DECIMAL FRACTIONS.
126. We have seen (112) (121) the necessity of reducing fractions
to a common denominator, in order to compare their magnitudes. We
have seen also how much more readily operations are performed upon
fractions which have the same, than upon those which have different,
denominators. On this account it has long been customary, in all those
parts of mathematics where fractions are often required, to use none
but such as either have, or can be easily reduced to others having, the
same denominators. Now, of all numbers, those which can be most easily
managed are such as 10, 100, 1000, &c., where 1 is followed by ciphers.
These are called DECIMAL NUMBERS; and a fraction whose denominator is
any one of them, is called a DECIMAL FRACTION, or more commonly, a
DECIMAL.
127. A whole number may be reduced to a decimal fraction, or one
decimal fraction to another, with the greatest ease. For example,
940 9400 94000
94 is ---, or ----, or ----- (106);
10 100 1000
3 30 300 3000
---- is ----, or ----, or ----- (108).
30 100 1000 10000
The placing of a cipher on the right hand of any number is the same
thing as multiplying that number by 10 (57), and this may be done as
often as we please in the numerator of a fraction, provided it be done
as often in the denominator (108).
128. The next question is, How can we reduce a fraction which is
not decimal to another which is, without altering its value? Take,
for example, the fraction ⁷/₁₆, multiply both the numerator and
denominator successively by 10, 100, 1000, &c., which will give a
series of fractions, each of which is equal to ⁷/₁₆ (108), viz. ⁷⁰/₁₆₀,
⁷⁰⁰/₁₆₀₀, ⁷⁰⁰⁰/₁₆₀₀₀, ⁷⁰⁰⁰⁰/₁₆₀₀₀₀, &c. The denominator of each of
these fractions can be divided without remainder by 16, the quotients
of which divisions form the series of decimal numbers 10, 100, 1000,
10000, &c. If, therefore, one of the numerators be divisible by 16,
the fraction to which that numerator belongs has a numerator and
denominator both divisible by 16. When that division has been made,
which (108) does not alter the value of the fraction, we shall have a
fraction whose denominator is one of the series 10, 100, 1000, &c.,
and which is equal in value to ⁷/₁₆. The question is then reduced to
finding the first of the numbers 70, 700, 7000, 70000, &c., which can
be divided by 16 without remainder.
Divide these numbers, one after the other, by 16, as follows:
16)70(4 16)700(43 16)7000(437 16)70000(4375
64 64 64 64
-- --- --- ---
6 60 60 60
48 48 48
-- --- ---
12 120 120
112 112
--- ---
8 80
80
--
0
It appears, then, that 70000 is the first of the numerators which is
divisible by 16. But it is not necessary to write down each of these
divisions, since it is plain that the last contains all which came
before. It will do, then, to proceed at once as if the number of
ciphers were without end, to stop when the remainder is nothing, and
then count the number of ciphers which have been used. In this case,
since 70000 is 16 × 4375,
70000 16 × 4375 4375
------, which is ----------, or -----,
160000 16 × 10000 10000
gives the fraction required.
Therefore, to reduce a fraction to a decimal fraction, annex ciphers
to the numerator, and divide by the denominator until there is no
remainder. The quotient will be the numerator of the required fraction,
and the denominator will be unity, followed by as many ciphers as were
used in obtaining the quotient.
EXERCISES.
Reduce to decimal fractions
½, ¼, ²/₂₅, ¹/₅₀, ³⁹²⁷/₁₂₅₀, and ⁴⁵³/₆₂₅.
_Answer_, ⁵/₁₀, ²⁵/₁₀₀, ⁸/₁₀₀, ²/₁₀₀, ³¹⁴¹⁶/₁₀₀₀₀, and ⁷²⁴⁸/₁₀₀₀₀.
129. It will happen in most cases that the annexing of ciphers to the
numerator will never make it divisible by the denominator without
remainder. For example, try to reduce ¹/₇ to a decimal fraction.
7)1000000000000000000, &c.
-------------------
142857142857142857, &c.
The quotient here is a continual repetition of the figures 1, 4, 2, 8,
5, 7, in the same order; therefore ¹/₇ cannot be reduced to a decimal
fraction. But, nevertheless, if we take as a numerator any number of
figures from the quotient 142857142857, &c., and as a denominator 1
followed by as many ciphers as were used in making that part of the
quotient, we shall get a fraction which differs very little from ¹/₇,
and which will differ still less from it if we put more figures in the
numerator and more ciphers in the denominator.
Thus, 1 {is less} 1 3 {which is not} 1
--- { } --- by --- { } ---
10 { than } 7 70 { so much as } 10
14 1 2 1
--- --- --- ---
100 7 700 100
142 1 6 1
---- --- ---- ----
1000 7 7000 1000
1428 1 4 1
----- --- ----- -----
10000 7 70000 10000
14285 1 5 1
------ --- ------ ------
100000 7 700000 100000
142857 1 1 1
------- --- ------- -------
1000000 7 7000000 1000000
&c. &c. &c. &c.
In the first column is a series of decimal fractions, which come nearer
and nearer to ¹/₇, as the third column shews. Therefore, though we
cannot find a decimal fraction which is exactly ¹/₇, we can find one
which differs from it as little as we please.
