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Title: Elements of arithmetic
Author: De Morgan, Augustus
Language: English
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ELEMENTS OF ARITHMETIC.
BY AUGUSTUS DE MORGAN,
OF TRINITY COLLEGE, CAMBRIDGE;
FELLOW OF THE ROYAL ASTRONOMICAL SOCIETY,
AND OF THE CAMBRIDGE PHILOSOPHICAL SOCIETY;
PROFESSOR OF MATHEMATICS IN UNIVERSITY COLLEGE, LONDON.
“Hominis studiosi est intelligere, quas utilitates
proprie afferat arithmetica his, qui solidam et
perfectam doctrinam in cæteris philosophiæ partibus
explicant. Quod enim vulgo dicunt, principium esse
dimidium totius, id vel maxime in philosophiæ partibus
conspicitur.”--MELANCTHON.
“Ce n’est point par la routine qu’on e’instruit, c’est par
sa propre réflexion; et il est essentiel de contracter
l’habitude de se rendre raison de ce qu’on fait: cette
habitude s’acquiert plus facilement qu’on ne pense; et une
fois acquise, elle ne se perd plus.”--CONDILLAC.
_SEVENTEENTH THOUSAND._
LONDON:
WALTON AND MABERLY,
UPPER GOWER STREET, AND IVY LANE, PATERNOSTER ROW.
M.DCCC.LVIII.
LONDON:
PRINTED BY J. WERTHEIMER AND CO.,
CIRCUS-PLACE, FINSBURY-CIRCUS.
PREFACE.
The preceding editions of this work were published in 1830, 1832,
1835, and 1840. This fifth edition differs from the three preceding,
as to the body of the work, in nothing which need prevent the four,
or any two of them, from being used together in a class. But it is
considerably augmented by the addition of eleven new Appendixes,[1]
relating to matters on which it is most desirable that the advanced
student should possess information. The first Appendix, on
_Computation_, and the sixth, on _Decimal Money_, should be read and
practised by every student with as much attention as any part of the
work. The mastery of the rules for instantaneous conversion of the
usual fractions of a pound sterling into decimal fractions, gives the
possessor the greater part of the advantage which he would derive from
the introduction of a decimal coinage.
At the time when this work was first published, the importance
of establishing arithmetic in the young mind upon reason and
demonstration, was not admitted by many. The case is now altered:
schools exist in which rational arithmetic is taught, and mere rules
are made to do no more than their proper duty. There is no necessity
to advocate a change which is actually in progress, as the works which
are published every day sufficiently shew. And my principal reason for
alluding to the subject here, is merely to warn those who want nothing
but routine, that this is not the book for their purpose.
A. DE MORGAN.
_London, May 1, 1846._
[1] Some separate copies of these Appendixes are printed, for those who
may desire to add them to the former editions.
TABLE OF CONTENTS.
BOOK I.
SECTION PAGE
I. Numeration 1
II. Addition and Subtraction 14
III. Multiplication 24
IV. Division 34
V. Fractions 51
VI. Decimal Fractions 65
VII. Square Root 89
VIII. Proportion 100
IX. Permutations and Combinations 118
BOOK II.
I. Weights and Measures, &c. 124
II. Rule of Three 144
III. Interest, &c. 150
APPENDIX.
I. On the mode of computing 161
II. On verification by casting out nines and elevens 166
III. On scales of notation 168
IV. On the definition of fractions 171
V. On characteristics 174
VI. On decimal money 176
VII. On the main principle of book-keeping 180
VIII. On the reduction of fractions to others of nearly
equal value 190
IX. On some general properties of numbers 193
X. On combinations 201
XI. On Horner’s method of solving equations 210
XII. Rules for the application of arithmetic to geometry 217
ELEMENTS OF ARITHMETIC.
BOOK I.
PRINCIPLES OF ARITHMETIC.
SECTION I.
NUMERATION.
1. Imagine a multitude of objects of the same kind assembled together;
for example, a company of horsemen. One of the first things that must
strike a spectator, although unused to counting, is, that to each man
there is a horse. Now, though men and horses are things perfectly
unlike, yet, because there is one of the first kind to every one of the
second, one man to every horse, a new notion will be formed in the mind
of the observer, which we express in words by saying that there is the
same _number_ of men as of horses. A savage, who had no other way of
counting, might remember this number by taking a pebble for each man.
Out of a method as rude as this has sprung our system of calculation,
by the steps which are pointed out in the following articles. Suppose
that there are two companies of horsemen, and a person wishes to
know in which of them is the greater number, and also to be able to
recollect how many there are in each.
2. Suppose that while the first company passes by, he drops a pebble
into a basket for each man whom he sees. There is no connexion between
the pebbles and the horsemen but this, that for every horseman there
is a pebble; that is, in common language, the _number_ of pebbles and
of horsemen is the same. Suppose that while the second company passes,
he drops a pebble for each man into a second basket: he will then have
two baskets of pebbles, by which he will be able to convey to any
other person a notion of how many horsemen there were in each company.
When he wishes to know which company was the larger, or contained most
horsemen, he will take a pebble out of each basket, and put them aside.
He will go on doing this as often as he can, that is, until one of the
baskets is emptied. Then, if he also find the other basket empty, he
says that both companies contained the same number of horsemen; if the
second basket still contain some pebbles, he can tell by them how many
more were in the second than in the first.
3. In this way a savage could keep an account of any numbers in which
he was interested. He could thus register his children, his cattle,
or the number of summers and winters which he had seen, by means
of pebbles, or any other small objects which could be got in large
numbers. Something of this sort is the practice of savage nations at
this day, and it has in some places lasted even after the invention
of better methods of reckoning. At Rome, in the time of the republic,
the prætor, one of the magistrates, used to go every year in great
pomp, and drive a nail into the door of the temple of Jupiter; a way of
remembering the number of years which the city had been built, which
probably took its rise before the introduction of writing.
4. In process of time, names would be given to those collections of
pebbles which are met with most frequently. But as long as small
numbers only were required, the most convenient way of reckoning them
would be by means of the fingers. Any person could make with his
two hands the little calculations which would be necessary for his
purposes, and would name all the different collections of the fingers.
He would thus get words in his own language answering to one, two,
three, four, five, six, seven, eight, nine, and ten. As his wants
increased, he would find it necessary to give names to larger numbers;
but here he would be stopped by the immense quantity of words which
he must have, in order to express all the numbers which he would be
obliged to make use of. He must, then, after giving a separate name
to a few of the first numbers, manage to express all other numbers by
means of those names.
5. I now shew how this has been done in our own language. The English
names of numbers have been formed from the Saxon: and in the following
table each number after ten is written down in one column, while
another shews its connexion with those which have preceded it.
One eleven ten and one[2]
two twelve ten and two
three thirteen ten and three
four fourteen ten and four
five fifteen ten and five
six sixteen ten and six
seven seventeen ten and seven
eight eighteen ten and eight
nine nineteen ten and nine
ten twenty two tens
twenty-one two tens and one fifty five tens
twenty-two two tens and two sixty six tens
&c. &c. &c. &c. seventy seven tens
thirty three tens eighty eight tens
&c. &c. ninety nine tens
forty four tens a hundred ten tens
&c. &c.
a hundred and one ten tens and one
&c. &c.
a thousand ten hundreds
ten thousand
a hundred thousand
a million ten hundred thousand
or one thousand thousand
ten millions
a hundred millions
&c.
[2] It has been supposed that _eleven_ and _twelve_ are derived
from the Saxon for _one left_ and _two left_ (meaning, after ten is
removed); but there seems better reason to think that _leven_ is a word
meaning ten, and connected with _decem_.
6. Words, written down in ordinary language, would very soon be too
long for such continual repetition as takes place in calculation. Short
signs would then be substituted for words; but it would be impossible
to have a distinct sign for every number: so that when some few signs
had been chosen, it would be convenient to invent others for the rest
out of those already made. The signs which we use areas follow:
0 1 2 3 4 5 6 7 8 9
nought one two three four five six seven eight nine
I now proceed to explain the way in which these signs are made to
represent other numbers.
7. Suppose a man first to hold up one finger, then two, and so on,
until he has held up every finger, and suppose a number of men to do
the same thing. It is plain that we may thus distinguish one number
from another, by causing two different sets of persons to hold up each
a certain number of fingers, and that we may do this in many different
ways. For example, the number fifteen might be indicated either by
fifteen men each holding up one finger, or by four men each holding up
two fingers and a fifth holding up seven, and so on. The question is,
of all these contrivances for expressing the number, which is the most
convenient? In the choice which is made for this purpose consists what
is called the method of _numeration_.
8. I have used the foregoing explanation because it is very probable
that our system of numeration, and almost every other which is used
in the world, sprung from the practice of reckoning on the fingers,
which children usually follow when first they begin to count. The
method which I have described is the rudest possible; but, by a little
alteration, a system may be formed which will enable us to express
enormous numbers with great ease.
