
the
Twenty
fifth
mean
proportional
in
the
foregoing
Table
,
viz.
6862238.
Is
to
its
respective
Logarithme
,
29802322.
So
are
the
significant
Figures
of
the
last
continual
mean
found
between
1
and
2
,
viz.
8262958.
To
its
Respective
Logarithme
35885571.
Now
if
you
prefix
before
the
Logarithme
last
found
8
Cyphers
,
it
will
be
0000000035885571
,
which
being
doubled
and
redoubled
23
times
,
(
because
there
were
23
continual
means
found
between
1
and
two
)
there
will
at
last
be
produced
0.30102998797568
,
which
is
the
logarithme
of
the
number
2
,
which
was
Required
,
but
because
we
intend
the
Table
of
logarithmes
to
consist
but
of
7
places
,
and
because
2
nines
follow
the
sixth
place
therefore
make
the
figure
2
to
be
3
and
so
shall
the
logarithme
of
2
be
0.301030
cancelling
the
following
Figures
as
superfluous
.
The
Logarithme
of
2
being
found
,
you
may
easily
find
the
logarithmes
of
4
,
5
,
8
,
16
,
20
,
25
,
32
,
40
,
50
,
64
,
&c.
by
Artificial
Multiplication
and
Division
,
which
is
by
adding
and
subtracting
of
logarithmes
;
for
if
you
take
the
logarithme
of
2
out
of
the
logarithme
of
10
,
there
will
remain
the
logarithme
of
5
and
the
logarithme
of
2
Doubled
gives
you
the
logarithme
of
4
,
then
add
the
logarithme
of
4
to
the
logarithme
of
2
,
and
you
have
the
logarithme
of
8
,
and
to
the
logarithme
of
8
add
the
logarithme
of
2
,
and
it
gives
you
the
logarithme
of
16
,
and
the
logarithme
of
5
added
to
the
logarithme
of
4
,
gives
he
logarimthe
of
20
,
and
the
logarithme
of
5
doubled
,
gives
the
Logarithme
of
25
,
&c.
In
the
next
place
you
are
to
get
the
Logarithmes
of
3
,
7
,
11
,
13
,
17
,
19
,
23
,
29
,
31
,
37
,
41
,
●…3
,
47
,
53
,
59
,
61
,
67
,
7●…
,
73
,
79
,
89
,
97
,
&c
by
help
of
which
all
the
rest
may
be
Calculated
.
IV.
The
first
figure
of
every
logarithme
,
which
is
separated
from
the
rest
by
a
point
or
prick
is
very
properly
called
the
Index
,
or
Char●…cteristick
of
the
logarithme
,
which
sheweth
the
Nature
of
the
Number
by
it
signified
,
viz.
whether
it
be
positive
,
or
negative
,
and
if
positive
,
of
what
number
of
places
it
doth
consist
,
and
if
negative
,
what
place
of
the
Decimal
Fraction
the
first
figure
of
the
number
by
it
signified
,
shall
possess
,
as
in
the
following
Table
,
So
the
logarithme
of
|
46768
|
will
be
|
4.670134
|
4676.8
|
3.670134
|
467.68
|
2.670134
|
46.768
|
1.670134
|
4.6768
|
0.670134
|
.
46768
|
−
1.670134
|
.
046768
|
−
2.670134
|
.
0046768
|
−
3.670134
|
.
00046768
|
−
4.670134
|
Whereby
you
may
perceive
that
the
logarithmes
of
absolute
and
defective
numbers
are
the
same
,
only
the
Characteristick
of
a
defective
number
is
marked
with
the
note
of
defection
,
for
the
logarithme
of
the
Absolute
number
46768
is
4.670134
,
the
Characteristick
4
,
shewing
the
number
by
it
signified
to
consist
of
5
places
,
as
is
already
said
in
the
fourth
Rule
of
the
second
Chapter
,
and
the
logarithme
of
the
mixt
number
46.768
is
1.670134
which
is
the
same
with
the
former
,
only
the
Characteristick
is
1
,
which
sheweth
the
integral
Number
by
it
signified
,
to
consist
of
2
places
,
the
●…est
being
a
decimal
Fraction
.
Likewise
the
logarithme
of
the
Decimal
.46768
is
−
1.670134
,
which
is
still
the
same
with
the
f●…rmer
,
only
its
Characteristick
being
marked
with
a
note
of
defection
,
sheweth
it
to
be
the
Logarithme
of
a
Decimal
Fraction
,
and
because
the
Characteristick
is
−
1
,
it
sheweth
that
the
first
figure
of
the
number
by
it
signified
doth
possess
the
first
place
of
the
Decimal
;
or
place
of
primes
:
Ag●…in
,
the
Logarithme
−
4
.
670●…
is
still
the
same
,
and
if
you
look
for
it
in
the
Table
of
Logarithms
,
not
regarding
the
Index
,
you
will
find
it
to
be
the
Logarithme
of
46768
,
but
because
its
Index
is
defective
,
I
conclude
it
to
be
the
Logarithme
of
a
Decimal
,
and
because
the
Index
or
Characteristick
is
−
4
,
therefore
I
conclude
that
the
first
Figure
of
the
number
signified
by
it
,
must
possess
the
fourth
place
of
the
Decimal
,
wherefore
place
3
Cyphers
before
it
,
and
you
have
.00046768
for
the
Decimal
s●…gnified
by
the
logarithme
−
4.670134
.
This
being
well
understood
,
the
rest
will
easily
be
attained
by
the
following
Directions
.
CHAP.
IV.
Of
the
Use
of
the
Table
of
Logarithmes
.
THE
use
of
the
Table
of
Logarithmes
is
twofold
,
viz.
First
,
To
find
therein
the
logarithme
of
any
given
number
,
or
to
find
the
number
appropriated
to
any
given
logarithme
.
Secondly
,
To
resolve
diverse
necessary
problems
in
Arithmetick
,
Geometry
,
Trigonometry
,
Astronomy
,
&c.
Concerning
the
first
of
these
,
I
shall
not
meddle
,
because
our
Limits
will
not
afford
sufficient
room
to
insert
a
Table
of
Logarithmes
,
and
the
Tables
already
published
by
others
are
sufficiently
explained
,
in
that
point
as
Mr.
Briggs
,
Mr.
Gunter
,
Dr.
Newton
,
Mr.
Wingate
,
Mr.
Norwood
,
Mr.
Phillips
,
&c.
Every
one
shewing
how
by
their
own
Tables
to
find
the
logarithme
of
any
number
,
or
the
number
to
any
logarithme
,
therefore
I
shall
proceed
to
shew
their
use
in
Arithmetick
,
viz.
how
to
Multiply
,
Divide
,
and
Extract
Roots
,
&c.
thereby
.
And
First
,
To
Multiply
by
the
Logarithmes
.
In
Multiplication
by
the
logarithmes
there
are
3
Cases
,
viz.
the
Characteristicks
of
the
logarithmes
of
the
Factors
are
either
both
affirmative
,
or
both
negative
;
or
else
they
are
the
one
affirmative
,
and
the
other
negative
?
I.
When
they
are
both
Affirmative
.
When
the
Characteristicks
of
the
logarithmes
of
the
Factors
are
both
Affirmative
,
then
the
sum
of
those
logaithmes
is
the
logarithme
of
the
fact
or
product
.
Examples
.
Note
that
if
you
carry
ten
to
the
Characteristicks
,
it
is
affirmative
,
as
in
the
last
Example
.
II.
When
they
are
both
Negative
.
When
the
logarithmes
of
the
Factors
have
their
Characteristicks
both
Negative
,
or
defective
,
then
the
sum
of
their
logarithmes
is
the
logarithme
of
their
product
,
the
sum
of
their
Characteristicks
being
also
negative
as
in
the
following
Examples
.
And
here
note
,
that
when
you
carry
ten
to
the
Characteristicks
it
is
affirmative
,
and
must
be
abated
out
of
their
sum
as
in
the
two
last
examples
.
III.
When
they
are
heterogeneal
,
viz.
the
one
Affirmative
,
and
the
other
Negative
.
When
the
Characteristicks
of
the
Factors
are
the
one
Negative
,
and
the
other
Affirmative
,
then
add
their
logarithms
together
,
and
when
you
come
to
the
Characteristicks
,
take
their
difference
,
and
place
it
for
the
Characteristick
of
the
Product
,
making
it
either
Affirmative
or
Negative
,
according
to
the
affection
of
that
wherein
lay
the
excess
;
and
here
note
,
that
if
you
carry
any
thing
to
the
Characteristicks
,
it
is
Affirmative
,
and
must
be
added
to
the
affirmative
characterist
.
As
in
the
following
Examples
.
CHAP.
V.
Division
by
the
Logarithmes
.
I.
TO
subtract
the
Logarithme
of
one
number
out
of
the
logarithme
of
another
,
is
the
same
(
and
produceth
the
same
effect
)
with
Division
in
Natural
Numbers
,
the
Logarithme
remaining
being
the
Logarithme
of
the
Quotient
.
II.
In
Division
by
the
Logarithmes
there
are
three
Cases
,
viz.
First
,
when
the
Characteristicks
of
the
Dividend
,
and
of
the
Divisor
are
both
Affirmative
:
Secondly
,
when
they
are
both
Negative
.
And
Thirdly
,
when
they
are
heterogeneal
,
viz.
the
one
Affirmative
,
and
the
other
Negative
.
Of
which
in
their
order
.
I.
When
they
are
both
Affirmative
.
III.
When
the
Characteristicks
of
the
Dividend
,
and
of
the
Divisor
,
are
both
Affirmative
,
then
if
you
subtract
the
Logarithme
of
the
Divisor
out
of
the
logarithme
the
Dividend
,
the
remainder
will
be
the
logarithme
of
the
Quotient
.
And
if
you
borrow
10
from
the
Characteristicks
,
it
is
Affirmative
.
Examples
.
These
Examples
are
so
plain
that
they
need
no
Explanation
.
II.
When
they
are
both
Negative
.
IV.
When
the
Characteristicks
of
the
Dividend
,
and
of
the
Divisor
,
are
both
Negative
,
subtract
the
Logarithme
of
the
Divisor
from
the
Logarithme
of
the
Dividend
,
and
the
Remainder
is
the
logarithme
of
the
Quotient
,
and
if
you
borrow
10
,
it
must
be
paid
to
the
Index
of
the
Divisor
affirmatively
.
Examples
.
The
first
and
second
of
the
foregoing
Examples
are
easily
understood
,
and
as
for
the
third
and
fourth
,
all
the
difficulty
therein
is
caused
by
borrowing
10
at
the
next
figure
to
the
Characteristicks
,
as
in
the
third
Example
,
in
subtracting
5
out
of
1.
Now
to
make
good
the
10
borrowed
,
I
pay
1
to
the
Characteristick
of
the
Divisor
,
and
because
the
said
1
is
affirmative
,
and
the
said
Characteristick
negative
,
therefore
subtract
it
from
the
Characteristick
of
the
Divisor
,
and
there
remains
nothing
;
wherefore
I
take
(o)
out
of
the
Characteristick
of
the
Dividend
,
and
there
remains
−
1
for
the
Characteristick
of
the
Quotient
.
The
same
is
to
be
understood
in
the
fourth
Example
,
and
in
all
other
of
the
same
Nature
.
A
General
Observation
drawn
from
the
third
and
fourth
Rules
foregoing
.
V.
If
when
the
Characteristicks
of
the
Dividend
and
Divisor
be
Homogeneal
,
(
that
is
,
both
affirmative
,
or
both
negative
)
the
Characteristick
of
the
Divisor
is
greater
than
the
Characteristick
of
the
Dividend
,
then
in
this
case
subtract
the
characteristick
of
the
Dividend
out
of
that
of
the
Divisor
,
placing
the
remainder
for
the
characteristick
of
the
Quotient
,
changing
its
sign
,
viz.
if
it
be
affirmative
,
make
it
negative
,
and
if
it
be
negative
,
make
it
affirmative
.
Remembring
the
Directions
under
the
last
Rule
when
you
borrow
from
the
Characteristicks
.
Observe
the
following
Examples
.
III.
When
they
are
Heterogeneal
,
viz.
the
one
Negative
,
the
other
Affirmative
.
VI.
When
the
Characteristicks
of
the
Dividend
and
the
Divisor
are
Heterogeneal
,
proceed
as
in
the
two
First
cases
,
till
you
come
to
the
characteristicks
,
and
then
instead
of
subtracting
the
one
characteristick
from
the
other
,
add
them
together
,
so
shall
their
sum
be
the
characteristick
of
the
Quotient
,
and
it
is
of
the
same
kind
with
the
Characteristick
of
the
Dividend
.
But
here
note
that
when
you
borrow
10
at
the
next
figure
to
the
characteristick
it
must
be
paid
to
the
characteristick
of
the
Divisor
affirmatively
viz.
If
the
characteristick
of
the
Divisor
be
affirmative
,
then
add
1
to
it
for
that
you
borrowed
,
and
if
it
be
negative
,
subtract
1
from
it
.
As
in
the
following
Examples
.
In
the
second
of
the
foregoing
Examples
I
borrow
1
(
in
the
place
next
the
characteristicks
)
by
subtracting
9
out
of
8
,
wherefore
to
make
it
good
,
I
subtract
1
from
−
2
(
the
characteristick
of
the
Divisor
,
)
because
it
is
Negative
,
and
the
remainder
which
is
−
1
I
add
to
1
(
the
characteristick
of
the
Dividend
)
and
their
sum
is
2
for
the
characteristick
of
the
Quotient
which
is
Affirmative
,
because
the
characteristick
of
the
Dividend
is
Affirmative
.
And
in
the
last
Example
,
I
likewise
borrow
1
from
the
characteristick
,
wherefore
to
make
it
good
,
I
add
1
to
the
characteristick
of
the
Divisor
,
(
because
it
is
affirmative
)
and
that
makes
it
3
,
which
added
to
−
1
(
the
characteristick
of
the
Dividend
)
makes
−
4
for
the
characteristick
of
the
Quotient
,
which
here
is
negative
,
because
the
characteristick
of
the
Dividend
is
negative
.
Other
Examples
for
Exercise
may
be
such
as
follow
.
CHAP.
VI.
To
raise
the
Powers
of
Numbers
,
viz.
to
find
the
Square
,
Cube
,
Biquadrate
,
or
Squared
Square
,
&c.
of
any
Number
.
Also
to
Extract
the
Square
,
Cube
,
Biquadrate
,
&c.
Roots
of
any
Number
by
the
Logarithmes
.
1.
BY
the
third
Section
of
the
Second
Chapter
of
this
Book
it
is
evident
,
that
if
you
add
the
Logarithmes
of
two
numbers
together
,
the
Sum
will
be
the
Logarithme
of
their
Product
;
And
by
the
first
and
fourth
Sections
of
the
9th
Chapter
of
my
Decimal
Arithmetick
,
it
appeareth
that
any
number
multiplyed
by
it self
,
produceth
its
Square
,
wherefore
if
you
double
(
or
multiply
by
2
)
the
logarithme
of
any
number
,
it
will
produce
the
logarithme
of
its
Square
,
which
if
duly
considered
you
will
find
that
to
Square
,
Cube
,
&c.
any
number
,
is
nothing
else
but
to
multiply
the
Logarithme
of
the
given
number
by
the
Index
of
the
Power
you
would
raise
it
to
,
viz.
If
you
would
find
the
Square
of
any
number
,
multiply
the
logarithme
thereof
by
2
,
so
shall
the
product
thereof
be
the
Logarithme
of
its
Square
;
and
if
you
would
find
the
Cube
of
any
number
,
multiply
its
logarithme
by
3
,
and
the
product
thereof
will
be
the
Logarithme
of
its
Cube
;
and
if
you
would
find
the
Biquadrate
of
any
number
,
multiply
its
Logarithme
by
4
,
and
it
will
produce
the
Logarithme
of
its
Biquadrate
,
&c.
As
in
the
following
Example
.
Let
it
be
required
to
find
the
Square
of
12
Which
being
multiplyed
by
2
,
produceth
2.158362
,
which
is
the
Logarithme
of
144
,
viz.
the
Square
of
12.
Again
let
it
be
required
to
find
the
Square
of
94.
Which
being
multiplyed
by
2
,
produceth
3.9462556
,
which
is
the
logarithme
of
8836
,
which
is
the
Square
of
94.
II.
But
if
the
characteristick
of
the
Logarithme
be
negative
,
that
is
,
if
the
given
number
,
whose
Square
,
Cube
,
Biquadrate
,
&c.
you
would
find
be
a
Decimal
Fraction
,
observe
,
that
in
multiplying
the
next
figure
to
the
characteristick
the
ten
,
or
tens
to
be
borne
in
mind
are
affirmative
,
and
are
to
be
deducted
out
of
the
Product
of
the
negative
characteristicks
.
Observe
the
several
Examples
following
The
like
is
to
be
observed
of
all
others
.
To
Extract
the
Square
,
Cube
,
Biquadrate
,
&c.
Roots
of
any
given
numbers
by
the
Logarithmes
.
III.
From
a
due
consideration
of
the
first
Section
of
this
Chapter
,
it
may
easily
be
perceived
,
that
To
Extract
the
Cube
Root
of
any
number
is
to
bipart
(
or
divide
by
2
)
its
logarithme
,
so
shall
that
biparted
logarithme
be
the
logarithme
of
the
Square
Root
desired
.
Examples
.
Let
it
be
required
to
Extract
the
Square
Root
of
75832.
Let
it
be
required
to
find
the
Square
Root
of
4489.
In
the
first
of
these
Examples
the
logarithme
of
75832
is
4.879852
which
being
biparted
(
or
divided
by
2
)
gives
1.826074
for
the
logarithme
of
(
275.37
)
the
Root
required
.
And
in
the
second
Example
3.652149
(
the
logarithme
of
(
4489
)
being
biparted
gives
1.826074
for
the
logarithme
of
(
67
)
its
Square
Root
.
So
will
the
Square
Root
of
36783
be
found
to
be
191.789
fere
,
and
the
Square
Root
of
3866
will
be
62.17717
fere
.
And
the
Square
Root
of
95
will
be
found
to
be
9.7468
,
&c.
To
Extract
the
Cube
Root
of
any
number
is
to
tripart
(
or
divide
by
3
)
its
logarithme
,
so
shall
this
triparted
logarithme
be
the
logarithme
of
the
Cube
Root
required
.
Examples
.
Let
it
be
required
to
Extract
the
Cube
Root
of
157464.
Let
it
be
required
to
find
the
Cube
Root
of
187237601580329.
In
the
first
of
these
Examples
where
it
is
required
to
extract
the
Cube
Root
of
157464
,
its
logar
.
is
5.197181
which
being
divided
by
3
,
hath
for
its
third
part
●…393
which
is
the
logarithme
of
(
54
)
the
Cube
Root
of
157464
,
which
was
required
.
And
the
same
is
to
be
observed
in
finding
the
Cube
Root
of
1872376015
80329
by
the
Logarithmes
;
or
it
any
other
positive
number
whatsoever
as
you
may
see
by
the
following
Examples
.
To
extract
the
Biquadrate
Root
of
any
given
number
,
do
thus
,
viz.
Take
its
logarithme
,
and
divide
To
extract
the
Biquadrat
Root
of
any
Number
.
it
by
4
,
so
shall
the
fourth
part
thereof
be
the
logarithme
of
the
biquad
.
Root
required
,
as
in
the
following
Example
.
Let
it
be
required
to
Extract
the
Biquadrate
Root
of
256.
Here
the
log
.
of
the
given
biquadrat
number
,
viz.
256
is
2.408239
which
being
divided
by
4
,
giveth
0.602059
for
the
logarithme
of
(4)
the
biquadrat
Root
required
.
In
like
manner
if
you
would
extract
the
Root
of
the
fifth
Power
of
any
number
given
,
Divide
its
logarithme
by
5
,
so
shall
the
Quotient
be
the
logarithme
of
its
Root
.
And
if
you
would
find
the
Root
from
the
sixth
power
of
any
Number
,
divide
its
logarithme
by
6
,
and
the
Quote
is
the
logarithme
of
the
Root
desired
,
&c.
IV.
But
here
you
are
to
observe
in
Extracting
the
Square
,
Cube
,
Biquadrat
,
or
any
other
Root
To
extract
the
Square
,
Cube
,
&c.
Roots
of
negative
numbers
by
the
Logarithmes
.
of
a
negative
number
,
or
decimal
by
the
logarithmes
,
that
if
you
cannot
evenly
divide
the
Index
or
Characteristick
of
the
logarithme
without
any
remainder
,
then
add
to
the
said
Characteristick
so
many
units
till
it
may
be
divided
without
any
remainder
,
and
place
the
Quotient
for
a
new
Characteristick
,
(
belonging
to
the
Root
;
)
Then
look
how
many
units
you
lent
to
the
Characteristick
,
and
esteem
them
so
many
tens
to
be
prefixed
to
the
logarithmetical
figure
immediately
following
the
Characteristick
,
then
proceed
to
finish
the
work
,
so
shall
this
new
logarithme
be
the
logarithme
of
the
Root
required
,
which
will
also
be
negative
.
Examples
follow
.
What
is
the
Square
Root
of
.144
What
is
the
Square
Root
of
.00324
What
is
the
Cube
Root
of
.000512
In
the
first
of
these
Examples
,
where
it
is
required
to
Extract
the
Square
Root
of
.144
,
its
logar
.
is
−
1.158362
,
which
(
according
to
the
third
Rule
)
I
should
bipart
,
(
or
divide
by
2
)
And
because
its
negative
Characteristick
(
−
1
)
cannot
be
evenly
divided
by
2
,
I
increase
it
by
an
unit
,
and
it
makes
2
,
then
will
the
Quote
be
−
1
for
the
Index
of
the
log
.
of
the
Root
,
then
do
I
proceed
to
the
next
Figure
,
to
the
Characteristick
which
is
1
,
and
because
I
added
1
to
the
Characteristick
,
therefore
I
increase
the
next
Figure
by
adding
10
to
it
.
(
or
prefixing
1
before
it
)
and
then
is
it
11
,
&c.
So
I
find
the
logarithme
of
the
Root
to
be
−
1.5755272
viz.
.37947
And
in
the
third
Example
,
where
it
is
required
to
find
the
Cube
Root
of
.000512
,
the
Index
of
its
logarithme
is
−
4
,
which
cannot
be
evenly
divided
by
3
,
therefore
I
add
2
to
it
to
make
it
6
,
and
the
Quotient
is
−
2
for
the
Index
of
the
loagarithme
of
the
Root
,
then
because
I
added
2
to
the
Index
−
4
,
therefore
I
increase
the
next
Figure
to
it
with
2
tens
,
making
it
27
,
&c.
So
is
the
Cube
Root
required
found
to
be
.05
.
Observe
the
like
in
extracting
of
any
other
Roots
:
Otherwise
,
you
may
make
use
of
the
following
Table
.
The
Use
of
the
foregoing
Table
.
In
the
foregoing
Table
the
Figures
2.3.4.5.6
placed
on
the
left
hand
,
are
the
Indices
of
Powers
,
whose
Roots
are
required
to
be
extracted
,
or
they
are
Divisors
by
which
to
divide
the
logarithme
of
any
given
power
,
in
order
to
find
out
its
Root
:
As
the
number
2
(
which
is
the
uppermost
)
is
the
Divisor
for
finding
the
logarithme
of
the
Square
Root
of
any
number
:
And
3
the
Divisor
for
the
Cube
;
and
4
for
the
Biquadrat
,
&c.
The
Figures
placed
between
the
perpendicular
line
,
and
the
several
lines
of
connection
,
and
under
A
are
the
Characteristicks
of
the
logarithmes
of
Negative
or
Decimal
Numbers
,
whose
Roots
are
required
to
be
extracted
;
And
the
Figures
placed
on
the
right
hand
of
the
perpendicular
line
under
B
,
are
the
Characteristicks
of
the
Logarithme
of
the
several
Roots
:
And
the
numbers
at
the
bottom
of
the
Table
,
viz.
50
,
40
,
30
,
&c.
are
the
numbers
to
be
added
,
or
rather
prefixed
to
the
first
Figure
of
the
logarithme
next
the
Characteristick
whose
Negative
Index
is
found
in
the
same
series
or
Collum
even
with
the
Divisor
,
&c.
Example
.
Let
it
be
required
to
extract
the
Cube
Root
of
.405224
.
The
Logarithme
of
the
given
Number
is
−
1.6076951
.
And
the
Divisor
whereby
to
extract
the
Cube
Root
is
3
,
which
I
find
in
the
foregoing
Table
on
the
left
hand
;
then
on
the
right
hand
of
its
line
of
connection
,
I
find
the
Characteristick
of
the
Logarithme
−
1.6076951
which
is
−
1
,
and
just
against
it
on
the
right
hand
under
B
I
find
−
1
for
the
Characteristick
of
the
Logarithme
of
the
Root
,
and
in
the
bottom
line
under
the
said
−
1
,
in
the
same
series
,
I
find
20
which
is
to
be
prefixed
to
the
Figure
in
the
Logarithme
next
the
Characteristick
,
&c.
and
having
finished
Division
,
I
find
the
Logarithme
of
its
Root
to
be
−
1.8692317
which
is
.74
.
So
if
the
Logarithme
−
6.418426
were
to
be
divided
by
4
,
First
find
the
Divisor
4
on
the
left
hand
of
the
Table
,
and
find
the
Characteristick
−
6
,
behind
its
line
of
connection
just
against
which
on
the
right
hand
of
the
perpendicular
line
you
have
−
2
for
the
Index
or
Characteristick
of
the
Quotient
;
and
at
the
bottom
,
just
underneath
−
6
you
have
20
,
which
being
added
to
(4)
the
first
figure
of
the
Dividend
next
the
Characteristick
makes
it
24
,
in
which
the
Divisor
3
is
contained
6
times
,
&c.
See
the
work
.
CHAP.
VII
.
Of
the
Use
of
Log.
in
Comparative
Arithmetick
.
PROP.
I.
Having
three
Numbers
given
,
to
find
a
fourth
proportional
.
THis
is
nothing
else
but
the
work
of
the
Rule
of
3
,
and
it
may
be
thus
performed
,
viz.
Add
the
Logarithme
of
the
second
and
third
Numbers
together
,
and
from
their
sum
subtract
the
Logarithme
of
the
first
,
so
shall
the
remainder
be
the
Logarithme
of
the
fourth
,
as
in
the
following
Example
.
The
3
given
numbers
are
3
,
24
,
and
108
unto
which
it
is
required
to
find
a
fourth
proportional
.
The
Operation
by
the
Logarithmes
.
Which
is
the
Logarithme
of
864.
the
fourth
proportional
required
.
The
former
work
may
be
somewhat
shortned
,
if
in
stead
of
the
Log.
of
the
first
you
take
its
Complement
Arithmetical
(
which
is
nothing
else
but
to
set
down
what
every
Figure
wants
of
9
,
till
you
come
to
that
next
the
right
hand
,
and
then
set
down
what
it
wants
of
10
)
and
then
add
them
all
3
together
,
and
cancel
the
first
Figure
of
the
sum
on
the
left
hand
,
and
then
will
the
sum
be
the
Log.
of
the
Answer
,
as
will
appear
by
working
the
foregoing
Examples
.
PROP.
II.
Between
two
Numbers
given
to
find
a
mean
proportional
.
When
the
Logarithmes
of
the
numbers
propounded
are
homogeneal
,
viz.
both
Affirmative
,
or
both
Negative
,
add
them
together
,
then
bipart
that
Logarithmetical
sum
,
so
have
you
the
Logarithme
of
the
mean
proportional
required
,
which
logarithme
so
found
is
of
the
same
kind
with
the
Logarithmes
of
the
number
given
.
Let
it
be
required
to
find
a
mean
proportional
between
18
and
6.
In
this
Example
the
Logrithme
of
18
,
and
6
being
added
together
make
2.02342
which
is
the
Logarithme
of
108
,
and
that
Logarithme
being
divided
by
2
(
which
is
the
same
with
extracting
the
Square
Root
of
its
significant
number
,
as
in
the
third
Rule
of
the
Sixth
Chapter
)
the
Quotient
is
2.02342
,
the
Logarithme
of
10.392
which
is
the
mean
proportional
required
.
Example
2.
Let
it
be
required
to
find
a
mean
proportional
between
.018
and
.006
.
II.
But
when
the
Characteristicks
of
the
Logarithmes
of
the
given
Numbers
be
heterogeneal
,
viz.
the
one
Affirmative
,
and
the
other
Negative
;
add
the
Logarithmes
together
as
before
,
till
you
come
to
the
Characteristicks
,
then
subtract
the
lesser
Characteristick
out
of
the
greater
,
(
according
to
the
third
Rule
of
the
fourth
Chapter
,
)
which
being
done
,
bipart
the
Logarithmetical
production
,
so
shall
the
Quote
be
the
Logarithme
of
the
mean
proportional
required
,
which
will
always
be
of
the
same
kind
with
that
Logarithme
of
the
given
Numbers
,
whose
Index
is
greatest
,
as
in
the
following
Examples
.
Example
1.
What
is
the
mean
proportional
between
36
and
●…5
?
Facit
4.2427
.
Example
2.
What
is
the
mean
proportional
between
12
and
.75
?
Facit
3.
PROP.
III.
Between
2
Numbers
given
,
to
find
two
mean
Proportionals
.
Whether
the
Numbers
given
be
Homogeneal
,
or
Heterogeneal
,
subtract
the
Logarithme
of
the
lesser
extream
from
the
Logarithme
of
the
greater
extream
,
then
take
1/3
of
the
difference
of
the
said
Logarithmes
,
and
add
it
to
the
Logarithme
of
the
lesser
extream
,
so
will
the
sum
be
the
Logarithme
of
the
lesser
mean
;
then
add
the
same
Difference
to
the
said
Logarithme
of
the
lesser
mean
,
and
the
sum
will
be
the
Logarithme
of
the
greater
mean
;
still
observing
the
Rules
delivered
in
the
fourth
and
sith
Chapters
of
this
book
,
in
adding
and
subtracting
of
Logarithmes
.
Examples
.
Ex.
1.
Let
it
be
required
to
find
2
mean
proportionals
between
144
and
12
?
Example
.
2.
Let
it
be
required
to
find
2
mean
proportionals
between
.75
and
.05
?
Example
.
3.
Let
it
be
required
to
find
2
mean
proportionals
,
between
125
and
.05
?
PROP.
IV.
Three
Numbers
given
to
find
a
fourth
in
a
Duplicate
Proportion
.
Take
the
Logarithmes
of
the
two
Numbers
which
have
one
and
the
same
Denomination
,
and
subtract
the
lesser
Logarithme
from
the
greater
,
and
double
the
remainder
,
(
that
is
multiply
it
by
2.
)
Then
if
the
first
number
be
less
than
the
second
,
add
the
said
doubled
difference
to
the
Logarithme
of
the
other
Number
,
so
will
the
sum
be
the
Logarithme
of
the
fourth
number
,
or
number
required
,
as
in
Example
.
The
superficial
content
of
a
Circle
whose
Diameter
is
14
Inches
is
154
Square
inches
,
I
demand
the
Content
of
another
Circle
,
whose
Diameter
is
25
Inches
?
Facit
491.07
square
Inches
.
See
the
operation
.
But
if
the
first
number
be
greater
than
the
second
,
then
instead
of
adding
the
doubled
difference
to
the
other
number
,
subtract
it
therefrom
,
so
shall
the
remainder
be
the
Logarithme
of
the
number
required
,
as
in
the
following
Example
.
There
is
a
Circle
whose
Diameter
is
28
Inches
,
and
its
superficial
Content
is
616
Square
Inches
,
I
demand
what
is
the
Superficial
Content
of
another
Circle
,
whose
Diameter
is
25
Inches
?
Facit
491.07
Square
Inches
,
as
in
the
former
Example
.
PROP.
V.
Having
3
Numbers
given
to
find
a
fourth
in
a
Triplicate
Proportion
.
Triple
the
Difference
of
the
Logarithmes
of
the
two
given
Terms
,
which
have
the
same
Denomination
.
Then
if
the
first
Term
be
less
than
the
second
,
add
the
said
Triple
Difference
to
the
Logarithme
of
the
other
Term
,
so
shall
the
sum
be
the
Logarithme
of
the
fourth
Term
required
,
as
in
the
following
Example
.
There
is
a
Bullet
whose
Diameter
is
4
Inches
,
and
its
weight
is
9
l.
I
demand
what
weight
a
Bullet
of
the
same
metal
will
be
,
whose
Diameter
is
8
Inches
?
Facit
72
l.
view
the
following
work
.
But
if
the
first
term
be
greater
than
the
second
,
then
subtract
the
said
tripled
Difference
from
the
logarithme
of
the
other
Term
,
so
shall
the
remainder
be
the
Logarithme
of
the
fourth
number
required
.
As
in
Example
.
There
is
a
Bullet
whose
Diameter
is
8
Inches
,
and
its
weight
is
72
pounds
,
I
demand
the
weight
of
another
Bullet
of
the
same
Mettal
,
whose
Diameter
is
4
Inches
?