This may also be illustrated thus: It is required to reduce ¹/₇ to
a decimal fraction without the error of say a millionth of a unit;
multiply the numerator and denominator of ¹/₇ by a million, and then
divide both by 7; we have then
1 1000000 1428571¹/₇
--- = ------- = -----------
7 7000000 1000000
If we reject the fraction ¹/₇ in the numerator, what we reject is
really the 7th part of the millionth part of a unit; or less than the
millionth part of a unit. Therefore ¹⁴²⁸⁵⁷/₁₀₀₀₀₀₀ is the fraction
required.
EXERCISES.
Make similar tables} 3 17 1
with } ---, ---, and ---.
these fractions } 91 143 247
} 3
The recurring} --- is 329670,329670, &c.
quotient of} 91
17
--- 118881,118881, &c.
143
1
--- 404858299595141700,4048582 &c.
247
130. The reason for the _recurrence_ of the figures of the quotient
in the same order is as follows: If 1000, &c. be divided by the
number 247, the remainder at each step of the division is less than
247, being either 0, or one of the first 246 numbers. If, then, the
remainder never become nothing, by carrying the division far enough,
one remainder will occur a second time. If possible, let the first
246 remainders be all different, that is, let them be 1, 2, 3, &c.,
up to 246, variously distributed. As the 247th remainder cannot be so
great as 247, it must be one of these which have preceded. From the
step where the remainder becomes the same as a former remainder, it is
evident that former figures of the quotient must be repeated in the
same order.
131. You will here naturally ask, What is the use of decimal
fractions, if the greater number of fractions cannot be reduced at
all to decimals? The answer is this: The addition, subtraction,
multiplication, and division of decimal fractions are much easier than
those of common fractions; and though we cannot reduce all common
fractions to decimals, yet we can find decimal fractions so near to
each of them, that the error arising from using the decimal instead
of the common fraction will not be perceptible. For example, if we
suppose an inch to be divided into ten million of equal parts, one of
those parts by itself will not be visible to the eye. Therefore, in
finding a length, an error of a ten-millionth part of an inch is of no
consequence, even where the finest measurement is necessary. Now, by
carrying on the table in (129), we shall see that
1428571 1 1
-------- does not differ from --- by --------;
10000000 7 10000000
and if these fractions represented parts of an inch, the first might
be used for the second, since the difference is not perceptible. In
applying arithmetic to practice, nothing can be measured so accurately
as to be represented in numbers without any error whatever, whether it
be length, weight, or any other species of magnitude. It is therefore
unnecessary to use any other than decimal fractions, since, by means of
them, any quantity may be represented with as much correctness as by
any other method.
EXERCISES.
Find decimal fractions which do not differ from the following fractions
by ¹/₁₀₀₀₀₀₀₀₀.
⅓ _Answer_, ³³³³³³³³/₁₀₀₀₀₀₀₀₀.
⁴/₇ ⁵⁷¹⁴²⁸⁵⁷/₁₀₀₀₀₀₀₀₀.
¹¹³/₃₅₅ ³¹⁸³⁰⁹⁸⁵/₁₀₀₀₀₀₀₀₀.
³⁵⁵/₁₁₃ ³¹⁴¹⁵⁹²⁹²/₁₀₀₀₀₀₀₀₀.
132. Every decimal may be immediately reduced to a quantity consisting
either of a whole number and more simple decimals, or of more simple
decimals alone, having one figure only in each of the numerators. Take,
for example,
147326 147326 326
------. By (115) ------- is 147----;
1000 1000 1000
and since 326 is made up of 300, and 20, and 6; by (112) ³²⁶/₁₀₀₀₀ =
³⁰⁰/₁₀₀₀ + ²⁰/₁₀₀₀ + ⁶/₁₀₀₀. But (108) ³⁰⁰/₁₀₀₀ is ³/₁₀, and ²⁰/₁₀₀₀
is ²/₁₀₀. Therefore, ¹¹⁴⁷³²6/₁₀₀₀ is made up of 147 + ³/₁₀ + ²/₁₀₀ +
6/₁₀₀₀. Now, take any number, for example, 147326, and form a number
of fractions having for their numerators this number, and for their
denominators 1, 10, 100, 1000, 10000, &c., and reduce these fractions
into numbers and more simple decimals, in the foregoing manner, which
will give the table below.
DECOMPOSITION OF A DECIMAL FRACTION.
147326
------ = 147326
1
147326 6
------ = 14732 + ---
10 10
147326 2 6
------ = 1473 + --- + ---
100 10 100
147326 3 2 6
------ = 147 + --- + --- + ----
1000 10 100 1000
147326 7 3 2 6
------ = 14 + --- + --- + ---- + -----
10000 10 100 1000 10000
147326 4 7 3 2 6
------ = 1 + --- + --- + ---- + ----- + ------
100000 10 100 1000 10000 100000
147326 1 4 7 3 2 6
------- = --- + --- + ---- + ----- + ------ + -------
1000000 10 100 1000 10000 100000 1000000
147326 1 4 7 3 2 6
-------- = --- + ---- + ----- + ------ + ------- + --------
10000000 100 1000 10000 100000 1000000 10000000
N.B. The student should write this table himself, and then proceed to
make similar tables from the following exercises.
EXERCISES.
Reduce the following fractions into a series of numbers and more simple
fractions:
31415926 31415926
--------, --------, &c.
10 100
2700031 2700031
-------, --------, &c.
10 100
2073000 2073000
-------, --------, &c.
10 100
3331303 3331303
-------, -------, &c.