9. Suppose that you are going to count some large number, for example,
to measure a number of yards of cloth. Opposite to yourself suppose a
man to be placed, who keeps his eye upon you, and holds up a finger for
every yard which he sees you measure. When ten yards have been measured
he will have held up ten fingers, and will not be able to count any
further unless he begin again, holding up one finger at the eleventh
yard, two at the twelfth, and so on. But to know how many have been
counted, you must know, not only how many fingers he holds up, but also
how many times he has begun again. You may keep this in view by placing
another man on the right of the former, who directs his eye towards his
companion, and holds up one finger the moment he perceives him ready
to begin again, that is, as soon as ten yards have been measured. Each
finger of the first man stands only for one yard, but each finger of
the second stands for as many as all the fingers of the first together,
that is, for ten. In this way a hundred may be counted, because the
first may now reckon his ten fingers once for each finger of the second
man, that is, ten times in all, and ten tens is one hundred (5).[3]
Now place a third man at the right of the second, who shall hold up
a finger whenever he perceives the second ready to begin again. One
finger of the third man counts as many as all the ten fingers of the
second, that is, counts one hundred. In this way we may proceed until
the third has all his fingers extended, which will signify that ten
hundred or one thousand have been counted (5). A fourth man would
enable us to count as far as ten thousand, a fifth as far as one
hundred thousand, a sixth as far as a million, and so on.
[3] The references are to the preceding articles.
10. Each new person placed himself towards your left in the rank
opposite to you. Now rule columns as in the next page, and to the right
of them all place in words the number which you wish to represent;
in the first column on the right, place the number of fingers which
the first man will be holding up when that number of yards has been
measured. In the next column, place the fingers which the second man
will then be holding up; and so on.
|7th.|6th.|5th.|4th.|3rd.|2nd.|1st.|
I.| | | | | | 5 | 7 | fifty-seven.
II.| | | | | 1 | 0 | 4 | one hundred and four.
III.| | | | | 1 | 1 | 0 | one hundred and ten.
IV.| | | | 2 | 3 | 4 | 8 | two thousand three hundred
| | | | | | | | and forty-eight.
V.| | | 1 | 5 | 9 | 0 | 6 | fifteen thousand nine
| | | | | | | | hundred and six.
VI.| | 1 | 8 | 7 | 0 | 0 | 4 | one hundred and
| | | | | | | | eighty-seven thousand
| | | | | | | | and four.
VII.| 3 | 6 | 9 | 7 | 2 | 8 | 5 | three million, six hundred
| | | | | | | | and ninety-seven
| | | | | | | | thousand, two hundred and
| | | | | | | | eighty-five.
11. In I. the number fifty-seven is expressed. This means (5) five tens
and seven. The first has therefore counted all his fingers five times,
and has counted seven fingers more. This is shewn by five fingers of
the second man being held up, and seven of the first. In II. the number
one hundred and four is represented. This number is (5) ten tens and
four. The second person has therefore just reckoned all his fingers
once, which is denoted by the third person holding up one finger;
but he has not yet begun again, because he does not hold up a finger
until the first has counted ten, of which ten only four are completed.
When all the last-mentioned ten have been counted, he then holds up
one finger, and the first being ready to begin again, has no fingers
extended, and the number obtained is eleven tens, or ten tens and one
ten, or one hundred and ten. This is the case in III. You will now find
no difficulty with the other numbers in the table.
12. In all these numbers a figure in the first column stands for only
as many yards as are written under that figure in (6). A figure in
the second column stands, not for as many yards, but for as many tens
of yards; a figure in the third column stands for as many hundreds of
yards; in the fourth column for as many thousands of yards; and so on:
that is, if we suppose a figure to move from any column to the one on
its left, it stands for ten times as many yards as before. Recollect
this, and you may cease to draw the lines between the columns, because
each figure will be sufficiently well known by the _place_ in which it
is; that is, by the number of figures which come upon the right hand of
it.
13. It is important to recollect that this way of writing numbers,
which has become so familiar as to seem the _natural_ method, is not
more natural than any other. For example, we might agree to signify one
ten by the figure of one with an accent, thus, 1′; twenty or two tens
by 2′; and so on: one hundred or ten tens by 1″; two hundred by 2″; one
thousand by 1‴; and so on: putting Roman figures for accents when they
become too many to write with convenience. The fourth number in the
table would then be written 2‴ 3′ 4′ 8, which might also be expressed
by 8 4′ 3″ 2‴, 4′ 8 3″ 2‴; or the order of the figures might be changed
in any way, because their meaning depends upon the accents which are
attached to them, and not upon the place in which they stand. Hence,
a cipher would never be necessary; for 104 would be distinguished
from 14 by writing for the first 1″ 4, and for the second 1′ 4. The
common method is preferred, not because it is more exact than this, but
because it is more simple.
14. The distinction between our method of numeration and that of the
ancients, is in the meaning of each figure depending partly upon the
place in which it stands. Thus, in 44444 each four stands for four of
_something_; but in the first column on the right it signifies only
four of the pebbles which are counted; in the second, it means four
collections of ten pebbles each; in the third, four of one hundred
each; and so on.
15. The things measured in (11) were yards of cloth. In this case one
yard of cloth is called the _unit_. The first figure on the right is
said to be in the _units’ place_, because it only stands for so many
units as are in the number that is written under it in (6). The second
figure is said to be in the _tens’_ place, because it stands for a
number of tens of units. The third, fourth, and fifth figures are in
the places of the _hundreds_, _thousands_, and _tens of thousands_, for
a similar reason.
16. If the quantity measured had been acres of land, an acre of land
would have been called the _unit_, for the unit is _one_ of the things
which are measured. Quantities are of two sorts; those which contain an
exact number of units, as 47 yards, and those which do not, as 47 yards
and a half. Of these, for the present, we only consider the first.
17. In most parts of arithmetic, all quantities must have the same
unit. You cannot say that 2 yards and 3 feet make 5 _yards_ or 5
_feet_, because 2 and 3 make 5; yet you may say that 2 _yards_ and 3
_yards_ make 5 _yards_, and that 2 _feet_ and 3 _feet_ make 5 _feet_.
It would be absurd to try to measure a quantity of one kind with a unit
which is a quantity of another kind; for example, to attempt to tell
how many yards there are in a gallon, or how many bushels of corn there
are in a barrel of wine.
18. All things which are true of some numbers of one unit are true of
the same numbers of any other unit. Thus, 15 pebbles and 7 pebbles
together make 22 pebbles; 15 acres and 7 acres together make 22 acres,
and so on. From this we come to say that 15 and 7 make 22, meaning that
15 things of the same kind, and 7 more of the same kind as the first,
together make 22 of that kind, whether the kind mentioned be pebbles,
horsemen, acres of land, or any other. For these it is but necessary to
say, once for all, that 15 and 7 make 22. Therefore, in future, on this
part of the subject I shall cease to talk of any particular units, such
as pebbles or acres, and speak of numbers only. A number, considered
without intending to allude to any particular things, is called an
_abstract_ number: and it then merely signifies repetitions of a unit,
or the _number of times_ a unit is repeated.
19. I will now repeat the principal things which have been mentioned in
this chapter.
I. Ten signs are used, one to stand for nothing, the rest for the first
nine numbers. They are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The first of these
is called a _cipher_.
II. Higher numbers have not signs for themselves, but are signified
by placing the signs already mentioned by the side of each other, and
agreeing that the first figure on the right hand shall keep the value
which it has when it stands alone; that the second on the right hand
shall mean ten times as many as it does when it stands alone; that the
third figure shall mean one hundred times as many as it does when it
stands alone; the fourth, one thousand times as many; and so on.
III. The right hand figure is said to be in the _units’ place_, the
next to that in the _tens’ place_, the third in the _hundreds’ place_,
and so on.
IV. When a number is itself an exact number of tens, hundreds, or
thousands, &c., as many ciphers must be placed on the right of it as
will bring the number into the place which is intended for it. The
following are examples:
Fifty, or five tens, 50: seven hundred, 700.
Five hundred and twenty-eight thousand, 528000.
If it were not for the ciphers, these numbers would be mistaken for 5,
7, and 528.
V. A cipher in the middle of a number becomes necessary when any one of
the denominations, units, tens, &c. is wanting. Thus, twenty thousand
and six is 20006, two hundred and six is 206. Ciphers might be placed
at the beginning of a number, but they would have no meaning. Thus 026
is the same as 26, since the cipher merely shews that there are no
hundreds, which is evident from the number itself.
20. If we take out of a number, as 16785, any of those figures which
come together, as 67, and ask, what does this sixty-seven mean? of what
is it sixty-seven? the answer is, sixty-seven of the same collections
as the 7, when it was in the number; that is, 67 hundreds. For the 6
is 6 thousands, or 6 ten hundreds, or sixty hundreds; which, with the
7, or 7 hundreds, is 67 hundreds: similarly, the 678 is 678 tens. This
number may then be expressed either as
1 ten thousand 6 thousands 7 hundreds 8 tens and 5;
or 16 thousands 78 tens and 5; or 1 ten thousand 678 tens and 5;
or 167 hundreds 8 tens and 5; or 1678 tens and 5, and so on.
21. EXERCISES.
I. Write down the signs for--four hundred and seventy-six; two thousand
and ninety-seven; sixty-four thousand three hundred and fifty; two
millions seven hundred and four; five hundred and seventy-eight
millions of millions.
II. Write at full length 53, 1805, 1830, 66707, 180917324, 66713721,
90976390, 25000000.