Facit
9
l.
See
the
operation
,
it
being
the
converse
of
the
former
.
CHAP.
VIII
.
Of
Anatocisme
,
or
Compound
Interest
,
wherein
is
shewed
how
by
the
Logarithmes
to
answer
all
Questions
concerning
the
Increase
,
or
present
worth
of
any
Sum
of
Money
or
Annuity
,
for
any
Term
of
Years
,
or
at
any
Rate
of
Interest
.
According
to
the
six
Fundamental
Theorems
invented
and
laid
down
by
Mr.
Oughtred
in
his
Treatise
De
Anatocismo
sive
Usura
Composita
,
annexed
to
his
Clavis
Mathematicae
.
I.
WHen
any
Question
in
Compound
Interest
is
proposed
,
it
willfall
under
one
of
the
six
Cases
following
,
viz.
-
1.
To
find
the
Increase
or
Amount
of
any
sum
of
money
put
out
at
Compound
Interest
for
any
number
of
years
,
and
at
any
Rate
of
Interest
propounded
.
-
2.
To
find
the
present
worth
of
any
sum
of
money
due
at
the
end
of
any
number
of
years
to
come
,
Rebate
being
allowed
at
any
Rate
of
Compound
Interest
.
-
3.
To
find
the
Increase
,
or
amount
of
an
Annuity
being
forborne
for
any
number
of
years
,
at
any
Rate
of
Compound
Interest
.
-
4.
To
find
what
Annuity
any
Sum
of
Money
due
at
the
end
of
any
Term
of
years
to
come
,
will
purchase
at
any
Rate
per
Cent.
-
5.
To
find
the
present
worth
of
any
Annuity
to
continue
any
number
of
years
,
allowing
Rebate
at
any
Rate
per
Cent.
-
6.
To
find
what
Annuity
any
Sum
of
Money
will
purchase
for
any
number
of
years
,
and
at
any
Rate
of
Interest
proposed
.
II.
When
any
Question
in
Compound
Interest
is
propounded
,
find
out
the
Interest
of
1
l.
and
let
1
l.
with
its
interest
be
the
Rate
of
Interest
implyed
in
the
Question
,
as
if
any
Question
were
proposed
at
8
per
Cent.
the
Int.
of
1
l.
for
a
year
is
.08
,
and
the
rate
of
Interest
is
1.08
,
if
at
6
per
Cent.
the
Rate
is
1.06
,
&c.
of
which
find
out
the
Logarithme
.
III.
When
in
any
Annuity
or
Debt
the
payments
be
half
Yearly
,
Quarterly
,
or
Monethly
,
&c.
you
are
to
divide
the
Logarithme
of
the
said
Rate
by
2
,
4
,
or
12
,
&c.
so
shall
that
Quotient
be
the
Logarithme
of
the
Rate
,
as
suppose
any
Question
were
propounded
at
8
per
Cent.
the
Rate
of
Interest
here
implyed
is
1.08
,
for
Which
said
Rate
is
for
yearly
payments
,
the
Logarithme
whereof
is
0.0374204
,
but
if
the
payments
are
to
be
half
yearly
,
then
if
you
divide
the
said
Logarithme
by
2
,
it
will
give
you
0.0187102
for
the
Logarithme
of
the
Rate
,
and
if
the
payments
be
Quarterly
,
divide
the
said
Logarithme
of
1.08
by
4
,
and
it
will
give
you
0.0093551
for
the
Log.
of
the
Rate
,
and
if
the
payments
be
monethly
,
then
if
you
divide
the
said
Log.
of
1.08
by
12
,
it
will
give
you
0.0031183
for
the
Log.
of
the
Rate
,
&c.
and
this
is
generally
the
first
thing
to
be
observed
in
every
Question
,
as
you
will
find
by
the
following
Examples
.
CASE
I.
To
find
the
Increase
or
Amount
of
any
Sum
of
Money
put
out
at
Compound
Interest
for
any
Term
of
years
,
and
at
any
Rate
of
Interest
propounded
.
Quest.
1.
If
50
l.
−
16
s.
be
put
out
at
8
per
Cent.
Compound
Interest
for
7
years
,
I
demand
how
much
will
then
be
due
to
the
Creditor
?
Facit
87
l.
−
01
s.
−
02
¾
d.
IV.
Multiply
the
Logarithme
of
the
Rate
by
the
number
of
years
,
and
the
product
will
give
you
the
Logarithme
of
the
Amount
of
1
l.
for
the
proposed
Time
,
to
which
if
you
add
the
Log.
of
the
Sum
propounded
,
the
sum
will
be
the
log
.
of
the
Answer
.
The
operation
by
the
Logarithmes
.
Quest.
2.
What
is
the
amount
of
76
l.
−
04
s.
for
3
4
years
at
8
per
Cent
?
Facit
97
l.
−
17
s.
−
01
d.
The
operation
by
the
Logarithmes
.
Quest.
3.
If
50
l.
be
put
forth
at
Interest
for
20
years
at
6
per
Cent.
I
demand
how
much
it
will
be
increased
to
at
the
end
of
the
said
time
.
Facit
168
l.
−
01
s.
−
10
d.
The
rate
of
Interest
here
proposed
is
6¼
=
6.25
per
Cent.
therefore
to
find
out
the
Rate
of
1
l.
for
a
year
,
say
by
the
Rule
of
proportion
.
So
that
the
Rate
of
Interest
implyed
in
the
Question
fit
for
Calculation
by
the
Log.
is
1.0625
according
to
the
second
Rule
of
this
Chapter
,
behold
The
Operation
.
Which
is
168
l.
−
01
s.
−
10
d.
very
near
,
and
so
much
will
50
l.
be
increased
to
in
20
years
at
6
l.
−
5
s.
per
Cent.
CASE
2.
To
find
the
present
worth
of
any
sum
of
money
due
at
the
end
of
any
number
of
years
to
come
,
Rebate
being
allowed
at
any
Rate
of
Compound
Interest
.
V.
When
it
is
required
to
find
the
present
worth
of
any
sum
of
Money
,
first
find
the
amount
of
1
l.
for
the
proposed
time
,
and
at
the
Rate
of
Interest
propounded
,
then
find
the
Logarithme
of
the
sum
proposed
to
be
Rebated
,
and
from
it
subtract
the
Logarithme
of
the
amount
of
1
l.
(
found
as
before
)
and
the
remainder
will
be
the
Logarithme
of
the
present
worth
of
the
sum
proposed
.
As
in
the
following
Example
.
Quest.
4.
What
is
30
l.
that
is
due
7
years
hence
worth
to
be
paid
presently
,
allowing
Rebate
at
8
per
Cent
?
Facit
17
l.
−
10
s.
−
01
½
d.
as
you
may
perceive
by
The
Operation
by
the
Logarithmes
.
which
is
17
l.
−
10
s.
−
01½
d.
and
so
much
is
the
present
worth
of
30
l.
due
7
years
hence
.
Quest.
5.
What
is
the
present
worth
of
120
l.
due
2
years
hence
,
allowing
Rebate
at
6
per
Cent
?
Facit
106
l.
−
15
s.
−
11
d.
The
Operation
by
the
Logarithmes
.
which
is
106
l.
−
15
s.
−
11
d.
and
so
much
is
the
present
worth
of
120
l.
due
2
years
hence
.
CASE
3.
To
find
the
Increase
,
or
Amount
of
an
Annuity
,
being
forborne
any
number
of
years
,
at
any
Rate
of
Compound
Interest
.
VI.
For
Resolving
Questions
concerning
the
forbearance
of
Annuities
,
you
are
(
by
the
fourth
Rule
)
first
to
find
out
the
Amount
of
1
l.
for
the
Time
,
and
at
the
Rate
of
Interest
propounded
.
Secondly
,
Find
out
the
Logarithme
of
the
said
amount
made
less
by
1
,
and
also
the
log
.
of
the
Rate
made
less
by
1
,
and
subtract
the
latter
from
the
former
,
so
shall
the
remainder
be
the
log
.
of
the
Amount
of
1
l.
Annuity
for
the
term
of
years
propounded
,
to
which
if
you
add
the
Logarithme
of
the
proposed
Annuity
,
the
sum
will
be
the
Logarithme
of
the
Amount
,
or
Increase
of
the
said
Annuity
.
As
in
the
following
Example
.
Quest.
6.
What
will
be
the
Amount
,
or
Increase
of
48
l.
−
16
s.
per
an
.
for
7
years
,
Compound
Interest
being
Computed
at
8
per
Cent.
Facit
435
l.
−
08
s.
−
05
d.
fere
.
See
The
Operation
by
the
Logarithmes
.
which
is
435
l.
−
8
s.
−
5½
d.
very
near
,
and
so
much
will
be
the
Increase
of
an
Annuity
of
48
l.
16
s.
in
7
years
at
8
per
Cent.
Compound
Interest
.
Quest.
7.
There
is
an
Annuity
of
50
l.
forborne
to
the
end
of
10
years
,
I
demand
how
much
is
then
due
,
Compound
Interest
being
computed
at
6
〈◊〉
per
Cent
?
Facit
666
l.
−
16
s.
as
you
will
find
by
The
Operation
by
the
Logarithmes
.
which
is
666
l.
−
16
s.
and
so
much
will
be
due
at
the
end
of
the
said
time
.
CASE
4.
To
find
what
Annuity
any
sum
due
at
any
time
to
come
will
purchase
to
continue
for
any
Time
,
and
at
any
Rate
of
Interest
proposed
.
VII
.
The
Operation
in
this
Case
is
the
same
in
every
respect
with
that
in
the
former
Case
,
only
whereas
in
the
last
case
you
subtracted
the
log
.
of
the
rate
less
1
from
the
log
.
of
the
increase
of
1
l.
less
1
,
so
in
this
you
must
subtract
the
log
.
of
the
increase
of
1
l.
less
1
,
from
the
log
.
of
the
rate
less
1
,
as
in
the
following
Example
.
Quest.
8.
There
is
705
l.
due
at
the
end
of
7
years
to
come
,
I
demand
what
Annuity
to
continue
7
years
,
the
same
will
purchase
,
Compound
Interest
being
allowed
at
8
per
Cent
?
Facit
79.015
l.
=
79
l.
−
●…
00
s.
as
you
may
find
if
you
observe
The
Operation
by
the
Logarithmes
.
Questions
of
this
Nature
may
be
solved
at
two
Operations
by
the
second
and
sixth
Cases
;
First
,
by
the
Rule
in
the
second
Case
find
the
present
worth
of
the
sum
propounded
,
then
by
the
sixth
,
find
what
Annuity
such
a
sum
will
purchase
.
CASE
5.
To
find
the
present
worth
of
an
Annuity
to
continue
any
Term
of
years
,
howsoever
payable
,
viz.
either
yearly
,
half
,
yearly
or
Quarterly
,
Rebate
being
allowed
at
any
rate
per
Cent.
VIII
.
Find
out
the
Logarithme
of
the
Rate
,
and
multiply
it
by
the
number
of
Years
or
Quarters
,
according
as
the
Annuity
is
payable
,
and
that
will
produce
the
Logarithme
of
the
increase
of
1
l.
for
the
proposed
time
,
to
which
add
the
Log.
of
the
Rate
made
less
by
1
,
and
subtract
that
sum
from
the
Log.
of
the
increase
of
1
l.
made
less
by
1
,
so
shall
the
remainder
be
the
Log
of
the
presentworth
of
1
l
annuity
for
the
time
proposed
to
which
add
the
logarithme
of
the
proposed
Annuity
,
and
the
sum
will
be
the
Logarithme
of
the
present
worth
of
the
given
Annuity
.
As
in
Example
.
Quest.
9.
What
is
the
present
worth
of
an
Annuity
of
30
l.
payable
by
yearly
payments
,
and
to
continue
30
years
,
allowing
Rebate
after
the
Rate
of
8
per
Cent.
per
Annum
?
Facit
337
l.
−
●…
14
s.
−
●…
09
½
d.
as
appears
in
The
Operation
by
the
Logarithmes
.
CASE
6.
To
find
out
what
Annuity
to
continue
any
term
of
years
any
given
sum
of
Money
will
purchase
at
any
Rate
of
Compound
Interest
.
IX
.
When
you
would
know
what
Annuity
any
given
sum
will
purchase
,
first
(
as
in
the
foregoing
Rules
)
find
out
the
Logarithme
of
the
Rate
,
which
multiply
by
the
proposed
time
,
so
will
that
product
be
the
Logarithme
of
the
increase
of
1
l.
to
which
add
the
Log.
of
the
rate
made
less
by
1
,
and
from
that
sum
subtract
the
Log.
of
the
said
increase
of
1
l.
made
less
by
1
,
so
will
the
remainder
be
the
Log.
of
what
1
l.
will
purchase
for
the
proposed
Time
,
to
which
if
you
add
the
log
.
of
the
given
purchase
money
,
the
sum
will
be
the
Log.
of
the
Annuity
that
the
given
sum
will
purchase
.
As
in
Example
.
Quest.
10.
What
Annuity
to
continue
7
years
,
and
payable
by
Quarterly
payments
will
246
l.
purchase
.
Allowing
Rebate
at
8
per
Cent
?
Facit
12.297
l.
The
Operation
by
the
Logarithmes
.
More
variety
of
Questions
might
be
stated
,
but
these
to
the
Ingenious
are
sufficient
.
FINIS
.
Cockers
ALGEBRAICAL
ARITHMETICK
,
CONTAINING
The
Doctrine
of
Composing
,
and
Resolving
an
EQUATION
,
With
all
other
Rules
requisite
for
the
understanding
of
that
Mysterious
Art
,
according
to
the
Method
used
by
Mr.
IOHN
KERSEY
,
in
his
incomparable
Treatise
of
ALGEBRA
.
Composed
by
EDWARD
COCKER
,
late
Practitioner
in
the
Arts
of
Writing
,
Arithmetick
,
and
Engraving
.
Perused
,
Corrected
,
and
Published
By
IOHN
HAWKINS
,
Writing-Master
at
St.
George's
Church
in
Southwark
.
Plato
foribus
Academiae
inscribi
jussit
;
Nemo
Arithmetices
Ignarus
hic
Ingrediatur
.
LONDON
,
Printed
in
the
Year
1684.
ALGEBRAICAL
DEFINITIONS
CHAP.
I.
Concerning
the
construction
of
Cossick
Powers
,
and
the
way
of
expressing
them
by
Letters
,
together
with
the
signification
of
all
such
Characters
or
Marks
as
are
used
in
the
ensuing
Treatise
.
1.
THE
Analytical
Art
generally
called
Algebra
,
is
that
by
which
,
when
a
Problem
,
or
hard
Question
is
propounded
,
we
assume
the
Quantity
or
Number
sought
,
as
if
it
were
really
known
;
and
,
with
this
assumed
Quantity
,
and
the
Quantity
or
Quantities
given
,
we
proceed
by
undeniable
Consequences
,
until
the
Quantity
first
assumed
is
found
to
be
equal
to
some
quantity
or
quantities
really
known
,
and
is
therefore
it self
also
known
.
II.
Algebra
is
either
Numeral
,
or
Literal
.
III.
Numeral
Algebra
is
so
called
,
because
all
the
given
Quantities
in
any
Question
are
expressed
by
Natural
Numbers
,
and
the
number
or
quantity
sought
is
solely
represented
by
some
Letter
or
Character
taken
at
the
pleasure
of
the
Artist
.
IV.
Literal
Algebra
is
so
called
,
because
when
a
question
is
resolved
after
this
method
,
the
known
or
given
quantities
as
well
as
the
unknown
,
are
all
expressed
by
Letters
of
the
Alphabet
,
or
some
other
Convenient
Marks
or
Characters
,
and
this
is
also
called
Specious
Algebra
;
and
when
a
question
is
resolved
after
this
manner
,
at
the
end
of
the
operation
,
there
is
discovered
a
Canon
,
directing
how
the
question
proposed
;
or
any
other
of
the
like
nature
may
be
solved
,
and
therefore
is
Literal
Algebra
,
accounted
more
excellent
than
Numeral
Algebra
,
for
that
produceth
not
a
Canon
without
extraordinary
difficulty
,
because
the
numbers
first
given
are
by
Arithmetical
operations
so
interwoven
and
confounded
,
that
it
may
seem
a
task
too
tedious
for
the
most
ingenious
Artist
to
trace
out
their
footsteps
.
V.
The
Doctrine
of
Algebra
consists
in
the
knowledge
of
certain
quantities
called
Cossick
Powers
,
which
we
shall
immediately
explain
.
VI.
In
a
series
,
or
rank
of
Geometrical
proportionals
continued
,
proceeding
from
Unity
,
or
one
,
whether
they
be
ascending
,
or
descending
,
all
the
Numbers
or
Terms
except
the
first
(
which
is
supposed
to
be
unity
)
are
called
Cossick
Numbers
,
or
Powers
,
as
for
Example
,
in
this
Rank
of
continual
proportionals
,
viz.
1
,
2
,
4
,
8
,
16
,
32
,
64
,
128
,
256
,
&c.
the
second
Term
(2)
is
called
the
Root
or
first
Power
,
the
third
term
(4)
is
called
the
Square
or
second
power
,
the
fourth
term
(8)
is
called
the
Cube
,
or
third
Power
;
the
fifth
(
16
)
is
called
the
Biquadrate
,
or
fourth
Power
;
32
is
the
fifth
Power
,
64
is
the
sixth
Power
,
128
is
the
seventh
Power
,
&c.
In
like
manner
if
you
take
a
Rank
of
Geometrical
proportionals
continued
,
and
descending
from
unity
,
viz.
1
,
½
,
¼
,
1/8
,
1/16
,
1/32
,
&c.
or
1
,
1/3
,
1/9
,
1/27
,
1/81
,
&c.
or
1
,
¼
,
1/16
,
1/64
,
&c.
The
second
term
is
called
the
Root
or
first
Power
,
the
third
term
is
called
the
second
power
,
the
fourth
term
is
called
the
third
Power
,
&c.
VII
.
Whence
it
is
evident
that
the
Square
or
Second
Power
is
generated
by
the
Multiplication
of
the
Root
or
first
Power
into
it self
,
and
the
Cube
or
third
power
is
generated
by
multiplying
the
second
Power
by
the
Root
,
or
by
multiplying
the
Root
3
times
into
it self
,
and
the
Biquadrate
or
fourth
power
is
produced
by
multiplying
of
the
third
power
by
the
Root
,
or
by
multiplying
the
Root
4
times
into
it self
,
and
the
fifth
Power
is
produced
by
multiplying
the
fourth
Power
by
the
Root
,
&c.
As
for
Example
.
If
you
take
2
for
a
Root
,
and
multiply
it
by
its
self
,
it
produceth
4
for
the
Square
or
second
power
of
the
Root
2
:
Again
,
multiply
4
by
the
Root
2
,
and
it
produceth
8
for
the
third
Power
,
or
Cube
of
the
Root
2
:
Again
,
if
you
multiply
the
Cube
8
by
the
Root
2
,
it
produceth
the
fourth
Power
,
or
Biquadrate
of
the
Root
2
,
&c.
In
like
manner
if
3
were
proposed
for
a
Root
,
it
being
multiplyed
by
it self
,
produceth
9
for
the
Square
,
or
Second
Power
of
3
,
and
9
being
multiplyed
by
the
Root
3
,
produceth
27
for
the
Cube
,
or
third
Power
of
the
Root
3
,
&c.
And
also
if
½
be
proposed
for
a
Root
,
and
it
be
multiplyed
by
it self
,
it
produceth
¼
for
the
Square
or
second
power
of
(
the
Root
)
¼
,
and
¼
(
the
Square
)
being
multiplyed
by
(
the
Root
)
½
,
it
produceth
1/8
for
the
Cube
,
or
third
Power
of
(
the
Root
)
½
&c.
Whence
it
is
evident
that
the
4
,
6
or
7
powers
of
any
Roots
may
be
found
out
,
without
any
respect
at
all
had
to
the
intermediate
Powers
between
the
Root
and
the
power
required
;
as
suppose
there
were
given
the
Root
3
,
and
it
were
required
to
find
the
fifth
power
of
it
.
I
take
3
,
and
set
it
down
5
times
in
order
thus
,
3
,
3
,
3
,
3
,
3
,
and
multiply
them
all
into
each
other
,
according
to
the
rule
of
continual
multiplication
,
and
the
last
product
(
which
is
243
)
is
the
fifth
power
of
the
Root
3
,
which
was
required
.
Again
,
let
it
be
required
to
find
the
fourth
power
of
5
,
I
take
5
,
and
set
it
down
4
times
thus
,
5
,
5
,
5
,
5
,
then
do
I
multiply
them
continually
,
and
find
the
last
product
to
be
625
,
which
is
the
fourth
power
(
of
the
given
Root
)
as
was
required
.
The
like
may
be
observed
in
the
finding
of
any
other
power
of
any
other
given
Root
.
VIII
.
If
there
be
a
series
of
Geometrical
proportionals
continued
,
and
against
each
power
there
be
placed
numbers
orderly
representing
the
number
or
degree
of
distance
of
each
power
from
the
Root
,
such
numbers
are
called
the
Indices
or
exponents
of
the
powers
,
because
they
shew
how
often
the
Root
is
involved
into
it self
for
the
production
of
such
a
power
,
as
in
the
Rank
,
or
Scale
of
Algebraical
powers
placed
in
the
margent
,
proceeding
from
the
root
2
,
to
the
tenth
power
thereof
,
which
is
1024
,
under
which
is
written
the
word
powers
,
and
then
against
each
particular
power
,
on
the
left
hand
thereof
,
is
expressed
Index
,
or
Exponent
of
that
Power
,
shewing
how
often
the
Root
is
involved
or
multiplyed
into
it self
to
produce
that
Power
:
As
for
Example
,
against
the
number
64
,
is
placed
the
number
6
,
which
sheweth
that
64
is
the
sixth
power
of
its
Root
,
or
that
its
Root
is
multiplyed
6
times
into
it self
to
produce
the
number
64.
The
like
is
to
be
understood
of
any
other
.
1
|
2
|
2
|
4
|
3
|
8
|
4
|
16
|
5
|
32
|
6
|
64
|
7
|
128
|
8
|
256
|
9
|
512
|
10
|
1024
|
Indices
.
|
Powers
.
|
Likewise
if
any
two
or
more
Indices
,
or
Exponents
be
added
together
,
their
sum
will
be
an
exponent
shewing
what
power
will
be
produced
by
the
multiplication
of
those
Powers
belonging
to
those
Exponents
or
Indices
which
you
add
together
;
As
in
the
foregoing
Table
let
it
be
required
to
find
out
what
power
of
the
Root
2
will
be
produced
by
multiplying
128
(
its
seventh
power
)
by
8
(
its
third
power
,
)
in
order
to
which
I
take
3
and
7
,
the
respective
Indices
of
the
given
powers
,
and
add
them
together
,
and
their
sum
is
10
,
which
sheweth
that
the
third
power
,
and
the
seventh
power
of
any
Number
,
or
Root
,
being
multiplyed
together
,
will
produce
the
tenth
power
of
that
Root
;
so
in
our
example
128
being
multiplyed
by
8
,
produceth
1024
,
which
is
the
tenth
power
of
the
Root
2.
In
like
manner
,
the
Indices
3
and
5
being
added
together
,
make
8
for
a
new
Exponent
,
which
sheweth
that
32
and
8
(
the
powers
belonging
to
those
Exponents
)
being
multiplyed
together
,
will
produce
the
eighth
power
,
viz.
256
,
as
appears
by
the
said
Table
,
the
like
of
any
other
.
So
that
you
see
that
the
addition
of
Indices
answers
to
the
multiplication
of
their
Correspondent
powers
.
And
in
like
manner
will
the
subtraction
of
Indices
,
or
Exponents
,
answer
to
the
Division
of
their
correspondent
powers
,
observing
always
to
make
the
power
corresponding
the
subtrahend
(
or
Index
to
be
subtracted
)
to
be
the
Divisor
.
IX
.
When
a
Question
is
propounded
,
and
its
solution
is
to
be
searched
out
by
the
Algebraick
Art
,
the
number
or
magnitude
sought
is
generally
called
a
Root
,
and
it
must
be
represented
or
signified
by
some
Character
or
Symbol
,
as
must
be
also
all
the
powers
proceeding
from
the
said
Root
according
to
the
tenure
of
the
Question
,
in
order
to
which
there
may
be
taken
some
letter
of
the
Alphabet
at
the
pleasure
and
discretion
of
the
Artist
,
as
a
,
b
,
c
,
or
d
,
&c.
to
express
the
said
Root
,
but
to
avoid
confusion
in
the
operation
,
by
the
commixture
of
known
with
unknown
Quantities
,
our
Modern
Analysts
have
been
accustomed
to
assume
vowels
to
represent
unknown
Quantities
,
and
to
put
Consonants
to
signifie
known
or
given
quantities
.
X.
If
for
the
number
or
quantity
sought
there
be
put
or
assumed
the
Vowel
a
,
then
its
Square
will
be
aa
,
that
is
,
a
being
multiplyed
by
it self
,
produceth
aa
,
that
is
a
squared
,
or
the
square
of
a
,
for
a
time
a
is
aa
,
and
the
Cube
or
third
power
raised
from
the
Root
a
,
is
aaa
,
that
is
,
a
times
aa
,
is
aaa
,
and
the
fourth
power
accordingly
is
aaaa
,
and
after
the
same
manner
may
any
higher
power
of
a
be
signified
:
In
like
manner
if
for
the
quantity
or
number
sought
there
be
assumed
,
the
letter
e
,
then
shall
the
Square
raised
therefrom
be
ee
,
and
the
third
power
eee
,
and
the
fourth
power
eeee
,
and
the
fifth
power
eeeee
,
&c.
Also
if
b
,
or
any
other
Consonant
,
be
put
for
a
given
or
known
Quantity
,
then
its
Square
will
be
bb
,
its
Cube
bbb
,
and
its
biquadrate
bbbb
,
&c.
But
by
some
Analysts
the
powers
of
a
,
or
any
other
letter
,
Vowel
,
or
Consonant
,
are
expressed
by
placing
the
Index
or
Exponent
of
the
power
in
a
small
Character
,
just
after
the
Symbol
,
even
with
the
head
thereof
,
viz.
a
,
a
²
,
a
³
,
a
,
&c.
signifie
the
Root
a
,
its
Square
,
its
Cube
,
and
its
Biquadrate
,
&c.
which
may
be
further
exemplified
by
the
following
Table
.
A
Table
shewing
the
Powers
of
Numbers
,
and
how
to
express
the
Simple
Powers
by
Alphabetical
Characters
.
The
Root
,
or
first
Power
.
|
1
|
2
|
3
|
4
|
a
a
|
The
Square
,
or
second
Power
.
|
1
|
4
|
9
|
16
|
aa
a
²
|
The
Cube
,
or
third
Power
.
|
1
|
8
|
27
|
64
|
aaa
a
³
|
The
Biquadrat
or
fourth
Power
,
|
1
|
16
|
81
|
256
|
aaaa
a
⁴
|
The
fifth
Power
|
1
|
32
|
243
|
1024
|
aaaaa
a
⁵
|
The
sixth
Power
|
1
|
64
|
729
|
4096
|
aaaaaa
a
⁶
|
The
seventh
Power
.
|
1
|
128
|
2187
|
16384
|
aaaaaaa
a
⁷
|
The
eighth
Power
.
|
1
|
256
|
6561
|
65536
|
aaaaaaaa
a
●
|
The
ninth
Power
|
1
|
512
|
19683
|
262144
|
aaaaaaaaa
a
⁹
|
The
tenth
Power
.
|
1
|
1024
|
59049
|
1048576
|
aaaaaaaaaa
a
¹⁰
|
XI
.
The
numbers
made
use
of
in
solving
of
Algebraical
Questions
,
are
either
absolute
Numbers
,
or
Numbers
prefixt
.
Absolute
numbers
are
those
which
are
disjunct
from
any
kind
of
Magnitude
or
Quantity
,
either
known
or
(
unknown
)
required
,
but
stand
simply
of
themselves
,
without
having
Relation
to
any
thing
else
,
as
5
,
10
,
20
,
100
,
¾
,
and
½
are
called
absolute
Numbers
.
Numbers
prefixt
are
such
as
are
immediately
prefixed
to
some
letter
or
letters
,
signifying
an
Algebraical
quantity
,
either
known
,
or
required
,
such
as
are
2a
,
4a
,
10a
,
100a
,
½
a
,
¾
a
,
3aa
,
5bbb
,
3a
●
,
5b
³
;
which
numbers
so
prefixed
,
shew
how
often
the
quantity
to
which
they
are
prefixed
,
is
to
be
taken
,
as
4a
signifieth
that
a
is
to
be
taken
4
times
,
and
5bbb
,
or
5b
³
,
signifieth
that
the
Cube
or
third
power
of
b
is
to
be
taken
5
Times
,
½
a
is
half
of
a
,
and
2/3b
is
two
thirds
of
b
;
The
like
is
to
be
understood
of
any
other
.
And
,
Note
,
that
when
you
have
any
Algebraical
Quantity
,
or
Letter
,
or
Character
,
not
having
any
number
prefixed
to
it
,
then
1
,
or
unity
must
be
imagined
to
be
prefixed
,
as
a
,
or
1a
,
b
,
or
1b
,
&c.
XII
.
As
in
Vulgar
and
Decimal
Arithmetick
,
so
in
Algebraical
Arithmetick
,
the
operations
are
performed
either
by
Absolute
Numbers
,
or
by
Alphabetical
characters
,
in
all
the
fundamental
rules
,
viz.
Addition
,
Subtraction
,
Multiplication
,
Division
,
and
the
Extraction
of
Roots
:
And
note
that
Where
it
is
required
to
performe
the
work
by
absolute
numbers
,
that
the
operation
is
in
every
respect
the
same
as
in
Common
Arithmetick
.
But
where
it
is
performed
by
Alphabetical
Letters
,
there
is
an
absolute
necessity
of
using
some
Characters
,
to
signifie
the
Operation
,
an
explanation
of
which
Characters
take
as
followeth
.
XIII
.
This
Character
(
+
)
is
a
sign
of
affirmation
or
Addition
,
when
it
is
placed
between
two
quantities
,
signifying
that
the
2
numbers
or
quantities
between
which
it
is
placed
,
are
to
be
added
together
,
and
is
as
much
as
to
say
plus
,
as
3+6
signifieth
the
sum
of
3
and
6
,
and
is
as
much
as
to
say
3
plus
6
,
or
3
more
6
,
which
is
9
,
and
4+7+9
signifieth
the
sum
of
4
,
7
,
and
9
,
which
is
20
;
so
a+b+c
signifieth
the
sum
of
a
,
b
,
and
c.
And
here
note
,
that
when
there
is
no
Mark
,
or
Character
before
any
Letter
or
quantity
,
then
is
it
Affirmative
,
and
the
Mark
(
+
)
is
supposed
to
stand
before
it
;
as
a
,
is
+a
,
or
+1a
,
and
b
is
+b
,
or
+1b
,
and
bcd
is
+bcd
the
like
of
others
.
XIV
.
This
Character
(
−
)
is
a
negative
sign
,
and
always
belongeth
to
the
Quantity
or
Number
which
followeth
it
,
denying
it
to
be
,
and
signifieth
a
fictitious
Number
,
or
Quantity
less
than
nothing
.
So
−
7
is
a
feigned
number
less
than
nothing
by
7
,
viz.
as
the
height
of
the
Sun
above
the
Horizon
may
be
affirmed
to
be
7
deg.
or
+7
deg.
so
when
it
is
depressed
7
degrees
below
the
Horison
,
its
height
may
be
said
to
be
−
7
deg.
that
is
,
7
deg.
less
than
nothing
.
But
when
the
said
sign
or
Character
is
placed
between
two
Numbers
or
Quantities
,
it
signifies
that
the
number
or
quantity
which
followeth
it
,
is
to
be
subtracted
out
of
some
Number
or
Quantity
going
before
it
,
as
12
−
8
signifieth
that
8
is
to
be
subtracted
out
of
12
,
or
it
signifieth
the
excess
of
12
above
8
,
or
the
Difference
between
12
and
8
,
which
is
4
,
so
a
−
b
signifieth
the
excess
of
a
above
b
,
and
it
is
as
much
as
to
say
(
a
less
b
,
)
so
a+b
−
c
signifieth
that
c
is
to
be
subtracted
from
the
sum
of
a
and
b.
XV.
This
Character
(
×
)
is
the
sign
of
Multiplication
,
and
signifieth
that
the
Numbers
or
Quantities
between
which
it
is
placed
,
are
to
be
multiplyed
together
,
as
4×5
signifieth
the
product
of
4
and
5
,
which
is
20
;
so
3×5×8
signifieth
the
product
of
the
continual
multiplication
of
3
,
5
,
and
8
,
viz.
120.
Likewise
b×c
signifieth
the
product
of
the
multiplication
of
b
by
c
,
and
b×c×d
signifieth
the
product
made
by
the
continual
multiplication
of
b
,
c
,
and
d
,
into
each
other
.