1000 10000
133. If, in this table, and others made in the same manner, you look at
those fractions which contain a whole number, you will see that they
may be made thus: Mark off, from the right hand of the numerator, as
many _figures_ as there are _ciphers_ in the denominator by a point, or
any other convenient mark.
This will give 14732·6 when the fraction is 147326
------
10
1473·26 147326
------
100
147·326 147326
------
1000
&c. &c.
The figures on the left of the point by themselves make the whole
number which the fraction contains. Of those on its right, the first
is the numerator of the fraction whose denominator is 10, the second
of that whose denominator is 100, and so on. We now come to those
fractions which do not contain a whole number.
134. The first of these is ¹⁴⁷³²⁶/₁₀₀₀₀₀₀ which the number of
_ciphers_ in the denominator is the same as the number of _figures_
in the numerator. If we still follow the same rule, and mark off all
the figures, by placing the point before them all, thus, ·147326,
the observation in (133) still holds good; for, on looking at
¹⁴⁷³²⁶/₁₀₀₀₀₀₀ in the table, we find it is
1 4 7 3 2 6
--- + --- + ---- + ----- + ------ + -------
10 100 1000 10000 100000 1000000
The next fraction is ¹⁴⁷³²⁶/₁₀₀₀₀₀₀₀, which we find by the table to be
1 4 7 3 2 6
--- + ---- + ----- + ------ + ------- + --------
100 1000 10000 100000 1000000 10000000
In this, 1 is not divided by 10, but by 100; if, therefore, we put a
point before the whole, the rule is not true, for the first figure on
the left of the point has the denominator which, according to the rule,
the second ought to have, the second that which the third ought to
have, and so on. In order to keep the same rule for this case, we must
contrive to make 1 the second figure on the right of the point instead
of the first. This may be done by placing a cipher between it and the
point, thus, ·0147326. Here the rule holds good, for by that rule this
fraction is
0 1 4 7 3 2 6
--- + --- + ---- + ----- + ------ + ------- + --------
10 100 1000 10000 100000 1000000 10000000
which is the same as the preceding line, since ⁰/₁₀ is 0, and need not
be reckoned.
Similarly, when there are two ciphers more in the denominator than
there are figures in the numerator, the rule will be true if we place
two ciphers between the point and the numerator. The rule, therefore,
stated fully, is this:
To reduce a decimal fraction to a whole number and more simple
decimals, or to more simple decimals alone if it do not contain a whole
number, mark off by a point as many figures from the numerator as there
are ciphers in the denominator. If the numerator have not places enough
for this, write as many ciphers before it as it wants places, and put
the point before these ciphers. Then, if there be any figures before
the point, they make the _whole number_ which the fraction contains.
The first figure after the point with the denominator 10, the second
with the denominator 100, and so on, are the _fractions_ of which the
first fraction is composed.
135. Decimal fractions are not usually written at full length. It is
more convenient to write the numerator only, and to cut off from the
numerator as many figures as there are ciphers in the denominator,
when that is possible, by a point. When there are more ciphers in the
denominator than figures in the numerator, as many ciphers are placed
before the numerator as will supply the deficiency, and the point is
placed before the ciphers. Thus, ·7 will be used in future to denote
⁷/₁₀, ·07 for ⁷/₁₀₀, and so on. The following tables will give the
whole of this notation at one view, and will shew its connexion with
the decimal notation explained in the first section. You will observe
that the numbers on the right of the units’ place stand for units
_divided_ by 10, 100, 1000, &c. while those on the left are units
_multiplied_ by 10, 100, 1000, &c.
The student is recommended always to write the decimal point in a line
with the top of the figures or in the middle, as is done here, and
never at the bottom. The reason is, that it is usual in the higher
branches of mathematics to use a point placed between two numbers or
letters which are multiplied together; thus, 15. 16, _a_. _b_, (_a_ +
_b_). (_c_ + _d_) stand for the products of those numbers or letters.
1234 4 4
I. 123·4 stands for ---- or 123--- or 123 + ---
10 10 10
1234 34 3 4
12·34 ---- or 12--- or 12 + --- + ---
100 100 10 100
1234 234 2 3 4
1·234 ---- or 1---- or 1 + --- + --- + ----
1000 1000 10 100 1000
1234 1 2 3 4
·1234 ----- or --- + --- + ---- + -----
10000 10 100 1000 10000
1234 1 2 3 4
·01234 ------ or --- + ---- + ----- + ------
100000 100 1000 10000 100000
1234 1 2 3 4
·001234 ------- or ---- + ----- + ------ + -------
1000000 1000 10000 100000 1000000
II. 1003 1 3
·01003 is ------ or --- + ------
100000 100 100000
1003 1 3
·1003 is ----- or --- + -----
10000 10 10000
1003 3
10·03 is ---- or 10 + ---
100 100
1003 3
100·3 is ---- or 100 + ---
10 10
III. 1 2 8 3
·1283 = --- + --- + ---- + -----
10 100 1000 10000
= ·1 + ·02 + ·008 + ·0003
= ·1 + ·0283 = ·12 + ·0083
= ·128 + ·0003 = ·108 + ·0203
= ·1003 + ·028 = ·1203 + ·008
{ 1 is 1000 inches
{ 2 is 200
{ 3 is 30
{ 4 is 4
IV. In 1234·56789 { 5 is ⁵/₁₀ of an inch
inches the { 6 is ⁶/₁₀₀
{ 7 is ⁷/₁₀₀₀
{ 8 is ⁸/₁₀₀₀₀
{ 9 is ⁹/₁₀₀₀₀₀
136. The ciphers on the right hand of the decimal point serve the same
purpose as the ciphers in (10). They are not counted as any thing
themselves, but serve to shew the place in which the accompanying
numbers stand. They might be dispensed with by writing the numbers in
ruled columns, as in the first section. They are distinguished from
the numbers which accompany them by calling the latter _significant
figures_. Thus, ·0003747 is a decimal of seven places with four
significant figures, ·346 is a decimal of three places with three
significant figures, &c.