III. What alteration takes place in a number made up entirely of nines,
such as 99999, by adding one to it?
IV. Shew that a number which has five figures in it must be greater
than one which has four, though the first have none but small figures
in it, and the second none but large ones. For example, that 10111 is
greater than 9879.
22. You now see that the convenience of our method of numeration arises
from a few simple signs being made to change their value as they
change the column in which they are placed. The same advantage arises
from counting in a similar way all the articles which are used in
every-day life. For example, we count money by dividing it into pounds,
shillings, and pence, of which a shilling is 12 pence, and a pound 20
shillings, or 240 pence. We write a number of pounds, shillings, and
pence in three columns, generally placing points between the columns.
Thus, 263 pence would not be written as 263, but as £1. 1. 11, where £
shews that the 1 in the first column is a pound. Here is a _system of
numeration_ in which a number in the second column on the right means
12 times as much as the same number in the first; and one in the third
column is twenty times as great as the same in the second, or 240 times
as great as the same in the first. In each of the tables of measures
which you will hereafter meet with, you will see a separate system of
numeration, but the methods of calculation for all will be the same.
23. In order to make the language of arithmetic shorter, some other
signs are used. They are as follow:
I. 15 + 38 means that 38 is to be added to 15, and is the same thing
as 53. This is the _sum_ of 15 and 38, and is read fifteen _plus_
thirty-eight (_plus_ is the Latin for _more_).
II. 64-12 means that 12 is to be taken away from 64, and is the same
thing as 52. This is the _difference_ of 64 and 12, and is read
sixty-four _minus_ twelve (_minus_ is the Latin for _less_).
III. 9 × 8 means that 8 is to be taken 9 times, and is the same thing
as 72. This is the _product_ of 9 and 8, and is read nine _into_ eight.
IV. 108/6 means that 108 is to be divided by 6, or that you must find
out how many sixes there are in 108; and is the same thing as 18. This
is the _quotient_ of 108 and 6; and is read a hundred and eight _by_
six.
V. When two numbers, or collections of numbers, with the foregoing
signs, are the same, the sign = is put between them. Thus, that 7
and 5 make 12, is written in this way, 7 + 5 = 12. This is called an
_equation_, and is read, seven _plus_ five _equals_ twelve. It is plain
that we may construct as many equations as we please. Thus:
12
7 + 9 - 3 = 12 + 1; --- - 1 + 3 × 2 = 11,
2
and so on.
24. It often becomes necessary to speak of something which is true not
of any one number only, but of all numbers. For example, take 10 and 7;
their sum[4] is 17, their difference is 3. If this sum and difference
be added together, we get 20, which is twice the greater of the two
numbers first chosen. If from 17 we take 3, we get 14, which is twice
the less of the two numbers. The same thing will be found to hold good
of any two numbers, which gives this general proposition,--If the sum
and difference of two numbers be added together, the result is twice
the greater of the two; if the difference be taken from the sum, the
result is twice the lesser of the two. If, then, we take _any_ numbers,
and call them the first number and the second number, and let the first
number be the greater; we have
[4] Any little computations which occur in the rest of this section may
be made on the fingers, or with counters.
(1st No. + 2d No.) + (1st No. - 2d No.) = twice 1st No.
(1st No. + 2d No.) - (1st No. - 2d No.) = twice 2d No.
The brackets here enclose the things which must be first done, before
the signs which join the brackets are made use of. Thus, 8-(2 + 1) × (1
+ 1) signifies that 2 + 1 must be taken 1 + 1 times, and the product
must be subtracted from 8. In the same manner, any result made from
two or more numbers, which is true whatever numbers are taken, may be
represented by using first No., second No., &c., to stand for them, and
by the signs in (23). But this may be much shortened; for as first No.,
second No., &c., may mean any numbers, the letters _a_ and _b_ may be
used instead of these words; and it must now be recollected that _a_
and _b_ stand for two numbers, provided only that _a_ is greater than
_b_. Let twice _a_ be represented by 2_a_, and twice _b_ by 2_b_. The
equations then become
(_a_ + _b_) + (_a_ - _b_) = 2_a_,
and (_a_ + _b_) - (_a_ - _b_) = 2_b_.
This may be explained still further, as follows:
25. Suppose a number of sealed packets, marked _a_, _b_, _c_, _d_, &c.,
on the outside, each of which contains a distinct but unknown number of
counters. As long as we do not know how many counters each contains, we
can make the letter which belongs to each stand for its number, so as
to talk of _the number a_, instead of the number in the packet marked
_a_. And because we do not know the numbers, it does not therefore
follow that we know nothing whatever about them; for there are some
connexions which exist between all numbers, which we call _general
properties_ of numbers. For example, take any number, multiply it by
itself, and subtract one from the result; and then subtract one from
the number itself. The first of these will always contain the second
exactly as many times as the original number increased by one. Take
the number 6; this multiplied by itself is 36, which diminished by one
is 35; again, 6 diminished by 1 is 5; and 35 contains 5, 7 times, that
is, 6 + 1 times. This will be found to be true of any number, and, when
proved, may be said to be true of the number contained in the packet
marked _a_, or of the number _a_. If we represent a multiplied by
itself by _aa_,[5] we have, by (23)
_aa_ - 1
------------- = _a_ + 1.
_a_ - 1
[5] This should be (23) _a_ × _a_, but the sign × is unnecessary here.
It is used with numbers, as in 2 × 7, to prevent confounding this,
which is 14, with 27.
26. When, therefore, we wish to talk of a number without specifying
any one in particular, we use a letter to represent it. Thus: Suppose
we wish to reason upon what will follow from dividing a number into
three parts, without considering what the number is, or what are the
parts into which it is divided. Let _a_ stand for the number, and _b_,
_c_, and _d_, for the parts into which it is divided. Then, by our
supposition,
_a_ = _b_ + _c_ + _d_.
On this we can reason, and produce results which do not belong to any
particular number, but are true of all. Thus, if one part be taken away
from the number, the other two will remain, or
_a_ - _b_ = _c_ + _d_.
If each part be doubled, the whole number will be doubled, or
2_a_ = 2_b_ + 2_c_ + 2_d_.
If we diminish one of the parts, as _d_, by a number _x_, we diminish
the whole number just as much, or
_a_ - _x_ = _b_ + _c_ + (_d_ - _x_).
27. EXERCISES.
What is _a_ + 2_b_ - _c_,
where _a_ = 12,
_b_ = 18,
_c_ = 7?--_Answer_, 41.
_aa_ - _bb_
What is ----------- ,
_a_ - _b_
where _a_ = 6 and _b_ = 2?--_Ans._ 8.
What is the difference between (_a_ + _b_)(_c_ + _d_)
and _a_ + _bc_ + _d_, for the following values of
_a_, _b_, _c_, and _d_?
_a_ | _b_ | _c_ | _d_ | _Ans._
1 | 2 | 3 | 4 | 10
2 | 12 | 7 | 1 | 25
1 | 1 | 1 | 1 | 1
SECTION II.
ADDITION AND SUBTRACTION.
28. There is no process in arithmetic which does not consist entirely
in the increase or diminution of numbers. There is then nothing which
might not be done with collections of pebbles. Probably, at first,
either these or the fingers were used. Our word _calculation_ is
derived from the Latin word _calculus_, which means a pebble. Shorter
ways of counting have been invented, by which many calculations, which
would require long and tedious reckoning if pebbles were used, are made
at once with very little trouble. The four great methods are, Addition,
Subtraction, Multiplication, and Division; of which, the last two are
only ways of doing several of the first and second at once.
29. When one number is increased by others, the number which is as
large as all the numbers together is called their _sum_. The process
of finding the sum of two or more numbers is called ADDITION, and, as
was said before, is denoted by placing a cross (+) between the numbers
which are to be added together.
Suppose it required to find the sum of 1834 and 2799. In order to add
these numbers, take them to pieces, dividing each into its units, tens,
hundreds, and thousands:
1834 is 1 thous. 8 hund. 3 tens and 4;
2799 is 2 thous. 7 hund. 9 tens and 9.
Each number is thus broken up into four parts. If to each part of the
first you add the part of the second which is under it, and then put
together what you get from these additions, you will have added 1834
and 2799. In the first number are 4 units, and in the second 9: these
will, when the numbers are added together, contribute 13 units to
the sum. Again, the 3 tens in the first and the 9 tens in the second
will contribute 12 tens to the sum. The 8 hundreds in the first and
the 7 hundreds in the second will add 15 hundreds to the sum; and
the thousand in the first with the 2 thousands in the second will
contribute 3 thousands to the sum; therefore the sum required is
3 thousands, 15 hundreds, 12 tens, and 13 units.
To simplify this result, you must recollect that--
13 units are 1 ten and 3 units.
12 tens are 1 hund. and 2 tens.
15 hund. are 1 thous. and 5 hund.
3 thous. are 3 thous.
Now collect the numbers on the right hand side together, as was done
before, and this will give, as the sum of 1834 and 2799,
4 thousands, 6 hundreds, 3 tens, and 3 units,
which (19) is written 4633.