But
for
the
most
part
Analysts
signifie
the
multiplication
of
literal
Quantities
by
setting
the
letters
together
like
letters
in
a
word
,
as
ab
is
the
same
with
a×b
,
and
abc
is
the
same
with
a×b×c
and
this
indeed
is
to
be
preferred
before
the
other
as
most
convenient
and
fittest
for
operation
.
XVI
.
This
Character
(
)
signifieth
the
Difference
btween
the
two
quantities
between
which
it
is
placed
,
when
it
is
not
known
in
which
of
them
the
excess
lyeth
.
So
b
c
signifieth
the
Difference
between
b
and
c
,
when
it
is
not
known
whether
b
be
greater
or
lesser
than
c.
XVII
.
The
said
4
Characters
defined
in
the
13
,
14
,
15
,
and
16
Sections
foregoing
,
viz.
+
,
−
,
×
.
and
,
may
oftentimes
have
Relation
to
such
a
Compound
Quantity
following
the
Character
,
as
hath
a
line
drawn
over
each
part
of
it
,
as
for
example
,
,
by
which
you
are
to
understand
that
the
Quantity
(
c
,
)
is
to
be
added
to
the
difference
between
the
Quantities
(
b
and
d
)
in
which
of
them
soever
the
excess
lyeth
.
Likewise
which
signifieth
that
the
difference
between
b
and
c
is
to
be
subtracted
from
the
Quantity
expressed
by
a.
Also
signifieth
that
the
sum
of
b
and
c
is
to
be
multiplyed
by
the
quantity
a
,
where
take
notice
that
in
regard
there
is
a
line
drawn
over
the
two
quantities
b
and
c
,
the
sign
×
hath
reference
to
the
multiplication
of
a
into
the
quantity
c
as
well
as
the
quantity
b
,
which
immediately
followeth
it
,
but
if
the
said
line
were
omitted
,
and
the
quantities
were
thus
expressed
,
a×b+c
,
it
would
signifie
the
quantity
e
to
be
added
to
the
product
of
the
multiplication
of
a
and
b.
Furthermore
signifieth
that
the
quantities
c
and
d
are
or
must
be
subtracted
from
the
Quantity
b
,
whereas
if
there
were
not
a
line
over
the
quantities
c
and
d
,
it
would
signifie
that
the
quantity
d
is
to
be
added
to
b
−
c.
And
(
)
signifieth
the
difference
between
the
quantity
c
,
and
the
sum
of
d
and
e
,
whereas
if
the
line
were
not
over
d
and
e
,
it
would
signifie
the
quantity
e
to
be
added
to
the
difference
between
c
and
d.
XVIII
.
This
Character
(
√
)
is
a
radical
sign
,
and
signifieth
that
the
Square
Root
of
the
quantity
or
quantities
following
it
,
is
to
be
extracted
as
,
signifieth
the
Square
Root
of
36
,
viz.
6.
So
signifieth
the
Square
Root
of
the
product
of
the
quantities
a
and
b
,
and
is
the
Square
Root
of
the
product
of
the
continual
multiplication
of
the
Quantities
,
a
,
b
,
and
c.
But
when
you
would
represent
the
Root
of
a
Power
that
is
higher
than
a
Square
;
then
immediately
after
the
said
Radical
sign
,
express
the
index
,
or
exponent
of
its
power
in
a
parenthesis
,
as
followeth
,
viz.
,
signifieth
the
Cube
Root
of
64
,
which
is
4
;
and
signifieth
the
biquadrate
Root
of
81
,
viz.
3.
Also
,
signifieth
the
Cube
Root
of
the
product
of
the
multiplication
of
the
quantities
,
a
,
and
b
,
and
signifieth
the
Biquadrate
Root
of
the
Product
of
the
multiplication
of
the
quantities
,
c
and
d.
And
the
said
Radical
sign
doth
oftentimes
belong
to
such
a
Compound
Quantity
following
it
,
as
hath
a
line
over
every
part
of
it
.
As
for
Example
,
signifieth
the
Square
Root
of
the
sum
of
the
Quantities
b
and
c.
So
signifieth
the
CubeRoot
of
theremainder
,
when
the
quantity
c
is
subtracted
from
the
sum
of
the
quantities
,
a
and
b
,
and
(
)
signifieth
the
Biquadrate
Root
of
the
remainder
,
when
the
quantity
c
is
subtracted
from
the
Sum
of
the
Square
of
a
added
to
b.
Likewise
signifieth
that
to
the
quantity
a
is
to
be
added
the
Square
Root
of
the
remainder
,
when
the
quantity
d
is
subtracted
from
the
sum
of
the
Square
of
the
quantity
b
,
and
the
quantity
c
:
And
these
and
such
like
are
by
Analysts
generally
called
universal
Roots
.
After
the
same
manner
may
be
expressed
the
universal
Square
Root
of
thus
,
viz.
which
signifieth
the
Square
Root
of
the
sum
when
b
is
added
to
the
Square
Root
of
.
XIX
.
This
Character
(
=
)
signifieth
an
Equation
,
or
equality
of
the
magnitudes
or
quantities
between
which
it
is
placed
,
and
imports
as
much
as
these
words
,
viz.
(
is
equal
to
)
as
in
the
following
Example
,
viz.
3+4
=
7
,
which
is
as
much
as
to
say
,
the
sum
of
3
and
4
,
or
3
plus
4
is
equal
to
7
;
so
7+9
=
12+4
=
16
imports
that
the
sum
of
7
and
9
is
equal
to
the
sum
of
12
,
and
4
which
is
equal
to
16
;
and
9
=
12
−
3
,
signifieth
that
9
is
equal
to
the
excess
of
12
above
3.
Also
4×5
=
2×10
=
16+4
=
20
signifieth
that
the
Rectangle
or
Product
of
4
by
5
is
equal
to
the
Rectangle
or
Product
of
10
by
2
,
which
is
equal
to
the
sum
of
16
and
4
,
equal
to
20.
Likewise
signifieth
that
the
Quotient
of
24
divided
by
6
,
is
equal
to
the
Quotient
of
8
divided
by
2.
Again
a+b
=
c
−
d
signifieth
that
the
sum
of
a
and
b
is
equal
to
the
excess
of
c
above
d
,
and
signifieth
that
the
sum
c
and
d
is
equal
to
the
Quotient
of
f
divided
by
g
;
and
b×c
=
r
−
s
signifieth
that
the
Rectangle
of
b
and
c
is
equal
to
the
excess
of
r
above
s
,
and
signifieth
that
a
is
equal
to
the
remainder
,
when
½
c
or
is
subtracted
from
the
universal
square
Root
of
this
will
be
made
plain
and
easie
to
the
ingenious
practitioner
by
the
ensuing
Examples
of
this
Treatise
.
XXI
.
This
Character
(
)
stands
for
the
word
(
greater
)
signifying
the
number
,
or
quantity
standing
on
the
left
hand
of
the
said
Character
to
be
greater
than
that
on
the
right
hand
thereof
;
as
signifieth
that
8
is
greater
than
3
;
also
signifieth
that
the
sum
of
a
and
b
is
greater
than
c
,
&c.
XXII
.
This
Character
(
)
stands
for
the
word
(
less
)
and
it
signifieth
that
the
number
or
quantity
standing
on
the
left
hand
thereof
,
is
lesser
than
that
on
the
right
hand
.
As
signifieth
that
the
sum
of
4
and
3
is
less
than
the
excess
of
20
above
8.
Likewise
is
thus
read
,
viz.
the
remainder
of
d
being
subtracted
from
c
is
lesser
than
the
sum
of
b
and
e.
XXIII
.
This
Character
(
)
is
always
placed
in
the
middle
between
4
Geometrical
proportionals
,
as
in
the
following
Examples
,
viz.
is
thus
to
be
read
,
viz.
as
2
is
to
4
,
so
9
is
to
18
;
or
after
the
manner
of
the
Rule
of
3
,
if
2
require
4
,
9
will
require
18.
Also
is
thus
read
,
as
b
is
to
c
,
so
is
d
to
e.
And
is
as
much
as
to
say
,
as
the
Compound
Quantity
a+e
is
to
the
quantity
b
,
so
is
the
Compound
Quantity
c+b
to
the
Quotient
of
the
Compound
Quantity
bc+bb
being
divided
by
a+e
.
XXIV
.
This
Character
(
)
placed
after
any
number
of
quantities
exceeding
two
,
declareth
the
said
numbers
or
quantities
after
which
it
is
placed
to
be
continual
Geometrical
proportionals
,
so
2
,
4
,
8
,
16
,
32
,
64
signifieth
the
said
numbers
to
be
continual
proportionals
Geometrical
,
for
,
as
2
is
to
4
,
so
is
4
to
8
,
and
so
is
8
to
16
,
and
so
is
16
to
32
,
and
so
is
32
to
64
,
&c.
Also
these
quantities
,
viz.
a.
b.
c.
d.
e.
are
continual
proportionals
Geometrical
,
for
,
as
a
is
to
b
,
so
is
c
to
d.
and
so
is
d
to
e.
CHAP.
II.
Addition
of
Algebraical
Integers
.
I.
AS
in
Common
Arithmetick
,
so
in
Algebraical
,
Addition
finds
out
the
aggregate
,
or
sum
of
two
or
more
given
quantities
however
expressed
numerally
or
literally
.
II.
When
the
quantities
given
to
be
added
are
alike
,
and
have
like
signs
,
collect
the
numbers
prefixed
to
each
quantity
into
one
sum
,
and
thereto
annex
the
letter
,
or
letters
of
any
one
of
the
given
quantities
,
and
then
prefix
the
sign
of
Affirmation
or
Negation
,
viz.
+
or
−
so
shall
the
quantity
thus
found
be
the
sum
desired
.
And
here
note
that
every
quantity
which
hath
no
number
prefixed
to
it
,
is
supposed
to
have
the
number
1
prefixed
,
so
is
a
=
1a
,
and
b
=
1b
.
Example
.
What
is
the
sum
of
3b+b+2b
?
Facit
6b
,
for
the
sum
of
the
numbers
prefixed
to
each
quantity
,
viz.
3
,
1
,
and
2.
is
6
,
to
which
if
I
annex
the
Character
b
,
it
will
be
6b
,
which
must
have
the
sign
+
prefixed
to
it
,
or
else
it
must
be
imagined
so
to
be
,
then
will
+6b
be
the
sum
of
the
given
Quantities
.
So
is
5ab
the
sum
of
3ab+2ab
.
And
−
4cd
the
sum
of
−
3cd
More
Examples
of
this
Rule
.
III.
When
the
quantities
given
to
be
added
together
are
alike
,
but
have
unlike
signes
,
then
subtract
the
lesser
number
prefixed
from
the
greater
,
and
to
the
remainder
annex
the
letter
or
letters
by
which
any
one
of
the
given
quantities
is
expressed
,
and
thereto
prefix
the
sign
of
+
or
−
according
to
the
sign
of
that
prefixed
number
wherein
lay
the
excess
,
so
shall
this
new
quantity
be
the
sum
of
the
quantities
propounded
.
Example
.
Let
it
be
required
to
add
+5cd
to
−
2cd
,
the
sum
will
be
found
to
be
+3cd
;
for
,
first
,
I
subtract
−
2
from
+5
and
there
remaines
3
to
which
I
annex
cd
so
will
there
be
3cd
,
to
which
I
prefix
the
sign
+
because
it
belongs
to
the
number
5
,
wherein
lay
the
excess
,
so
have
I
+3cd
for
the
sum
required
.
See
the
work
.
Again
if
it
were
required
to
add
+4aa
to
−
7aa
the
sum
would
be
found
to
be
−
3aa
,
because
the
sign
−
belongs
to
the
number
7
wherein
lay
theexcess
,
see
the
work
in
the
margent
.
More
Examples
of
this
last
Rule
.
And
here
note
,
that
if
the
numbers
prefixed
to
the
given
quantity
beequal
,
and
they
have
different
signs
,
then
their
sum
will
be
0
,
so
i●…
it
were
required
to
add
+8bcd
to
−
8bcd
their
sum
will
be
0
,
the
negative
sign
destroying
the
affirmative
.
IV.
When
the
quantities
given
to
be
added
are
more
than
2
,
and
having
different
signs
,
then
according
to
the
second
Rule
of
this
Chapter
,
bring
the
quantities
having
like
signs
into
one
sum
,
that
is
the
affirmative
quantities
into
one
sum
,
and
the
negative
into
another
,
then
by
the
foregoing
third
Rule
add
those
two
quantities
together
,
so
shall
their
sum
be
the
number
sought
.
Example
Let
it
be
required
to
and
the
sum
of
3aa+7aa
−
2aa
−
5aa
.
First
,
by
the
said
second
Rule
I
find
the
sum
of
3aa+7aa
to
be
10aa
;
and
the
sum
of
−
2aa
−
5aa
to
be
−
7aa
,
then
by
the
said
third
Rule
I
find
the
sum
of
10aa
−
7aa
to
be
3aa
,
or
+3aa
so
that
I
conclude
the
sum
of
3aa+7aa
−
2aa
−
5aa
to
be
+3aa
.
More
Examples
of
this
Rule
follow
.
V.
When
the
Simple
quantities
given
to
be
added
together
,
be
unlike
,
then
(
how
many
soever
there
be
)
set
them
one
after
another
in
the
same
line
without
altering
their
signs
.
Example
.
What
is
the
sum
of
4b
added
to
3cd
?
Facit
4b+3cd
for
the
sum
.
More
Examples
of
this
Rule
follow
.
Algebraical
Addition
of
Compound
Integers
.
VI.
The
Addition
of
Compound
Algebraical
Integers
is
easily
performed
by
the
help
of
the
foregoing
Rules
of
this
Chapter
,
whether
the
Compound
quantities
to
be
added
are
alike
,
or
unlike
;
as
you
may
easily
perceive
by
the
work
of
the
following
Examples
.
Let
it
be
required
to
find
the
sum
of
3a+b
,
and
5a+3b
.
their
sum
will
be
8a+4b
for
3a+5a
=
8a
,
and
3b+b
=
4b
,
whose
sum
is
8a+4b
by
the
second
Rule
of
this
Chapter
.
Also
the
sum
of
6cd+3bb
and
2cd
−
5bb
will
be
found
to
be
8cd
−
2bb
for
(
by
the
second
Rule
of
this
Chapt.
(
6cd+2cd
=
8cd
,
and
by
the
third
Rule
the
sum
of
3bb
−
5bb
=
−
2bb
which
2
sums
added
together
by
the
fifth
Rule
of
this
Chap.
will
be
8cd
−
2bb
.
Moreover
if
it
were
required
to
find
the
sum
of
these
Compound
Quantities
,
viz.
15gg+8a
−
20
and
3gg
−
3a+12
it
will
be
18gg+5a
−
8
for
15gg+3gg
=
18gg
by
the
second
Rule
,
and
the
sum
of
8a
−
3a
=
5a
by
the
third
Rule
,
and
by
the
same
12
−
20
=
−
8
,
the
sum
of
which
3
sums
is
18gg+5a
−
8
by
the
fifth
Rule
of
this
Chapter
.
And
the
sum
of
8b
−
16+2cd
and
24
−
5b+3cd
is
3b+8+5cd
.
And
here
note
that
in
setting
down
of
Compound
quantities
to
be
added
together
,
it
matters
not
which
of
them
you
set
first
,
so
that
to
every
quantity
there
is
prefixed
its
proper
sign
;
as
3a+b
−
cc
is
the
same
with
b+3a
−
cc
and
with
−
cc+b+3a
,
&c.
More
Examples
of
the
Addition
of
Compound
Algebraical
Integers
.
CHAP.
III.
Subtraction
of
Algebraick
Integers
.
I
Shall
not
here
need
to
give
you
a
definition
of
the
nature
of
subtraction
,
but
shall
only
give
you
a
general
Rule
for
the
finding
out
of
the
remainder
,
excess
,
or
difference
of
any
two
quantities
,
and
that
in
all
cases
whatsoever
.
I.
When
a
Quantity
Single
or
Compound
,
is
given
to
be
subtracted
from
another
,
then
change
the
sign
,
or
signs
of
the
quantity
to
be
subtracted
,
into
the
contrary
signs
,
that
is
+
into
−
,
and
−
into
+
;
which
being
done
,
add
the
two
given
quantities
together
by
the
Rules
of
the
foregoing
second
Chapter
,
so
shall
their
sum
be
the
difference
,
or
remainder
sought
.
Example
1.
Let
it
be
required
to
subtract
3a
from
8a
.
The
quantity
here
given
to
be
subtracted
is
3a
,
which
according
to
the
second
rule
of
the
second
chap.
is
+3a
,
therefore
must
its
sign
+
be
changed
into
−
,
so
will
it
be
−
3a
,
which
being
added
to
8a
(
by
the
third
Rule
of
the
second
Chap.
their
sum
will
be
5a
,
for
,
8a
−
3a
=
5a
,
and
8a
and
is
the
difference
between
the
quantities
so
much
3a
.
Example
2.
Let
it
be
required
to
subtract
−
3bc
from
4bc
.
Here
because
−
3bc
is
the
quantity
to
be
subtracted
,
therefore
must
its
sign
−
be
changed
into
+
,
so
will
it
be
+3bc
,
which
being
added
to
4bc
,
by
the
second
Rule
of
the
second
Chapter
,
their
sum
is
7bc
,
for
,
3bc+4bc
=
7bc
,
and
so
much
is
the
remainder
when
−
3bc
is
subtracted
from
+4bc
.
Example
3.
Let
it
be
required
to
subtract
−
3bde
from
−
9bde
.
Here
because
−
3bde
is
the
quantity
to
be
subtracted
,
therefore
must
its
sign
−
be
changed
into
+
,
so
will
it
be
+3bde
which
being
added
to
−
9bde
according
to
the
third
Rule
of
the
second
Chapter
their
sum
will
be
−
6bde
for
+3bde
−
9bde
=
−
6bde
,
and
so
much
is
the
remainder
when
−
3bde
is
subtracted
from
−
9bde
.
Example
4.
Subtract
3cd
from
8de
.
The
sign
of
3cd
being
changed
,
it
will
be
−
3cd
,
which
being
added
to
8de
by
the
fifth
Rule
of
the
second
Chapter
,
their
sum
will
be
8de
−
3cd
which
is
the
remainder
when
3cd
is
subtracted
from
●…
de
.
Example
5.
What
is
the
remainder
when
−
3bc
is
subtracted
from
2cd
?
Facit
3bc+2cd
,
or
2cd+3bc
.
In
all
which
Examples
you
see
that
the
sign
of
the
quantity
given
to
be
subtracted
is
changed
into
the
contrary
sign
.
More
Examples
of
Subtraction
of
Simple
Algebraick
Integers
.
Example
6.
Example
7.
Example
8.
Example
9.
And
when
it
is
required
to
subtract
a
Compound
Integer
from
a
Compound
Integer
,
the
operation
will
not
in
any
wise
differ
from
the
former
,
observing
alway
to
change
+
into
−
,
and
−
into
+
as
will
appear
by
the
following
Examples
.
Example
10.
From
3a+4b
let
it
be
required
to
subtract
2a
−
b.
Here
2a
−
b
being
the
quantity
to
be
subtracted
from
the
other
;
its
signs
must
be
changed
into
the
contrary
signs
.
And
then
instead
of
2a
−
b
you
will
have
−
2a+b
,
which
being
added
to
3a+4b
the
sum
will
be
a+5b
=
3a+4b
−
2a+b
,
and
so
much
is
the
remainder
,
when
subtraction
is
performed
according
to
the
tenure
of
the
Question
.
See
the
work
laid
down
as
followeth
.
Example
11.
From
3aa
−
2dc+ab
let
it
be
required
to
subtract
aa+3ab
−
3dc
.
The
quantity
here
given
to
be
subtracted
is
aa+3ab
−
3dc
,
whose
signs
being
changed
,
it
will
then
be
−
aa
−
3ab+3dc
,
which
being
added
to
3aa
−
2dc+ab
,
the
sum
will
be
3aa
−
2dc+ab
−
aa
−
3ab+3dc
,
which
according
to
the
sixth
Rule
of
the
second
Chapter
is
equal
to
2aa+dc
−
2ab
.
See
the
following
operation
.
But
when
the
given
quantities
are
unlike
,
then
place
the
quantity
to
be
subtracted
immediately
after
the
quantity
out
of
which
it
is
to
be
subtracted
in
the
same
line
changing
its
signs
,
which
new
quantity
when
the
said
quantity
is
so
annexed
,
is
the
remainder
required
,
which
will
admit
of
no
Contraction
,
because
the
quantities
are
unlike
.
Example
12.
Let
it
de
required
to
subtract
3ab+2aa
from
7bc+6cd
,
the
quantity
given
to
be
subtracted
is
3ab+2aa
which
annexed
to
the
other
given
quantity
,
changing
its
signs
,
will
give
7bc+6cd
−
3ab
−
2aa
,
which
is
the
remainder
required
.
See
the
following
work
.
More
Examples
of
subtraction
in
Compound
Algebraick
Integers
.
As
in
Natural
Arithmetick
,
the
remainder
and
the
sum
subtracted
being
added
together
,
will
be
equal
to
the
number
from
which
the
subtraction
is
made
,
so
is
it
likewise
in
Algebraical
Arithmetick
;
for
if
you
add
the
remaining
Quantity
to
the
Quantity
subtracted
,
the
sum
will
be
equal
to
the
Quantity
out
of
which
the
subtraction
is
made
.
As
in
the
first
Example
of
this
Chapter
,
where
it
is
required
to
subtract
3a
from
8a
,
and
the
remainder
is
5a
;
now
if
to
5a
you
add
3a
,
the
sum
will
be
5a+3a
=
8a
;
And
in
the
twelfth
Example
,
where
it
is
required
to
subtract
3ab+2aa
from
7bc+6cd
;
the
remainder
is
found
to
be
7bc+6cd
−
3ab
−
2aa
,
to
which
if
you
add
the
number
subtracted
,
viz.
3ab+
2aa
,
the
sum
will
be
7bc+6cd
equal
to
the
given
quantity
out
of
which
subtraction
is
made
,
for
2ab+1aa
being
added
to
−
3ab
−
2aa
they
destroy
each
other
,
because
their
signs
are
unlike
,
and
this
may
serve
for
a
sufficient
(
and
indeed
the
only
)
proof
of
the
work
.
CHAP.
IV.
Multiplication
in
Algebraick
Integers
.
I.
IN
Multiplication
of
Algebraical
Quantities
,
there
are
always
two
quantities
given
,
to
find
out
a
third
.
II.
The
two
quantities
given
are
called
the
Factors
,
and
the
third
quantity
invented
,
or
found
by
the
said
Factors
,
is
called
the
Product
,
Fact
,
or
Rectangle
.
III.
When
the
given
Factors
are
single
quantities
alike
,
or
unlike
,
if
they
have
not
natural
numbers
prefixed
to
them
,
the
fact
is
discovered
at
first
sight
,
and
is
performed
by
joyning
both
the
quantities
together
in
one
,
without
any
Character
between
them
,
like
letters
in
a
word
.
But
special
regard
must
be
had
to
the
signs
of
the
given
quantities
,
in
all
kinds
of
Multiplication
.
whether
by
Simple
or
Compound
Quantities
.
and
whether
with
,
or
without
numbers
prefixed
to
them
;
the
nature
of
the
product
wholly
depending
thereupon
,
viz.
If
the
signs
of
the
quantities
to
be
multiplyed
together
be
alike
,
that
is
,
both
+
or
both
−
,
then
the
sign
of
the
product
or
fact
will
be
+
,
but
if
they
be
of
different
kinds
,
viz.
the
one
+
,
and
the
other
−
,
then
the
sign
of
the
product
will
be
−
,
as
you
will
find
by
the
several
Examples
following
.
Example
1.
What
is
the
Product
of
a
multiplyed
by
b
?
Facit
ab
.
Here
because
both
the
Factors
are
signed
with
+
,
therefore
the
sign
of
the
product
is
+
.
In
like
manner
,
if
the
given
Factors
had
been
−
a
and
−
b
the
product
would
have
been
(
ab
or
ba
)
the
same
as
before
,
because
the
signs
of
the
Factors
are
both
alike
,
viz.
both
−
.
But
if
the
given
Factors
had
been
+a
and
−
b
,
or
−
a
and
+b
then
the
product
or
fact
would
have
been
−
ba
or
−
ab
,
because
the
signs
of
the
Factors
are
unlike
viz.
the
one
+
and
the
other
−
,
observe
the
like
in
all
cases
whatsoever
.
Example
.
2.
What
is
the
Product
of
abc
multiplyed
by
cd
?
Facit
abccd
or
+abccd
.
And
if
you
had
been
to
multiply
−
abc
by
−
cd
,
the
product
would
have
been
the
same
,
viz.
abccd
,
or
+abccd
.
But
if
the
Factors
had
been
−
abc
by
+cd
,
or
+abc
by
−
cd
,
then
the
Fact
would
have
been
−
abccd
,
because
the
signs
of
the
Factors
are
unlike
.
More
Examples
of
the
like
Nature
.
IV.
When
the
Quantities
given
to
be
subtracted
are
Single
,
or
Simple
Quantities
,
(
whether
alike
,
or
unlike
)
having
natural
numbers
prefixed
to
them
,
then
in
such
cases
let
the
natural
numbers
be
multiplyed
together
,
and
to
their
product
annex
the
product
of
the
given
Algebraical
Quantities
,
they
being
multiplyed
together
as
in
the
last
Rule
,
so
shall
this
new
quantity
found
be
the
product
required
.
As
in
the
following
Examples
.
Example
1.
Let
it
be
required
to
multiply
3a
by
9a
.
First
,
I
multiply
the
numbers
prefixed
to
both
quantities
,
the
one
by
the
other
,
viz.
9
by
3
,
and
their
product
is
27
,
to
which
I
annex
the
Letters
contained
in
both
quantities
,
viz.
aa
,
and
they
make
27aa
,
which
is
the
product
,
or
Fact
required
.
Example
2.
Let
it
be
required
to
multiply
3aa
by
4b
.
Here
first
I
multiply
the
numbers
prefixed
together
,
viz.
3
and
4
,
and
they
make
12
;
to
which
product
I
annex
the
Letters
of
both
quantities
given
,
viz.
aa
and
b
,
and
they
make
12aab
for
the
product
required
.
Example
3.
What
is
the
product
of
−
3abc
by
−
5cd
?
Here
first
I
multiply
the
given
numbers
prefixed
,
viz.
3
and
5
,
and
they
produce
15
,
to
which
I
annex
the
Letter
in
both
the
quantities
,
viz.
abc
and
cd
,
and
they
make
15abccd
,
to
which
I
prefix
the
sign+
,
because
the
signs
of
the
given
quantities
were
both
alike
,
viz.
−
,
and
then
will
the
product
or
fact
be
+15abccd
.
Example
4.
What
is
the
product
of
+6ab
multiplyed
by
−
3cd
?
First
,
multiply
the
numbers
prefixed
,
viz.
6
and
3
,
and
the
product
is
18
,
to
which
I
annex
the
Letters
in
both
the
given
quantities
,
ab
and
cd
,
and
it
makes
18abcd
,
to
which
I
prefix
the
sign
−
(
because
the
signs
of
the
Factors
were
unlike
,
viz.
the
one+
,
and
the
other
−
)
and
then
the
product
will
be
−
18abcd
.
More
Examples
of
the
like
Nature
.
V.
When
Compound
quantities
are
to
be
multiplyed
,
the
operation
(
in
effect
)
is
the
same
with
multiplication
of
Simple
Quantities
delivered
in
the
foregoing
Rules
,
for
you
are
to
multiply
every
particular
quantity
in
the
multiplicand
by
each
particular
quantity
in
the
multiplyar
,
(
not
regarding
whether
you
begin
the
work
at
the
right
hand
or
the
left
,
)
and
then
let
the
several
products
be
joyned
together
according
to
the
Rules
of
Algebraical
Addition
,
and
that
sum
will
be
the
product
required
.
The
following
Examples
will
make
the
Rule
plain
.
Example
1.
In
the
first
place
let
it
be
required
to
multiply
(
a
Compound
Quantity
by
a
Simple
,
viz.
)
ab+d
by
a.
And
in
order
thereto
,
First
,
I
multiply
a
into
ab
,
and
the
product
is
aab
,
and
then
into
d
,
and
it
produceth
ad
,
so
is
aab+ad
the
product
required
,
each
member
of
the
product
being
affirmative
,
because
all
parts
of
the
Factors
were
Affirmative
.
Example
2.
Let
it
be
required
to
multiply
aa+ab
−
c
by
b.
The
Product
of
aa
by
b
is
+aab
,
and
the
product
of
ab
by
b
is
+abb
,
and
the
product
of
−
c
by
b
is
−
cb
,
all
which
particular
products
being
joyned
,
and
one
Compound
quantity
composed
thereof
,
it
will
give
aab+abb
bc
for
the
product
required
.
See
the
work
in
the
margent
.
Example
3.
Let
it
be
required
to
multiply
the
Compound
quantity
ac+dg
by
the
Compound
quantity
c+d
.
First
,
Multiply
each
member
of
the
multiplicand
by
c
,
and
the
product
is
acc+cdg
,
then
multiply
each
member
of
the
multiplicand
by
d
,
and
the
product
is
acd+ddg
,
which
two
products
being
joyned
together
by
the
Rules
of
Algebraical
Addition
,
the
sum
is
acc+cdg+acd+ddg
,
which
is
the
product
required
,
as
appears
by
the
operation
.
Example
.
4.
What
is
the
product
of
da+bc
multiplyed
by
da
−
ab
?
First
,
Multiply
the
Multiplicand
da+bc
by
da
(
the
first
member
of
the
multiplyar
)
and
it
produceth
ddaa+dabc
,
then
multiply
the
said
Multiplicand
by
−
ab
(
the
second
member
of
the
multiplyar
)
and
it
produceth
−
aadb
−
abbc
,
which
two
quantities
being
joyned
together
,
give
ddaa+dabc
−
aadb
−
abbc
for
the
product
required
.
As
you
may
see
in
the
Operation
.
Example
5.
What
is
the
Product
of
a+b
−
c
multiplyed
by
a+b
−
c
?
First
,
Multiply
each
member
of
the
Multiplicand
by
a
(
the
first
member
of
the
multiplyar
)
and
it
produceth
aa+ab
−
ac
,
then
multiply
each
said
member
in
the
multiplicand
by
b
,
(
the
second
member
of
the
multiplyar
)
and
it
produceth
ab+bb
−
bc
,
then
multiply
each
member
of
the
said
multiplicand
by
−
c
(
the
third
and
last
member
of
the
multiplyar
)
and
the
product
is
−
ac
−
bc+c●…
;
which
said
three
products
being
joyned
together
according
to
the
Rules
of
Algebraical
Addition
,
will
give
aa+2ab
−
2ac+bb
−
2bc
+cc
which
is
the
Square
of
a+b
−
c
or
product
required
,
as
appears
by
the
whole
operation
.
And
if
there
are
natural
numbers
prefixed
to
any
of
the
Compound
Quantities
,
the
operation
will
not
be
different
from
the
foregoing
Examples
of
this
Rule
,
regard
being
had
to
the
fourth
Rule
of
this
Chapter
,
as
will
appear
by
the
following
Example
.
Example
.
6.
What
is
the
Product
of
3b+2c
multiplyed
by
4b
−
3c
?
First
,
By
the
fourth
Rule
multiply
each
member
of
(
3b+2c
)
the
multiplicand
by
4b
,
and
the
product
is
12bb+8bc
;
then
multiply
the
said
multiplicand
by
−
3c
,
and
the
product
is
−
9bc
−
6cc
,
which
being
added
to
12bb+8bc
,
the
sum
will
be
12bb
−
bc
−
6cc
which
is
the
product
required
.
Example
7.
What
is
the
product
of
2a+2c
−
8
multiplyed
by
2a
−
5
?
First
,
2a+2c
−
8
being
multiplyed
by
2a
,
produceth
4aa+4ac
−
16a
,
for
2a×2a
=
4aa
and
2a×
2e
=
4ae
and
2a×
−
8
=
−
16a
which
is
marked
with
−
,
because
the
signs
of
the
Factors
are
unlike
,
viz.
−
8
and
+2a
.
Secondly
2a+2e
−
8
being
multiplyed
by
−
5
,
produceth
−
10a
−
10e
+40
;
for
−
5×
+2a
=
−
10a
and
−
5×
+2e
=
−
10e
,
and
−
5×
−
8
=
+40
,
all
which
quantities
being
joyned
together
by
the
Rules
of
Algebraical
addition
will
give
+4aa+4ae
−
26a
−
10e
+40
which
is
the
product
required
.
See
the
work
.
More
Examples
in
Multiplication
of
Compound
Algebraical
Integers
.
CHAP.
V.
Division
in
Algebraick
Integers
.
I.