137. The value of a decimal is not altered by putting any number of
ciphers on its right. Take, for example, ·3 and ·300. The first (135)
is ³/₁₀, and the second ³⁰⁰/₁₀₀₀, which is made from the first by
multiplying both its numerator and denominator by 100, and (108) is the
same quantity.
138. To reduce two decimals to a common denominator, put as many
ciphers on the right of that which has the smaller number of places
as will make the number of places in both fractions the same. Take,
for example, ·54 and 4·3297. The first is ⁵⁴/₁₀₀, and the second
⁴³²⁹⁷/₁₀₀₀₀. Multiply the numerator and denominator of the first by 100
(108), which reduces it to ⁵⁴⁰⁰/₁₀₀₀₀, which has the same denominator
as ⁴³²⁹⁷/₁₀₀₀₀. But ⁵⁴⁰⁰/₁₀₀₀₀ is ·5400 (135). In whole numbers, the
decimal point should be placed at the end: thus, 129 should be written
129·. It is, however, usual to omit the point; but you must recollect
that 129 and 129·000 are of the same value, since the first is 129 and
the second ¹²⁹⁰⁰⁰/₁₀₀₀.
139. The rules which were given in the last chapter for addition,
subtraction, multiplication, and division, apply to all fractions, and
therefore to decimal fractions among the rest. But the way of writing
decimal fractions, which is explained in this chapter, makes the
application of these rules more simple. We proceed to the different
cases.
Suppose it required to add 42·634, 45·2806, 2·001, and 54. By (112)
these must be reduced to a common denominator, which is done (138) by
writing them as follows: 42·6340, 45·2806, 2·0010, and 54·0000. These
are decimal fractions, whose numerators are 426340, 452806, 20010, and
540000, and whose common denominator is 10000. By (112) their sum is
426340 + 452806 + 20010 + 540000 1439156
--------------------------------, which is -------
10000 10000
or 143·9156. The simplest way of doing this is as follows: write the
decimals down under one another, so that the decimal points may fall
under one another, thus:
42·634
45·2806
2·001
54
--------
143·9156
Add the different columns together as in common addition, and place the
decimal point under the other decimal points.
EXERCISES.
What are 1527 + 64·732094 + 2·0013 + ·00001974;
2276·3 + ·107 + ·9 + 26·3172 + 56732·001;
and 1·11 + 7·7 + ·0039 + ·00142 + ·8838?
_Answer_, 1593·73341374, 59035·6252, 9·69912.
140. Suppose it required to subtract 91·07324 from 137·321. These
fractions when reduced to a common denominator are 91·07324 and
137·32100 (138). Their difference is therefore
13732100 - 9107324 4624776
------------------, which is -------
100000 100000
or 46·24776. This may be most simply done as follows: write the less
number under the greater, so that its decimal point may fall under that
of the greater, thus:
137·321
91·07324
---------
46·24776
Subtract the lower from the upper line, and wherever there is a figure
in one line and not in the other, proceed as if there were a cipher in
the vacant place.
EXERCISES.
What is 12362 - 274·22107 + ·5;
9976·2073942 - ·00143976728;
and 1·2 + ·03 + ·004 - ·0005?
_Answer_, 12088·27893, 9976·20595443272; and 1·2335.
141. The multiplication of a decimal by 10, 100, 1000, &c., is
performed by merely moving the decimal point to the right. Suppose,
for example, 13·2079 is to be multiplied by 100. The decimal is
¹³²⁰⁷⁹/₁₀₀₀₀, which multiplied by 100 is (117) ¹³²⁰⁷⁹/₁₀₀, or 1320·79.
Again, 1·309 × 100000 is ¹³⁰⁹/₁₀₀₀ × 100000, or (116) ¹³⁰⁹⁰⁰⁰⁰⁰/₁₀₀₀ or
130900. From these and other instances we get the following rule: To
multiply a decimal fraction by a decimal number (126), move the decimal
point as many places to the right as there are ciphers in the decimal
number. When this cannot be done, annex ciphers to the right of the
decimal (137) until it can.
142. Suppose it required to multiply 17·036 by 4·27. The first of these
decimals is ¹⁷⁰³⁶/₁₀₀₀, and the second ⁴²⁷/₁₀₀. By (118) the product
of these fractions has for its numerator the product of 17036 and 427,
and for its denominator the product of 1000 and 100; therefore this
product is ⁷²⁷⁴³⁷²/₁₀₀₀₀₀, or 72·74372. This may be done more shortly
by multiplying the two numbers 17036 and 427, and cutting off by the
decimal point as many places as there are decimal places both in 17·036
and 4·27, because the product of two decimal numbers will contain as
many ciphers as there are ciphers in both.