30. The former process, written with the signs of (23) is as follows:
1834 = 1 × 1000 + 8 × 100 + 3 × 10 + 4
2799 = 2 × 1000 + 7 × 100 + 9 × 10 + 9
Therefore,
1834 + 2799 = 3 × 1000 + 15 × 100 + 12 × 10 + 13
But 13 = 1 × 10 + 3
12 × 10 = 1 × 100 + 2 × 10
15 × 100 = 1 × 1000 + 5 × 100
3 × 1000 = 3 × 1000
Therefore,
1834 + 2799 = 4 × 1000 + 6 × 100 + 3 × 10 + 3
= 4633.
31. The same process is to be followed in all cases, but not at the
same length. In order to be able to go through it, you must know how to
add together the simple numbers. This can only be done by memory; and
to help the memory you should make the following table three or four
times for yourself:
+----+----+----+----+----+----+----+----+----+----+
| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
+----+----+----+----+----+----+----+----+----+----+
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
+----+----+----+----+----+----+----+----+----+----+
| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
+----+----+----+----+----+----+----+----+----+----+
| 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
+----+----+----+----+----+----+----+----+----+----+
| 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
+----+----+----+----+----+----+----+----+----+----+
| 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
+----+----+----+----+----+----+----+----+----+----+
| 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
+----+----+----+----+----+----+----+----+----+----+
| 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
+----+----+----+----+----+----+----+----+----+----+
| 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
+----+----+----+----+----+----+----+----+----+----+
| 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
+----+----+----+----+----+----+----+----+----+----+
The use of this table is as follows: Suppose you want to find the sum
of 8 and 7. Look in the left-hand column for either of them, 8, for
example; and look in the top column for 7. On the same line as 8, and
underneath 7, you find 15, their sum.
32. When this table has been thoroughly committed to memory, so that
you can tell at once the sum of any two numbers, neither of which
exceeds 9, you should exercise yourself in adding and subtracting two
numbers, one of which is greater than 9 and the other less. You should
write down a great number of such sentences as the following, which
will exercise you at the same time in addition, and in the use of the
signs mentioned in (23).
12 + 6 = 18 22 + 6 = 28 19 + 8 = 27
54 + 9 = 63 56 + 7 = 63 22 + 8 = 30
100 - 9 = 91 27 - 8 = 19 44 - 6 = 38, &c.
33. When the last two articles have been thoroughly studied, you will
be able to find the sum of any numbers by the following process,[6]
which is the same as that in (29).
[6] In this and all other processes, the student is strongly
recommended to look at and follow the first Appendix.
RULE I. Place the numbers under one another, units under units, tens
under tens, and so on.
II. Add together the units of all, and part the _whole_ number thus
obtained into units and tens. Thus, if 85 be the number, part it into
8 tens and 5 units; if 136 be the number, part it into 13 tens and 6
units (20).
III. Write down the units of this number under the units of the rest,
and keep in memory the number of tens.
IV. Add together all the numbers in the column of tens, remembering
to take in (or carry, as it is called) the tens which you were told
to recollect in III., and divide this number of tens into tens and
hundreds. Thus, if 335 tens be the number obtained, part this into 33
hundreds and 5 tens.
V. Place the number of tens under the tens, and remember the number of
hundreds.
VI. Proceed in this way through every column, and at the last column,
instead of separating the number you obtain into two parts, write it
all down before the rest.
EXAMPLE.--What is
1805 + 36 + 19727 + 3 + 1474 + 2008
1805
36
19727
3
1474
2008
-----
25053
The addition of the units’ line, or 8 + 4 + 3 + 7 + 6 + 5, gives
33, that is, 3 tens and 3 units. Put 3 in the units’ place, and add
together the line of tens, taking in at the beginning the 3 tens which
were created by the addition of the units’ line. That is, find 3 + 0
+ 7 + 2 + 3 + 0, which gives 15 for the number of tens; that is, 1
hundred and 5 tens. Add the line of hundreds together, taking care to
add the 1 hundred which arose in the addition of the line of tens;
that is, find 1 + 0 + 4 + 7 + 8, which gives exactly 20 hundreds,
or 2 thousands and no hundreds. Put a cipher in the hundreds’ place
(because, if you do not, the next figure will be taken for hundreds
instead of thousands), and add the figures in the thousands’ line
together, remembering the 2 thousands which arose from the hundreds’
line; that is, find 2 + 2 + 1 + 9 + 1, which gives 15 thousands, or 1
ten thousand and 5 thousand. Write 5 under the line of thousands, and
collect the figures in the line of tens of thousands, remembering the
ten thousand which arose out of the thousands’ line; that is, find 1 +
1, or 2 ten thousands. Write 2 under the ten thousands’ line, and the
operation is completed.
34. As an exercise in addition, you may satisfy yourself that what I
now say of the following square is correct. The numbers in every row,
whether reckoned upright, or from right to left, or from corner to
corner, when added together give the number 24156.
+----+----+----+----+----+----+----+----+----+----+----+
|2016|4212|1656|3852|1296|3492| 936|3132| 576|2772| 216|
+----+----+----+----+----+----+----+----+----+----+----+
| 252|2052|4248|1692|3888|1332|3528| 972|3168| 612|2412|
+----+----+----+----+----+----+----+----+----+----+----+
|2448| 288|2088|4284|1728|3924|1368|3564|1008|2808| 648|
+----+----+----+----+----+----+----+----+----+----+----+
| 684|2484| 324|2124|4320|1764|3960|1404|3204|1044|2844|
+----+----+----+----+----+----+----+----+----+----+----+
|2880| 720|2520| 360|2160|4356|1800|3600|1440|3240|1080|
+----+----+----+----+----+----+----+----+----+----+----+
|1116|2916| 756|2556| 396|2196|3996|1836|3636|1476|3276|
+----+----+----+----+----+----+----+----+----+----+----+
|3312|1152|2952| 792|2592| 36|2232|4032|1872|3672|1512|
+----+----+----+----+----+----+----+----+----+----+----+
|1548|3348|1188|2988| 432|2628| 72|2268|4068|1908|3708|
+----+----+----+----+----+----+----+----+----+----+----+
|3744|1584|3384| 828|3024| 468|2664| 108|2304|4104|1944|
+----+----+----+----+----+----+----+----+----+----+----+
|1980|3780|1224|3420| 864|3060| 504|2700| 144|2340|4140|
+----+----+----+----+----+----+----+----+----+----+----+
|4176|1620|3816|1260|3456| 900|3096| 540|2736| 180|2376|
+----+----+----+----+----+----+----+----+----+----+----+
35. If two numbers must be added together, it will not alter the sum if
you take away a part of one, provided you put on as much to the other.
It is plain that you will not alter the whole number of a collection
of pebbles in two baskets by taking any number out of one, and putting
them into the other. Thus, 15 + 7 is the same as 12 + 10, since 12 is 3
less than 15, and 10 is three more than 7. This was the principle upon
which the whole of the process in (29) was conducted.
36. Let _a_ and _b_ stand for two numbers, as in (24). It is impossible
to tell what their sum will be until the numbers themselves are
known. In the mean while _a_ + _b_ stands for this sum. To say, in
algebraical language, that the sum of _a_ and _b_ is not altered by
adding _c_ to _a_, provided we take away _c_ from _b_, we have the
following equation:
(_a_ + _c_) + (_b_ - _c_) = _a_ + _b_;
which may be written without brackets, thus,
_a_ + _c_ + _b_ - _c_ = _a_ + _b_.
For the meaning of these two equations will appear to be the same, on
consideration.
37. If _a_ be taken twice, three times, &c., the results are
represented in algebra by 2_a_, 3_a_, 4_a_, &c. The sum of any two of
this series may be expressed in a shorter form than by writing the sign
+ between them; for though we do not know what number _a_ stands for,
we know that, be it what it may,
2_a_ + 2_a_ = 4_a_, 3_a_ + 2_a_ = 5_a_, 4_a_ + 9_a_ = 13_a_;
and generally, if _a_ taken _m_ times be added to _a_ taken _n_ times,
the result is _a_ taken _m_ + _n_ times, or
_ma_ + _na_ = (_m_ + _n_)_a_.
38. The use of the brackets must here be noticed. They mean, that the
expression contained inside them must be used exactly as a single
letter would be used in the same place. Thus, _pa_ signifies that _a_
is taken _p_ times, and (_m_ + _n_)_a_, that _a_ is taken _m_ + _n_
times. It is, therefore, a different thing from _m_ + _na_, which means
that _a_, after being taken _n_ times, is added to _m_. Thus (3 + 4) ×
2 is 7 × 2 or 14; while 3 + 4 × 2 is 3 + 8, or 11.
39. When one number is taken away from another, the number which is
left is called the _difference_ or _remainder_. The process of finding
the difference is called SUBTRACTION. The number which is to be taken
away must be of course the lesser of the two.
40. The process of subtraction depends upon these two principles.
I. The difference of two numbers is not altered by adding a number to
the first, if you add the same number to the second; or by subtracting
a number from the first, if you subtract the same number from the
second. Conceive two baskets with pebbles in them, in the first of
which are 100 pebbles more than in the second. If I put 50 more
pebbles into each of them, there are still only 100 more in the first
than in the second, and the same if I take 50 from each. Therefore, in
finding the difference of two numbers, if it should be convenient, I
may add any number I please to both of them, because, though I alter
the numbers themselves by so doing, I do not alter their difference.