IN
Division
Algebraical
(
as
in
Division
in
Common
Arithmetick
)
there
are
two
Quantities
given
to
find
out
a
third
;
which
Quantity
sought
is
called
the
Quote
,
or
Quotient
;
and
of
the
Quantities
given
,
that
which
is
to
be
divided
,
is
called
the
Dividend
,
and
the
Quantity
by
which
it
is
to
be
divided
,
is
called
the
Divisor
?
II.
When
it
is
required
to
divide
one
number
or
quantity
by
another
,
if
you
place
the
Dividend
for
the
Numerator
,
and
the
Divisor
for
the
Denominator
of
a
Fraction
,
that
Fraction
so
composed
is
equal
to
the
Quotient
that
would
arise
by
the
real
Division
of
the
one
by
the
other
.
For
if
it
were
required
to
divide
4
by
5
,
the
quotient
would
be
4/5
,
or
if
it
were
required
to
divide
12
by
7
,
the
Quotient
would
be
.
The
reason
of
which
is
the
ground
of
the
general
part
of
Division
in
Algebra
;
for
when
one
quantity
is
to
be
divided
by
another
,
set
the
quantity
that
is
Dividend
for
the
Numerator
of
a
Fraction
,
and
the
quantity
that
is
the
Divisor
set
for
the
Denominator
.
So
if
it
were
required
to
divide
the
quantity
b
by
the
quantity
a
,
I
would
place
them
thus
,
viz.
which
signifieth
the
Quotient
of
b
divided
by
a.
In
like
manner
if
it
were
required
to
Divide
abe
by
cd
,
the
Quotient
would
be
.
And
if
5ac
were
to
be
divided
by
3cd
,
the
Quotient
would
be
.
And
if
bed
were
to
be
divided
by
7
,
the
Quotient
would
be
.
The
same
is
to
be
observed
in
Division
of
Compound
Algebraick
Integers
,
for
if
it
were
required
to
divide
a+b
by
c
,
the
Quotient
would
be
,
and
if
5b
were
to
be
divided
by
3a+bed
,
the
Quotient
would
be
.
More
Examples
of
Division
according
to
the
foregoing
Rule
.
III.
When
in
any
Quotient
that
is
expressed
according
to
the
foregoing
Rule
,
there
are
the
same
Letter
or
Letters
repeated
in
every
part
or
member
of
the
Numerator
and
Denominator
,
you
may
cancel
such
Letter
or
Letters
,
but
be
sure
that
what
you
cancel
in
one
part
,
to
cancel
the
very
same
in
all
the
rest
.
So
shall
this
new
Quantity
be
a
true
Quotient
,
equivalent
to
what
it
was
before
the
said
Letters
were
cancelled
.
Example
.
What
is
the
Quotient
of
bd
divided
by
b
?
According
to
the
foregoing
Rule
the
Quotient
is
;
but
because
the
letter
b
is
found
both
in
the
Numerator
and
in
the
Denominator
,
therefore
cancel
b
in
both
of
them
,
and
then
you
will
find
the
Quotient
to
be
d
,
for
.
Again
let
it
be
required
to
divide
ab+ad
by
acd
,
the
Quotient
is
,
and
because
the
letter
a
is
found
in
every
member
of
the
Numerator
and
Denominator
,
cast
it
out
of
every
one
,
and
then
you
will
have
for
the
Quotient
.
Likewise
if
you
were
to
divide
ab+abc+abe
by
abd+abf
the
Quotient
you
would
find
to
be
which
being
contracted
by
cancelling
ab
in
each
member
of
the
Numerator
and
Denominator
,
there
will
be
found
for
the
Quotient
.
And
bb+b
being
to
be
divided
by
b
,
the
Quotient
will
be
and
by
cancelling
b
in
every
part
,
there
will
be
b+1
for
the
Quotient
abbreviated
,
for
and
by
cancelling
b
in
every
part
,
there
will
be
and
.
The
same
is
to
be
observed
whether
the
signs
be
+
or
−
;
so
if
it
be
required
to
divide
ab●…+
cbe
,
by
bde
−
bce
the
Quotient
will
be
And
because
be
is
found
in
each
Quantity
,
I
cancel
it
,
and
then
the
Quotient
contracted
,
or
abbreviated
will
be
found
to
be
.
More
Examples
of
Contractions
or
Abbreviations
in
Division
of
Algebraick
Integers
according
to
the
foregoing
Rule
.
IV.
If
when
it
is
required
to
divide
a
Simple
or
Compound
Algebraick
Quantity
by
a
Simple
Quantity
,
there
be
prefixed
to
every
member
a
number
,
or
numbers
,
that
may
be
divided
by
any
other
number
without
any
remainder
,
then
instead
of
the
given
prefixed
numbers
prefix
the
Quotient
of
each
of
the
said
numbers
divided
by
the
said
common
measurer
,
not
neglecting
to
cancel
any
letter
that
may
be
found
in
each
part
of
the
Numerator
and
Denominator
,
according
to
the
foregoing
third
Rule
.
As
for
Example
.
Divide
16bc
by
4bd
.
Here
according
to
the
foregoing
Rule
,
the
Quotient
is
,
but
because
the
prefixed
numbers
16
and
4
will
admit
of
4
for
a
common
measure
,
therefore
I
divide
them
both
by
4
,
and
the
Quotients
are
4
and
1
,
which
I
prefix
to
the
given
Quantities
instead
of
16
and
4
,
and
then
the
Quotient
will
be
,
or
,
and
because
b
is
contained
both
in
the
Numerator
,
and
the
Denominator
,
cancel
it
,
so
have
you
for
the
Quotient
required
.
Moreover
15abc+12bd
being
given
to
be
divided
by
3bg
,
the
Quotient
will
be
found
to
be
.
For
,
I
first
discover
that
the
prefixed
numbers
15
,
12
;
and
3
,
have
3
for
their
common
measure
,
by
which
they
being
severally
divided
,
give
5
,
4
,
and
1
,
which
being
prefixed
to
the
said
Quantities
after
b
,
(
which
is
found
in
every
quantity
is
cancelled
)
there
will
be
for
the
true
Quotient
required
.
More
Examples
of
Contraction
in
Division
,
according
to
the
two
last
Rules
.
V.
When
in
Compound
quantities
one
or
more
letter
or
letters
is
repeated
in
every
member
,
then
will
the
remaining
Letters
in
each
quantity
evenly
divide
the
said
Compound
quantity
without
any
remainder
,
and
the
quotient
will
be
the
Letter
or
letters
repeated
in
each
member
as
aforesaid
.
As
for
Example
.
1.
What
is
the
Quotient
of
ba+ca
divided
by
b+c
?
Here
it
is
evident
that
the
Quotient
will
be
a
;
for
proof
whereof
take
the
Divisor
b+c
,
and
multiply
by
the
Quotient
a
,
according
to
the
fifth
Rule
of
the
fourth
Chap.
and
the
product
will
be
ba+ca
equal
to
the
Dividend
.
2.
Likewise
if
it
were
required
to
divide
bad+cad
by
b+c
,
the
Quotient
will
be
found
to
be
ad
.
3.
Also
by
the
same
reason
if
you
divide
2ab
−
2acd
−
2adb
by
b
−
cd
−
db
,
the
Quotient
will
be
2a
,
and
if
you
divide
the
same
dividend
by
2b
−
2cd
−
−
2db
,
the
Quotient
will
be
a.
More
Examples
of
the
like
nature
.
The
reason
why
−
1
is
the
last
member
of
the
last
example
is
,
because
−
c
,
or
−
1c
is
the
last
member
of
the
Dividend
,
for
according
to
the
thirteenth
Rule
of
the
first
Chapter
,
when
a
quantity
hath
no
number
prefixed
to
it
;
it
is
supposed
to
have
the
number
1
before
it
.
And
here
note
,
that
as
in
Multiplication
of
Algebraick
Integers
+
by
+
,
and
−
by
−
produceth
+
,
and
+
by
−
produceth
−
,
so
in
Division
,
if
you
divide
+
by
+
,
or
−
by
−
,
the
sign
of
the
Quote
will
be
+
,
but
if
you
divide
+
by
−
,
or
−
,
by
+
,
the
sign
of
the
Quote
will
be
−
;
so
if
3ab
be
divided
by
3a
,
the
Quote
will
be
b
,
or
+
b
,
and
if
−
3ab
be
divided
by
−
3a
,
the
Quote
will
be
+b
,
for
if
you
multiply
−
3a
by
+b
,
the
product
will
be
−
3ab
by
the
third
Rule
of
the
fourth
Chapter
foregoing
.
Also
if
you
divide
+3ab
by
−
3a
,
or
−
3ab
by
+3a
,
the
Quotient
will
be
−
b
,
for
+3a
being
multiplyed
by
−
b
,
produceth
−
3ab
,
and
−
3a
being
multiplyed
by
−
b
,
produceth
+3ab
.
VI.
From
a
due
consideration
of
the
manner
of
operating
the
Examples
of
the
last
Rule
,
a
way
may
be
discovered
to
divide
a
Compound
Quantity
by
a
Simple
,
or
Compound
Quantity
,
and
to
find
out
the
true
Quotient
when
it
likewise
will
be
a
Compound
Quantity
,
the
practice
of
which
will
be
made
plain
by
the
following
Examples
.
Example
1.
Let
it
be
required
to
divide
ba+ca
,
by
a.
Having
placed
the
Dividend
and
Divisor
as
is
usual
in
vulgar
Arithmetick
,
and
as
you
see
in
the
following
operation
.
The
do
I
seek
how
often
a
is
contained
in
ba
,
(
the
first
member
of
the
Dividend
)
and
the
answer
is
b
times
,
therefore
I
put
b
in
the
Quotient
,
and
thereby
I
multiply
(
a
)
the
Divisor
,
and
the
product
is
+ba
,
which
must
be
subtracted
from
ba
in
the
Dividend
,
(
and
therefore
I
change
its
sign
into
−
ba
by
the
first
Rule
of
the
third
Chapter
,
and
there
remaineth
0
,
then
do
I
bring
down
+ca
the
next
member
of
the
Dividend
,
and
divide
it
by
a
,
and
the
Quotient
is
+c
,
by
which
I
again
multiply
(
a
)
the
Divisor
,
and
the
Product
is
ca
,
which
subtracted
from
ca
there
remaineth
0
,
and
so
the
work
of
Division
is
ended
,
and
I
find
the
Quotient
of
ba+ca
divided
by
a
be
b+c
;
for
proof
whereof
if
you
multiply
b+c
by
a
(
the
Divisor
)
the
product
will
be
ba+ca
equal
to
the
given
Dividend
.
Example
2.
Let
it
be
required
to
divide
ba+ca+be+ce
by
b+c
.
Having
disposed
of
the
Dividend
and
Divisor
in
order
to
the
work
with
a
crooked
line
behind
which
to
place
the
Quotient
,
as
in
Common
Arithmetick
;
then
first
I
seek
how
often
b
(
the
first
member
of
the
Divisor
is
contained
in
ba
,
(
the
first
member
of
the
Dividend
)
and
there
ariseth
a
,
which
I
put
in
the
Quotient
,
and
thereby
multiply
each
member
of
the
Divisor
,
viz.
b+c
,
and
the
product
is
ba+ca
,
which
place
under
the
two
first
quantities
of
the
dividend
towards
the
left
hand
,
viz
,
under
ba+be
,
and
by
the
first
Rule
of
the
second
Chapter
subtract
it
therefrom
,
so
will
the
remainder
be
0
;
to
which
I
bring
down
the
remaining
part
of
the
dividend
,
viz.
be+ce
,
and
divide
be
by
e
,
and
there
ariseth
in
the
Quotient
e
,
by
which
I
multiply
the
whole
Divisor
b+c
,
and
the
Product
is
be+ce
,
which
subtracted
from
the
Dividend
be+ce
,
the
Remainder
is
0.
See
the
whole
work
as
followeth
.
So
that
the
quotient
is
a+e
,
now
to
prove
the
work
,
multiply
the
Divisor
b+c
by
the
Quotient
(
a+e
)
according
to
the
fifth
Rule
of
the
fourth
Chapter
,
and
the
Product
you
will
find
to
be
ba+ca+be+ce
which
is
equal
to
the
given
Dividend
;
and
therefore
I
conclude
the
operation
to
be
truly
performed
.
Example
3.
In
like
manner
,
if
you
divide
ba+bd+ca+cd
−
ae
−
de
by
a+d
,
the
Quotient
will
be
found
to
be
b+c
−
e
according
to
the
following
work
.
The
work
of
the
last
Example
explained
.
In
the
foregoing
Example
,
First
,
I
divide
ba
(
the
first
member
of
the
Dividend
)
by
a
the
first
quantity
or
member
of
the
Divisor
,
and
there
ariseth
a
in
the
Quotient
,
which
is
+a
,
(
because
the
signs
of
the
Dividend
and
Divisor
are
+
)
and
thereby
I
multiply
the
Divisor
a+d
.
and
the
Product
is
ba+bd
,
which
I
place
under
the
two
first
members
of
the
Dividend
as
you
see
in
the
work
,
and
subtract
it
therefrom
,
and
the
remainder
is
0
,
to
which
I
bring
down
the
two
next
quantities
,
viz.
+ca+cd
.
Then
do
I
divide
+ca
by
a
,
and
there
ariseth
in
the
Quotient
+c
,
(
because
the
Dividend
and
Divisor
are
both
signed
+
,
)
by
which
I
multiply
the
said
Divisor
,
and
the
Product
is
+ca+cd
which
I
placeunder
the
dividend
,
and
subtract
it
therefrom
,
and
there
remaineth
0
,
to
which
I
annex
the
two
next
and
last
members
of
the
Dividend
,
viz.
−
ae
−
de
;
and
divide
−
ae
by
+a
,
and
the
Quotient
is
−
e
,
(
because
the
signs
of
the
Dividend
and
Divisor
are
different
,
viz.
the
one
+
,
and
the
other
−
)
and
thereby
I
multiply
the
whole
Divisor
,
and
the
Product
is
−
ae
−
de
which
subtracted
from
(
−
ae
−
de
)
the
Dividend
,
the
remainder
is
0
,
and
so
the
work
is
finished
,
and
the
Quotient
arising
by
this
Division
is
b+c
−
e
,
as
you
may
prove
at
your
leisure
.
If
the
quantities
or
members
of
the
Dividend
of
the
foregoing
Example
are
not
placed
in
the
same
order
that
is
there
expressed
the
,
effect
of
the
operation
will
be
the
same
,
as
you
may
see
by
the
following
work
.
First
,
I
divide
ba
by
a
,
and
the
Quote
is
b
,
by
which
I
multiply
the
Divisor
(
a+d
)
and
the
product
is
ba+bd
which
I
subtract
from
ba+ca
and
the
remainder
is
(
by
the
Rule
of
the
third
Chapter
(
ba+ca
−
ba
−
bd
which
being
contracted
by
the
Rules
of
Addition
is
ca
−
bd
,
(
for
+ba
and
−
ba
expunge
each
other
,
)
then
to
this
remainder
do
I
bring
down
the
two
next
quantities
of
the
Dividend
,
viz.
−
ae+bd
which
being
annexed
to
the
said
remainder
ca
−
bd
it
then
(
makes
for
a
new
dividual
)
ca
−
bd
−
ae+bd
,
but
−
bd
and
+bd
destroy
each
other
,
and
therefore
the
dividual
contracted
is
ca
−
ce
,
which
I
divide
by
a+d
as
before
,
and
the
Quotient
is
+c
,
by
which
I
multiply
the
Divisor
,
and
the
Product
is
ca+cd
,
which
being
subtracted
from
the
said
dividual
ca
−
ae
,
the
remainder
is
ca
−
ae
−
ca
−
cd
which
being
contracted
,
is
−
ae●…cd
,
to
which
I
joyn
the
two
next
quantities
in
the
Dividend
,
viz.
−
cd
−
de
,
and
it
makes
(
−
ae
−
cd+cd
−
de
)
=
−
ae
−
de
(
for
−
cd
and
+cd
destroy
each
other
)
for
a
new
dividual
,
which
I
divide
by
the
said
Divisor
a+d
,
and
the
Quote
is
−
e
,
by
which
I
multiply
the
Divisor
,
and
the
Product
is
−
ae
−
de
,
which
subtracted
from
−
ae
−
de
(
the
dividual
)
the
remainder
is
−
ae
−
de+ae+de
=
0
,
and
so
the
work
is
finished
,
and
I
find
the
Quotient
to
be
b+c
−
e
as
before
.
But
here
note
by
the
way
that
it
doth
not
alwaies
fall
out
that
you
are
to
divide
the
first
member
of
the
dividual
by
Note
the
first
of
the
Divisor
,
but
by
some
other
member
which
you
can
discover
will
do
the
work
without
making
a
Fraction
.
As
in
the
following
Example
:
Example
.
4.
Let
it
be
required
to
divide
aa
−
ee
by
a+e
?
First
,
I
divide
aa
by
a
,
and
there
ariseth
in
the
Quotient
a
,
by
which
I
multiply
the
Divisor
a+e
,
and
the
Product
is
aa+ae
,
which
subtracted
from
the
Dividend
(
aa
−
ee
)
the
remainder
is
−
ee
−
ae
for
a
dividual
,
and
then
I
do
not
seek
how
often
a
is
contained
in
ee
,
for
then
the
answer
would
be
a
Fraction
,
but
I
divide
ee
by
its
correspondent
Divisor
+e
,
and
there
ariseth
−
e
,
to
be
written
in
the
Quotient
next
after
a
,
but
not
+e
,
because
−
divided
by
+
quotes
−
.
Then
I
multiply
the
whole
Divisor
a+e
by
−
e
,
and
the
product
is
−
ae
−
ee
,
which
subtracted
from
the
said
dividual
−
ee
−
ae
,
the
remainder
is
0
,
so
is
the
work
ended
,
and
I
find
the
Quotient
to
be
a
−
e.
See
the
operation
.
Example
5.
If
it
were
required
to
divide
aaa+abd+baa+bbd
by
aa+bd
by
the
Quotient
would
be
found
to
be
a+b
,
as
appears
by
the
work
.
Example
6.
If
aaa
−
abb+abd+baa
−
bbb+bbd
be
divided
by
a+b
the
Quote
will
be
aa
−
bb+bd
,
as
by
the
operation
.
Example
7
,
Let
it
be
required
to
divide
18abbc+9bbcc+24bcc+24abc+16cc
by
3bc+4c
,
the
quotient
will
be
found
to
be
6ab+3bc+4c
.
See
the
following
operation
.
When
Algebraical
Division
according
to
the
Rules
before
delivered
,
will
not
exactly
performe
the
work
without
any
remainder
,
then
you
may
place
the
Dividend
and
Divisor
Fraction
wise
,
which
is
indeed
the
most
general
practice
amongst
Algebrists
;
or
else
proceed
in
Division
as
far
as
you
can
by
the
preceeding
method
;
and
then
place
the
remainder
for
a
Numerator
over
the
Divisor
,
as
in
the
following
Example
,
where
aa
−
bb+ac
is
divided
by
a+b
,
and
the
quotient
is
,
the
remainder
being
+ac
.
CHAP.
VI.
The
Doctrine
of
Algebraical
Fractions
.
And
First
,
Of
Reduction
.
1.
HE
that
intends
a
considerable
proficiency
in
this
mysterious
Art
,
must
be
very
well
acquainted
with
the
Doctrine
of
Vulgar
Fractions
,
a
mean
knowledge
therein
not
being
sufficient
;
for
all
operations
whatsoever
in
Algebraick
Fractions
have
their
dependance
thereupon
,
being
wrought
in
every
respect
as
vulgar
Fractions
they
are
by
the
help
of
the
Rules
contained
in
the
several
Chapters
foregoing
,
and
there
are
very
few
questions
solved
Algebraically
,
but
what
have
one
or
more
Fractions
concerned
in
its
operation
.
To
reduce
Fractions
,
having
unequal
Denominators
to
Fractions
of
the
same
value
,
having
a
common
Denominator
.
II.
When
you
would
reduce
Algebraical
Fractions
to
a
common
Denominator
,
multiply
the
Numerator
of
the
first
Fraction
into
the
Denominator
or
denominators
of
the
rest
,
so
shall
the
●…ct
be
a
Numerator
equal
to
the
Numerator
of
the
first
Fraction
,
likewise
multiply
the
Numerators
of
the
second
,
third
,
&c.
Fractions
into
all
the
Denominators
except
its
own
,
and
the
several
products
shall
be
so
many
new
Numerators
,
then
multiply
all
the
Denominators
continually
,
so
shall
the
product
be
a
common
Denominator
to
all
the
Numerators
found
out
as
before
.
Example
1.
Reduce
and
to
a
common
Denominator
.
Multiply
the
Numerator
a
(
of
the
first
Fraction
)
into
the
Denominator
(
c
)
of
the
second
Fraction
,
and
the
product
is
ac
,
for
a
Numerator
=
a
,
then
multiply
(
d
)
the
Numerator
of
the
second
Fraction
into
(
b
)
the
Denominator
of
the
first
,
and
the
product
is
db
,
for
a
Numerator
=
d
,
then
multiply
the
Denominators
together
,
viz.
b
into
c
,
and
the
product
bc
is
the
Denominator
common
to
both
the
Numerators
,
so
will
the
2
new
Fractions
be
and
for
=
and
.
Example
2.
What
Fractions
are
=
,
and
,
having
an
equal
or
common
Denominator
?
Facit
,
and
,
for
a×
a×
d
=
aad
,
=
the
Numerator
a
,
and
b×
c×
d
=
bcd
=
the
Numerator
b
,
and
c×
c×
a
=
cca
=
the
Numerator
c
,
and
c×
a×
d
=
cad
which
is
the
common
Denominator
.
To
reduce
an
Algebraical
Fraction
to
its
lowest
Terms
equivalent
.
III.
When
in
the
Numerator
and
Denominator
of
an
Algebraical
Fraction
,
the
same
letter
or
letters
is
contained
,
then
cancel
the
same
in
both
,
and
if
there
be
any
numbers
prefixt
,
if
you
can
discover
any
number
that
will
divide
them
both
without
any
remainder
,
then
prefix
those
quotients
instead
of
the
numbers
prefixed
before
;
so
shall
this
new
Fraction
be
of
the
same
value
with
the
Fraction
proposed
.
So
will
be
reduced
to
,
by
cancelling
cb
in
the
Numerator
and
Denominator
.
Also
being
reduced
to
its
lowest
terms
will
be
by
cancelling
bd
in
every
part
,
and
dividing
the
prefixed
numbers
by
8.
More
Examples
follow
.
Or
if
you
can
(
in
a
Compound
Algebraical
Fraction
)
discover
a
quantity
that
will
divide
the
Numerator
and
Denominator
without
any
remainder
,
(
according
to
the
sixth
Rule
of
the
fifth
Chapter
)
then
shall
the
quotients
be
a
new
Numerator
and
a
new
Denominator
equal
to
the
Fraction
in
its
given
Termes
.
As
in
the
following
Examples
.
To
reduce
an
Integral
quantity
to
an
Algebraical
Fraction
.
IV.
Multiply
the
given
Quantity
by
the
intended
Denominator
,
so
shall
the
Product
be
the
Numerator
required
.
As
in
the
following
Examples
.
Let
it
be
required
to
reduce
the
quantity
b
to
a
Fraction
,
having
ad
for
its
Denominator
.
To
do
which
I
multiply
the
given
quantity
b
by
ad
,
and
the
product
is
the
Numerator
,
viz.
bad
,
so
shall
be
the
Fraction
required
,
for
.
Also
if
it
were
required
to
reduce
the
quantity
e
to
a
Fraction
,
whose
Denominator
should
be
b+c
,
it
would
be
.
V.
If
it
be
required
to
reduce
a
mixt
quantity
to
a
Fraction
,
multiply
the
Integral
quantity
by
the
Denominator
of
the
Fractional
part
,
and
join
the
product
to
the
Numerator
of
the
Fractional
part
,
so
shall
the
sum
be
the
Numerator
.
As
in
Example
.
Reduce
to
an
improper
Fraction
.
Mult
.
the
Integral
part
a+b
by
the
Denominator
d
,
and
the
Product
da+db
which
being
added
to
the
numerator
c
,
makes
da+db+c
for
the
numerator
,
to
which
placing
d
for
a
Denominator
,
it
gives
for
the
answer
.
VI.
When
you
are
to
express
an
Algebraick
Integer
Fraction
wise-without
an
Assigned
Denominator
,
then
make
the
given
quantity
the
Numerator
,
and
1
the
Denominator
.
So
will
ab
be
and
cd
will
be
and
a+b
will
be
,
&c.
These
things
are
so
plain
that
they
need
no
farther
explanation
by
examples
.
CHAP.
VII
.
Of
Addition
and
Subtraction
of
Algebraical
Fractions
.
1.
WHen
the
Fractions
given
to
be
added
together
have
an
equal
or
common
Denominator
,
add
the
Numerators
together
,
and
place
their
sum
for
a
Numerator
over
the
common
Denominator
,
which
new
Fraction
shall
be
the
sum
of
the
given
Fractions
,
but
if
they
have
not
a
common
Denominator
,
reduce
them
by
the
second
Rule
of
the
sixth
Chapter
,
and
then
proceed
as
before
.
Example
1.
What
is
the
sum
of
and
?
Facit
,
the
sum
of
the
Numerators
,
viz.
a
and
ac
is
a+ac
which
placed
over
the
Denominator
b
,
gives
for
the
sum
required
.
Example
2.
So
also
the
sum
of
and
will
be
.
Example
3.
And
the
sum
of
and
will
be
found
to
be
for
the
sum
of
the
Numerators
is
a+b
−
c+a+b+c+2a
−
2b
=
4a
.
Example
4.
What
is
the
sum
of
and
?
Facit
the
given
Fraction
being
reduced
to
a
common
Denominator
by
the
second
Rule
of
the
sixth
Chapter
,
are
and
whose
sum
is
.
Example
.
5.
What
is
the
sum
of
and
?
Facit
.
II.
When
it
is
required
to
gather
mixed
quantities
into
one
sum
,
then
add
the
fractional
parts
together
by
the
foregoing
Rule
,
and
likewise
bring
the
Integers
into
one
sum
,
and
the
sum
of
these
two
sums
will
be
the
sum
required
.
Example
.
What
is
the
sum
of
and
?
The
Sum
of
the
Fractions
being
added
by
the
foregoing
Rule
is
|
|
To
which
sum
if
you
add
the
Integral
parts
of
the
propounded
mixed
Quantities
,
the
sum
required
will
be
|
|
Subtraction
of
Algebraical
Fractions
.
III.
If
the
2
given
Fractions
have
not
a
common
Denominator
,
then
reduce
them
to
such
by
the
second
Rule
of
the
sixth
Chapter
,
then
(
by
the
Rule
of
the
third
Chapter
)
subtract
the
Numerator
of
the
Fraction
to
be
subtracted
from
the
Numerator
of
the
other
Fraction
,
and
place
the
remainder
for
a
Numerator
over
the
common
Denominator
,
which
new
Fraction
shall
be
the
remainder
sought
.
As
in
the
following
Examples
.
If
you
would
subtract
from
,
Take
the
Numerator
ab
from
the
Numerator
bc
,
and
the
remainder
is
bc
−
ab
which
being
placed
over
the
Denominator
e
,
it
will
give
for
the
remainder
,
or
difference
sought
.
Also
let
it
be
required
to
subtract
from
The
Difference
of
the
Numerators
of
the
given
Fractions
is
|
|
Which
remainder
or
difference
being
made
a
Numerator
to
the
common
Denominator
will
give
the
difference
sought
which
is
|
|
And
if
it
were
required
to
subtract
from
.
The
given
mixt
quantities
will
(
by
the
fifth
Rule
of
the
sixth
Chapter
)
be
reduced
to
and
Which
will
be
reduced
to
these
Fractions
of
the
same
value
,
having
a
common
Denominator
,
viz.
|
|
And
if
from
the
Numerator
caae+abe
you
subtract
the
Numerator
cae+ccd
−
ccb
there
will
be
given
for
the
remainder
sought
,
viz.
|
|
In
like
manner
if
from
a
it
be
required
to
subtract
.
First
by
the
fourth
Rule
of
the
sixth
Chapter
,
reduce
the
quantity
a
to
the
improper
Fractional
Quantity
and
therefrom
subtract
the
given
Fraction
so
will
you
have
the
remainder
soughtwhich
is
CHAP.
VIII
.
Multiplication
and
Division
of
Algebraical
Fractions
.
I.
WHen
it
is
required
to
multiply
two
Algebraical
Fractions
the
one
by
the
other
,
the
work
is
the
same
as
in
Vulgar
Fractions
,
for
if
you
multiply
the
Numerators
of
the
given
Fractions
together
,
and
likewise
their
Denominators
together
,
and
place
their
Respective
Products
for
a
new
Numerator
,
and
a
new
Denominator
,
that
new
Fraction
shall
be
the
Product
required
.
Example
1.
What
is
the
Product
of
multiplyed
by
?
Facit
,
for
9a
ab
=
9aab
which
is
the
Numerator
,
and
c
b
=
cb
the
Denominator
.
Example
2.
What
is
the
Product
of
multiplyed
by
?
Facit
;
for
4cd●…
3a+2c
=
12cda+8ccd
which
is
the
Numerator
.
Example
3.
What
is
the
Product
of
multiplyed
by
?
Facit
;
for
and
by
the
Rule
of
the
sixth
Chapter
,
and
which
is
the
Product
required
.
Example
4.
What
is
the
Product
of
ab
multiplyed
by
?
Facit
;
for
and
which
is
the
Product
required
.
II.
If
it
so
chance
that
you
have
a
Fraction
to
be
multiplyed
by
an
Integer
that
is
equal
to
the
Denominator
of
the
Fraction
,
then
take
the
Numerator
for
the
Product
Example
.
What
is
the
Product
of
being
multiplyed
by
a
−
e
?
Facit
aa
−
2ae+ee
.
The
reason
of
which
is
plain
,
for
the
Numerator
being
multiplyed
by
the
Integer
,
and
the
same
Integer
being
put
as
a
Denominator
to
the
Product
,
the
Quotient
arising
by
the
Division
of
the
Numerator
by
the
said
Denominator
,
will
be
equal
to
the
Numerator
of
the
given
Fraction
;
so
.
III.
When
it
is
required
to
divide
one
Algebraical
Fraction
by
another
,
if
they
have
a
common
Denominator
,
cancel
the
Denominator
,
and
divide
the
Numerator
of
the
Dividend
by
that
of
the
Divisor
,
so
shall
that
Quote
be
the
Quotient
sought
.
So
if
it
were
required
to
divide
by
the
Quotient
will
be
found
to
be
,
for
having
cast
away
the
common
Denominator
d
,
and
divided
(
abc
)
the
Numerator
of
the
Dividend
by
(
bcc
)
the
Numerator
of
the
Divisor
,
the
Quotient
will
be
which
is
by
cancelling
bc
in
the
Nemerator
and
Denominator
.
IV.
When
the
given
Algebraical
Fractions
have
not
a
common
denominator
,
then
multiply
the
denominator
of
the
divisor
into
the
Numerator
of
the
dividend
,
and
the
product
is
a
new
numerator
,
also
multiply
the
numerator
of
the
divisor
into
the
denominator
of
the
dividend
,
and
the
product
is
a
new
denominator
;
which
new
Fraction
is
the
quotient
sought
,
and
this
is
a
general
Rule
in
all
cases
whatsoever
,
and
is
the
same
with
division
in
Vulgar
Fractions
,
only
keeping
to
the
Algebraick
Rules
.
Example
.
What
is
the
quotient
of
being
divided
by
?
Facit
▪
for
c×a
=
ca
,
(
the
new
Numerator
,
)
and
a●…b
=
ba
(
the
new
Denominator
.
Likewise
,
If
it
were
required
to
divide
by
the
Quotient
would
be
for
the
numerator
of
the
dividend
is
3a+b
,
and
the
denominator
of
the
divisor
is
a+b
,
and
which
is
the
Numerator
;
and
the
numerator
of
the
divisor
is
2bb
,
and
the
denominator
of
the
dividend
is
c
,
and
2bb×c
=
2bbc
,
which
is
the
Denominator
.
The
like
is
to
be
observed
in
all
cases
both
in
multiplication
,
and
Division
of
Algebraical
Fractions
.
CHAP.
IX
.
The
Rule
of
Three
in
Algebraick
Quantities
.
I.
THE
Rule
of
Three
in
Algebraical
quantities
represented
by
Letters
,
(
whether
it
be
Direct
or
Inverse
)
differs
not
from
the
Rule
of
Three
in
Vulgar
Arithmetick
,
Respect
being
had
to
the
Rules
of
Algebraical
Multiplication
and
Division
,
before
delivered
in
this
Book
,
for
(
in
a
direct
proportion
)
if
you
multiply
the
second
term
by
the
third
,
and
divide
the
Product
thereof
by
the
first
,
the
quotient
will
be
the
fourth
quantity
sought
in
proportion
.
Example
1.
If
b
gives
c
,
what
will
d
give
?
Facit
.