143. This question now arises: What if there should not be as many
figures in the product as there are decimal places in the multiplier
and multiplicand together? To see what must be done in this case,
multiply ·172 by ·101, or ¹⁷²/₁₀₀₀ by ¹⁰¹/₁₀₀₀. The product of these
two is ¹⁷³⁷²/₁₀₀₀₀₀₀, or ·017372 (135). Therefore, when the number of
places in the product is not sufficient to allow the rule of the last
article to be followed, as many ciphers must be placed at the beginning
as will make up the deficiency.
ADDITIONAL EXAMPLES.
·001 × ·01 is ·00001
56 × ·0001 is ·0056.
EXERCISES.
Shew that
3·002 × 3·002 = 3 × 3 + 2 × 3 × ·002 + ·002 × ·002
11·5609 × 5·3191 = 8·44 × 8·44 - 3·1209 × 3·1209
8·217 × 10·001 = 8 × 10 + 8 × ·001 + 10 × ·217 + ·001 × ·217.
Fraction. Square. Cube.
82·92 6875·7264 570135·233088
·0173 ·00029929 ·000005177717
1·43 2·0449 2·924207
·009 ·000081 ·000000729
15·625 × 64 = 1000
1·5625 × ·64 = 1
·015625 × ·0064 = ·0001
·15625 × ·64 = ·1
1562·5 × ·064 = 100
15625000 × ·064 = 1000000
144. The division of a decimal by a decimal number, such as 10, 100,
1000, &c., is performed by moving the decimal point as many places to
the left as there are ciphers in the decimal number. If there are not
places enough in the dividend to allow of this, annex ciphers to the
beginning of it until there are. For example, divide 1734·229 by 1000:
the decimal fraction is ¹⁷³⁴²²⁹/₁₀₀₀, which divided by 1000 (123) is
¹⁷³⁴²²⁹/₁₀₀₀₀₀₀, or 1·734229. If, in the same way, 1·2106 be divided by
10000, the result is ·00012106.
145. Before proceeding to shorten the rule for the division of one
decimal fraction by another, it will be necessary to resume what was
said in (128) upon the reduction of any fraction to a decimal fraction.
It was there shewn that ⁷/₁₆ is the same fraction as ⁴³⁷⁵/₁₀₀₀₀ or
·4375. As another example, convert ³/₁₂₈ into a decimal fraction.
Follow the same process as in (128), thus:
128)300000000000(234375
256
----
440
384
----
560
512
----
480
384
----
960
896
----
640
640
---
0
Since 7 ciphers are used, it appears that 30000000 is the first of the
series 30, 300, &c., which is divisible by 128; and therefore ³/₁₂₈
or, which is the same thing (108), ³⁰⁰⁰⁰⁰⁰⁰/₁₂₈₀₀₀₀₀₀₀ is equal to
²³⁴³⁷⁵/₁₀₀₀₀₀₀₀ or ·0234375 (135).
From these examples the rule for reducing a fraction to a decimal is:
Annex ciphers to the numerator; divide by the denominator, and annex
a cipher to each remainder after the figures of the numerator are all
used, proceeding exactly as if the numerator had an unlimited number
of ciphers annexed to it, and was to be divided by the denominator.
Continue this process until there is no remainder, and observe how many
ciphers have been used. Place the decimal point in the quotient so as
to cut off as many figures as you have used ciphers; and if there be
not figures enough for this, annex ciphers to the beginning until there
are places enough.
146. From what was shewn in (129), it appears that it is not every
fraction which can be reduced to a decimal fraction. It was there
shewn, however, that there is no fraction to which we may not find a
decimal fraction as near as we please. Thus, ¹/₁₀, ¹⁴/₁₀₀, ¹⁴²/₁₀₀₀,
¹⁴²⁸/₁₀₀₀₀, ¹⁴²⁸⁵/₁₀₀₀₀₀, &c., or ·1, ·14, ·142, ·1428, ·14285, were
shewn to be fractions which approach nearer and nearer to ¹/₇. To find
either of these fractions, the rule is the same as that in the last
article, with this exception, that, I. instead of stopping when there
is no remainder, which never happens, stop at any part of the process,
and make as many decimal places in the quotient as are equal in number
to the number of ciphers which have been used, annexing ciphers to the
beginning when this cannot be done, as before. II. Instead of obtaining
a fraction which is exactly equal to the fraction from which we set
out, we get a fraction which is very near to it, and may get one still
nearer, by using more of the quotient. Thus, ·1428 is very near to ¹/₇,
but not so near as ·142857; nor is this last, in its turn, so near as
·142857142857, &c.
147. If there should be ciphers in the numerator of a fraction, these
must not be reckoned with the number of ciphers which are necessary in
order to follow the rule for changing it into a decimal fraction. Take,
for example, ¹⁰⁰/₁₂₅; annex ciphers to the numerator, and divide by the
denominator. It appears that 1000 is divisible by 125, and that the
quotient is 8. One cipher only has been annexed to the numerator, and
therefore 100 divided by 125 is ·8. Had the fraction been ¹/₁₂₅, since
1000 divided by 125 gives 8, and three ciphers would have been annexed
to the numerator, the fraction would have been ·008.
148. Suppose that the given fraction has ciphers at the right of its
denominator; for example, ³¹/₂₅₀₀. Then annexing a cipher to the
numerator is the same thing as taking one away from the denominator;
for, (108) ³¹⁰/₂₅₀₀ is the same thing as ³¹/₂₅₀, and ³¹⁰/₂₅₀ as ³¹/₂₅.
The rule, therefore, is in this case: Take away the ciphers from the
denominator.