II. Since 6 exceeds 4 by 2,
and 3 exceeds 2 by 1,
and 12 exceeds 5 by 7,
6, 3, and 12 together, or 21, exceed 4, 2, and 5 together, or 11, by
2, 1, and 7 together, or 10: the same thing may be said of any other
numbers.
41. If _a_, _b_, and _c_ be three numbers, of which _a_ is greater than
_b_ (40), I. leads to the following,
(_a_ + _c_) - (_b_ + _c_) = _a_ - _b_.
Again, if _c_ be less than _a_ and _b_,
(_a_ - _c_) - (_b_ - _c_) = _a_ - _b_.
The brackets cannot be here removed as in (36). That is, _p_- (_q_-_r_)
is not the same thing as _p_-_q_- _r_. For, in the first, the
difference of _q_ and _r_ is subtracted from _p_; but in the second,
first _q_ and then _r_ are subtracted from _p_, which is the same as
subtracting as much as _q_ and _r_ together, or _q_ + _r_. Therefore
_p_-_q_-_r_ is _p_-(_q_ + _r_). In order to shew how to remove the
brackets from _p_ -(_q_-_r_) without altering the value of the result,
let us take the simple instance 12-(8-5). If we subtract 8 from 12, or
form 12-8, we subtract too much; because it is not 8 which is to be
taken away, but as much of 8 as is left after diminishing it by 5. In
forming 12-8 we have therefore subtracted 5 too much. This must be set
right by adding 5 to the result, which gives 12-8 + 5 for the value
of 12-(8-5). The same reasoning applies to every case, and we have
therefore,
_p_ - (_q_ + _r_) = _p_ - _q_ - _r_.
_p_ - (_q_ - _r_) = _p_ - _q_ + _r_.
By the same kind of reasoning,
_a_ - (_b_ + _c_ - _d_ - _e_) = _a_ - _b_ - _c_ + _d_ + _e_.
2_a_ + 3_b_ - (_a_ - 2_b_) = 2_a_ + 3_b_ - _a_ + 2_b_ = _a_ + 5_b_.
4_x_ + _y_ - (17_x_ - 9_y_) = 4_x_ + _y_ - 17_x_ + 9_y_
= 10_y_ - 13_x_.
42. I want to find the difference of the numbers 57762 and 34631. Take
these to pieces as in (29) and
57762 is 5 ten-th. 7 th. 7 hund. 6 tens and 2 units.
34631 is 3 ten-th. 4 th. 6 hund. 3 tens and 1 unit.
Now 2 units exceed 1 unit by 1 unit.
6 tens 3 tens 3 tens.
7 hundreds 6 hundreds 1 hundred.
7 thousands 4 thousands 3 thousands.
5 ten-thousands 3 ten-thous. 2 ten-thous.
Therefore, by (40, Principle II.) all the first column _together_
exceeds all the second column by all the third column, that is, by
2 ten-th. 3 th. 1 hund. 3 tens and 1 unit,
which is 23131. Therefore the difference of 57762 and 34631 is 23131,
or 57762-34631 = 23131.
43. Suppose I want to find the difference between 61274 and 39628.
Write them at length, and
61274 is 6 ten-th. 1 th. 2 hund. 7 tens and 4 units.
39628 is 3 ten-th. 9 th. 6 hund. 2 tens and 8 units.
If we attempt to do the same as in the last article, there is a
difficulty immediately, since 8, being greater than 4, cannot be
taken from it. But from (40) it appears that we shall not alter the
difference of two numbers if we add the same number to _both_ of them.
Add ten to the first number, that is, let there be 14 units instead of
four, and add ten also to the second number, but instead of adding ten
to the number of units, add one to the number of tens, which is the
same thing. The numbers will then stand thus,
6 ten-thous. 1 thous. 2 hund. 7 tens and 14 _units_.[7]
3 ten-thous. 9 thous. 6 hund. 3 _tens_ and 8 units.
[7] Those numbers which have been altered are put in italics.
You now see that the units and tens in the lower can be subtracted from
those in the upper line, but that the hundreds cannot. To remedy this,
add one thousand or 10 hundred to both numbers, which will not alter
their difference, and remember to increase the hundreds in the upper
line by 10, and the thousands in the lower line by 1, which are the
same things. And since the thousands in the lower cannot be subtracted
from the thousands in the upper line, add 1 ten thousand or 10 thousand
to both numbers, and increase the thousands in the upper line by 10,
and the ten thousands in the lower line by 1, which are the same
things; and at the close the numbers which we get will be,
6 ten-thous. 11 _thous._ 12 _hund._ 7 tens and 14 _units_.
4 _ten-thous._ 10 _thous._ 6 hund. 3 _tens_ and 8 units.
These numbers are not, it is true, the same as those given at the
beginning of this article, but their difference is the same, by (40).
With the last-mentioned numbers proceed in the same way as in (42),
which will give, as their difference,
2 ten-thous. 1 thous. 6 hund. 4 tens, and 6 units, which is 21646.
44. From this we deduce the following rules for subtraction:
I. Write the number which is _to be subtracted_ (which is, of course,
the lesser of the two, and is called the _subtrahend_) under the other,
so that its units shall fall under the units of the other, and so on.
II. Subtract each figure of the lower line from the one above it, if
that can be done. Where that cannot be done, add ten to the upper
figure, and then subtract the lower figure; but recollect in this case
always to increase the next figure in the lower line by 1, before you
begin to subtract it from the upper one.
45. If there should not be as many figures in the lower line as in the
upper one, proceed as if there were as many ciphers at the beginning
of the lower line as will make the number of figures equal. You do not
alter a number by placing ciphers at the beginning of it. For example,
00818 is the same number as 818, for it means
0 ten-thous. 0 thous. 8 hunds. 1 ten and 8 units;
the first two signs are nothing, and the rest is
8 hundreds, 1 ten, and 8 units, or 818.
The second does not differ from the first, except in its being said
that there are no thousands and no tens of thousands in the number,
which may be known without their being mentioned at all. You may ask,
perhaps, why this does not apply to a cipher placed in the middle of
a number, or at the right of it, as, for example, in 28007 and 39700?
But you must recollect, that if it were not for the two ciphers in the
first, the 8 would be taken for 8 tens, instead of 8 thousands; and if
it were not for the ciphers in the second, the 7 would be taken for 7
units, instead of 7 hundreds.
46. EXAMPLE.
What is the difference between 3708291640030174
and 30813649276188
----------------
Difference 3677477990753986
EXERCISES.
I. What is 18337 + 149263200 - 6472902?--_Answer_ 142808635.
What is 1000 - 464 + 3279 - 646?--_Ans._ 3169.
II. Subtract
64 + 76 + 144 - 18 from 33 - 2 + 100037.--_Ans._ 99802.
III. What shorter rule might be made for subtraction when all the
figures in the upper line are ciphers except the first? for example, in
finding
10000000 - 2731634.
IV. Find 18362 + 2469 and 18362-2469, add the second result to the
first, and then subtract 18362; subtract the second from the first, and
then subtract 2469.--_Answer_ 18362 and 2469.
V. There are four places on the same line in the order A, B, C, and D.
From A to D it is 1463 miles; from A to C it is 728 miles; and from B
to D it is 1317 miles. How far is it from A to B, from B to C, and from
C to D?--_Answer._ From A to B 146, from B to C 582, and from C to D
735 miles.
VI. In the following table subtract B from A, and B from the remainder,
and so on until B can be no longer subtracted. Find how many times B
can be subtracted from A, and what is the last remainder.
No. of
A B times. Remainder.
23604 9999 2 3606
209961 37173 5 24096
74712 6792 11 0
4802469 654321 7 222222
18849747 3141592 6 195
987654321 123456789 8 9
SECTION III.
MULTIPLICATION.
47. I have said that all questions in arithmetic require nothing but
addition and subtraction. I do not mean by this that no rule should
ever be used except those given in the last section, but that all
other rules only shew shorter ways of finding what might be found,
if we pleased, by the methods there deduced. Even the last two rules
themselves are only short and convenient ways of doing what may be done
with a number of pebbles or counters.
48. I want to know the sum of five seventeens, or I ask the following
question: There are five heaps of pebbles, and seventeen pebbles in
each heap; how many are there in all? Write five seventeens in a
column, and make the addition, which gives 85. In this case 85 is
called the _product_ of 5 and 17, and the process of finding the
product is called MULTIPLICATION, which gives nothing more than the
addition of a number of the same quantities. Here 17 is called the
_multiplicand_, and 5 is called the _multiplier_.
17
17
17
17
17
----
85
49. If no question harder than this were ever proposed, there would be
no occasion for a shorter way than the one here followed. But if there
were 1367 heaps of pebbles, and 429 in each heap, the whole number is
then 1367 times 429, or 429 multiplied by 1367. I should have to write
429 1367 times, and then to make an addition of enormous length. To
avoid this, a shorter rule is necessary, which I now proceed to explain.
50. The student must first make himself acquainted with the products of
all numbers as far as 10 times 10 by means of the following table,[8]
which must be committed to memory.