In
this
Example
the
second
and
third
quantities
are
c
and
d
,
which
being
multiplyed
together
,
produce
cd
by
the
Third
Rule
of
the
fourth
Chapter
,
which
being
divided
by
(
b
)
the
first
quantity
,
the
quotient
is
which
is
the
fourth
proportion
sought
for
.
Which
may
be
proved
according
to
the
proof
of
the
Rule
of
Three
Direct
laid
down
in
the
Tenth
Chapter
of
my
Vulgar
Arithmetick
:
For
,
The
Product
of
the
second
and
third
Terms
is
cd
.
And
The
Product
of
the
first
and
fourth
Terms
is
And
by
3d
the
Rule
of
the
6th
Chap.
(
the
first
Term
)
which
was
to
be
proved
.
Example
2.
If
b+c
require
d
,
what
will
d+e
require
?
Facit
.
For
,
Example
3.
If
12
require
36
,
what
will
4ab
require
?
Facit
.
For
,
II.
Nor
will
the
operation
be
different
from
the
former
,
if
any
of
the
3
given
quantities
be
a
Fraction
,
or
if
they
be
all
Fractions
,
observing
the
Rules
of
multiplication
and
Division
in
Algebraick
quantities
;
or
when
any
of
the
given
Terms
is
a
mixt
quantity
,
let
it
be
reduced
to
the
form
of
a
Fraction
,
by
multiplying
the
Integral
part
by
the
denominator
,
and
joyning
the
product
to
the
numerator
of
the
Fractional
part
and
then
multiply
and
divide
as
before
.
Example
4.
If
require
d
,
what
will
require
?
Facit
?
For
if
you
first
reduce
to
the
form
of
a
Fraction
,
it
will
be
and
the
second
Term
d
being
set
Fraction-wise
,
will
be
,
then
if
you
multiply
(
)
the
third
Term
by
(
)
the
second
Term
,
the
Product
will
be
,
which
being
divided
by
(
)
the
first
Term
,
the
Quotient
will
be
,
which
is
the
fourth
proportional
sought
.
For
,
I
shall
not
need
here
to
give
any
Examples
in
the
Inverse
Rule
of
proportion
in
Algebraick
quantities
,
the
manner
of
the
operation
being
the
same
with
the
former
,
only
the
proportion
flows
backward
,
as
in
the
Rule
of
Three
Inverse
in
Vulgar
Arithmetick
.
CHAP.
X.
A
Collection
of
some
easy
Questions
wherein
the
Rules
hitherto
delivered
are
Exercised
,
taken
out
of
Mr.
Oughtred's
Clavis
Mathematicae
,
Chap.
11.
Sir
Ionas
Moors
Arithmetick
in
Species
,
Chap.
10
,
and
Mr.
Kersy's
Elements
of
Algebra
,
Cap.
10.
of
the
First
Book
.
I.
THere
are
two
Quantities
or
numbers
,
whereof
the
greater
is
a
(
=
4
)
and
the
lesser
is
e
(
=
2
)
What
is
their
sum
?
What
their
difference
?
What
the
product
of
their
multiplication
?
What
the
Quotient
of
the
greater
divided
by
the
lesser
?
What
the
Quotient
of
the
lesser
divided
by
the
greater
?
What
the
sum
of
their
Squares
?
What
the
difference
of
their
Squares
?
What
is
the
sum
of
their
sum
,
and
difference
?
What
is
the
difference
of
their
sum
and
differences
?
What
is
the
Product
of
their
sum
and
difference
?
What
the
Square
of
their
sum
?
What
the
Square
of
their
difference
?
What
the
Square
of
their
Product
?
1
The
sum
of
the
quantities
proposed
is
|
|
2
Their
difference
is
|
|
3
Their
Product
by
multiplication
|
|
4
The
quote
of
the
greater
divided
by
the
lesser
.
|
|
5
The
quote
of
the
lesser
by
the
greater
.
|
|
6
The
Sum
of
their
Squares
.
|
|
7
The
difference
of
their
Squares
.
|
|
8
The
sum
of
their
sum
and
difference
.
|
|
9
The
difference
of
their
sum
and
difference
.
|
|
10
The
Product
of
their
sum
and
difference
.
|
|
10
The
Square
of
their
sum
.
|
|
12
The
Square
of
the
difference
.
|
|
13
The
Square
of
their
product
.
|
|
II.
There
are
two
quantities
whose
sum
is
b
(
=
12
)
and
the
greater
of
them
is
a
(
=
8.
)
I
demand
what
is
the
lesser
?
What
their
difference
?
What
is
the
product
of
their
multiplication
?
What
is
the
sum
of
their
Squares
?
What
the
difference
of
their
Squares
?
1
The
lesser
is
|
|
2
Their
difference
is
|
|
3
The
Product
is
|
|
4
The
sum
of
their
Squares
is
|
|
5
The
difference
of
their
Squares
is
|
|
III.
There
are
two
Quantities
or
Numbers
whose
difference
is
d
,
(
=
4
)
and
the
greater
of
them
is
a
(
=
8
)
I
demand
what
is
the
lesser
?
What
is
their
sum
?
What
their
Rectangle
or
Product
?
What
the
sum
of
their
Squares
?
What
the
difference
of
their
Squares
?
1
The
difference
or
excess
being
subtracted
from
the
greater
gives
the
lesser
|
|
2
Their
sum
is
|
|
3
Their
Product
or
Rectangle
is
|
|
4
The
sum
of
their
Squares
is
|
|
5
The
difference
of
their
Squares
is
|
|
IV.
There
are
two
Numbers
,
Magnitudes
,
or
Quantities
;
whereof
the
Ratio
of
the
greater
to
the
lesser
is
as
r
to
s
,
(
or
as
3
to
2
)
and
the
greater
of
them
is
a
(
=
12.
)
I
demand
what
is
the
lesser
?
What
is
their
Sum
?
What
their
difference
?
What
their
Rectangle
,
or
Product
?
What
the
sum
of
their
Squares
?
And
what
the
difference
of
their
Squares
?
1
The
lesser
is
by
the
Rule
of
3
|
|
2
Their
sum
is
|
|
3
Their
difference
is
|
|
4
Their
Rectangle
or
Product
is
|
|
5
The
sum
of
their
Squares
is
|
|
6
The
difference
of
their
Squares
is
|
|
But
if
the
Ratio
between
the
lesser
and
the
greater
had
been
given
as
s
to
r
,
(
or
as
2
to
3
)
and
the
lesser
had
been
given
e
(
=
8
)
then
,
1
The
greater
by
the
Rule
of
3
would
be
|
|
2
Their
sum
|
|
3
Their
difference
|
|
4
Their
Rectangle
,
or
Product
|
|
5
The
sum
of
their
Squares
|
|
6
The
difference
of
their
Squares
,
|
|
V.
There
are
two
numbers
or
Quantities
whereof
the
Rectangle
or
Product
is
b
(
=
96
)
and
the
greater
quantity
is
a
(
=
12
)
What
is
the
lesser
?
What
their
sum
?
What
their
difference
?
What
the
sum
of
their
Squares
?
And
what
the
differencé
of
their
Squares
?
1
The
Product
given
being
divided
by
(
a
)
the
lesser
quantity
is
|
|
2
Their
sum
|
|
3
Their
difference
|
|
4
The
sum
of
their
Squares
|
|
5
The
difference
of
their
Squares
.
|
|
But
if
the
Rectangle
had
been
given
b
,
as
before
,
and
the
lesser
quantity
had
been
given
e
(
=
8
)
Then
1
The
greater
would
have
been
found
by
Division
to
be
|
|
2
Their
Sum
|
|
3
Their
difference
|
|
4
The
Sum
of
their
Squares
.
|
|
5
The
difference
of
their
Squares
.
|
|
CHAP.
XI
.
Reduction
of
Equations
.
I.
AN
Equation
is
an
equality
between
two
quantities
of
different
names
,
whether
the
comparison
of
Equality
be
between
Simple
,
or
Compound
Quantities
,
or
both
;
between
which
two
Quantities
there
is
always
this
Charactor
viz.
=
.
So
〈◊〉
●…his
following
Equation
,
viz.
a
=
3c
,
a
is
said
to
be
the
first
part
,
and
3c
the
second
part
of
the
Equation
,
and
signifieth
that
some
Number
or
Quantity
represented
by
a
is
equal
to
three
times
another
Number
or
Quantity
represented
by
c.
So
a
=
b+c
signifieth
that
some
Quantity
represented
by
a
is
equal
to
the
sum
of
two
other
Numbers
or
quantities
represented
by
b
and
c.
The
manner
of
composing
an
Equation
will
be
understood
by
solving
of
the
several
Questions
contained
in
this
and
other
following
Chap.
But
when
known
,
are
mingled
with
unknown
Quantities
,
in
an
equation
they
must
be
so
separated
or
reduced
that
the
unknown
Quantity
or
Quantities
may
remain
intire
on
the
one
side
,
or
part
,
and
the
known
or
given
Quantities
on
the
other
side
or
part
of
the
Equation
,
which
to
performe
is
the
work
of
Reduction
,
and
which
is
contained
in
the
several
following
Rules
of
this
Chap.
Here
note
that
the
Quantity
unknown
or
sought
in
every
Equation
is
represented
by
the
Letter
a
,
or
some
other
Vowel
,
and
the
Quantity
or
quantities
known
or
given
are
represented
by
Consonants
,
as
b
,
c
,
d
,
f
,
&c.
Reduction
by
Addition
.
II.
If
equal
numbers
or
quantities
be
added
to
equal
numbers
or
quantities
,
the
sums
or
totals
will
be
equal
,
and
therefore
If
it
be
granted
that
|
|
Then
by
adding
+8
to
each
part
of
the
Equation
there
ariseth
|
|
Then
because
in
the
first
part
of
the
Equation
there
is
+8
and
−
8
,
they
destroy
each
other
by
the
Third
Rule
of
the
Second
Chap.
and
it
followeth
that
|
|
Again
let
this
Equation
be
proposed
to
be
reduced
,
viz.
|
|
Then
by
adding
b
to
each
part
of
the
Equation
,
there
ariseth
|
|
And
because
−
b
and
+b
are
in
the
first
part
of
the
equation
,
they
destroy
each
other
,
and
the
Equation
is
|
|
Likewise
if
|
|
Then
by
adding
b+c
to
each
part
of
the
equation
there
ariseth
|
|
Now
from
a
due
consideration
of
the
premises
it
followeth
,
that
if
in
an
Equation
there
be
any
Number
or
Quantity
proposed
with
the
sign
−
before
it
,
then
if
it
be
transferred
to
the
other
side
of
the
equation
,
and
cancelled
on
the
side
or
part
where
it
now
standeth
,
the
effect
will
be
the
same
as
the
adding
of
that
Quantity
to
each
part
of
the
Equation
,
and
this
by
Artists
is
called
Transposition
.
As
in
the
first
of
the
foregoing
Examples
,
where
it
is
granted
That
|
|
And
by
transposing
−
8
on
the
other
side
of
the
Equation
,
making
it
there
+8
it
giveth
|
|
And
in
the
second
Example
where
|
|
By
transposing
−
b
,
cancelling
it
on
the
first
side
of
the
equation
,
and
making
it
+b
on
the
other
,
it
is
|
|
And
let
it
be
granted
that
|
|
Then
by
transposing
of
−
bb
and
−
d
there
ariseth
|
|
Reduction
by
Subtraction
.
III.
If
in
any
Equation
there
be
any
number
or
quantity
signed
with
+
(
on
which
side
of
the
equation
soever
)
if
it
be
cancelled
on
that
side
,
and
placed
on
the
other
side
with
the
sign
−
prefixed
to
it
,
the
work
of
Reduction
is
truly
performed
,
and
this
is
also
called
Transposition
,
and
is
only
the
converse
of
the
foregoing
Rule
.
Examples
.
Let
it
be
granted
that
|
|
Then
if
+8
be
cancelled
,
and
placed
on
the
other
part
of
the
equation
with
the
sign
−
it
will
give
|
|
Which
equation
being
contracted
is
|
|
Again
let
be
given
|
|
By
the
Transposition
of
+b
on
the
first
side
the
Equation
it
is
|
|
And
by
Transposition
of
aa
on
the
second
side
of
the
equation
it
is
|
|
Also
if
|
|
By
Transposition
of
b+c
to
the
second
side
of
the
equation
it
is
|
|
And
by
the
Transposition
of
ba
to
the
first
side
of
the
equation
it
is
|
|
Which
method
(
in
reducing
of
the
premised
Equation
)
is
deduced
from
this
general
Axiom
,
viz.
If
from
equal
Numbers
or
Quantities
,
equal
Numbers
or
Quantities
are
subtracted
,
the
remainder
shall
be
equal
.
So
in
the
second
Example
there
is
given
this
equation
,
viz.
|
|
First
by
subtracting
b
from
each
part
of
the
Equation
,
there
is
|
|
Then
I
subtract
aa
from
each
part
,
and
there
remaineth
|
|
Reduction
by
Multiplication
.
IV.
When
in
an
Equation
one
or
both
parts
are
Fractions
,
then
let
them
be
reduced
to
a
common
denominator
by
the
,
2d
,
4th
,
and
5th
Rules
of
the
sixth
Chapter
,
and
then
casting
away
the
Denominator
,
use
only
the
Numerators
,
so
shall
Equations
exprest
by
Algebraical
Fractions
be
reduced
to
other
Equations
,
consisting
altogether
of
Integers
.
As
in
the
following
Examples
.
If
|
|
Then
by
reducing
9
in
the
second
part
of
the
equation
to
a
Fraction
,
having
8
for
its
denominator
,
it
is
|
|
And
by
casting
away
the
denominator
which
is
common
to
both
it
is
|
|
Again
if
|
|
Then
by
reducing
a
,
on
the
first
side
of
the
equation
to
to
a
Fraction
haveing
a+b
for
its
denominator
it
is
|
|
And
by
casting
away
the
common
denominator
a+b
the
equation
is
|
|
Likewise
if
|
|
The
quantities
being
reduced
to
a
common
denominator
are
|
|
And
the
common
denominator
cb
being
cast
away
the
equation
is
|
|
V.
When
either
part
of
an
Equation
is
Composed
of
a
mixed
Quantity
or
Quantities
,
let
the
Integral
part
,
or
parts
be
reduced
to
a
Fraction
or
Fractions
,
and
then
proceed
as
in
the
last
Example
.
It
is
granted
that
|
|
First
it
is
reduced
to
|
|
Which
Fractional
equation
being
reduced
according
to
the
foregoing
Rule
is
|
|
VI.
When
some
power
or
degree
of
the
number
or
quantity
sought
,
is
multiplyed
into
each
part
,
and
each
member
of
an
Equation
,
then
let
that
degree
or
power
be
cancelled
in
each
part
and
member
so
will
it
quite
vanish
,
and
the
Equation
will
be
reduced
to
more
Simple
Terms
.
As
for
Example
.
Let
it
be
granted
that
|
|
Forasmuch
as
a
is
a
Factor
in
each
part
,
and
member
of
the
equation
,
therefore
it
being
expunged
in
each
,
there
ariseth
this
equation
.
|
|
VII
.
When
(
according
to
the
second
,
third
,
fourth
,
and
fifth
Rules
,
)
an
Equation
is
reduced
,
and
that
some
known
Number
or
Quantity
is
multiplyed
into
the
quantity
sought
;
then
divide
each
part
of
the
Equation
by
that
known
Quantity
,
to
the
end
that
the
Quantity
sought
may
have
no
quantity
multiplyed
into
it
but
1
(
or
unity
.
)
As
in
Example
.
If
it
be
granted
that
|
|
Then
because
the
Quantity
sought
is
(
a
)
multiplyed
by
b
,
I
divide
each
part
of
the
equation
by
b
,
and
there
ariseth
|
|
VIII
.
When
any
one
part
of
an
Equation
is
composed
of
a
surd
Quantity
,
(
viz.
such
as
hath
the
radical
sign
√
prefixed
to
it
,
)
and
the
other
part
is
a
rational
Quantity
;
then
let
that
rational
quantity
be
raised
to
the
power
signified
by
the
Radical
sign
,
and
then
cast
away
the
said
Radical
sign
,
so
shall
both
parts
of
the
Equation
be
a
rational
quantity
.
As
,
If
it
be
proposed
that
|
|
Square
8
,
and
place
its
Square
in
the
room
of
it self
,
casting
away
the
radical
sign
from
the
first
part
of
the
equation
,
and
then
it
will
be
|
|
Likewise
if
|
|
Then
by
raising
the
second
part
of
the
equation
,
to
its
Square
,
and
casting
away
the
radical
sign
from
the
first
part
there
ariseth
this
equation
,
viz.
|
|
Again
,
if
|
|
The
second
part
of
the
equation
being
squared
,
and
the
radical
sign
cancelled
in
the
first
,
there
ariseth
|
|
Reduction
by
Division
.
IX
.
If
equal
Quantities
be
divided
by
equal
Quantities
,
the
Quotients
thence
arising
will
be
equal
.
For
,
If
|
|
Then
by
dividing
each
part
of
the
equation
by
a
,
there
ariseth
this
equation
.
|
|
And
if
|
|
Then
by
dividing
each
part
of
the
equation
by
a
there
ariseth
|
|
And
da
in
the
second
part
of
the
equation
being
transposed
by
the
third
Rule
of
this
Chapter
there
ari
seth
this
equation
,
viz.
|
|
And
if
|
|
Then
by
dividing
each
part
of
the
equation
by
b
−
c
,
it
is
|
|
CHAP.
XII
.
To
Convert
Analogies
into
Equations
,
and
Equations
into
Analogies
.
I.
THis
is
deduced
from
this
universal
Theorem
,
viz.
That
if
four
quantities
are
proportionals
,
the
product
of
the
two
means
is
equal
to
the
product
of
the
two
Extreams
,
and
if
three
numbers
are
proportionals
,
the
product
of
the
two
extreams
is
equal
to
the
Square
of
the
means
.
1.
Let
there
be
proposed
these
four
proportionals
.
|
|
Then
by
the
said
Theorem
this
equation
will
follow
,
viz.
|
|
2
Let
there
be
proposed
these
3
continual
proportionals
,
viz.
|
|
That
is
to
say
|
|
whence
there
followeth
this
equation
,
viz.
|
|
II.
From
a
due
consideration
of
the
premises
it
is
evident
that
Equations
may
oftentimes
be
resolved
into
proportionals
,
viz.
when
the
Product
of
two
Quantities
is
found
equal
to
the
product
of
two
other
quantities
:
Then
as
any
one
of
the
Factors
in
the
first
side
of
the
equation
is
to
any
one
of
the
Factors
in
the
second
part
of
the
Equation
,
so
is
the
remaining
Factor
of
the
second
part
,
to
the
remaining
Factor
in
the
first
part
:
And
the
Converse
,
Suppose
that
|
|
From
thence
may
be
drawn
this
Analogie
.
|
|
The
truth
of
which
may
be
proved
by
the
first
Rule
of
this
Chapter
,
for
thereby
the
said
Analogy
may
be
reduced
to
the
given
equation
,
viz.
bc
=
ad
.
Again
if
|
|
Then
from
thence
may
be
deduced
this
Analogy
,
viz.
|
|
or
|
|
or
|
|
Likewise
if
|
|
Then
may
that
equation
be
resolved
into
these
proportionals
.
|
|
And
if
|
|
Then
it
will
be
found
that
|
|
III.
When
it
happens
that
there
is
an
Equation
between
an
Algebraical
Fraction
,
and
an
Integer
,
if
the
Numerator
of
the
said
Fraction
can
be
resolved
into
two
such
Quantities
,
as
being
multiplyed
the
one
by
the
other
,
will
produce
the
said
Numerator
,
then
will
the
said
equation
produce
this
proportion
,
viz.
As
the
Denominator
of
the
Fraction
is
to
one
of
the
Factors
of
which
the
Numerator
is
produced
,
so
is
the
other
Factor
to
the
Integer
,
unto
which
the
said
Fraction
is
equal
.
Examples
.
If
it
be
granted
that
|
|
Then
may
that
equation
be
resolved
into
this
Analogy
.
|
|
For
,
|
|
Again
if
|
|
Then
may
that
equation
be
resolved
into
this
Analogy
.
|
|
And
also
if
.
|
|
Then
may
the
said
equation
be
resolved
into
this
Analogy
,
viz.
|
|
The
Practice
of
the
two
last
Rules
will
be
plainly
discovered
in
the
next
Chapter
(
in
the
resolution
of
Questions
producing
simple
equations
)
to
be
of
most
excellent
use
in
discovering
or
laying
down
of
Theorems
for
the
ready
solution
of
the
Question
proposed
,
or
any
other
of
the
same
nature
,
which
Theorems
are
to
be
reserved
in
store
for
the
finding
out
of
new
,
and
the
confirmation
of
old
Truths
.
CHAP.
XIII
.
The
Resolution
of
Arithmetical
Questions
(
Algebraically
)
which
produce
Simple
Equations
I.
AN
Equation
is
two-fold
,
viz.
First
,
Simple
,
and
secondly
,
Adfected
,
or
Compounded
.
II.
A
Simple
Equation
is
when
the
Quantity
sought
(
solely
possessing
one
part
of
the
Equation
)
is
either
expressed
by
a
Single
or
Simple
Root
,
as
a
,
or
by
a
Single
or
Simple
Power
as
aa
,
or
aaa
,
&c.
as
in
these
Equations
,
viz.
a
=
32
,
and
aa
=
64
,
or
4aaa
=
256
,
and
such
like
.
III.
When
a
Question
is
propounded
,
and
to
be
resolved
Algebraically
,
Then
for
the
Answer
put
a
,
and
for
each
of
the
given
Numbers
put
Consonants
,
then
proceed
according
to
the
Tenure
of
the
Question
,
by
Addition
,
Subtraction
,
Multiplication
,
or
Division
,
until
an
Equation
is
Composed
;
and
when
the
Equation
is
composed
,
then
proceed
to
reduce
it
,
(
according
to
the
Rules
contained
in
the
11th
Chapter
)
until
the
Quantity
unknown
(
being
a
or
some
power
of
a
)
do
solely
possess
one
part
of
the
Equation
,
and
the
known
or
given
Quantities
the
other
part
,
and
then
will
the
Quantity
sought
be
also
known
.
IV.
I
shall
in
the
Resolution
of
every
Question
proceed
(
gradatim
)
step
by
step
,
according
to
the
method
used
by
Mr.
Kersey
,
each
step
being
numbred
orderly
in
the
Margent
,
from
the
beginning
to
the
end
by
1
,
2
,
3
,
4
,
&c.
And
I
shall
only
proceed
in
the
operation
literally
,
because
otherwise
this
Treatise
would
swell
to
a
bigger
Volume
than
is
at
present
intended
;
but
I
shall
give
the
Learner
a
taste
of
Numeral
Algebra
,
in
the
solution
of
two
or
three
of
the
first
Questions
thereby
.
Quest.
1.
There
are
two
numbers
whose
sum
is
48
(
or
b
)
and
the
excess
of
the
greater
above
the
lesser
is
14
(
or
c
)
I
demand
what
are
the
Numbers
.
The
Solution
literally
.
1
For
the
greater
number
put
|
|
2
From
which
if
you
subtract
the
difference
(
c
)
you
will
have
the
lesser
,
which
is
|
|
3
The
greater
and
lesser
being
added
together
,
will
be
equal
to
(
b
)
the
sum
whence
this
equation
|
|
4
And
by
the
Transposition
of
−
c
the
equation
is
|
|
5
Then
dividing
each
part
of
the
equation
by
2
,
it
is
|
|
And
if
from
you
subtract
(
c
)
the
excess
of
the
greater
above
the
lesser
,
the
lesser
will
be
|
|
So
that
the
numbers
sought
are
31
and
17
,
for
by
the
fifth
step
(
a
)
the
greater
is
found
to
be
=
to
and
b
is
given
48
,
and
c
is
given
14
,
the
sum
of
which
is
62
,
which
divided
by
2
,
gives
3●…
,
for
the
greater
,
and
by
the
sixth
step
,
if
from
the
greater
you
subtract
the
difference
(
c
)
the
remainder
will
give
the
lesser
,
which
is
17
,
for
.
Now
if
the
fifth
and
sixth
steps
are
duly
considered
,
they
will
present
you
with
this
Theorem
.
The
sum
of
the
sum
and
difference
of
any
two
Numbers
being
divided
by
2
,
will
give
the
greater
Number
,
and
the
difference
of
any
two
Numbers
being
subtracted
from
half
the
sum
of
the
sum
and
difference
,
the
remainder
will
give
the
lesser
number
.
The
Solution
Numerally
.
1
For
the
greater
number
put
|
|
2
From
which
if
you
subtract
the
difference
(
14
)
the
lesser
is
|
|
3
Which
added
together
,
will
be
the
sum
,
whence
this
Equation
.
|
|
4
And
by
transposition
of
of
14
it
will
be
|
|
5
And
both
parts
of
the
Equation
being
divided
by
2
,
will
give
the
value
of
(
a
)
the
greater
.
|
|
6
From
which
if
you
subtract
(
14
)
the
Difference
,
the
remainder
will
give
the
lesser
by
the
2
step
.
|
|
So
that
the
Numbers
sought
are
31
and
17
,
which
will
satisfie
the
conditions
of
the
Question
.
Quest.
2.
There
are
two
Numbers
whose
Sum
is
56
(
or
b
)
and
the
lesser
hath
such
proportion
to
the
greater
,
as
2
to
5
,
(
or
c
to
d
)
I
demand
what
are
the
Numbers
?
1
For
the
lesser
number
put
|
|
2
Then
by
the
Rule
of
Three
find
the
greater
,
viz.
|
|
3
Wherefore
the
sum
of
the
2
numbers
sought
is
|
|
4
Which
sum
must
be
equal
to
the
given
sum
,
whence
this
equation
.
|
|
5
Which
equation
being
reduced
by
the
fourth
and
fifth
Rules
of
the
eleventh
Chap.
the
value
of
a
will
be
found
to
be
|
|
6
And
by
the
first
,
second
,
and
fifth
steps
the
greater
number
will
be
discovered
to
be
|
|
So
that
the
numbers
sought
are
40
and
16
for
(
a
)
the
lesser
is
found
to
be
by
the
fifth
step
viz.
the
Product
of
(
cb
)
56
by
2
divided
by
(
c+d
)
the
sum
of
2
and
5
,
viz.
7
,
which
is
16
,
&c.
And
if
(
according
to
the
third
Rule
of
the
twelfth
Chap.
)
the
two
last
steps
be
turned
into
proportionals
,
it
will
give
this
Theorem
.
As
the
sum
of
the
Terms
which
represent
the
ratio
of
two
Numbers
,
is
to
the
sum
of
the
numbers
themselves
,
so
is
the
lesser
term
to
the
lesser
number
;
and
so
is
the
greater
Term
to
the
greater
Number
.
Therefore
if
the
sum
of
two
Numbers
is
given
,
and
also
their
Ratio
,
the
Numbers
themselves
are
also
given
by
this
Theorem
.
The
same
Question
solved
Numerally
.
1
For
the
lesser
number
put
|
|
2
Then
by
the
Rule
of
Three
the
greater
number
is
found
|
|
3
Then
will
their
sum
be
|
|
4
And
according
to
the
tenure
of
the
Question
,
their
sum
must
be
equal
to
the
given
sum
whence
this
Equation
|
|
5
And
that
Equation
being
reduced
by
the
fifth
and
sixth
Rules
of
the
eleventh
Chap.
the
value
of
a
will
be
found
to
be
|
|
6
Which
being
subtracted
from
the
given
sum
,
the
greater
number
is
|
|
Quest.
3.
A
Gentleman
asked
his
Friend
(
that
had
4
purses
in
his
hand
)
what
money
he
had
in
each
Purse
?
To
whom
he
answered
that
he
knew
not
but
(
quoth
he
)
this
I
know
,
that
in
the
second
purse
there
are
8
(
or
b
)
Crowns
more
than
in
the
first
or
least
purse
,
and
in
the
third
8
Crowns
more
than
the
second
,
and
in
the
fourth
or
biggest
purse
there
are
8
Crowns
more
than
in
the
third
,
and
twice
as
many
as
in
the
first
or
least
,
I
demand
what
number
of
Crowns
he
had
in
each
purse
?
1
For
the
number
of
Crowns
in
the
first
purse
put
|
|
2
Then
in
the
second
there
is
|
|
3
And
in
the
third
there
is
|
|
4
And
in
the
fourth
|
|
5
Which
according
to
the
Tenure
of
the
Question
is
double
to
that
in
the
first
,
whence
this
Equation
|
|
6
Then
by
the
Transposition
of
a
from
the
first
side
of
the
equation
,
it
is
|
|
which
discovereth
the
value
of
a
to
be
3b
,
or
3
times
8
,
which
is
24
,
&c.
which
is
the
number
of
Crowns
in
the
first
purse
,
and
consequently
the
number
of
Crowns
in
each
purse
is
24
,
32
,
40
,
and
48
,
which
will
satisfie
the
conditions
of
the
Question
.
The
same
Question
solved
Numerally
.
1
For
the
Crowns
in
the
first
purse
put
|
|
2
Then
in
the
second
there
is
|
|
3
And
in
the
third
|
|
4
And
in
the
fourth
|
|
5
Which
is
double
to
the
number
of
Crowns
in
the
first
,
whence
this
equation
|
|
6
Which
equation
being
reduced
by
the
transposition
of
a
discovers
the
value
of
a
,
viz.
|
|
Quest.
4.
Three
men
build
a
Ship
which
cost
them
2700l
.
(
or
b
)
Pounds
,
of
which
B
must
pay
double
to
what
A
must
pay
,
and
C
must
pay
three
times
as
much
as
B
,
I
demand
the
share
that
each
must
pay
?
1
For
the
sum
to
be
paid
by
A
,
put
|
|
2
Then
B
must
pay
|
|
3
And
C
must
pay
|
|
4
The
sum
of
these
3
quantities
are
equal
to
the
total
charge
,
whence
this
Equation
|
|
5
Which
being
reduced
,
discovers
the
value
of
a
,
viz.
|
|
which
is
the
sume
that
A
must
pay
,
viz.
300l
.
Therefore
B
must
pay
.
which
is
twice
as
much
as
A
,
and
C
must
pay
.
which
is
three
times
as
much
as
B.
Quest.
5.
There
is
a
Fish
whose
head
is
supposed
to
be
9
(
or
b
)
Inches
,
and
his
Tail
is
as
long
as
his
Head
and
half
his
Body
,
and
his
Body
is
as
long
as
his
Head
and
his
Tail
,
I
demand
the
length
of
such
a
Fish
?
1
For
the
length
of
the
body
put
|
|
2
Then
will
the
Tail
be
|
|
3
Then
if
to
the
Tail
you
add
the
length
of
the
head
,
viz.
b
,
the
sum
is
|
|
4
Which
according
to
the
tenure
of
the
Question
is
equal
to
the
length
of
the
body
,
whence
this
equation
|
|
5
And
the
second
part
of
the
equation
being
cleared
of
the
unknown
quantity
a
by
Reduction
,
gives
the
value
of
a
the
length
of
the
body
,
viz.
|
|
6
Then
according
to
the
Tenure
of
the
question
,
if
therefrom
you
subtract
(
b
)
the
length
of
the
head
,
the
remainder
will
be
the
length
of
the
Tail
,
which
is
|
|
By
the
fifth
step
the
length
of
the
Body
is
found
to
be
4b
=
36
,
And
by
the
sixth
step
the
length
of
the
Tail
is
discovered
to
be
3b
=
3×9
=
27.
So
that
the
length
of
the
head
is
(
given
)
9
inches
,
the
length
of
the
Tail
27
Inches
,
and
the
length
of
the
Body
36
Inches
,
which
numbers
will
satisfie
the
conditions
of
the
question
,
So
that
the
whole
length
of
the
Fish
is
9+27+36
=
72
Inches
.
Quest.
6.
A
Father
lying
at
the
point
of
Death
,
left
to
his
3
Sons
A
,
B
,
and
C
all
his
Estate
in
Money
,
and
divided
it
thus
,
viz.
to
A
he
gave
½
,
wanting
44
(
or
b
)
pounds
,
and
to
B
he
gave
1/3
and
14
,
(
or
c
)
pounds
over
,
and
to
C
he
gave
the
rest
,
which
was
82
(
or
d
)
pounds
less
than
the
share
of
B
,
Now
I
demand
what
was
the
Fathers
Estate
?
1
For
the
Fathers
Estate
put
|
|
2
Then
will
the
share
left
to
A
be
|
|
3
And
the
share
of
B
|
|
4
And
by
the
third
step
the
share
of
C
is
|
|
5
The
Quantities
in
the
3
last
steps
being
added
together
,
give
|
|
6
Which
must
be
equal
to
the
Fathers
Estate
,
whence
this
Equation
.
|
|
Which
equation
after
due
reduction
and
Transposition
of
quantities
the
value
of
a
is
discovered
to
be
|
|
And
6b
=
6×44
=
264
,
and
6d
=
6×82
=
492
,
and
12c
=
12×14
=
168
,
now
264+492
−
168
=
588
,
so
that
the
Fathers
estate
was
588
pounds
,
of
which
A
had
250
l.