EXERCISES.
Reduce the following fractions to decimal fractions:
1 36 297 1
---, ----, ----, and ---.
800 1250 64 128
_Answer_, ·00125, ·0288, 4·640625, and ·0078125.
Find decimals of 6 places very near to the following fractions:
27 156 22 194 2637 1 1 3
--, ---, -----, ---, ----, ----, ---, and ---.
49 33 37000 13 9907 2908 466 277
_Answer_, ·551020, 4·727272, ·000594, 14·923076, ·266175,
·000343, ·002145, and ·010830.
149. From (121) it appears, that if two fractions have the same
denominator, the first may be divided by the second by dividing the
numerator of the first by the numerator of the second. Suppose it
required to divide 17·762 by 6·25. These fractions (138), when reduced
to a common denominator, are 17·762 and 6·250, or ¹⁷⁷⁶²/₁₀₀₀ and
⁶²⁵⁰/₁₀₀₀. Their quotient is therefore ¹⁷⁷⁶²/₆₂₅₀, which must now be
reduced to a decimal fraction by the last rule. The process at full
length is as follows: Leave out the cipher in the denominator, and
annex ciphers to the numerator, or, which will do as well, to the
remainders, when it becomes necessary, and divide as in (145).
625)17762(284192
1250
-----
5262
5000
-----
2620
2500
-----
1200
625
-----
5750
5625
-----
1250
1250
----
0
Here four ciphers have been annexed to the numerator, and one has been
taken from the denominator. Make five decimal places in the quotient,
which then becomes 2·84192, and this is the quotient of 17·762 divided
by 6·25.
150. The rule for division of one decimal by another is as follows:
Equalise the number of decimal places in the dividend and divisor,
by annexing ciphers to that which has fewest places. Then, further,
annex as many ciphers to the dividend[18] as it is required to have
decimal places, throw away the decimal point, and operate as in common
division. Make the required number of decimal places in the quotient.
[18] Or remove ciphers from the divisor; or make up the number of
ciphers partly by removing from the divisor and annexing to the
dividend, if there be not a sufficient number in the divisor.
Thus, to divide 6·7173 by ·014 to three decimal places, I first write
6·7173 and ·0140, with four places in each. Having to provide for three
decimal places, I should annex three ciphers to 6·7173; but, observing
that the divisor ·0140 has one cipher, I strike that one out and annex
two ciphers to 6·7173. Throwing away the decimal points, then divide
6717300 by 014 or 14 in the usual way, which gives the quotient 479807
and the remainder 2. Hence 479·807 is the answer.
The common rule is: Let the quotient contain as many decimal places
as there are decimal places in the dividend more than in the divisor.
But this rule becomes inoperative except when there are more decimals
in the dividend than in the divisor, and a number of ciphers must
be annexed to the former. The rule in the text amounts to the same
thing, and provides for an assigned number of decimal places. But the
student is recommended to make himself familiar with the rule of the
_characteristic_ given in the Appendix, and also to accustom himself
to _reason out_ the place of the decimal point. Thus, it should be
visible, that 26·119 ÷ 7·2436 has one figure before the decimal point,
and that 26·119 ÷ 724·36 has one cipher after it, preceding all
significant figures.
Or the following rule may be used: Expunge the decimal point of the
divisor, and move that of the dividend as many places to the right as
there were places in the divisor, using ciphers if necessary. Then
proceed as in common division, making one decimal place in the quotient
for every decimal place of the final dividend which is used. Thus
17·314 divided by 61·2 is 173·14 divided by 612, and the decimal point
must precede the first figure of the quotient. But 17·314 divided by
6617·5 is 173·14 by 66175; and since three decimal places of 173·14000
... must be used before a quotient figure can be found, that quotient
figure is the third decimal place, or the quotient is ·002.....
EXAMPLES.
3·1 ·00062
----- = 1240, ------ = ·00096875
·0025 ·64
EXERCISES.
15·006 × 15·006 - ·004 × ·004
Shew that ----------------------------- = 15·002,
15·01
and that
·01 × ·01 × ·01 + 2·9 × 2·9 × 2·9
--------------------------------- = 2·9 × 2·9 - 2·9 × ·01 + ·01 × ·01
2·91
1 1 365
What are -------, ---------, and ------, as far as 6 places
3·14159 2·7182818 ·18349
of decimals?--_Answer_, ·318310, ·367879, and 1989·209221.
Calculate 10 terms of each of the following series, as far as 5 places
of decimals.
1 1 1 1
1 + --- + ----- + --------- + ------------- + &c. = 1·71824.
2 2 × 3 2 × 3 × 4 2 × 3 × 4 × 5
1 1 1 1
1 + --- + --- + --- + --- + &c. = 2·92895.
2 3 4 5
80 81 82 83 84
---- + ---- + ---- + ---- + ---- + &c. = 9·88286.