+----+----+----+----+----+----+----+----+----+----+----+----+
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 3 | 6 | 9 | 12 | 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 | 36 | 40 | 44 | 48 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 | 55 | 60 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 | 66 | 72 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 70 | 77 | 84 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 8 | 16 | 24 | 32 | 40 | 48 | 56 | 64 | 72 | 80 | 88 | 96 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 9 | 18 | 27 | 36 | 45 | 54 | 63 | 72 | 81 | 90 | 99 |108 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 |100 |110 |120 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 11 | 22 | 33 | 44 | 55 | 66 | 77 | 88 | 99 |110 |121 |132 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 12 | 24 | 36 | 48 | 60 | 72 | 84 | 96 |108 |120 |132 |144 |
+----+----+----+----+----+----+----+----+----+----+----+----+
[8] As it is usual to learn the product of numbers up to 12 times 12, I
have extended the table thus far. In my opinion, all pupils who shew a
tolerable capacity should slowly commit the products to memory as far
as 20 times 20, in the course of their progress through this work.
If from this table you wish to know what is 7 times 6, look in the
first upright column on the left for either of them; 6 for example.
Proceed to the right until you come into the column marked 7 at the
top. You there find 42, which is the product of 6 and 7.
51. You may find, in this way, either 6 times 7, or 7 times 6, and
for both you find 42. That is, six sevens is the same number as seven
sixes. This may be shewn as follows: Place seven counters in a line,
and repeat that line in all six times. The number of counters in the
whole is 6 times 7, or six sevens, if I reckon the rows from the top
to the bottom; but if I count the rows that stand side by side, I find
seven of them, and six in each row, the whole number of which is 7
times 6, or seven sixes. And the whole number is 42, whichever way I
count. The same method may be applied to any other two numbers. If the
signs of (23) were used, it would be said that 7 × 6 = 6 × 7.
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
52. To take any quantity a number of times, it will be enough to take
every one of its parts the same number of times. Thus, a sack of corn
will be increased fifty-fold, if each bushel which it contains be
replaced by 50 bushels. A country will be doubled by doubling every
acre of land, or every county, which it contains. Simple as this
may appear, it is necessary to state it, because it is one of the
principles on which the rule of multiplication depends.
53. In order to multiply by any number, you may multiply separately
by any parts into which you choose to divide that number, and add the
results. For example, 4 and 2 make 6. To multiply 7 by 6 first multiply
7 by 4, and then by 2, and add the products. This will give 42, which
is the product of 7 and 6. Again, since 57 is made up of 32 and 25, 57
times 50 is made up of 32 times 50 and 25 times 50, and so on. If the
signs were used, these would be written thus:
7 × 6 = 7 × 4 + 7 × 2.
50 × 57 = 50 × 32 + 50 × 25.
54. The principles in the last two articles may be expressed thus: If
_a_ be made up of the parts _x_, _y_, and _x_, _ma_ is made up of _mx_,
_my_, and _mz_; or,
if _a_ = _x_ + _y_ + _z_.
_ma_ = _mx_ + _my_ + _mz_,
or, _m_(_x_ + _y_ + _z_) = _mx_ + _my_ + _mz_.
A similar result may be obtained if _a_, instead of being made up of
_x_, _y_, and _z_, is made by combined additions and subtractions, such
as _x_ + _y_-_z_, _x_- _y_ + _z_, _x_-_y_-_z_, &c. To take the first as
an instance:
Let _a_ = _x_ + _y_ - _z_,
then _ma_ = _mx_ + _my_ - _mz_.
For, if _a_ had been _x_ + _y_, _ma_ would have been _mx_ + _my_. But
since _a_ is less than _x_ + _y_ by _z_, too much by _z_ has been
repeated every time that _x_ + _y_ has been repeated;--that is, _mz_
too much has been taken; consequently, _ma_ is not _mx_ + _my_, but
_mx_ + _my_-_mz_. Similar reasoning may be applied to other cases, and
the following results may be obtained:
_m_(_a_ + _b_ + _c_ - _d_) = _ma_ + _mb_ + _mc_ - _md_.
_a_(_a_ - _b_) = _aa_ - _ab_.
_b_(_a_ - _b_) = _ba_ - _bb_.
3(2_a_ - 4_b_) = 6_a_ - 12_b_.
7_a_(7 + 2_b_) = 49_a_ + 14_ab_.
(_aa_ + _a_ + 1)_a_ = _aaa_ + _aa_ + _a_.
(3_ab_ - 2_c_)4_abc_ = 12_aabbc_ - 8_abcc_.
55. There is another way in which two numbers may be multiplied
together. Since 8 is 4 times 2, 7 times 8 may be made by multiplying 7
and 4, and then multiplying that _product_ by 2. To shew this, place 7
counters in a line, and repeat that line in all 8 times, as in figures
I. and II.
I.
+---------------+
| ● ● ● ● ● ● ● |
A | ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+
+---------------+
| ● ● ● ● ● ● ● |
B | ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+
II.
+---------------+
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+
+---------------+
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+
+---------------+
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+
+---------------+
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+
The number of counters in all is 8 times 7, or 56. But (as in fig. I.)
enclose each four rows in oblong figures, such as A and B. The number
in each oblong is 4 times 7, or 28, and there are two of those oblongs;
so that in the whole the number of counters is twice 28, or 28 x 2, or
7 first multiplied by 4, and that product multiplied by 2. In figure
II. it is shewn that 7 multiplied by 8 is also 7 first multiplied by
2, and that product multiplied by 4. The same method may be applied
to other numbers. Thus, since 80 is 8 times 10, 256 times 80 is 256
multiplied by 8, and that product multiplied by 10. If we use the
signs, the foregoing assertions are made thus:
7 × 8 = 7 × 4 × 2 = 7 × 2 × 4.
256 × 80 = 256 × 8 × 10 = 256 × 10 × 8.
EXERCISES.
Shew that 2 × 3 × 4 × 5 = 2 × 4 × 3 × 5 = 5 × 4 × 2 × 3, &c.
Shew that 18 × 100 = 18 × 57 + 18 × 43.
56. Articles (51) and (55) may be expressed in the following way, where
by _ab_ we mean _a_ taken _b_ times; by _abc_, _a_ taken _b_ times, and
the result taken _c_ times.
_ab_ = _ba_.
_abc_ = _acb_ = _bca_ = _bac_, &c.
_abc_ = _a_ × (_bc_) = _b_ × (_ca_) = _c_ × (_ab_).
If we would say that the same results are produced by multiplying by
_b_, _c_, and _d_, one after the other, and by the product _bcd_ at
once, we write the following:
_a_ × _b_ × _c_ × _d_ = _a_ × _bcd_.
The fact is, that if any numbers are to be multiplied together, the
product of any two or more may be formed, and substituted instead
of those two or more; thus, the product _abcdef_ may be formed by
multiplying
_ab_ _cde_ _f_
_abf_ _de_ _c_
_abc_ _def_ &c.
57. In order to multiply by 10, annex a cipher to the right hand of the
multiplicand. Thus, 10 times 2356 is 23560. To shew this, write 2356 at
length which is
2 thousands, 3 hundreds, 5 tens, and 6 units.
Take each of these parts ten times, which, by (52), is the same as
multiplying the whole number by 10, and it will then become
2 tens of thou. 3 tens of hun. 5 tens of tens, and 6 tens,
which is
2 ten-thou. 3 thous. 5 hun. and 6 tens.
This must be written 23560, because 6 is not to be 6 units, but 6 tens.
Therefore 2356 × 10 = 23560.
In the same way you may shew, that in order to multiply by 100 you
must affix two ciphers to the right; to multiply by 1000 you must
affix three ciphers, and so on. The rule will be best caught from the
following table:
13 × 10 = 130
13 × 100 = 1300
13 × 1000 = 13000
13 × 10000 = 130000
142 × 1000 = 142000
23700 × 10 = 237000
3040 × 1000 = 3040000
10000 × 100000 = 1000000000
58. I now shew how to multiply by one of the numbers, 2, 3, 4, 5, 6, 7,
8, or 9. I do not include 1, because multiplying by 1, or taking the
number once, is what is meant by simply writing down the number. I want
to multiply 1368 by 8. Write the first number at full length, which is
1 thousand, 3 hundreds, 6 tens, and 8 units.
To multiply this by 8, multiply each of these parts by 8 (50) and (52),
which will give
8 thousands, 24 hundreds, 48 tens, and 64 units.
Now 64 units are written thus 64
48 tens 480
24 hundreds 2400
8 thousands 8000
Add these together, which gives 10944 as the product of 1368 and 8, or
1368 × 8 = 10944. By working a few examples in this way you will see
for following rule.
59. I. Multiply the first figure of the multiplicand by the multiplier,
write down the units’ figure, and reserve the tens.
II. Do the same with the second figure of the multiplicand, and add
to the product the number of tens from the first; put down the units’
figure of this, and reserve the tens.
III. Proceed in this way till you come to the last figure, and then
write down the whole number obtained from that figure.
IV. If there be a cipher in the multiplicand, treat it as if it were a
number, observing that 0 × 1 = 0, 0 × 2 = 0, &c.
60. In a similar way a number can be multiplied by a figure which is
accompanied by ciphers, as, for example, 8000. For 8000 is 8 × 1000,
and therefore (55) you must first multiply by 8 and then by 1000, which
last operation (57) is done by placing 3 ciphers on the right. Hence
the rule in this case is, multiply by the simple number, and place the
number of ciphers which follow it at the right of the product.