B
210
l
,
and
C
128
,
which
Numbers
do
answer
the
conditions
of
the
Question
.
Quest.
7.
Two
persons
thus
discoursed
together
concerning
their
Money
,
quoth
A
to
B
,
give
me
3
(
or
b
)
of
your
Crowns
,
and
I
shall
have
as
many
as
you
;
nay
quoth
B
to
A
,
but
if
you
will
give
me
3
of
your
Crowns
,
I
shall
have
5
times
as
many
as
you
.
Now
I
demand
how
many
Crowns
had
each
person
?
1
For
the
number
of
Crowns
which
A
had
put
|
|
2
Then
forasmuch
as
adding
3
(
or
b
)
Crowns
to
A
will
be
equal
to
the
Crowns
remaining
to
B
after
he
had
given
3
Crowns
to
A
,
therefore
B
will
then
have
left
|
|
3
And
consequently
if
you
add
thereto
the
3
(
or
b
)
Crowns
which
he
gave
to
A
the
sum
will
be
the
number
of
Crowns
which
B
had
at
first
,
which
is
|
|
4
Then
if
from
the
number
of
Crowns
A
had
at
first
(
a
)
you
subtract
3
(
or
b
)
crowns
,
there
will
remain
to
A
a
−
b
crowns
,
and
giving
the
same
to
B
,
he
will
then
have
|
|
5
Which
according
to
the
tenure
of
the
Question
is
five
times
as
much
as
what
A
had
left
,
whence
there
ariseth
this
Equation
.
|
|
6
Which
equation
being
reduced
by
the
second
and
seventh
Rules
of
the
eleventh
Chapter
,
the
value
of
a
is
discovered
to
be
|
|
7
And
by
the
sixth
and
third
steps
the
number
of
Crowns
which
B
had
at
first
are
found
to
be
|
|
So
that
it
is
found
that
A
had
6
Crowns
,
and
B
had
12
Crowns
,
which
numbers
will
satisfie
the
conditions
of
the
Question
.
For
,
Quest.
8.
A
Labourer
had
576
(
or
b
)
pence
for
threshing
60
(
or
c
)
Quarters
of
Corn
,
viz.
Wheat
and
Barly
;
for
the
wheat
he
had
12
(
or
d
)
pence
per
Quarter
,
and
for
the
Barly
he
had
6
(
or
f
)
pence
per
Quarter
,
I
demand
how
many
Quarters
of
each
he
threshed
?
1
For
the
quarters
of
wheat
which
he
threshed
put
|
|
2
Then
the
quarters
of
Barley
will
be
|
|
3
The
quantity
of
Wheat
in
the
first
step
being
multiplyed
by
its
price
produceth
|
|
4
The
quantity
of
Barley
in
the
second
step
being
multiplyed
by
its
price
,
produceth
|
|
5
The
sum
of
the
quantities
in
the
two
last
steps
must
be
equal
to
the
given
price
of
the
60
quarters
,
whence
this
equation
|
|
6
Which
being
reduced
by
the
second
,
third
,
and
fifth
Rules
of
the
eleventh
Chap.
the
quantity
of
Wheat
will
be
discovered
to
be
|
|
7
And
by
the
second
and
fifth
steps
the
quantity
of
Barly
is
discovered
to
be
|
|
So
that
the
quarters
of
Wheat
which
he
threshed
were
36
,
and
the
quarter
of
Barly
24.
The
Proof
.
Quest.
9.
A
Gentleman
bought
a
Cloak
of
a
Sales-man
,
which
cost
him
3
l.
−
10
s.
or
70
(
or
b
)
shillings
,
and
desiring
the
Sales-man
to
tell
him
what
he
gained
thereby
,
he
said
he
gained
¼
(
or
c
)
of
what
it
cost
him
,
the
question
is
what
the
Cloak
cost
the
first
penny
?
1
Suppose
the
Cloak
cost
|
|
2
Then
he
gained
|
|
3
The
first
and
Second
steps
being
added
together
,
their
sum
will
be
equal
to
the
sum
which
the
Gentleman
gave
for
it
,
whence
this
equation
|
|
4
Which
Equation
being
reduced
by
the
ninth
Rule
of
the
eleventh
Chap.
the
value
of
a
will
be
discovered
to
be
|
|
So
that
it
cost
56
shillings
,
¼
of
which
is
14
shillings
,
and
56+14
=
70.
And
if
the
quantity
in
the
fourth
step
be
duly
considered
,
you
will
find
that
if
the
gain
had
been
any
other
part
or
parts
of
the
first
cost
,
if
the
price
it
was
sold
for
had
been
divided
by
the
Fraction
representing
the
part
of
gain
,
increased
by
1
,
the
quote
would
have
been
the
answer
.
Quest.
10.
A
Gentleman
hired
a
Labourer
to
work
for
him
for
40
(
or
b
)
days
,
and
made
this
agreement
with
him
that
for
every
day
he
wrought
he
should
have
20
(
or
c
)
pence
,
and
for
every
day
that
he
played
he
should
forfeit
8
(
or
d
)
pence
,
and
at
the
end
of
the
said
40
dayes
he
received
184
(
or
f
)
pence
,
which
was
his
full
due
.
Now
I
demand
how
many
dayes
he
wrought
,
and
how
many
dayes
he
played
?
1
For
the
number
of
days
he
wrought
,
put
|
|
2
Then
the
number
of
days
he
played
will
be
|
|
3
And
if
the
time
he
wrought
(
in
the
first
step
)
be
multiplyed
by
20
(
c
)
it
will
produce
the
total
he
gained
by
work
,
viz.
|
|
4
And
if
the
time
he
played
(
in
the
2d
step
)
be
drawn
into
8
(
d
)
the
product
will
be
what
he
lost
by
play
|
|
5
And
if
the
total
loss
(
in
the
fourth
step
)
be
subtracted
from
the
gain
(
in
the
third
step
)
the
remainder
will
be
what
he
received
,
whence
this
Equation
|
|
6
Which
being
reduced
by
the
second
and
ninth
Rules
of
the
eleventh
Chapter
,
it
will
discover
the
value
of
a
to
be
eighteen
which
is
the
days
that
he
wrought
.
|
|
7
And
from
the
sixth
and
second
steps
the
number
of
days
he
played
are
discovered
to
be
22
dayes
,
viz.
|
|
So
that
by
the
sixth
step
it
appears
he
wrought
18
dayes
,
and
by
the
seventh
step
it
appears
that
he
played
22
dayes
.
The
proof
.
Quest.
11.
A
person
(
in
the
Afternoon
)
being
asked
what
a
Clock
it
was
,
answered
that
3/5
(
or
b
)
parts
of
the
time
from
Noon
was
equal
to
5/8
(
or
c
)
parts
of
the
time
remaining
to
midnight
,
now
,
(
supposing
the
time
from
Noon
to
Midnight
to
be
divided
in
12
(
or
d
)
equal
parts
or
hours
)
I
demand
what
was
the
present
hour
of
the
day
?
1
For
the
hour
sought
put
|
|
2
Then
the
time
to
midnight
will
be
|
|
3
Then
will
3/5
(
or
b
)
parts
of
the
hour
from
Noon
be
|
|
4
And
5/8
(
or
c
)
parts
of
the
time
remaining
till
midnight
will
be
|
|
5
Therefore
from
the
third
and
fourth
steps
there
ariseth
this
equation
.
|
|
6
Which
equation
being
reduced
according
to
the
second
and
ninth
Rules
of
the
eleventh
Chap.
gives
the
value
of
a
(
to
be
the
hour
sought
)
viz.
|
|
So
that
the
hour
sought
was
,
and
consequently
the
time
remaining
till
midnight
was
hours
,
which
two
numbers
will
answer
the
conditions
of
the
question
,
for
,
3/5
parts
of
,
which
is
is
equal
to
5/8
parts
of
,
as
you
may
prove
at
your
leisure
.
Moreover
,
If
the
last
step
be
converted
into
proportionals
by
the
third
Rule
of
the
twelfth
Chap.
it
will
give
this
Theorem
.
As
the
sum
of
the
parts
of
any
two
Numbers
(
wherein
there
is
an
equality
)
is
to
the
sum
of
those
Numbers
,
so
is
the
given
parts
of
any
one
of
those
Numbers
,
to
the
other
Number
.
As
suppose
it
were
required
to
find
out
two
Numbers
,
whose
sum
is
27
,
and
such
,
that
¾
of
the
one
may
be
equal
to
3/5
of
the
other
,
the
same
may
be
found
out
by
the
said
Theorem
.
For
,
which
number
so
found
is
the
number
sought
,
whereof
3/5
is
to
be
taken
;
and
the
other
is
27
−
15
=
12
,
or
it
may
be
found
by
the
following
proportion
,
viz.
Quest.
12.
One
asked
a
Shepherd
what
was
the
price
of
his
hundred
Sheep
,
quoth
he
,
I
have
not
an
hundred
,
but
if
I
had
as
many
more
,
and
half
as
many
more
,
and
7½
(
or
b
)
sheep
,
then
I
should
have
just
100
(
or
c
)
I
demand
how
many
sheep
he
had
?
1
For
the
number
of
sheep
he
had
,
put
|
|
2
Which
being
doubled
is
|
|
3
And
if
to
the
second
step
you
add
half
the
first
,
it
is
|
|
4
And
if
to
the
third
step
there
be
added
7½
(
or
b
)
the
sum
is
|
|
5
Which
quantity
in
the
fourth
step
is
equal
to
100
(
or
c
)
whence
this
equation
|
|
6
Which
equation
being
reduced
by
the
5th
and
7th
Rules
of
the
11th
.
Chap.
the
value
of
a
will
be
discovered
to
be
37
,
viz.
|
|
So
that
the
number
of
sheep
he
had
were
37
for
37+37+37/2+7½
=
100
CHAP.
XIV
.
How
to
Extract
the
Root
of
a
Square
formed
from
a
Binomial
,
and
how
by
having
any
two
of
the
Members
of
such
a
Square
given
to
find
out
the
third
.
I.
A
Binomial
is
a
quantity
consisting
of
two
names
or
parts
,
as
a+b
,
or
a
−
b
,
aa+ee
,
b+d
,
&c.
And
when
a
Square
is
formed
from
such
a
Root
,
it
will
consist
of
three
members
or
parts
,
viz.
two
Affirmative
Squares
of
the
parts
of
which
the
Binomial
is
composed
,
and
the
double
Rectangle
of
those
parts
,
which
double
Rectangle
is
sometimes
affirmative
,
and
sometimes
negative
,
viz.
Affirmative
,
when
the
parts
of
the
Binomial
are
both
affirmative
,
or
both
negative
,
that
is
,
when
they
are
both
signed
with
+
,
or
both
with
−
;
and
negative
,
when
one
of
the
parts
of
the
Binomial
Root
is
signed
with
+
,
and
the
other
with
−
.
So
if
a+b
were
given
for
a
Root
,
its
Square
would
be
aa+2ab+bb
which
is
composed
of
(
aa
and
bb
)
the
Squares
of
the
parts
of
which
the
Root
is
composed
,
and
of
(
2ab
)
the
double
Product
,
or
Rectangle
made
by
the
multiplication
of
the
said
parts
(
a
and
b
)
one
by
the
other
.
See
the
work
.
So
if
it
were
required
to
find
the
Square
of
the
Binomial
a
−
b
,
or
b
−
a
it
(
being
multiplyed
by
it self
)
would
be
aa
−
2ab+bb
,
which
is
composed
of
(
aa
and
bb
)
the
sum
of
the
Squares
of
the
parts
,
and
their
double
Rectangle
,
as
before
,
but
(
2ab
)
the
Double
Rectangle
of
the
parts
is
signed
with
−
,
so
that
the
Squares
of
the
difference
of
any
two
numbers
or
quantities
is
equal
to
the
sum
of
the
Squares
of
the
said
quantities
or
numbers
made
less
by
their
double
Rectangle
.
As
by
the
work
.
So
if
the
Number
10
were
divided
into
8
and
2
,
viz.
8+2
,
its
Square
would
be
64+32+4
=
10×10
=
100.
And
the
Square
of
8
−
2
is
64
−
32+4
=
6×6
=
36
for
8
−
2
=
6
and
6×6
=
36.
Note
,
That
a
Binomial
Root
having
one
of
its
parts
signed
with
−
,
is
by
some
Authors
called
a
Residual
Root
,
as
a
−
b
,
and
c
−
d
&c.
are
Residuals
.
II.
From
what
hath
been
said
concerning
the
Square
of
a
Binomial
,
m●…
be
inferred
this
Theorem
.
If
a
Compound
quantity
consisting
of
3
members
,
whereof
two
are
Squares
of
different
names
,
with
the
sign+prefixed
to
them
,
and
the
third
is
the
double
Rectangle
of
the
Roots
of
those
Squares
,
having
also
the
sign+prefixed
to
it
,
then
shall
the
Square
Root
of
such
a
compound
quantity
be
the
sum
of
the
Square
Roots
of
the
said
two
simple
Squares
;
but
if
the
said
double
Rectangle
hath
the
sign
−
prefixed
to
it
,
the
Square
Root
of
the
said
Compound
quantity
,
shall
be
the
difference
of
the
said
Roots
.
So
the
Square
Root
of
aa+2ab+bb
will
be
found
to
be
a+b
,
for
the
Square
Root
of
aa
is
a
,
and
the
Square
Root
of
bb
is
b
,
which
two
Roots
added
together
,
give
a+b
.
Also
the
Square
Root
of
aa+8a+16
will
be
found
to
be
a+4
,
the
2
Squares
in
the
given
quantity
are
aa
and
16
,
and
8a
is
the
double
product
of
(
a
and
4
)
the
said
Roots
being
multiplyed
the
one
by
the
other
.
Likewise
the
Square
Root
of
aa
−
2ab+bb
is
a
−
b
,
or
b
−
a
,
not
a+b
,
because
the
double
Rectangle
(
2ab
)
is
signed
with
−
.
Furthermore
the
Square
Root
of
9aa+12ba+4bb
is
3a+2b
:
The
two
Square
quantities
in
the
said
Compound
Square
are
9aa
,
and
4bb
,
whose
Roots
are
3a
and
2b
,
and
12bb
is
the
double
Product
of
3a
and
2b
being
multiplyed
together
.
And
the
Square
Root
of
aa
−
20a+100
is
a
−
10
,
for
the
two
Squares
in
this
Compound
Square
Quantity
are
aa
and
100
,
whose
Square
Roots
are
a
and
10
,
and
20a
is
the
double
rectangle
of
10
and
a
they
being
multiplyed
together
.
The
foregoing
Theorem
being
well
understood
will
be
of
excellent
use
in
the
Resolution
of
Questions
,
producing
Quadratick
Equations
,
as
you
will
find
by
the
Questions
contained
in
the
next
Chapter
.
III.
When
it
is
required
to
extract
the
square
Root
of
a
quantity
whose
Root
cannot
be
exactly
extracted
,
then
prefix
the
radical
sign
to
it
,
which
shall
represent
its
Square
Root
.
So
the
Square
Root
of
bc
is
,
or
,
and
the
Square
Root
of
is
thus
represented
,
viz.
,
or
,
&c.
IV.
From
a
due
consideration
of
the
foregoing
Theorem
,
a
way
is
discovered
how
by
having
any
two
of
the
members
of
a
Square
formed
from
a
binomial
Root
,
the
third
member
may
be
found
out
.
For
,
When
two
Affirmative
Square
Quantities
are
given
for
two
of
the
members
of
a
Square
formed
from
a
binomial
Root
,
then
take
the
Roots
of
those
two
Squares
and
multiply
them
the
one
by
the
other
,
and
double
the
Product
,
so
shall
that
Product
doubled
be
the
third
member
,
which
being
annexed
to
the
two
given
Squares
,
either
by
+
,
or
−
,
it
will
make
an
exact
Compound
Square
,
whose
Root
shall
be
a
Binomial
.
So
if
aa+bb
were
given
for
two
of
the
members
of
a
Square
,
first
,
I
find
their
Roots
to
be
a
and
b
,
which
being
multiplyed
the
one
by
the
other
,
produce
ab
,
and
that
Product
being
doubled
gives
2ab
,
for
the
middle
Term
of
the
Compound
Square
Quantity
to
make
it
a
compleat
square
,
the
Root
whereof
is
a
Binomial
,
viz.
aa+2ab+bb
,
if
the
said
double
product
be
joyned
to
the
said
sum
of
the
Squares
by
the
sign
−
,
it
will
give
the
Compound
Square
Quantity
aa
−
2ab+bb
whose
Root
is
a
−
b.
Also
if
25aa+16bb
were
given
for
two
of
the
members
of
a
Square
,
whose
Root
is
a
Binomial
.
The
said
Square
being
compleated
,
will
be
25aa+40ab+16bb
,
or
25aa
−
40ab+16bb
,
whose
Root
is
either
5a+4b
,
or
5a
−
4b
.
V.
When
the
two
given
members
of
a
Compound
Square
Quantity
,
whose
Root
is
a
Binomial
,
are
the
double
product
or
rectangle
,
and
one
of
the
two
affirmative
squares
,
divide
half
the
said
double
product
by
the
Root
of
the
given
square
,
and
square
the
Quotient
,
so
shall
that
square
be
the
third
member
sought
,
which
being
joyned
to
the
two
given
Quantities
with
the
sign
+
,
it
will
give
you
a
compleat
square
having
for
its
root
a
Binomial
.
As
for
Example
.
Let
aa+2ba
be
proposed
for
2
of
the
members
of
a
square
,
whose
Root
is
a
Binomial
:
First
,
I
take
half
of
(
2ba
)
the
said
double
product
and
it
is
ba
,
which
being
divided
by
(
a
)
the
Root
of
(
aa
)
the
given
Square
,
the
Quotient
is
b
,
whose
square
is
bb
for
the
third
member
sought
.
Again
,
Let
25aa+40a
be
the
two
proposed
terms
of
such
a
square
,
whose
Root
is
a
Binomial
,
and
let
it
be
required
to
find
the
other
square
which
shall
make
it
a
compleat
square
,
raised
from
a
Binomial
Root
;
in
order
to
which
,
first
,
I
take
half
(
40a
)
the
double
Product
,
viz.
20a
,
and
divide
it
by
the
Root
of
(
25aa
)
the
given
square
,
which
is
5a
,
and
the
Quotient
is
4
,
which
being
squared
,
gives
16
for
the
third
member
required
,
which
being
joyned
to
the
rest
,
gives
25aa+40a+16
for
the
square
compleated
.
VI.
When
the
two
given
members
of
a
square
raised
from
a
Binomial
Root
,
are
such
that
one
of
them
is
a
square
affirmative
without
any
Number
or
Quantity
prefixed
to
it
,
and
the
other
is
the
Root
of
the
said
square
multiplyed
by
some
other
Quantity
,
then
is
that
other
Quantity
by
Artists
called
the
Coefficient
,
and
if
you
square
half
the
said
coefficient
,
or
,
(
which
is
all
one
)
take
¼
of
the
square
of
the
coefficient
,
that
shall
be
the
third
member
required
,
which
being
joyned
to
the
two
given
quantities
by
the
sign
+
,
it
will
give
you
a
compleat
square
raised
from
a
Binomial
Root
.
Example
.
Let
the
two
given
members
of
a
square
be
aa+2ba
,
and
let
it
be
required
to
find
out
the
third
member
.
Here
the
coefficient
is
2b
,
half
of
which
is
b
,
which
being
squared
,
gives
bb
for
the
third
member
which
was
sought
,
so
is
the
square
compleated
aa+2ab+bb
.
In
like
manner
,
if
the
two
given
members
of
a
square
were
aa+ba
,
and
it
were
required
to
find
out
the
third
member
.
Here
the
coefficient
is
b
,
half
of
which
is
½
b
,
or
,
whose
square
is
¼
bb
,
or
for
the
member
sought
.
Also
let
the
two
given
members
of
a
square
be
aa+8a
,
and
let
it
be
required
to
find
out
the
third
member
.
Here
the
coefficient
is
8
,
half
of
which
is
4
,
whose
square
is
16
,
for
the
third
member
required
,
so
is
aa+8a+16
,
a
compleat
square
,
whose
Root
is
a+4
.
Again
,
if
the
two
given
members
of
a
square
be
aa
−
ca
,
and
the
third
is
required
;
First
,
I
take
half
the
coefficient
c
,
viz.
½c
,
and
then
square
it
,
and
it
gives
¼
cc
,
or
for
the
member
sought
,
and
so
is
the
square
compleated
aa
−
ca+¼cc
,
whose
Root
is
a
−
½c
.
In
like
manner
,
if
it
were
required
to
make
aa+3ba
a
compleat
square
,
take
half
the
coefficient
(
3b
)
which
is
3/2b
,
or
,
whose
square
is
9/4bb
,
or
,
which
being
joyned
to
the
two
given
Terms
with
the
sign
+
,
it
gives
aa+3ba+
9/4bb
,
whose
Root
is
a+3/2b
.
The
same
Rule
is
to
be
observed
for
the
squaring
of
half
the
coefficient
when
it
is
a
Fraction
.
As
for
Example
.
Let
the
two
members
of
a
square
raised
from
a
Binomial
given
be
aa+
,
and
let
.
it
be
required
to
find
the
third
member
.
Here
half
the
coefficient
is
which
being
squared
,
gives
for
the
member
sought
,
and
so
the
square
being
compleated
,
is
,
whose
Root
is
.
VII
.
When
the
Root
of
the
given
square
hath
no
coefficient
then
the
number
1
is
supposed
to
be
the
co-efficient
,
half
whereof
,
(
viz.
½
)
being
squared
,
gives
(
)
the
third
member
sought
to
make
it
a
compleat
square
.
So
aa+a
being
given
for
2
of
the
members
of
a
square
raised
from
a
Binomial
,
its
third
member
to
make
the
square
compleat
will
be
¼
,
for
aa+a
=
aa+●…a
,
where
the
Coefficient
is
1
,
whose
half
is
½
,
which
being
squared
,
gives
¼
for
the
third
member
sought
,
so
the
Square
being
compleated
,
is
aa+a+¼
,
whose
Root
is
a+½
.
This
Chapter
ought
to
be
well
understood
before
any
further
progress
be
made
,
for
the
manner
how
to
resolve
Questions
which
produce
Quadratick
(
or
square
)
Equations
doth
principally
depend
thereupon
.
CHAP.
XV.
Concerning
the
Resolution
of
Questions
producing
Quadratick
Equations
.
1.
QUadratick
(
or
square
)
Equations
,
are
such
adfected
or
▪
Compounded
Equations
as
consist
of
three
terms
,
the
highest
of
which
is
a
square
,
and
is
called
the
highest
term
in
the
Equation
,
of
which
3
Terms
,
2
are
always
unknown
,
and
the
third
is
always
known
;
the
first
of
the
three
is
the
square
of
the
Quantity
,
or
Number
sought
,
and
the
second
Term
is
the
Product
of
the
Quantity
sought
,
being
multiplyed
by
some
known
Number
or
Quantity
,
and
is
called
the
middle
Term
of
an
Equation
,
and
the
third
Term
is
a
Number
,
or
Quantity
purely
known
.
So
in
this
Equation
,
viz.
aa+ba
=
d
,
the
first
and
highest
Term
or
member
is
aa
,
which
is
the
square
of
the
Quantity
or
Number
sought
,
and
ba
is
the
middle
term
of
the
Equation
which
is
the
Product
of
the
Quantity
sought
,
it
being
drawn
into
b
(
which
is
known
,
)
and
the
third
term
or
member
of
this
Equation
is
b
,
which
is
really
known
,
and
is
usually
called
the
Absolute
Number
or
Quantity
given
.
II.
The
Equations
of
this
kind
are
of
3
Forms
,
which
are
laid
down
by
Mr.
Kersey
in
the
15th
Chapter
of
the
first
Book
of
his
Elements
of
Algebra
,
as
followeth
,
viz.
Equations
of
the
first
Form
.
Equations
of
the
second
Form
.
Equations
of
the
third
Form
.
III.
The
Resolution
of
Equations
which
fall
under
the
first
Form
.
When
an
Equation
is
composed
after
any
of
the
three
foregoing
Forms
,
and
any
known
Quantities
are
mixed
with
unknown
,
let
it
be
so
reduced
by
transposition
(
according
to
the
Rules
of
the
Eleventh
Chapter
)
as
that
the
known
quantities
may
possess
one
side
,
and
the
unknown
Quantities
the
other
side
of
the
Equation
.
Example
.
Let
this
Equation
be
given
,
viz.
aa+b
=
−
ba+bdc
.
By
the
transposition
of
B
on
the
first
part
of
the
Equation
,
and
−
ba
on
the
second
part
,
it
will
be
reduced
to
this
Equation
,
viz.
aa+ba
=
bdc
−
b
,
which
is
an
Equation
of
the
first
Form
:
And
when
your
Equation
is
so
reduced
,
add
to
each
part
of
the
Equation
the
square
of
half
the
coefficient
,
and
so
will
the
first
part
of
the
Equation
be
an
Exact
and
compleat
square
,
then
according
to
the
2d
and
3d
Rule
of
the
Fourteenth
Chapter
extract
the
square
Root
of
both
parts
of
the
Equation
,
and
from
the
Square
Roots
of
both
parts
of
the
Equation
subtract
half
the
coefficient
,
and
then
you
will
discover
the
value
of
a.
As
in
the
following
Examples
.
Quest.
1.
What
number
is
that
which
being
squared
,
and
multiplyed
by
8
(
or
b
)
the
sum
of
the
said
Square
and
Product
is
equal
to
384
(
or
c
)
?
Resolution
.
1
For
the
number
sought
put
|
|
2
Whose
Square
is
|
|
3
Its
Product
by
8
(
or
b
)
is
|
|
4
The
sum
of
the
second
and
third
steps
must
be
equal
to
384
(
or
c
)
whence
this
Equation
.
|
|
5
To
each
part
of
the
equation
add
the
square
of
(
●…
b
)
half
the
coefficient
,
then
will
it
be
|
|
6
Then
by
extracting
the
square
root
of
both
parts
of
the
equation
by
the
second
and
third
Rules
of
the
14
Chap.
it
will
be
reduced
to
|
|
7
By
the
transposition
of
½
b
to
the
second
part
of
the
Equation
the
value
of
a
is
discovered
to
be
|
|
Which
Equation
is
thus
expressed
in
words
,
viz.
the
number
sought
is
equal
to
the
remainder
,
when
(
½
b
)
4
is
subtracted
from
the
square
root
of
the
sum
of
(
c
)
384
and
¼
of
the
Square
of
(
b
)
8
(
added
together
)
which
is
16
,
so
that
the
value
of
a
is
16.
For
c+¼bb
=
400
and
.
and
20
−
4
=
16.
Quest.
2.
What
Number
is
that
whose
Square
being
multiplyed
by
4
(
or
b
)
and
its
Biquadrate
(
or
fourth
Power
)
multiplyed
by
6
(
or
c
)
and
the
Products
added
together
,
the
sum
is
3850
,
(
or
d
)
Resolution
.
1
For
the
number
sought
put
|
|
2
It
s
Square
multiplyed
by
b
is
|
|
3
It
s
Biquadrat
multiplyed
by
c
is
|
|
4
The
sum
of
the
second
and
third
steps
must
be
equal
to
3850
(
or
d
)
whence
this
Equation
,
viz.
|
|
5
And
because
the
highest
power
of
the
equation
is
multiplyed
by
c
,
therefore
each
part
being
divided
by
c
the
equation
is
|
|
6
To
each
part
of
the
equation
add
half
the
square
of
the
coefficient
(
)
and
the
equation
will
be
|
|
7
Then
the
square
Root
of
each
part
of
the
equation
in
the
sixth
step
,
being
extracted
by
the
second
and
third
Rules
of
the
14
Chapter
,
the
equation
then
will
be
|
|
8
And
by
the
transposition
of
to
the
second
part
of
the
equation
the
value
of
aa
is
found
to
be
|
|
9
And
because
the
equation
in
the
8
step
is
the
value
of
aa
,
therefore
if
the
square
Root
of
each
part
of
that
equation
be
extracted
,
the
value
of
a
it self
will
be
discovered
to
be
|
|
which
in
words
is
as
much
as
to
say
the
Number
sought
(
or
a
)
is
equal
to
the
square
Root
of
the
remainder
when
(
)
1/3
is
subtracted
from
the
square
Root
of
the
sum
of
3850/6
(
or
)
and
16/144
(
or
)
being
added
together
,
so
that
the
value
of
a
,
(
or
the
Number
sought
)
is
5.
For
,
and
76/3
−
1/3
(
and
,
which
is
the
number
sought
.
The
Proof
.
You
must
remember
always
to
reduce
a
Fraction
to
its
lowest
Terms
before
you
extract
its
Root
.
III.
The
Resolution
of
Equations
which
fall
under
the
second
of
the
three
Formes
before
mentioned
.
Quest.
1.
What
number
is
that
which
having
8
(
or
b
)
times
its
self
subtracted
from
its
square
,
the
remainder
▪
is
48
(
or
c
)
?
Resolution
.
1
For
the
number
sought
put
|
|
2
Then
will
its
square
be
|
|
3
The
first
step
multiplyed
by
b
is
|
|
4
If
the
third
step
be
subtracted
from
the
second
,
the
remainder
will
be
48
(
or
c
)
whence
this
equation
|
|
5
To
each
part
of
that
equation
add
the
square
of
(
½
b
)
half
the
coefficient
,
and
then
it
will
be
|
|
6
Extract
the
square
Root
of
each
part
of
the
last
equation
by
the
second
and
third
Rules
of
the
14th
Chapter
,
and
it
is
|
|
7
And
by
the
transposition
of
½
b
to
the
second
part
of
the
equation
,
the
value
of
a
is
discovered
to
be
12.
|
|
which
in
words
is
as
much
as
to
say
,
the
Number
sought
(
or
a
)
is
equal
to
the
sum
of
the
universal
Square
Root
of
the
sum
of
48
(
or
c
)
and
a
fourth
part
of
the
square
of
b
(
or
¼
bb
)
being
added
to
4
(
or
½
)
which
is
12
for
c
=
48
and
¼
bb
=
16
,
and
48+16
=
64
and
,
and
8+4
(
½
b
)
=
12.
The
Proof
.
Quest.
2.
What
number
is
that
which
having
12
(
or
b
)
times
its
square
subtracted
from
its
Biquadrate
,
or
fourth
power
,
the
remainder
is
3328
(
or
c
)
?
Resolution
.
1
For
the
number
sought
put
|
|
2
Then
its
biquadrat
is
|
|
2
And
its
square
multiplyed
by
12
(
or
b
)
is
|
|
4
The
difference
of
the
second
and
third
steps
must
be
equal
to
3328
(
or
c
,
)
whence
this
equation
,
viz.
|
|
5
Square
half
the
coefficient
,
and
add
it
to
each
part
of
the
equation
,
and
then
it
will
be
|
|
6
Extract
the
square
Root
of
both
parts
of
the
equation
by
the
second
and
third
Rules
of
the
14
Chap.
and
then
the
equation
will
be
|
|
7
By
the
transposition
of
−
½
b
to
the
contrary
Coast
the
value
of
(
a
)
the
number
sought
will
be
discovered
to
be
|
|
8
By
extracting
the
square
root
of
both
parts
of
the
equation
in
the
7th
step
the
value
of
a
is
found
to
be
8.
|
|
which
is
as
much
as
to
say
,
that
the
Number
sought
,
(
or
a
)
is
equal
to
the
universal
square
root
of
the
sum
of
6
(
or
½
b
)
being
added
to
the
universal
square
Root
of
the
sum
of
3328
,
(
or
c
)
and
36
(
or
¼
bb
)
which
upon
tryal
you
will
find
to
be
8.
For
,
c
=
3328
,
and
b
=
12
,
and
¼
bb
=
36
,
wherefore
3328+36
=
3364
,
and
,
and
58+
(
½
b
)
6
=
64
,
and
,
which
is
the
Number
sought
.
IV.
The
manner
of
resolving
Equations
,
which
fall
under
the
last
of
the
three
forms
before-mentioned
.
Let
the
equation
proposed
(
if
it
falls
under
the
third
and
last
form
)
be
reduced
to
an
equation
of
the
second
form
,
by
the
transposition
of
its
terms
,
as
in
the
following
questions
,
viz.
What
Number
is
that
whose
square
being
subtracted
from
12
(
or
b
)
times
it self
the
remainder
is
32
(
or
c
)
?
Resolution
.
1
For
the
number
sought
put
|
|
2
Its
product
by
12
(
or
b
)
is
|
|
3
If
from
the
second
step
you
subtract
(
aa
)
the
square
of
the
first
step
,
the
remainder
is
|
|
4
The
remainder
in
the
third
step
is
equal
to
32
(
or
c
)
whence
this
equation
|
|
Now
by
transposition
I
reduce
to
an
equation
of
the
second
of
the
foresaid
forms
.