81 82 83 84 85
151. We now enter upon methods by which unnecessary trouble is saved in
the computation of decimal quantities. And first, suppose a number of
miles has been measured, and found to be 17·846217 miles. If you were
asked how many miles there are in this distance, and a rough answer
were required which should give miles only, and not parts of miles,
you would probably say 17. But this, though the number of whole miles
contained in the distance, is not the nearest number of miles; for,
since the distance is more than 17 miles and 8 tenths, and therefore
more than 17 miles and a half, it is nearer the truth to say, it is 18
miles. This, though too great, is not so much too great as the other
was too little, and the error is not so great as half a mile. Again,
if the same were required within a tenth of a mile, the correct answer
is 17·8; for though this is too little by ·046217, yet it is not so
much too little as 17·9 is too great; and the error is less than half
a tenth, or ¹/₂₀. Again, the same distance, within a hundredth of a
mile, is more correctly 17·85 than 17·84, since the last is too little
by ·006217, which is greater than the half of ·01; and therefore 17·84
+ ·01 is nearer the truth than 17·84. Hence this general rule: When a
certain number of the decimals given is sufficiently accurate for the
purpose, strike off the rest from the right hand, observing, if the
first figure struck off be equal to or greater than 5, to increase the
last remaining figure by 1.
The following are examples of a decimal abbreviated by one place at a
time.
3·14159, 3·1416, 3·142, 3·14, 3·1, 3·0
2·7182818, 2·718282, 2·71828, 2·7183, 2·718, 2·72, 2·7, 3·0
1·9919, 1·992, 1·99, 2·00, 2·0
152. In multiplication and division it is useless to retain more
places of decimals in the result than were certainly correct in
the multiplier, &c., which gave that result. Suppose, for example,
that 9·98 and 8·96 are distances in inches which have been measured
correctly to two places of decimals, that is, within half a hundredth
of an inch each way. The real value of that which we call 9·98 may be
any where between 9·975 and 9·985, and that of 8·96 may be any where
between 8·955 and 8·965. The product, therefore, of the numbers which
represent the correct distances will lie between 9·975 × 8·955 and
9·985 × 8·965, that is, taking three decimal places in the products,
between 89·326 and 89·516. The product of the actual numbers given
is 89·4208. It appears, then, that in this case no more than the
whole number 89 can be depended upon in the product, or, at most,
the first place of decimals. The reason is, that the error made in
measuring 8·96, though only in the third place of decimals, is in
the multiplication increased at least 9·975, or nearly 10 times; and
therefore affects the second place. The following simple rule will
enable us to judge how far a product is to be depended upon. Let _a_ be
the multiplier, and _b_ the multiplicand; if these be true only to the
first decimal place, the product is within (_a_ + _b_)/20[19] of the
truth; if to two decimal places, within (_a_ + _b_)/200; if to three,
within (_a_ + _b_)/2000; and so on. Thus, in the above example, we have
9·98 and 8·96, which are true to two decimal places: their sum divided
by 200 is ·0947, and their product is 89·4208, which is therefore
within ·0947 of the truth. If, in fact, we increase and diminish
89·4208 by ·0947, we get 89·5155 and 89·3261, which are very nearly
the limits found within which the product must lie. We see, then, that
we cannot in this case depend upon the first place of decimals, as
(151) an error of ·05 cannot exist if this place be correct; and here
is a possible error of ·09 and upwards. It is hardly necessary to say,
that if the numbers given be exact, their product is exact also, and
that this article applies where the numbers given are correct only to
a certain number of decimal places. The rule is: Take half the sum
of the multiplier and multiplicand, remove the decimal point as many
places to the left as there are correct places of decimals in either
the multiplier or multiplicand; the result is the quantity within which
the product can be depended upon. In division, the rule is: Proceed
as in the last rule, putting the dividend and divisor in place of the
multiplier and multiplicand, and divide by the _square_ of the divisor;
the quotient will be the quantity within which the division of the
first dividend and divisor may be depended upon. Thus, if 17·324 be
divided by 53·809, both being correct to the third place, their half
sum will be 35·566, which, by the last rule, is made ·035566, and is to
be divided by the square of 53·809, or, which will do as well for our
purpose, the square of 50, or 2500. The result is something less than
·00002, so that the quotient of 17·324 and 53·809 can be depended on to
four places of decimals.
[19] These are not quite correct, but sufficiently so for every
practical purpose.
153. It is required to multiply two decimal fractions together, so as
to retain in the product only a given number of decimal places, and
dispense with the trouble of finding the rest. First, it is evident
that we may write the figures of any multiplier in a contrary order
(for example, 4321 instead of 1234), provided that in the operation we
move each line one place to the right instead of to the left, as in the
following example:
2221 2221
1234 4321
---- ----
8884 2221
6663 4442
4442 6663
2221 8884
------- -------
2740714 2740714
Suppose now we wish to multiply 348·8414 by 51·30742, reserving only
four decimal places in the product. If we reverse the multiplier, and
proceed in the manner just pointed out, we have the following:
3488414
2470315 |
---------+
17442070 |
3488414|
1046524|2
24418|898
1395|3656
69|76828
----------+------
17898·1522|23188
Cut off, by a vertical line, the first four places of decimals, and
the columns which produced them. It is plain that in forming our
abbreviated rule, we have to consider only, I. all that is on the left
of the vertical line; II. all that is carried from the first column on
the right of the line. On looking at the first column to the left of
the line, we see 4, 4, 8, 5, 9, of which the first 4 comes from 4 ×
1′,[20] the second 4 from 1 × 3′, the 8 from 8 × 7′, the 5 from 8 × 4′,
and the 9 from 4 × 2′. If, then, we arrange the multiplicand and the
reversed multiplier thus,
3488414
2470315
each figure of the multiplier is placed under the first figure of
the multiplicand which is used with it in forming the first _four_
places of decimals. And here observe, that the units’ figure in the
multiplier 51·30742, viz. 1, comes under 4, the _fourth_ decimal
place in the multiplicand. If there had been no carrying from the
right of the vertical line, the rule would have been: Reverse the
multiplier, and place it under the multiplicand, so that the figure
which was the units’ figure in the multiplier may stand under the last
place of decimals in the multiplicand which is to be preserved; place
ciphers over those figures of the multiplier which have none of the
multiplicand above them, if there be any: proceed to multiply in the
usual way, but begin each figure of the multiplier with the figure of
the multiplicand which comes above it, taking no account of those on
the right: place the first figures of all the lines under one another.