EXAMPLE.
Multiply 1679423800872
by 60000
------------------
100765428052320000
61. EXERCISES.
What is 1007360 × 7? _Answer_, 7051520.
123456789 × 9 + 10 and 123 × 9 + 4?--_Ans._ 1111111111 and 1111.
What is 136 × 3 + 129 × 4 + 147 × 8 + 27 × 3000?--_Ans._ 83100.
An army is made up of 33 regiments of infantry, each containing 800
men; 14 of cavalry, each containing 600 men; and 2 of artillery, each
containing 300 men. The enemy has 6 more regiments of infantry, each
containing 100 more men; 3 more regiments of cavalry, each containing
100 men less; and 4 corps of artillery of the same magnitude as those
of the first: two regiments of cavalry and one of infantry desert from
the former to the latter. How many men has the second army more than
the first?--_Answer_, 13400.
62. Suppose it is required to multiply 23707 by 4567. Since 4567 is
made up of 4000, 500, 60, and 7, by (53) we must multiply 23707 by each
of these, and add the products.
Now (58) 23707 × 7 is 165949
(60) 23707 × 60 is 1422420
23707 × 500 is 11853500
23707 × 4000 is 94828000
---------
The sum of these is 108269869
which is the product required.
It will do as well if, instead of writing the ciphers at the end of
each line, we keep the other figures in their places without them. If
we take away the ciphers, the second line is one place to the left of
the first, the third one place to the left of the second, and so on.
Write the multiplier and the multiplicand over these lines, and the
process will stand thus:
23707
4567
------
165949
142242
118535
94828
---------
108269869
63. There is one more case to be noticed; that is, where there is a
cipher in the middle of the multiplier. The following example will shew
that in this case nothing more is necessary than to keep the first
figure of each line in the column under the figure of the multiplier
from which that line arises. Suppose it required to multiply 365 by
101001. The multiplier is made up of 100000, 1000 and 1. Proceed as
before, and
365 × 1 is 365
(57) 365 × 1000 is 365000
365 × 100000 is 36500000
--------
The sum of which is 36865365
and the whole process with the ciphers struck off is:
365
101001
------
365
365
365
--------
36865365
64. The following is the rule in all cases:
I. Place the multiplier under the multiplicand, so that the units of
one may be under those of the other.
II. Multiply the whole multiplicand by each figure of the multiplier
(59), and place the unit of each line in the column under the figure of
the multiplier from which it came.
III. Add together the lines obtained by II. column by column.
65. When the multiplier or multiplicand, or both, have ciphers on the
right hand, multiply the two together without the ciphers, and then
place on the right of the product all the ciphers that are on the right
both of the multiplier and multiplicand. For example, what is 3200 ×
3000? First, 3200 is 32 × 100, or one hundred times as great as 32.
Again, 32 × 13000 is 32 × 13, with three ciphers affixed, that is 416,
with three ciphers affixed, or 416000. But the product required must
be 100 times as great as this, or must have two ciphers affixed. It is
therefore 41600000, having as many ciphers as are in both multiplier
and multiplicand.
66. When any number is multiplied by itself any number of times, the
result is called a _power_ of that number. Thus:
6 is called the first power of 6
6 × 6 second power of 6
6 × 6 × 6 third power of 6
6 × 6 × 6 × 6 fourth power of 6
&c. &c.
The second and third powers are usually called the _square_ and
_cube_, which are incorrect names, derived from certain connexions of
the second and third power with the square and cube in geometry. As
exercises in multiplication, the following powers are to be found.
Number proposed. Square. Cube.
972 944784 918330048
1008 1016064 1024192512
3142 9872164 31018339288
3163 10004569 31644451747
5555 30858025 171416328875
6789 46090521 312908547069
The fifth power of 36 is 60466176
fourth 50 6250000
fourth 108 136048896
fourth 277 5887339441
67. It is required to multiply _a_ + _b_ by _c_ + _d_, that is, to take
_a_ + _b_ as many times as there are units in _c_ + _d_. By (53) _a_
+ _b_ must be taken _c_ times, and _d_ times, or the product required
is (_a_ + _b_)_c_ + (_a_ + _b_)_d_. But (52) (_a_ + _b_)_c_ is _ac_ +
_bc_, and (_a_ + _b_)_d_ is _ad_ + _bd_; whence the product required is
_ac_ + _bc_ + _ad_ + _bd_; or,
(_a_ + _b_)(_c_ + _d_) = _ac_ + _bc_ + _ad_ + _bd_.
By similar reasoning
(_a_ - _b_)(_c_ + _d_) is (_a_ - _b_)_c_ + (_a_ - _b_)_d_; or,
(_a_ - _b_)(_c_ + _d_) = _ac_ - _bc_ + _ad_ - _bd_.
To multiply _a_-_b_ by _c_-_d_, first take _a_-_b_ _c_ times, which
gives _ac_-_bc_. This is not correct; for in taking it _c_ times
instead of _c_-_d_ times, we have taken it _d_ times too many; or have
made a result which is (_a_-_b_)_d_ too great. The real result is
therefore _ac_-_bc_-(_a_ -_b_)_d_. But (_a_-_b_)_d_ is _ad_- _bd_, and
therefore
(_a_ - _b_)(_c_ - _d_) = _ac_ - _bc_ - _ad_ - _bd_
= _ac_ - _bc_ - _ad_ + _bd_ (41)
From these three examples may be collected the following rule for the
multiplication of algebraic quantities: Multiply each term of the
multiplicand by each term of the multiplier; when the two terms have
both + or both-before them, put + before their product; when one has
+ and the other-, put-before their product. In using the first terms,
which have no sign, apply the rule as if they had the sign +.
68. For example, (_a_ + _b_)(_a_ + _b_) gives _aa_ + _ab_ + _ab_ +
_bb_. But _ab_ + _ab_ is 2_ab_; hence the _square_ of _a_ + _b_ is
_aa_ + 2_ab_ + _bb_. Again (_a_- _b_)(_a_-_b_) gives _aa_-_ab_-_ab_
+ _bb_. But two subtractions of _ab_ are equivalent to subtracting
2_ab_; hence the _square_ of _a_- _b_ is _aa_-2_ab_ + _bb_. Again, (_a_
+ _b_)(_a_-_b_) gives _aa_ + _ab_-_ab_ -_bb_. But the addition and
subtraction of _ab_ makes no change; hence the product of _a_ + _b_ and
_a_- _b_ is _aa_-_bb_.
Again, the square of _a_ + _b_ + _c_ + _d_ or (_a_ + _b_ + _c_ +
_d_)(_a_ + _b_ + _c_ + _d_) will be found to be _aa_ + 2_ab_ + 2_ac_
+ 2_ad_ + _bb_ + 2_bc_ + 2_bd_ + _cc_ + 2_cd_ + _dd_; or the rule for
squaring such a quantity is: Square the first term, and multiply all
that come _after_ by twice that term; do the same with the second, and
so on to the end.
SECTION IV.
DIVISION.
69. Suppose I ask whether 156 can be divided into a number of parts
each of which is 13, or how many thirteens 156 contains; I propose a
question, the solution of which is called DIVISION. In this case, 156
is called the _dividend_, 13 the _divisor_, and the number of parts
required is the _quotient_; and when I find the quotient, I am said to
divide 156 by 13.
70. The simplest method of doing this is to subtract 13 from 156,
and then to subtract 13 from the remainder, and so on; or, in common
language, to _tell off_ 156 by thirteens. A similar process has already
occurred in the exercises on subtraction, Art. (46). Do this, and
mark one for every subtraction that is made, to remind you that each
subtraction takes 13 once from 156, which operations will stand as
follows:
156
13 1
------
143
13 1
------
130
13 1
------
117
13 1
------
104
13 1
------
91
13 1
------
78
13 1
------
65
13 1
------
52
13 1
------
39
13 1
------
26
13 1
------
13
13 1
------
0
Begin by subtracting 13 from 156, which leaves 143. Subtract 13 from
143, which leaves 130; and so on. At last 13 only remains, from which
when 13 is subtracted, there remains nothing. Upon counting the number
of times which you have subtracted 13, you find that this number is 12;
or 156 contains twelve thirteens, or contains 13 twelve times.
This method is the most simple possible, and might be done with
pebbles. Of these you would first count 156. You would then take 13
from the heap, and put them into one heap by themselves. You would
then take another 13 from the heap, and place them in another heap by
themselves; and so on until there were none left. You would then count
the number of heaps, which you would find to be 12.
71. Division is the opposite of multiplication. In multiplication you
have a number of heaps, with the same number of pebbles in each, and
you want to know how many _pebbles_ there are in all. In division you
know how many there are in all, and how many there are to be in each
heap, and you want to know how many _heaps_ there are.
72. In the last example a number was taken which contains an exact
number of thirteens. But this does not happen with every number. Take,
for example, 159. Follow the process of (70), and it will appear that
after having subtracted 13 twelve times, there remains 3, from which
13 cannot be subtracted. We may say then that 159 contains twelve
thirteens and 3 _over_; or that 159, when divided by 13, gives a
_quotient_ 12, and a _remainder_ 3. If we use signs,
159 = 13 × 12 + 3.