And
First
,
5
By
transposition
of
aa
to
the
contrary
part
,
the
equation
is
|
|
6
Then
by
transposition
of
c
in
the
fifth
step
,
the
equation
is
|
|
7
And
by
transposition
of
ba
in
the
sixth
step
,
the
equation
will
be
|
|
So
that
from
a
due
consideration
of
the
method
used
in
reducing
Equations
of
the
third
form
to
Equations
of
the
second
form
you
may
easily
perceive
that
the
work
of
transposition
in
the
fifth
,
sixth
,
and
seventh
steps
is
performed
only
by
changing
the
signs
of
all
the
Terms
of
the
Equation
in
the
fourth
step
,
viz.
by
changing
+
into
−
,
and
−
into
+
.
So
the
Equation
in
the
fourth
step
is
ba
−
aa
=
c.
And
by
changing
the
signs
of
ba
−
aa
on
the
first
part
of
the
Equation
,
and
of
c
in
the
second
part
into
−
ba+aa
,
and
−
c
,
the
Equation
will
then
be
−
ba+aa
=
−
c
,
or
aa
−
ba
=
−
c
,
which
is
the
same
with
that
in
the
seventh
step
,
and
it
is
now
an
Equation
of
the
second
of
the
three
foregoing
forms
,
so
that
I
now
proceed
to
the
solution
of
the
Equation
8
The
square
of
half
the
coefficient
(
●…
b
)
in
the
seventh
step
to
each
part
of
the
equation
,
it
will
then
be
|
|
9
The
square
root
of
each
part
of
the
last
equation
being
extracted
by
the
second
and
third
Rules
of
the
14
Chapter
the
equation
will
then
be
|
|
10
And
by
the
transposition
of
●…
b
in
the
ninth
step
to
the
contrary
part
,
the
value
of
a
will
then
be
found
to
be
(8)
|
|
which
is
as
much
as
to
say
that
the
number
sought
(
or
a
)
is
equal
to
the
sum
of
6
(
or
●…
b
)
being
added
to
the
Square
Root
of
the
remainder
,
when
32
(
or
c
)
is
subtracted
from
36
(
or
¼
bb
)
which
is
8
:
For
,
¼
bb
−
c
=
4
whose
square
Root
is
2
,
and
2+6
(
or
●…
b
)
=
8
which
is
the
number
sought
.
The
proof
.
And
96
−
64
(
aa
)
=
32
(
or
c
)
which
was
propounded
.
V.
The
Resolution
of
various
Questions
producing
Quadratick
Equations
.
Quest.
1.
There
are
two
Numbers
whose
sum
is
12
(
or
b
)
and
the
sum
of
their
squares
is
80
(
or
c
)
I
demand
what
are
those
numbers
?
Resolution
.
1
For
one
of
the
numbers
sought
put
|
|
2
Then
the
other
will
be
|
|
3
Then
the
sum
of
their
Squares
will
be
|
|
4
Which
quantity
in
the
third
step
is
equal
to
80
(
or
c
)
whence
this
equation
.
|
|
5
Which
equation
being
duly
reduced
by
the
Rules
of
the
eleventh
Chap.
giveth
this
equation
.
|
|
6
Which
equation
being
solved
according
to
the
third
Rule
of
this
Chapter
,
the
value
of
a
is
discovered
to
be
|
|
7
Wherefore
I
conclude
the
numbers
sought
are
8
and
4
,
for
their
sum
is
12
,
and
the
sum
of
their
squares
is
80.
8
Moreover
the
Equation
in
the
sixth
step
will
give
this
Canon
,
If
from
half
the
given
sum
of
the
squares
you
subtract
half
the
square
of
the
given
sum
,
and
to
the
remainder
you
add
half
the
given
sum
,
the
square
root
thereof
being
added
to
the
said
half
sum
of
the
numbers
,
the
sum
of
this
addition
will
give
you
the
greater
number
sought
,
and
the
greater
number
being
subtracted
from
the
given
sum
of
the
numbers
,
will
give
the
lesser
number
sought
.
Quest.
2.
There
are
two
numbers
,
the
product
of
whose
multiplication
is
96
(
or
b
)
and
the
sum
of
their
squares
is
208
(
or
c
)
I
demand
what
are
those
numbers
?
Resolution
.
1
For
one
of
the
numbers
sought
put
|
|
2
Then
by
dividing
96
(
or
b
)
by
a
,
the
Quotient
will
give
the
other
which
is
|
|
3
The
square
of
the
number
in
the
first
step
is
|
|
4
The
square
of
the
other
number
in
the
second
step
is
|
|
5
And
the
sum
of
their
squares
in
the
third
and
fourth
steps
is
|
|
6
Which
sum
in
the
fifth
step
must
be
equal
to
the
given
sum
of
the
Squares
208
(
or
c
)
whence
followeth
this
equation
,
viz.
|
|
7
Which
Equation
in
the
last
step
beingduly
reduced
by
the
Rules
of
theeleventh
Chapter
the
value
of
a
will
be
discovered
to
be
8
So
that
I
conclude
the
numbers
sought
to
be
12
and
8
,
for
their
product
is
96
,
and
the
sum
of
their
squares
is
208.
9
Moreover
the
Equation
in
the
seventh
step
giveth
this
CANON
,
From
¼
of
the
Square
of
the
given
sum
of
the
Squares
subtract
the
Square
of
the
given
Product
of
the
Multiplication
of
the
numbers
sought
,
and
extract
the
square
Root
of
the
remainder
,
and
to
the
said
Square
Root
add
half
the
given
sum
of
the
said
squares
,
and
then
extract
the
square
Root
of
the
sum
of
that
Addition
,
so
shall
that
square
Root
be
one
of
the
Numbers
sought
,
by
which
if
you
divide
the
given
Product
,
the
Quotient
will
be
the
other
Number
sought
.
QUEST.
3.
There
are
two
Numbers
whose
sum
is
12
(
or
b
)
and
the
Product
of
their
Multiplication
is
20
(
or
c
)
what
are
the
Numbers
?
RESOLUTION
.
1
For
one
of
the
Numbers
sought
put
|
|
2
Which
if
you
subtract
from
(
12
)
b
the
given
sum
,
the
remainder
will
be
the
other
number
,
viz.
|
|
3
And
if
the
first
and
second
steps
be
multiplyed
the
one
by
the
other
,
the
Product
will
be
|
|
4
Which
Product
the
Question
requires
to
be
equal
to
20
(
or
c
)
from
whence
this
equation
|
|
5
Which
equation
is
of
the
third
and
last
form
,
mentioned
in
the
beginning
of
this
Chapter
,
which
being
duly
reduced
by
the
Rules
of
the
eleventh
Chapter
,
it
will
be
6
Which
Equation
being
solved
according
to
the
method
used
in
the
fourth
Rule
of
this
Chapter
,
the
value
of
a
will
be
discovered
to
be
7
So
that
I
conclude
the
Numbers
sought
to
be
10
and
2
,
whose
sum
is
12
,
and
their
product
20
,
according
to
the
conditions
of
the
Question
,
moreover
the
Equation
in
the
sixth
step
,
will
present
you
with
this
CANON
,
From
the
Square
of
half
the
given
sum
of
the
Numbers
sought
,
subtract
their
given
product
,
and
extract
the
square
Root
of
the
remainder
,
and
to
its
square
Root
add
half
the
given
sum
of
the
numbers
sought
,
so
shall
the
sum
of
that
Addition
be
the
greater
Number
sought
,
which
being
subtracted
from
the
said
given
sum
will
leave
the
lesser
.
QUEST.
4.
There
are
three
Numbers
which
are
Geometrical
proportionals
continued
,
the
mean
whereof
is
12
(
or
b
)
and
the
two
extreams
are
such
,
that
their
difference
is
18
(
or
c
)
I
demand
what
are
those
three
Numbers
?
RESOLUTION
.
1
For
the
lesser
extream
put
|
|
2
Then
the
greater
will
be
|
|
3
Then
will
the
product
made
by
the
multiplication
of
the
extreams
in
the
first
and
second
steps
be
|
|
4
Which
Product
(
or
Rectangle
)
in
the
third
step
must
be
equal
to
the
square
(
of
(
12
or
b
)
the
mean
whence
this
Equation
.
|
|
5
Which
Equation
being
solved
by
the
second
Rule
of
this
Chapter
,
the
value
of
a
will
be
found
to
be
|
|
6
I
say
the
extream
proportionals
sought
are
6
and
24
,
whose
difference
is
18
:
For
,
7
The
Equation
in
the
fifth
step
being
well
considered
,
will
present
you
with
this
CANON
,
If
to
the
Square
of
the
given
mean
you
add
the
square
of
half
the
difference
of
the
extreams
,
or
(
which
is
all
one
)
¼
part
of
the
square
of
the
given
difference
of
the
extreams
,
and
extract
the
square
root
of
the
sum
of
that
Addition
,
and
then
from
that
square
root
subtract
half
the
said
difference
,
the
remainder
will
be
the
lesser
extream
,
and
if
thereto
you
add
the
given
difference
,
that
sum
will
be
the
greater
extream
.
QUEST.
5.
A
Draper
sold
a
piece
of
Cloth
for
24
l.
(
or
b
)
and
gained
as
much
per
Cent.
(
or
c
)
as
the
cloth
cost
him
,
I
demand
how
much
it
cost
him
?
RESOLUTION
.
1
For
the
price
which
the
cloth
cost
put
|
|
2
Then
will
the
gain
by
its
sale
be
|
|
3
Then
by
the
Rule
of
three
find
how
much
is
gained
per
Cent.
saying
,
,
so
that
his
gain
per
cent
.
was
|
|
4
Which
quantity
in
the
3d
step
according
to
the
tenure
of
the
Question
must
be
equal
to
what
the
cloth
cost
in
the
first
step
whence
this
Equation
,
viz.
|
|
5
Which
being
reduced
by
the
Rules
of
the
eleventh
Chapter
,
it
will
be
|
|
6
Which
(
being
an
Equation
of
the
first
of
the
3
forms
delivered
in
the
beginning
in
the
fifteenth
Chap.
)
being
solved
by
the
2d
and
3d
rule
the
14
Chapter
,
the
value
of
a
is
discovered
to
be
|
|
I
say
the
cloth
cost
20
l.
which
is
the
value
of
a
,
for
and
,
so
that
he
gained
4
l.
in
laying
out
20
:
For
,
and
so
the
conditions
of
the
Question
are
satisfied
.
QUEST.
7.
A
Merchant
bought
a
certain
number
of
pieces
of
cloth
,
and
paid
30
pounds
(
or
b
)
per
Cloth
,
and
sold
them
again
at
such
a
rate
per
Cloth
,
that
if
the
pounds
he
sold
a
Cloth
for
be
multiplyed
by
the
pounds
he
gained
per
Cloth
,
the
product
will
be
equal
to
the
Cube
of
the
number
of
pounds
gained
per
Cloth
,
I
demand
what
he
gained
per
Cloth
,
and
what
he
sold
each
Cloth
for
?
RESOLUTION
.
1
For
the
number
of
pounds
gained
per
piece
,
put
|
|
2
To
which
if
you
add
30
(
or
b
)
the
sum
will
be
the
number
of
pounds
it
was
sold
for
per
piece
,
viz.
|
|
3
And
if
(
according
to
the
tenure
of
the
Question
)
the
second
step
be
multiplyed
by
the
first
,
the
product
will
be
|
|
4
Which
Product
in
the
third
step
,
must
(
according
to
the
nature
of
the
Question
)
be
equal
to
the
Cube
of
the
pounds
gained
per
Cloth
in
the
first
step
whence
this
equation
,
viz.
|
|
5
Which
equation
being
reduced
by
the
third
and
sixth
Rules
of
the
eleventh
Chap.
it
will
then
be
|
|
6
Which
equation
in
the
5th
step
being
solved
by
the
7th
Rule
of
the
14th
Chap.
and
the
sixth
Rule
of
this
Chap.
the
value
of
a
will
be
discovered
to
be
|
|
which
is
as
much
as
to
say
in
words
,
a
(
or
the
gain
per
Cloth
)
is
equal
to
the
sum
when
½
is
added
the
Square
Root
of
the
sum
of
30
,
and
½
added
together
,
(
viz.
the
Square
Root
of
30¼
)
which
is
.
I
say
he
gained
6
pounds
per
Cloth
,
and
he
sold
it
for
36
pounds
per
Cloth
,
which
two
numbers
will
satisfie
the
conditions
of
the
Question
.
The
proof
.
QUEST.
8.
A
Brick-layer
,
and
a
Labourer
wrought
together
at
the
building
of
a
certain
house
42
days
,
(
or
b
)
and
the
labourer
he
wrought
4
(
or
c
)
days
more
than
the
Brick-layer
did
to
gain
one
pound
,
and
at
the
end
of
the
42
dayes
the
Brick-layer
received
for
his
work
1¾
pounds
(
or
d
)
more
than
the
Labourer
,
I
demand
how
many
days
each
of
them
wrought
for
1l
.
RESOLUTION
.
1
For
the
number
of
days
which
the
Bricklayer
wrought
for
1l
.
put
|
|
2
Then
according
to
the
conditions
of
the
question
,
the
number
of
days
that
the
labourer
wrought
for
1l
.
will
be
|
|
3
By
the
Rule
of
proportion
find
how
many
pounds
the
Brick
layer
received
for
the
work
of
42
days
,
as
followeth
,
which
is
|
|
4
Then
find
out
by
the
Rule
of
3
howmany
pounds
the
Labourer
received
for
his
work
in
42
days
thus
which
is
|
|
5
But
the
Brick-layer
received
11/3
(
or
b
)
pounds
more
than
the
Labourer
for
42
days
work
,
wherefore
if
to
the
4th
step
you
add
d
(
or
11/3
)
it
will
be
equal
to
what
the
Brick-layer
received
in
the
third
step
,
whence
this
equation
,
viz.
|
|
6
Which
Equation
being
reduced
by
the
fourth
,
second
,
and
seventh
Rules
of
the
eleventh
Chapter
,
it
will
then
be
|
|
7
The
Equation
in
the
fifth
step
being
solved
by
the
Rule
of
this
Chapter
,
the
value
of
a
will
be
discovered
,
viz.
|
|
which
is
as
much
as
to
say
a
(
or
the
number
of
dayes
which
the
Brick-layer
wrought
)
is
equal
to
the
difference
when
(
〈◊〉
)
2
is
subtracted
from
the
square
root
of
the
sum
of
or
96
&
which
is
100
=
10
−
2
=
8.
I
say
the
Brick-layer
wrought
8
days
for
twenty
shillings
,
and
the
Labourer
wrought
8+4
=
12
days
,
which
two
numbers
will
satisfie
the
conditions
of
the
Question
as
will
appear
by
the
Proof
.
First
by
the
Rule
of
Three
find
what
the
Brick-layer
received
for
the
42
dayes
,
saying
,
Then
find
how
much
the
Labourer
received
for
his
42
days
work
by
the
Rule
of
Three
,
saying
,
So
that
I
find
the
Brick-layer
for
his
42
days
work
received
5l
.
−
5s
.
and
the
Labourer
3l
.
−
10s
.
which
is
1l
.
−
15
,
or
1¾
l.
less
than
the
Bricklayer
received
.
QUEST.
9.
A
Gentleman
bought
a
House
,
and
sold
it
again
for
280
pounds
(
or
b
)
and
by
its
sale
he
gained
so
many
pounds
,
that
their
Square
being
added
to
the
square
of
the
number
of
pounds
it
cost
him
,
the
sum
will
amount
to
52000
(
or
c
)
pounds
,
now
I
demand
how
much
the
house
cost
him
?
RESOLUTION
.
1
For
the
number
of
pounds
which
the
house
cost
,
put
|
|
2
Then
will
the
gain
by
its
sale
be
|
|
3
The
Square
of
(
a
)
its
first
cost
is
|
|
4
The
Square
of
the
gain
by
Sale
is
|
|
5
The
sum
of
the
two
quantities
in
the
third
and
fourth
steps
is
|
|
6
Which
quantity
in
the
fifth
step
is
equal
to
52000
(
or
c
)
whence
this
Equation
|
|
7
Which
equation
being
reduced
by
the
third
and
seventh
Rules
of
the
eleventh
Chapter
it
will
be
|
|
8
Which
Equation
being
solved
by
the
third
Rule
of
this
Chapter
,
the
value
of
a
will
be
discovered
,
viz.
|
|
which
is
as
much
as
to
say
in
words
,
(
a
)
the
price
which
the
house
cost
is
equal
to
the
sum
when
half
what
he
sold
it
for
is
added
to
the
Square
Root
of
the
sum
of
half
the
given
sum
of
the
Squares
added
to
a
fourth
part
of
the
Square
of
what
it
was
sold
for
,
that
sum
being
made
less
by
half
the
Square
of
what
it
was
sold
for
;
which
was
220
l.
and
he
gained
by
the
sale
60
l.
For
,
,
and
,
and
,
now
26000+19600
=
45600
,
and
45600
−
39200
=
6400
,
and
,
and
80+
(
●…
b
)
140
=
220
,
which
is
the
number
of
pounds
the
house
cost
,
and
280
−
220
=
60
,
which
is
the
number
of
pounds
he
gained
by
the
Sale
of
the
house
,
as
you
will
find
by
The
proof
.
whereby
the
conditions
of
the
Question
are
answered
.
QUEST.
10.
A
Draper
sold
2
pieces
of
Cloth
(
whereof
one
containeth
6
(
or
b
)
yards
more
than
the
other
)
for
two
equal
numbers
of
shillings
,
the
lesser
piece
he
selleth
for
2
(
or
c
)
shillings
per
yard
more
than
the
other
,
and
the
number
of
shillings
which
one
piece
was
sold
for
,
did
exceed
the
number
of
yards
in
both
pieces
by
186
(
or
d
)
the
Question
is
what
was
the
Number
of
yards
in
each
piece
,
and
what
each
piece
was
sold
for
per
yard
?
RESOLUTION
.
1
For
the
number
of
yards
in
the
least
piece
put
|
|
2
Then
will
the
yards
in
the
greater
piece
be
|
|
3
Then
will
the
sum
of
the
yards
in
both
pieces
be
|
|
4
Then
if
(
according
to
the
nature
of
the
Question
)
to
the
sum
of
the
yards
in
the
third
step
,
you
add
186
(
or
d
)
the
sum
will
be
the
number
of
shillings
which
each
piece
was
sold
for
,
viz.
|
|
5
And
if
the
quantity
in
the
fourth
step
be
divided
by
(
a
)
the
quantity
in
the
first
step
,
the
Quotient
will
give
the
number
of
shillings
that
1
yard
of
the
least
piece
was
sold
for
|
|
6
And
if
the
said
Quantity
in
in
the
fourth
step
be
divided
by
the
number
of
yards
in
the
biggest
piece
,
(
which
is
the
Quantity
in
the
second
step
)
the
Quotient
will
give
the
number
of
shillings
that
a
yard
of
the
biggest
piece
was
sold
for
,
which
is
|
|
7
If
to
the
Quantity
in
the
sixth
step
you
add
2
(
or
c
)
shillings
,
it
will
then
be
|
|
8
Which
Quantity
in
the
seventh
step
(
as
the
Question
requires
)
is
equal
to
the
Quantity
in
the
fifth
step
,
whence
this
Equation
,
viz.
9.
Which
equation
in
the
eighth
step
being
reduced
by
the
Rules
of
the
Eleventh
Chapter
,
it
will
then
be
10
The
Equation
in
the
last
step
being
solved
by
the
third
Rule
of
this
Chapter
,
the
value
of
a
will
be
discovered
to
be
11
But
if
you
consider
well
the
Equation
in
the
ninth
step
,
you
will
find
the
coefficient
to
be
0
,
for
,
and
therefore
,
whence
the
middle
term
in
that
Equation
is
0
,
and
therefore
the
middle
term
being
removed
,
the
Equation
will
be
which
is
a
simple
Equation
,
and
if
the
Square
root
of
both
parts
of
that
equation
be
extracted
,
the
value
of
a
will
be
discovered
to
be
,
which
is
the
same
with
the
value
of
a
in
the
tenth
step
,
as
you
may
easily
find
upon
Tryal
,
wherefore
I
say
,
The
number
of
yards
in
the
least
piece
is
24.
And
the
number
of
yards
in
the
biggest
piece
is
24+6
=
30
,
which
two
numbers
will
satisfie
the
conditions
of
the
Questions
as
will
appear
by
The
Proof
.
The
number
of
yards
in
both
pieces
is
24+30
=
54
,
which
if
added
to
186
(
as
the
Question
requires
)
will
give
the
number
of
shillings
which
one
piece
was
sold
for
,
which
is
54+186
=
240
,
and
the
least
piece
was
sold
at
10
shillings
per
yard
.
And
the
price
of
a
yard
of
the
biggest
piece
was
8
s.
For
,
which
is
two
shillings
per
yard
less
than
the
lesser
piece
was
sold
for
per
yard
,
and
therefore
the
answer
is
true
,
and
the
conditions
of
the
Question
are
satisfied
.
CHAP.
XVI
.
The
Doctrine
of
Surd
Quantities
.
I.
ALL
Quantities
or
Numbers
whatsoever
,
whether
Integral
,
or
Fractional
,
are
called
Rational
,
but
when
the
Root
of
any
power
cannot
be
exactly
extracted
,
such
Root
is
called
Irrati
)
nal
or
Surd
,
and
is
expressed
by
putting
the
Radical
sign
before
the
number
out
of
which
the
Root
proposed
ought
to
be
extracted
;
as
√
or
placed
before
any
Number
or
Quantity
signifieth
the
Square
Root
of
the
Quantity
or
Number
,
and
the
Cube
Root
,
and
the
Biquadrate
Root
,
&c
,
So
,
or
or
signifieth
the
Square
Root
of
12
,
and
its
Cube
Root
,
&c.
II.
Surd
Numbers
are
two-fold
,
viz.
Simple
,
and
Compound
;
A
Simple
Surd
Quantity
is
when
the
Radical
sign
is
prefixed
to
a
Simple
Quantity
,
as
or
.
A
Compound
Surd
Quantitie
consists
of
several
Simple
Surds
,
which
are
connected
together
by
+
or
−
,
as
,
and
,
and
which
last
Compound
Surd
is
usually
called
an
universal
Root
.
III.
To
Reduce
Simple
Surd
Quantities
that
have
different
radical
signs
to
a
common
radical
sign
.
Let
the
Indices
of
the
given
Powers
be
reduced
to
their
lowest
Terms
by
their
common
measurer
,
and
set
the
Quotients
under
their
respective
Dividends
,
and
multiply
cross-wise
,
so
shall
the
product
be
the
Index
required
,
before
which
placing
√
,
it
shall
then
be
the
common
radical
sign
required
;
Then
raise
the
Powers
of
the
given
Roots
to
the
powers
signified
by
the
said
altern
Quotients
,
before
which
said
Powers
place
the
common
radical
sign
found
as
before
,
so
will
you
have
new
surd
Quantities
equal
to
the
given
Quantities
,
and
having
equal
Radical
signs
.
Example
.
Let
it
be
required
to
reduce
,
and
to
two
other
Roots
equivalent
to
the
former
,
having
a
common
radical
sign
.
First
,
the
exponents
6
and
8
are
reduced
to
3
and
4
,
which
being
placed
under
the
given
exponents
6
and
8
as
you
see
,
and
having
multiplyed
Cross-wise
.
viz.
3×8
,
or
4×6
,
you
have
24
for
a
new
index
,
to
which
prefix
√
,
and
it
is
for
the
common
radical
sign
,
and
then
raising
12
to
the
third
power
thereof
,
and
8
to
the
fourth
,
you
have
and
equal
to
,
and
.
So
if
it
were
required
to
reduce
,
and
to
Surd
Roots
equivalent
thereto
,
having
a
common
Radical
sign
,
it
will
be
as
followeth
.
IV.
Multiplication
in
Simple
Surd
Quantities
.
1.
If
the
Quantities
given
to
be
multiplyed
have
a
common
radical
sign
,
then
multiply
them
together
without
any
regard
to
the
sign
,
and
to
the
product
prefix
the
given
Radical
sign
,
which
new
quantity
shall
be
the
Product
sought
.
So
if
be
to
be
multiplyed
by
,
the
Product
will
be
,
and
by
produceth
.
and
by
,
produceth
and
by
produceth
.
&c.
2.
But
if
the
Quantities
given
to
be
multiplyed
have
not
a
common
Radical
sign
,
let
them
be
reduced
to
such
by
the
third
Rule
foregoing
,
and
then
proceed
as
before
.
Example
.
What
is
the
Product
of
by
?
The
said
Quantities
being
reduced
to
a
common
radical
sign
,
will
be
and
which
being
multiplyed
together
,
produce
which
is
the
product
sought
.
So
the
being
multiplyed
by
they
being
Reduced
to
a
common
radical
sign
,
are
,
and
which
being
multiplyed
produce
.
3.
When
a
surd
Quantity
is
to
be
multiplyed
by
a
rational
quantity
,
then
first
raise
the
given
rational
quantity
to
the
power
of
the
given
quantity
,
whose
Roo●…
is
irrational
or
surd
;
and
then
proceed
as
before
.
So
if
it
were
required
to
multiply
by
5
,
the
rational
number
5
being
raised
to
the
second
power
is
25
,
and
then
you
will
have
to
multiply
by
,
whose
Product
is
.
Likewise
being
to
be
multiplyed
by
a
,
the
Product
will
be
,
for
a
being
raised
to
the
third
power
is
aaa
,
and
by
produceth
as
before
.
V.
Division
in
Simple
Surd
Quantities
.
1.
Reduce
the
Surd
Quantities
given
to
be
divided
to
a
common
Radical
sign
by
the
third
Rule
of
this
Chapter
,
and
then
divide
the
Quantity
following
the
Radical
sign
of
the
Dividend
by
the
quantity
sollowing
the
radical
sign
of
the
Divisor
,
and
to
the
Quotient
prefix
the
said
common
Radical
sign
,
so
shall
that
Surd
Quantity
be
the
Quo-tient
sought
.
Example
.
There
being
given
to
be
divided
by
,
the
quotient
will
be
.
And
being
to
be
divided
by
,
the
quotient
will
be
and
being
given
to
be
divided
by
,
the
Quotient
will
be
,
for
the
given
quantities
being
reduced
to
a
common
radical
sign
,
are
and
.
VI.
Addition
and
Subtraction
of
simple
Surd
quantities
.
1.
When
the
Surd
Roots
to
be
added
together
,
are
equal
,
multiply
any
one
of
them
by
the
given
number
of
Surd
quantities
,
so
shall
that
product
be
the
sum
required
,
before
which
prefix
the
radical
sign
given
,
so
the
sum
of
and
is
,
for
the
given
number
of
roots
is
2
,
whose
square
is
4
,
and
,
so
being
to
be
added
to
,
their
sum
is
and
being
to
be
added
to
,
and
their
sum
will
be
for
the
given
Number
of
Surds
is
3
,
and
being
multiplyed
by
3
,
viz.
(
by
the
third
part
of
the
fourth
Rule
)
the
Product
is
which
is
the
sum
of
,
,
and
,
which
was
required
.
2.
When
two
unequal
Surd
Roots
which
have
the
same
Radical
sign
prefixed
to
each
of
them
,
be
to
be
added
together
,
or
when
the
lesser
of
them
is
to
be
subtracted
from
the
greater
,
Then
you
must
first
try
whether
they
be
commensurable
,
or
not
;
that
is
,
if
after
they
have
been
divided
by
their
greatest
common
measurer
,
the
Quotients
be
rational
Quantities
,
then
multiply
the
sum
of
those
rational
quantities
by
the
said
common
Divisor
,
and
the
Product
shall
be
the
sum
of
the
Surd
Quantities
propounded
;
and
i●…
the
difference
of
those
Rational
Quotients
be
multiplyed
by
the
said
common
measurer
,
then
will
the
Product
be
the
difference
of
the
Surd
Quantities
propounded
.
Example
.
Let
it
be
required
to
find
the
sum
and
difference
of
,
and
,
their
greatest
common
measurer
is
,
by
which
they
being
divided
,
the
Quotients
are
5
and
,
viz.
5
and
1
,
whose
sum
is
7
,
which
being
multiplyed
by
,
the
Product
is
7
or
,
which
is
the
desired
sum
of
the
Surd
Quantities
propounded
.
And
if
the
difference
of
the
said
Rational
Quotients
,
viz.
5
−
2
(
or
3
)
be
multiplyed
by
the
said
common
Divisor
(
)
the
Product
will
be
,
which
is
the
difference
of
the
Surd
Quantities
given
,
the
lesser
being
subtracted
from
the
greater
.
But
if
the
simple
Surd
quantities
given
to
be
added
,
or
subtracted
,
be
incommensurable
,
neither
their
sum
nor
difference
can
be
exprest
by
any
simple
Term
,
or
Root
,
but
their
sum
and
difference
must
be
exprest
by
+
and
−
;
as
suppose
you
were
to
add
and
together
,
their
sum
would
be
,
and
their
difference
.
The
like
of
other
quantities
exprest
by
letters
.
CHAP.
XVII
.
The
Parts
of
Numeration
in
Compound
Surd
Quantities
.
I.
Addition
and
Subtraction
in
Compound
Surd
Quantities
.
THE
Addition
and
Subtraction
of
Compound
Surd
quantities
is
the
same
with
the
simple
Surds
,
having
respect
to
the
signs
of
Affirmation
and
Negation
,
viz.
+
and
−
.
So
if
to
you
add
the
sum
will
be
and
if
from
you
subtract
the
difference
will
be
.
Likewise
if
to
you
are
to
add
the
sum
is
and
if
you
subtract
the
latter
from
the
former
,
the
remainder
will
be
These
two
examples
are
of
Compound
Surd
quantities
which
are
commensurable
,
and
the
next
is
of
Compound
Surd
quantities
,
partly
commensurable
,
and
partly
incommensurable
.
As
Let
it
be
required
to
add
,
to
the
sum
will
be
,
and
if
the
former
be
subtracted
from
the
latter
,
the
remainder
will
be
.
The
same
is
to
be
observed
in
Addition
and
Subtraction
of
Compound
Surd
quantities
altogether
incommensurable
.
As
in
the
following
Examples
.
II.
Multiplication
in
Compound
Surd
Quantities
.
III.
Division
in
Compound
Surd
Quantities
.
These
Examples
will
not
seem
difficult
to
the
ingenious
,
if
what
is
before
delivered
concerning
Surd
quantities
be
duly
considered
.
CHAP.
XVII
.
The
parts
of
Numeration
in
Universal
Surd
Roots
.
WHen
it
is
required
to
extract
the
Root
of
any
Compound
quantity
,
whether
Square
,
Cube
,
Biquadrat
,
&c.
if
they
cannot
be
exactly
extracted
without
any
remainder
;
then
if
to
such
given
compound
quantity
you
prefix
the
Radical
sign
,
such
Roots
are
called
Universal
Surd
Roots
,
and
first
,
concerning
I.
Multiplication
in
Universal
Surds
.
1.
When
any
Universal
Root
is
to
be
multiplyed
by
a
Rational
quantity
,
or
by
any
Surd
,
multiply
the
Square
of
the
Multiplicand
by
the
Square
of
the
Multiplyar
,
when
the
Universal
Radical
sign
is
quadratick
,
or
the
Cube
of
the
Multiplicand
by
the
Cube
of
the
Multiplyar
,
when
the
Universal
Radical
sign
is
Cubical
,
and
before
that
Product
prefix
the
given
universal
Radical
sign
,
so
shall
that
new
universal
Root
be
the
Product
sought
.
Example
.
Let
it
be
required
to
multiply
by
2
this
universal
Square
Root
,
viz.
I
take
the
square
of
2
,
which
is
4
,
and
the
square
of
,
which
is
,
and
multiply
it
by
4
,
and
the
Product
is
,
whose
universal
square
Root
is
the
Product
sought
,
viz.
Also
if
were
to
be
multiplyed
by
2
,
or
doubled
,
take
the
Cube
of
the
universal
Root
given
,
which
is
,
and
multiply
the
same
by
the
Cube
of
2
,
which
is
8
,
and
the
Product
is
,
the
Cube
Root
of
which
is
the
Product
sought
,
viz.
,
and
it
is
double
to
the
Surd
Root
given
.
In
like
manner
,
if
it
were
required
to
multiply
into
its
self
,
or
to
find
.
it
s
square
;
the
squares
of
the
parts
are
and
the
sum
of
which
is
24
,
and
the
Product
made
by
the
Multiplication
of
the
parts
one
into
the
other
,
viz.
into
is
,
(
for
the
difference
of
the
Squares
of
12
and
is
138
whose
square
Root
is
,
and
the
double
of
the
said
product
is
,
which
added
to
24
(
the
sum
of
the
squares
of
the
parts
)
makes
,
which
is
the
square
of
.
Likewise
if
is
to
be
multiplyed
by
,
the
Product
will
be
found
to
be
20
,
for
if
(
which
is
the
square
of
)
be
subtracted
from
36
(
the
square
of
6
)
there
will
remain
which
is
20
,
the
Product
sought
.