To correct this rule, so as to allow for what is carried from the right
of the vertical line, observe that this consists of two parts, 1st,
what is carried directly in the formation of the different lines, and
2dly, what is carried from the addition of the first column on the
right. The first of these may be taken into account by beginning each
figure of the multiplier with the one which comes on its right in the
multiplicand, and carrying the tens to the next figure as usual, but
without writing down the units. But both may be allowed for at once,
with sufficient correctness, on the principle of (151), by carrying
1 from 5 up to 15, 2 from 15 up to 25, &c.; that is, by carrying the
nearest ten. Thus, for 37, 4 would be carried, 37 being nearer to 40
than to 30. This will not always give the last place quite correctly,
but the error may be avoided by setting out so as to keep one more
place of decimals in the product than is absolutely required to be
correct. The rule, then, is as follows:
[20] The 1′ here means that the 1 is in the multiplier.
154. To multiply two decimals together, retaining only _n_ decimal
places.
I. Reverse the multiplier, strike out the decimal points, and place the
multiplier under the multiplicand, so that what was its units’ figure
shall fall under the _n_ᵗʰ decimal place of the multiplicand, placing
ciphers, if necessary, so that every place of the multiplier shall have
a figure or cipher above it.
II. Proceed to multiply as usual, beginning each figure of the
multiplier with the one which is in the place to its right in the
multiplicand: do not set down this first figure, but carry its
_nearest_ ten to the next, and proceed.
III. Place the first figures of all the lines under one another; add as
usual; and mark off _n_ places from the right for decimals.
It is required to multiply 136·4072 by 1·30609, retaining 7 decimal
places.
1364072000
906031
----------
1364072000
409221600
8184432
122766
-----------
178·1600798
In the following examples the first two lines are the multiplicand and
multiplier; and the number of decimals to be retained will be seen from
the results.
·4471618 33·166248 3·4641016
3·7719214 1·4142136 1732·508
========= ========== ============
37719214 033166248 346410160
8161744 63124141 8052371
-------- ---------- ------------
15087686 3316625 346410160
1508768 1326650 242487112
264034 33166 10392305
3772 13266 692820
2263 663 173205
38 33 2771
30 10 ------------
--------- 2 6001·58373
1·6866591 --------
46·90415
Exercises may be got from article (143).
155. With regard to division, take any two numbers, for example,
16·80437921 and 3·142, and divide the first by the second, as far as
any required number of decimal places, for example, five. This gives
the following:
3·142)16·80437921(5·34830
15·710
-------
1·0943 |
9426 |
-----|
15177|
(A) 12568|
---- -----|-
2609 2609|9
2514 2513|6
---- ----|--
95 96|32
94 94|26
-- --|---
1 2|061
Now cut off by a vertical line, as in (153), all the figures which
come on the right of the first figure 2, in the last remainder 2061.
As in multiplication, we may obtain all that is on the left of the
vertical line by an abbreviated method, as represented at (A). After
what has been said on multiplication, it is useless to go further
into the detail; the following rule will be sufficient: To divide one
decimal by another, retaining only _n_ places: Proceed one step in
the ordinary division, and determine, by (150), in what place is the
quotient so obtained; proceed in the ordinary way, until the number of
figures remaining to be found in the quotient is less than the number
of figures in the divisor: if this should be already the case, proceed
no further in the ordinary way. Instead of annexing a figure or cipher
to the remainder, cut off a figure from the divisor, and proceed one
step with this curtailed divisor as usual, remembering, however, in
multiplying this divisor, to carry the _nearest ten_, as in (154), from
the figure which was struck off; repeat this, striking off another
figure of the divisor, and so on, until no figures are left. Since we
know from the beginning in what place the first figure of the quotient
is, and also how many decimals are required, we can tell from the
beginning how many figures there will be in the whole quotient. If the
divisor contain more figures than the quotient, it will be unnecessary
to use them: and they may be rejected, the rest being corrected as in
(151): if there be ciphers at the beginning of the divisor, if it be,
for example,
·3178
·003178, since this is -----,
100
divide by ·3178 in the usual way, and afterwards multiply the quotient
by 100, or remove the decimal point two places to the right. If,
therefore, six decimals be required, eight places must be taken in
dividing by ·3178, for an obvious reason. In finding the last figure
of the quotient, the nearest should be taken, as in the second of the
subjoined examples.
Places required, 2 8
Divisor, ·41432 3·1415927
Dividend, 673·1489 2·71828180
41432 2·51327416
-------- ----------
258828 20500764
248592 18849556
------- --------
10237[21] 1651208
8286 1570796
----- -------
1951 80412
1657 62832
----- -----
294 17580
290 15708
--- -----
4 1872
4 1571
- ----
0 301
283
---
18
19
--
Quotient, 1624·71 ·86525596
[21] This is written 7 instead of 6, because the figure which is
abandoned in the dividend is 9 (151).
Examples may be obtained from (143) and (150).