Also
,
and
and
2×10
=
20
as
before
.
Again
,
if
it
be
required
to
multiply
by
a
,
the
squares
of
the
given
quantities
are
aa+bb
and
aa
,
which
being
multiplyed
the
one
into
the
other
,
the
Product
will
be
aaaa+bbaa
,
the
universal
Square
Root
of
which
is
the
Product
ought
,
viz.
which
may
be
more
sompendiously
exprest
thus
,
.
II.
Division
in
Universal
Surds
.
As
in
Multiplication
you
multiplyed
the
Square
of
the
Multiplicand
by
the
square
of
the
Multiplyar
,
the
given
Radical
sign
being
Quadratick
,
&c.
So
in
Division
of
Universal
Surd
Roots
you
are
to
divide
the
square
of
the
Dividend
by
the
square
of
the
Divisor
,
when
the
universal
Radical
sign
is
Quadratick
,
and
Divide
the
Cube
of
the
Dividend
by
the
Cube
of
the
Divisor
,
when
the
universal
Radical
sign
is
Cubical
,
&c.
so
shall
the
Quotient
,
when
the
universal
radical
sign
given
is
prefixed
thereto
,
be
the
Quotient
required
.
Example
.
What
is
the
Quotient
when
is
divided
by
2
?
Here
I
divide
(
which
is
the
square
of
the
dividend
)
by
4
(
the
square
of
the
given
Divisor
)
and
there
ariseth
the
universal
square
Root
of
which
,
viz.
is
the
Quotient
required
.
Also
if
it
were
required
to
divide
by
2
,
the
Quotient
would
be
found
to
be
here
the
Cube
of
the
given
Dividend
is
which
being
divided
by
8
(
the
Cube
of
2
)
there
will
arise
,
to
which
if
you
prefix
the
universal
radical
sign
of
its
Cube
Root
,
it
will
be
which
is
the
Quotient
sought
.
Likewise
ifit
be
required
to
divide
by
a
,
the
Quotient
will
be
found
to
be
for
,
the
square
of
the
Dividend
is
aaaa+bbaa
,
and
the
square
of
the
Divisor
is
aa
,
and
when
Division
is
ended
,
there
will
arise
aa+bb
the
universal
square
Root
of
which
is
which
is
the
quotient
sought
.
But
when
the
work
of
Division
in
universal
Surd
Quantities
happens
to
be
intricate
,
and
its
operation
canno
be
finished
without
a
remainder
,
you
may
set
the
power
of
the
Dividend
for
a
Numerator
,
and
the
power
of
the
Divisor
for
a
Denominator
,
and
against
the
line
of
Separation
,
place
,
or
prefix
the
universal
radical
sign
,
which
universal
Root
so
signified
shall
be
the
Quotient
sought
.
As
if
it
were
required
to
divide
by
the
quotient
will
be
.
III.
Addition
and
Subtraction
in
Universal
Surd
Quantities
.
1.
If
two
Universal
Surd
quantities
that
are
commensurable
are
proposed
to
be
added
together
,
or
subtracted
,
the
operation
may
be
performed
like
simple
Surds
.
As
for
Example
.
If
the
sum
and
difference
of
and
were
required
.
Here
each
of
the
said
quantities
being
divided
by
their
greatest
common
measurer
,
the
Quotients
are
and
,
viz.
2
and
1
,
which
are
rational
Numbers
expressing
the
proportion
of
the
Surds
propounded
,
therefore
if
their
common
Divisor
be
multiplyed
2+1
(
viz.
3
)
it
giveth
for
the
sum
required
,
and
the
said
common
Divisor
being
multiplyed
by
(
2
−
1
)
the
difference
of
the
said
2
and
1
,
it
will
produce
for
the
difference
of
the
Roots
proposed
.
Likewise
if
it
were
required
to
find
the
sum
and
difference
of
and
.
The
said
Quantities
being
reduced
,
are
and
Therefore
is
their
sum
and
their
difference
is
.
2.
When
the
Root
of
a
residual
is
to
be
added
to
,
or
subtracted
,
from
the
Root
of
its
correspondent
Binomial
,
then
may
those
Roots
be
connected
together
by
the
signs
+
and
−
;
and
then
the
whole
being
multiplyed
by
it self
,
the
universal
Root
of
the
Product
shall
be
the
sum
or
difference
of
the
Roots
propounded
.
As
suppose
were
propounded
to
be
added
to
the
given
Roots
being
connected
together
by
+
,
make
which
composed
Quantity
being
multiplyed
by
it self
,
produceth
,
whose
universal
Square
Root
(
)
shall
be
the
sum
of
the
Quantities
proposed
to
be
added
.
But
if
be
multiplyed
into
it self
,
the
product
will
be
,
whose
universal
square
Root
is
the
difference
of
the
two
given
Roots
.
3.
But
if
the
universal
roots
to
be
added
or
subtracted
are
not
commensurable
,
&c
,
then
they
are
to
be
added
by
+
,
and
subtracted
by
−
.
So
if
it
were
required
to
add
to
their
sum
would
be
And
the
latter
being
subtracted
from
the
former
,
the
remainder
would
be
And
the
sum
of
being
added
to
will
be
and
the
latter
being
subtracted
from
the
former
,
the
remainder
will
be
IV.
The
Extraction
of
the
Square
Root
out
of
Binomials
,
and
Reseduals
.
Subtract
the
Square
of
the
lesser
part
of
the
given
Binomial
,
from
the
Square
of
the
greater
part
,
and
add
the
Square
root
of
the
remainder
to
the
greater
part
,
and
also
subtract
it
therefrom
,
and
then
extract
the
square
roots
of
the
Sum
and
remainder
,
and
joyn
them
together
by
+
if
the
quantity
proposed
be
a
Binomial
,
but
by
−
if
it
be
a
Residual
,
which
roots
so
joyned
,
are
the
square
root
of
the
given
Binomial
,
or
Residual
.
Example
1.
Extract
the
Square
root
of
this
Binomial
,
viz.
|
|
1
From
the
square
of
the
greater
,
part
38
,
viz.
from
|
|
2
Subtract
the
square
of
the
lesser
part
,
viz.
which
is
|
|
3
The
remainder
is
|
|
4
The
square
root
of
the
remainder
is
|
|
5
To
which
root
if
you
add
the
greater
part
38
,
the
sum
is
|
|
6
The
half
of
which
sum
is
|
|
7
The
square
root
of
the
said
half
sum
is
the
greater
part
of
the
root
sought
,
which
is
|
|
8
From
the
greater
part
of
the
given
Binomial
,
viz.
38
,
subtract
the
square
root
in
the
fourth
step
,
viz.
12
,
the
remainder
is
|
|
9
The
half
of
which
is
|
|
10
The
square
root
of
the
said
half
remainder
is
the
lesser
part
of
the
root
sought
.
|
|
11
To
which
if
you
add
the
quantity
in
the
seventh
step
,
the
sum
will
be
the
square
root
sought
,
viz.
|
|
which
is
the
square
root
of
the
given
Binomial
,
but
if
the
given
surd
quantity
had
been
a
Residual
,
viz.
if
it
had
been
required
to
extract
the
square
root
of
,
then
the
root
would
have
been
.
Example
2.
Extract
the
square
root
of
this
Binomial
,
viz.
|
|
1
The
square
of
the
greater
part
7
,
is
|
|
2
From
which
subtract
the
square
of
the
lesser
part
,
(
viz.
0
,
)
which
is
|
|
3
The
remainder
is
|
|
4
The
square
root
of
that
remainder
is
|
|
5
To
which
square
root
add
the
greater
part
of
the
given
Binomial
,
viz.
7
,
and
the
sum
is
|
|
6
The
half
of
which
sum
is
|
|
7
The
square
root
of
the
said
sum
is
the
greater
part
of
the
root
sought
,
which
is
|
|
8
From
the
greater
part
of
the
given
Binomial
,
viz.
from
7
)
subtract
in
the
fourth
step
,
and
the
remainder
is
|
|
9
The
half
of
which
remainder
is
|
|
10
The
square
root
of
the
said
half
remainder
is
the
lesser
part
of
the
root
sought
,
which
is
|
|
11
Which
being
joyned
to
the
greater
part
of
the
root
sought
in
the
seventh
step
by
the
sign
+
,
the
sum
will
be
the
square
root
sought
,
which
is
But
if
the
lesser
part
of
the
said
root
found
in
the
tenth
step
be
joyned
to
the
greater
part
found
in
the
seventh
step
by
interposing
the
sign
−
instead
of
+
,
it
will
then
be
the
square
root
of
the
residual
.
Example
3.
Let
it
be
required
to
extract
the
square
root
of
this
Binomial
,
viz.
aa+d
added
to
,
supposing
the
greater
part
of
the
given
Binomial
,
to
be
aa+d
.
Then
,
1
The
square
of
the
greater
part
is
|
|
2
From
which
subtract
the
square
of
the
lesser
part
(
)
viz.
4daa
,
and
the
remainder
is
|
|
3
The
square
root
of
that
remainder
is
|
|
4
To
which
square
root
add
the
greater
part
of
the
given
Binomial
,
viz.
aa+d
,
and
the
sum
is
|
|
5
The
half
of
which
sum
is
|
|
6
The
square
root
of
which
half
sum
is
the
greater
part
of
the
root
sought
,
which
is
|
|
7
From
(
aa+d
)
the
greater
part
of
the
given
binomial
subtract
the
square
root
found
in
the
third
step
(
aa
−
d
)
and
the
remainder
is
|
|
8
The
half
of
which
remainder
is
|
|
9
The
square
root
of
which
half
remainder
,
is
the
lesser
part
of
the
root
sought
,
viz.
|
|
10
Which
said
root
being
joyned
to
the
greater
part
found
in
the
sixth
step
by
the
sign
+
,
it
will
be
the
root
sought
,
viz.
|
|
but
if
the
quantity
in
the
ninth
step
be
joyned
to
the
quantity
in
the
sixth
step
,
by
interposing
the
sign
−
,
it
will
then
be
the
square
root
of
the
residual
,
aa+d
less
.
Example
4.
Let
it
be
required
to
extract
the
square
root
of
supposing
the
greater
part
to
be
|
|
|
1
The
square
of
the
greater
part
is
|
|
2
From
which
subtract
the
square
of
the
lesser
part
,
which
is
|
|
3
And
the
remainder
is
|
|
4
The
square
root
of
that
remainder
is
|
|
5
To
which
if
you
add
the
greater
part
of
the
given
binomial
,
the
sum
is
|
|
6
The
half
of
which
sum
is
|
|
7
The
square
root
of
the
said
half
sum
is
the
greater
part
of
the
root
sought
,
which
is
|
|
8
From
the
greater
part
of
the
given
binomial
subtract
the
square
root
found
in
the
fourth
step
,
and
the
remainder
is
|
|
9
The
half
of
which
remainder
is
|
|
10
The
square
root
of
the
said
half
remainder
is
the
lesser
part
of
the
root
sought
,
which
is
|
|
11
If
to
the
greater
part
of
the
root
sought
in
the
seventh
step
,
you
joyn
the
lesser
part
in
the
eleventh
step
,
by
interposing
the
sign
+
,
it
will
then
be
the
root
sought
,
which
is
But
if
the
two
said
quantities
are
joyned
together
by
the
interposition
of
the
sign
−
,
it
will
then
be
the
square
root
of
the
residual
less
2ed
.
V.
Some
Questions
to
exercise
the
Rules
of
this
and
the
foregoing
Chapters
.
QUEST.
1.
Let
it
be
required
to
divide
100
(
or
c
)
into
two
such
unequal
parts
,
that
100
multiplyed
by
the
lesser
part
may
be
equal
to
the
square
of
the
greater
.
RESOLUTION
.
1
For
the
greater
number
put
|
|
2
Then
will
the
lesser
be
|
|
3
By
which
if
you
multiply
100
(
or
c
)
the
product
will
be
|
|
4
Which
quantity
in
the
3d
step
must
be
equal
to
the
square
of
the
Quantity
in
the
first
step
,
whence
this
equation
.
|
|
5
Which
Equation
being
reduced
by
the
rules
of
the
11
Chap.
and
solved
the
value
of
a
will
be
discovered
to
be
|
|
which
Equation
in
the
last
step
being
duly
considered
,
will
present
you
with
this
Theorem
.
To
the
Square
of
the
given
line
or
number
add
a
fourth
part
of
its
Square
,
and
extract
the
Square
root
of
that
sum
,
then
from
the
said
square
root
subtract
half
the
given
line
,
so
shall
the
remainder
be
the
greater
segment
,
or
number
sought
.
QUEST.
2.
What
Number
is
that
whose
square
being
made
less
by
the
Rectangle
of
it self
drawn
into
12
(
or
b
)
the
remainder
is
equal
to
f
?
1
For
the
number
sought
put
|
|
2
The
square
of
which
is
|
|
3
The
Rectangle
of
a
in
b
is
|
|
4
If
the
quantity
in
the
third
step
be
subtracted
from
the
quantity
in
the
second
step
,
the
remainder
is
equal
to
f
,
whence
this
Equation
.
|
|
5
Which
Equation
being
solved
by
the
rules
of
the
11th
Chapter
the
value
of
a
will
be
found
to
be
|
|
The
Proof
.
6
If
|
|
7
Then
by
subtracting
½
b
from
each
part
of
the
equation
there
remaineth
|
|
8
Then
by
squaring
each
part
of
the
equation
you
have
|
|
9
And
by
subtracting
¼
bb
from
both
sides
of
the
equation
there
remaineth
|
|
which
was
to
be
proved
QUEST.
3.
1.
Let
c
and
d
be
put
for
two
such
known
Quantities
that
d
not
,
and
let
a
be
put
for
a
quantity
unknown
,
and
let
it
be
granted
that
ca
−
aa
=
d
what
is
the
value
of
a
?
2
The
given
equation
in
the
first
step
is
one
of
the
third
form
mentioned
in
the
beginning
of
the
fifteenth
Chapter
,
and
it
will
be
found
that
the
2
values
of
a
are
By
either
of
which
values
of
a
the
Equation
propounded
in
the
first
step
may
be
expounded
,
as
will
appear
by
the
DEMONSTRATION
,
1
If
|
|
2
Then
by
the
transposition
of
½
c
to
the
contrary
coast
,
it
is
|
|
3
And
by
squaring
each
part
of
the
last
equation
,
it
is
|
|
4
And
by
subtracting
¼
cc
from
each
part
of
the
equation
,
it
is
|
|
5
And
by
changing
the
signs
on
the
quantities
on
each
side
of
the
equation
it
is
|
|
which
was
to
be
demonstrated
.
6
Again
if
|
|
7
Then
by
transposition
of
to
the
other
side
it
is
|
|
8
And
by
transposition
of
a
it
is
|
|
9
And
by
squaring
each
part
of
the
equation
it
will
then
be
|
|
10
And
by
subtracting
¼
cc
from
both
parts
of
the
equation
,
it
is
|
|
11
And
the
quantities
on
both
sides
of
the
equation
being
transposed
to
the
contrary
coast
,
and
the
signs
of
each
thereby
changed
,
the
equation
will
then
be
|
|
which
was
likewise
to
be
proved
.
QUEST.
4.
Let
it
be
required
to
divide
100
into
two
such
parts
that
if
each
part
be
divided
by
the
other
,
the
sum
of
the
Quotients
may
be
3.
This
is
Quest.
1.
of
the
ninth
Chapter
of
the
second
Book
of
Kersey's
Elements
of
Algebra
,
and
it
is
thus
wrought
,
viz.
1
For
one
of
the
parts
sought
put
|
|
2
Then
will
the
other
be
|
|
3
Each
of
which
quantities
in
the
first
and
second
steps
being
mutually
divided
by
each
other
(
according
to
the
import
of
the
question
)
this
equation
ariseth
|
|
4
Which
equation
being
duly
reduced
,
gives
|
|
5
Which
is
an
equation
of
the
third
form
mentioned
in
Chap.
15.
and
being
solved
according
to
the
method
there
given
the
two
values
of
a
will
be
found
to
be
|
|
which
you
may
easily
prove
at
your
leisure
.
CHAP.
XVIII
.
Algebraical
Questions
Resolved
by
various
Positions
.
MR.
Kersey
in
the
Twelfth
Chapter
of
the
second
Book
of
his
Elements
of
Algebra
,
hath
laid
down
Rules
for
the
solution
of
Questions
Algebraically
by
various
Positions
;
assuming
a
peculiar
letter
to
represent
every
one
of
the
Quantities
sought
,
viz.
a
for
one
unknown
Quantity
,
e
for
another
,
and
y
for
a
third
,
&c.
and
for
the
performance
of
the
work
he
hath
laid
down
3
Rules
which
are
as
followeth
,
viz.
RULE
1.
When
many
Quantities
are
sought
in
a
Question
,
let
them
be
represented
by
various
letters
,
and
let
the
tenor
of
the
Question
be
represented
by
Equations
,
which
done
by
Transposition
find
what
any
single
letter
in
the
first
equation
is
equal
to
;
Then
wheresoever
that
Letter
is
found
in
the
other
equations
,
instead
thereof
,
take
what
it
is
found
equal
to
,
so
will
that
letter
quite
vanish
out
of
the
following
Equations
;
Then
by
Transposition
set
a
second
letter
alone
in
one
of
those
equations
out
of
which
the
first
letter
was
cancelled
and
poceed
as
before
,
so
at
length
one
of
the
letters
will
be
made
known
'
by
help
of
which
the
rest
will
be
easily
discovered
.
RULE
2.
When
the
same
Quantity
,
(
suppose
a
)
is
found
in
two
several
Equations
,
and
equal
numbers
are
prefixed
to
those
Quantities
,
then
if
their
signs
be
both
+
or
both
−
,
subtract
the
lesser
Equation
from
the
greater
,
but
if
the
signs
be
one
+
,
and
the
other
−
,
then
add
those
two
Equations
together
,
so
will
the
said
Quantity
a
quite
vanish
.
RULE
3.
When
the
same
Quantity
,
(
suppose
a
)
is
found
in
two
several
Equations
,
but
the
numbers
prefixed
to
those
equal
Quantities
are
unequal
,
those
two
Equations
may
be
reduced
to
two
others
which
shall
have
equal
numbers
prefixed
to
the
said
Quantity
a
thus
,
viz.
Multiply
all
the
Quantities
in
the
first
Equation
by
the
number
prefixed
to
a
in
the
second
equation
;
and
also
multiply
all
the
quantities
in
the
second
Equation
by
the
number
prefixed
to
the
same
quantity
a
in
the
first
Equation
,
so
by
such
alternate
multiplication
two
new
equations
will
be
produced
,
wherein
the
numbers
prefixed
to
the
said
quantity
a
will
be
equal
to
one
another
,
and
then
proceed
according
to
the
second
Rule
,
and
expel
the
same
quantity
out
of
the
Rest
of
the
Equations
;
proceed
in
like
manner
with
a
second
Quantity
,
until
at
length
some
one
Quantity
,
be
made
known
;
by
which
all
the
rest
may
be
found
out
.
The
three
foregoing
Rules
will
be
exercised
in
the
Resolution
of
the
following
Questions
.
QUEST.
1.
Divide
100
(
or
c
)
into
two
such
numbers
,
(
viz.
a
and
e
)
that
may
be
equal
to
30
(
or
d
)
I
demand
the
numbers
a
and
e
?
RESOLUTION
.
1
If
|
|
2
And
|
|
3
Then
by
transposition
of
e
in
the
first
step
,
you
will
have
|
|
4
By
reducing
the
Equation
in
the
second
step
,
so
as
a
may
solely
possess
one
side
thereof
,
you
will
have
|
|
5
If
instead
of
a
in
the
4th
step
you
take
what
a
is
equal
to
in
the
third
step
,
you
will
have
this
Equation
,
viz.
|
|
6
The
first
part
of
the
Equation
in
the
fifth
step
being
multiplyed
by
5
,
will
give
|
|
7
By
the
transposition
of
−
5e
it
is
|
|
8
And
by
the
transposition
of
15d
in
the
last
step
,
you
have
|
|
9
Each
part
of
the
Equation
in
the
last
step
being
divided
by
2
,
will
give
the
value
of
e
,
viz.
|
|
I
say
the
value
of
e
is
25
,
and
the
value
of
a
is
100
−
25
=
75
which
will
answer
the
conditions
of
the
Question
.
As
appears
by
The
Proof
.
QUEST.
2.
There
are
two
Numbers
(
a
the
greater
,
and
e
the
lesser
)
whose
difference
is
4
(
or
b
)
and
the
difference
of
their
Squares
is
64
(
or
c
)
what
are
the
Numbers
?
RESOLUTION
.
1
If
|
|
2
Then
by
the
transposition
of
e
you
have
|
|
3
And
if
|
|
4
Then
by
transposition
of
ee
it
is
|
|
5
If
both
parts
of
the
equation
in
the
second
step
be
squared
,
it
will
be
|
|
6
And
if
instead
of
aa
in
the
fifth
Equation
,
you
place
what
it
is
equal
to
in
the
fourth
step
,
the
equation
will
then
be
|
|
7
By
subtracting
ee
from
both
parts
of
the
last
equation
you
will
have
|
|
8
By
dividing
both
parts
of
the
last
equation
by
b
,
you
will
then
have
|
|
9
By
transposition
of
b
in
the
last
step
,
the
Equation
will
then
be
|
|
10
And
if
both
parts
of
the
equation
in
the
last
step
be
divided
by
2
,
the
value
of
e
will
then
be
discovered
to
be
|
|
I
say
the
value
of
e
(
the
lesser
number
)
is
6
,
and
by
the
second
step
(
a
)
the
greater
number
is
e+b
=
6+4
=
10
,
which
two
numbers
,
(
viz.
10
and
6
)
will
satisfie
the
Conditions
of
the
Question
,
as
will
appear
by
The
Proof
.
QUEST.
3.
A
Maid
being
at
Market
sold
10
dozen
of
eggs
,
and
twelve
Pounds
of
Butter
for
thirteen
shilling
,
and
at
another
time
,
and
at
the
same
rate
,
she
selleth
Eight
dozen
of
Eggs
,
and
18
pounds
of
Butter
for
16
shillings
,
I
demand
how
she
sold
her
Eggs
per
dozen
,
and
her
Butter
per
pound
?
Let
a
represent
the
desired
value
of
a
dozen
of
Eggs
,
and
for
the
price
of
a
pound
of
Butter
put
e
,
and
then
may
the
Question
being
abstracted
from
words
be
stated
thus
,
viz.
What
are
the
values
of
a
and
e
?
RESOLUTION
.
3
By
transposition
of
12e
in
the
first
step
,
that
equation
will
be
|
|
4
And
both
parts
of
the
last
equation
being
divided
by
10
it
is
|
|
5
By
transposition
of
●…8e
in
the
second
step
that
equation
will
be
|
|
6
Each
part
of
the
last
equation
being
divided
by
8
will
give
|
|
7
If
instead
of
a
in
the
sixth
step
,
you
place
what
it
is
equal
to
in
the
fourth
step
the
equation
will
then
be
|
|
8
Both
parts
of
the
last
equation
being
reduced
to
integers
will
give
|
|
9
By
transposition
of
180e
and
104
in
the
last
equation
each
to
the
contrary
coast
the
equation
will
then
be
|
|
10
If
each
part
of
the
last
equation
be
divided
by
84
the
value
of
e
will
be
discovered
to
be
|
|
which
is
8d
for
the
price
of
1
pound
of
butter
.
|
|
11
By
the
10th
step
the
value
of
e
is
discovered
to
be
2/3s
.
by
which
means
the
value
of
a
(
by
the
quantities
in
the
fourth
and
sixth
steps
)
is
found
to
be
6d
.
for
the
4th
step
is
|
|
And
it
hath
been
found
before
that
so
that
and
so
the
maid
sold
her
Eggs
at
6d
.
per
dozen
,
and
her
Butter
at
8d
.
per
pound
,
which
will
answer
the
conditions
of
the
Question
.
QUEST.
4.
Three
Men
,
viz.
A
,
B
,
and
C
discourse
thus
together
concerning
their
Age
,
quoth
B
to
A
,
your
age
added
to
mine
is
54
,
(
or
b
years
)
quoth
C
to
B
,
and
my
age
added
to
yours
makes
78
(
or
c
)
years
,
and
quoth
A
to
C
my
age
added
to
yours
is
72
(
or
d
)
years
,
I
demand
the
age
of
each
person
?
Let
the
Age
of
each
person
be
represented
by
the
letters
a
e
y
,
viz.
for
the
age
of
A
put
a
,
for
the
age
of
B
put
e
,
and
for
the
age
of
C
put
y
;
and
the
Question
being
abstracted
from
words
,
will
be
as
followeth
,
viz.
What
are
the
values
of
a
e
and
y
?
RESOLUTION
.
4
By
transposition
of
e
in
the
first
step
there
will
arise
|
|
5
Ifinstead
of
a
in
the
third
step
you
put
what
a
is
equal
to
in
the
fourth
step
,
there
will
arise
|
|
6
By
the
transposition
of
d
and
e
in
the
last
step
,
there
ariseth
|
|
7
And
if
instead
of
e
in
the
second
step
you
take
the
latter
part
of
the
sixth
step
,
there
will
then
arise
|
|
8
In
which
last
equation
there
is
no
unknown
quantity
but
y
,
and
therefore
the
equation
being
duly
reduced
,
will
discover
the
value
of
y
to
be
|
|
9
If
in
the
sixth
step
in
stead
of
y
you
take
the
latter
part
of
the
equation
in
the
eighth
step
,
the
value
of
e
will
be
found
to
be
|
|
10
And
if
instead
of
e
in
the
fourth
step
,
you
take
the
latter
part
of
the
Equation
in
the
ninth
step
,
the
value
of
a
will
be
discovered
,
viz.
|
|
And
thus
the
work
is
finished
,
and
the
equations
in
the
Eighth
,
ninth
,
and
tenth
steps
present
you
with
this
CANON
,
From
the
sum
of
every
two
of
the
three
given
Numbers
subtract
the
third
number
remaining
,
so
shall
the
three
remaining
numbers
being
divided
by
2
be
the
numbers
sought
.
So
the
numbers
sought
in
the
Question
,
viz.
a
,
e
,
and
y
,
are
found
to
be
24
,
30
,
and
48
,
viz.
the
age
of
A
is
24
,
the
age
of
B
is
30
,
and
the
age
of
C
is
48
,
which
three
numbers
will
satisfie
the
conditions
of
the
Question
for
24+30
=
54
,
and
30+48
=
78
,
and
48+24
=
72
:
QUEST.
5.
What
two
numbers
are
those
whose
sum
is
20
,
(
or
b
)
and
their
difference
4
(
or
c
)
?
Let
a
be
put
for
the
greater
number
sought
,
and
e
for
the
lesser
,
and
then
the
Question
being
extracted
from
words
may
be
stated
thus
,
viz.
What
are
a
and
e
?
RESOLUTION
.
3
Forasmuch
as+1●…
is
found
in
each
of
the
equations
in
the
first
and
second
steps
,
therefore
(
by
the
second
Rule
)
they
being
subtracted
,
do
give
this
Equation
,
viz.
|
|
4
And
by
dividing
both
parts
of
the
equation
in
the
third
step
by
2
,
the
value
of
e
will
〈…〉
|
|
5
And
if
instead
of
〈◊〉
the
second
step
you
put
what
▪
is
equal
to
in
the
fourth
step
,
you
will
have
this
equation
,
viz.
|
|
6
By
the
transposition
of
−
●…b
and+
c
to
the
contrary
coast
,
the
value
of
a
will
be
discovered
,
viz.
|
|
From
the
fourth
and
sixth
steps
is
raised
this
CANON
,
If
from
half
the
sum
of
two
numbers
you
subtract
half
their
difference
,
the
remainder
will
be
the
lesser
number
,
and
if
to
half
their
sum
you
add
half
their
Difference
that
sum
will
be
the
greater
number
,
whereby
the
two
numbers
sought
in
this
Question
are
found
to
be
12
and
8
;
for
,
12+8
=
20
and
12
−
8
=
4.
QUEST.
6.
What
3
Numbers
are
those
that
if
to
the
first
there
be
added
121
(
or
b
)
the
sum
will
be
equal
to
the
sum
of
the
first
and
second
,
and
if
to
the
second
there
be
added
121
,
the
sum
will
be
equal
to
double
the
sum
of
the
first
and
third
,
and
if
to
the
third
there
be
added
121
,
their
sum
will
be
triple
the
sum
of
the
first
and
second
?
If
for
the
number
sought
you
put
a
,
e
,
and
y
,
viz.
for
the
first
number
a
,
for
the
second
e
,
and
for
the
third
y
,
then
the
Question
being
abstracted
from
words
may
be
stated
thus
,
viz.
What
are
the
numbers
a
e
y
?
RESOLUTION
.
4
By
the
transposition
of
y
in
the
first
equation
,
there
ariseth
|
|
5
And
if
instead
of
e
in
the
second
equation
you
take
what
is
equal
thereto
in
the
fourth
equation
there
ariseth
.
|
|
6
The
last
equation
after
due
reduction
will
be
|
|
7
And
if
instead
of
3e
in
the
third
equation
you
take
the
triple
of
what
e
is
found
equal
to
,
in
the
fourth
step
,
you
will
find
the
following
Equation
to
arise
,
viz.
|
|
8
Which
equation
after
due
reduction
by
transposition
the
quantities
will
be
found
to
be
|
|
9
And
both
parts
of
the
last
equation
being
divided
by
4
there
ariseth
.
|
|
10
Then
if
instead
of
3y
in
the
sixth
equation
there
be
taken
the
triple
of
the
latter
part
of
the
ninth
equation
there
ariseth
|
|
11
After
due
Reduction
of
the
equation
in
the
tenth
step
,
the
value
of
a
will
be
discovered
,
viz.
|
|
12
Again
,
if
instead
of
a
the
sixth
Equation
you
put
the
latter
part
of
the
Eleventh
Equation
,
there
ariseth
|
|
13
After
due
reduction
of
the
equation
in
the
twelfth
step
,
the
value
of
y
will
be
discovered
to
be
|
|
14
And
if
for
a
and
y
in
the
fourth
step
there
be
put
their
equals
in
the
11th
and
13th
steps
there
will
arise
|
|
15
The
Equation
in
the
last
step
being
duly
reduced
,
will
discover
the
value
of
e
,
viz.
|
|
From
the
Eleventh
,
thirteenth
,
and
fifteenth
steps
is
gathered
this
CANON
.
If
the
number
given
to
be
added
to
the
three
numbers
required
be
divided
by
11
,
the
Quotient
will
give
the
first
number
,
and
its
Quintuple
(
or
Product
by
5
)
being
divided
by
11
,
will
give
the
second
number
,
and
its
septuple
(
or
product
by
7
)
being
divided
by
11
,
will
give
the
third
number
.
By
which
Canon
the
numbers
required
in
the
Question
are
11
,
55
,
and
77
,
(
the
second
being
5
times
as
much
as
the
first
,
and
the
third
is
7
times
as
much
as
the
first
)
which
said
numbers
will
satisfie
the
conditions
of
the
Question
,
as
will
appear
by
The
Proof
.
which
was
to
be
done
.
QUEST.
7.
What
two
numbers
are
those
that
if
to
10
times
the
greater
there
be
added
six
times
the
lesser
,
the
sum
will
be
228
(
or
b
)
and
if
from
4
times
the
greater
you
subtract
2
times
the
lesser
,
the
remainder
will
be
56
(
or
c
)
?
for
the
two
numbers
put
a
and
e
,
and
then
the
foregoing
question
being
abstracted
from
words
,
may
be
stated
thus
.
viz.
What
are
the
numbers
a
and
e
?
RESOLUTION
.
3
The
first
equation
(
accordding
to
the
third
Rule
)
being
multiplyed
by
4
,
which
is
prefixed
to
a
in
the
second
equation
,
produceth
|
|
4
And
the
second
Equation
being
multiplyed
by
10
,
which
is
prefixed
to
a
in
the
first
,
it
produceth
|
|
5
And
if
from
the
equation
in
the
third
step
you
subtract
the
equation
in
the
fourth
step
,
because
40a
is
found
in
both
(
according
to
the
second
Rule
)
there
ariseth
this
equation
,
viz.
|
|
6
Both
parts
of
the
equation
in
the
fifth
step
,
being
divided
by
44
,
the
value
of
e
will
be
discovered
to
be
|
|
7
If
in
stead
of
−
2e
in
the
second
step
,
you
put
double
the
latter
part
of
the
equation
in
the
sixth
step
you
will
have
this
equation
.
|
|
8
The
seventh
equation
being
duly
reduced
,
the
value
of
a
will
be
discovered
to
be
|
|
By
the
sixth
and
eighth
steps
the
numbers
sought
are
●…8
and
8
,
which
will
answer
the
conditions
of
the
Question
as
you
may
perceive
by
The
Proof
.
Soli
Deo
Gloria
.
FINIS
.