EXERCISES.
146 = 24 × 6 + 2, or 146 contains six twenty-fours and 2 over.
146 = 6 × 24 + 2, or 146 contains twenty-four sixes and 2 over.
300 = 42 × 7 + 6, or 300 contains seven forty-twos and 6 over.
39624 = 7277 × 5 + 3239.
73. If _a_ contain _b_ _q_ times with a remainder _r_, _a_ must be
greater than _bq_ by _r_; that is,
_a_ = _bq_ + _r_.
If there be no remainder, _a_ = _bq_. Here _a_ is the dividend, _b_ the
divisor, _q_ the quotient, and _r_ the remainder. In order to say that
_a_ contains _b_ _q_ times, we write,
_a_/_b_ = _q_, or _a_ : _b_ = _q_,
which in old books is often found written thus:
_a_ ÷ _b_ = _q_.
74. If I divide 156 into several parts, and find how often 13 is
contained in each of them, it is plain that 156 contains 13 as often as
all its parts together. For example, 156 is made up of 91, 39, and 26.
Of these
91 contains 13 7 times,
39 contains 13 3 times,
26 contains 13 2 times;
therefore 91 + 39 + 26 contains 13 7 + 3 + 2 times, or 12 times.
Again, 156 is made up of 100, 50, and 6.
Now 100 contains 13 7 times and 9 over,
50 contains 13 3 times and 11 over,
6 contains 13 0 times[9] and 6 over.
[9] To speak always in the same way, instead of saying that 6 does not
contain 13, I say that it contains it 0 times and 6 over, which is
merely saying that 6 is 6 more than nothing.
Therefore 100 + 50 + 6 contains 13 7 + 3 + 0 times and 9 + 11 + 6 over;
or 156 contains 13 10 times and 26 over. But 26 is itself 2 thirteens;
therefore 156 contains 10 thirteens and 2 thirteens, or 12 thirteens.
75. The result of the last article is expressed by saying, that if
_a_ = _b_ + _c_ + _d_, then _a_/_m_ = _b_/_m_ + _c_/_m_ + _d_/_m_
76. In the first example I did not take away 13 more than once at a
time, in order that the method might be as simple as possible. But
if I know what is twice 13, 3 times 13, &c., I can take away as many
thirteens at a time as I please, if I take care to mark at each step
how many I take away. For example, take away 13 ten times at once from
156, that is, take away 130, and afterwards take away 13 twice, or take
away 26, and the process is as follows:
156
130 10 times 13.
---
26
26 2 times 13.
---
0
Therefore 156 contains 13 10 + 2, or 12 times.
Again, to divide 3096 by 18.
3096
1800 100 times 18.
----
1296
900 50 times 18.
----
396
360 20 times 18.
----
36
36 2 times 18.
----
0
Therefore 3096 contains 18 100 + 50 + 20 + 2, or 172 times.
77. You will now understand the following sentences, and be able to
make similar assertions of other numbers.
450 is 75 × 6; it therefore contains any number, as 5, 6 times as often
as 75 contains it.
135 contains 3 more than 26 times; therefore,
Twice 135 ” 3 ” 52 or twice 26 times.
10 times 135 ” 3 ” 260 or 10 times 26
50 times 135 ” 3 ” 1300 or 50 times 26
472 contains 18 more than 21 times; therefore,
4720 contains 18 more than 210 times,
47200 contains 18 more than 2100 times,
472000 contains 18 more than 21000 times,
32 contains 12 more than 2 times, and less than 3 times.
320 ” 12 ” 20 30
3200 ” 12 ” 200 300
32000 ” 12 ” 2000 3000
&c. &c. &c.
78. The foregoing articles contain the principles of division. The
question now is, to apply them in the shortest and most convenient way.
Suppose it required to divide 4068 by 18, or to find 4068/18 (23).
If we divide 4068 into any number of parts, we may, by the process
followed in (74), find how many times 18 is contained in each of these
parts, and from thence how many times it is contained in the whole.
Now, what separation of 4068 into parts will be most convenient?
Observe that 4, the first figure of 4068, does not contain 18; but that
40, the first and second figures together, _does contain 18 more than
twice, but less than three times_.[10] But 4068 (20) is made up of 40
hundreds, and 68; of which, 40 hundreds (77) contains 18 more than 200
times, and less than 300 times. Therefore, 4068 also contains more than
200 times 18, since it must contain 18 more times than 4000 does. It
also contains 18 less than 300 times, because 300 times 18 is 5400, a
greater number than 4068. Subtract 18 200 times from 4068; that is,
subtract 3600, and there remains 468. Therefore, 4068 contains 18 200
times, and as many more times as 468 contains 18.
[10] If you have any doubt as to this expression, recollect that it
means “contains more than two eighteens, but not so much as three.”
It remains, then, to find how many times 468 contains 18. Proceed
exactly as before. Observe that 46 contains 18 more than twice, and
less than 3 times; therefore, 460 contains it more than 20, and less
than 30 times (77); as does also 468. Subtract 18 20 times from 468,
that is, subtract 360; the remainder is 108. Therefore, 468 contains
18 20 times, and as many more as 108 contains it. Now, 108 is found to
contain 18 6 times exactly; therefore, 468 contains it 20 + 6 times,
and 4068 contains it 200 + 20 + 6 times, or 226 times. If we write down
the process that has been followed, without any explanation, putting
the divisor, dividend, and quotient, in a line separated by parentheses
it will stand, as in example(A).
Let it be required to divide 36326599 by 1342 (B).
A. B.
18)4068(200 + 20 + 6 1342)36326599(20000 + 7000 + 60 + 9
3600 26840000
---- --------
468 9486599
360 9394000
--- -------
108 92599
108 80520
--- -----
0 12079
12078
-----
1
As in the previous example, 36326599 is separated into 36320000 and
6599; the first four figures 3632 being separated from the rest,
because it takes four figures from the left of the dividend to make
a number which is greater than the divisor. Again, 36320000 is found
to contain 1342 more than 20000, and less than 30000 times; and 1342
× 20000 is subtracted from the dividend, after which the remainder is
9486599. The same operation is repeated again and again, and the result
is found to be, that there is a quotient 20000 + 7000 + 60 + 9, or
27069, and a remainder 1.
Before you proceed, you should now repeat the foregoing article at
length in the solution of the following questions. What are
10093874 66779922 2718218
-------- , -------- , ------- ?
3207 114433 13352
the quotients of which are 3147, 583, 203; and the remainders 1445,
65483, 7762.
79. In the examples of the last article, observe, 1st, that it is
useless to write down the ciphers which are on the right of each
subtrahend, provided that without them you keep each of the other
figures in its proper place: 2d, that it is useless to put down the
right hand figures of the dividend so long as they fall over ciphers,
because they do not begin to have any share in the making of the
quotient until, by continuing the process, they cease to have ciphers
under them: 3d, that the quotient is only a number written at length,
instead of the usual way. For example, the first quotient is 200 + 20
+ 6, or 226; the second is 20000 + 7000 + 60 + 9, or 27069. Strike
out, therefore, all the ciphers and the numbers which come above them,
except those in the first line, and put the quotient in one line; and
the two examples of the last article will stand thus:
18)4068(226 1342)36326599(27069
36 2684
--- -----
46 9486
36 9394
--- -----
108 9259
108 8052
--- -----
0 12079
12078
-----
1
80. Hence the following rule is deduced:
I. Write the divisor and dividend in one line, and place parentheses on
each side of the dividend.
II. Take off from the left-hand of the dividend the least number of
figures which make a number greater than the divisor; find what number
of times the divisor is contained in these, and write this number as
the first figure of the quotient.
III. Multiply the divisor by the last-mentioned figure, and subtract
the product from the number which was taken off at the left of the
dividend.
IV. On the right of the remainder place the figure of the dividend
which comes next after those already separated in II.: if the remainder
thus increased be greater than the divisor, find how many times the
divisor is contained in it; put this number at the right of the first
figure of the quotient, and repeat the process: if not, on the right
place the next figure of the dividend, and the next, and so on until it
is greater; but remember to place a cipher in the quotient for every
figure of the dividend which you are obliged to take, except the first.
V. Proceed in this way until all the figures of the dividend are
exhausted.
In judging how often one large number is contained in another, a first
and rough guess may be made by striking off the same number of figures
from both, and using the results instead of the numbers themselves.
Thus, 4,732 is contained in 14,379 about the same number of times
that 4 is contained in 14, or about 3 times. The reason is, that 4
being contained in 14 as often as 4000 is in 14000, and these last
only differing from the proposed numbers by lower denominations, viz.
hundreds, &c. we may expect that there will not be much difference
between the number of times which 14000 contains 4000, and that which
14379 contains 4732: and it generally happens so. But if the second
figure of the divisor be 5, or greater than 5, it will be more accurate
to increase the first figure of the divisor by 1, before trying the
method just explained. Nothing but practice can give facility in this
sort of guess-work.
81. This process may be made more simple when the divisor is not
greater than 12, if you have sufficient knowledge of the multiplication
table (50). For example, I want to divide 132976 by 4. At full length
the process stands thus:
4)132976(33244
12
---
12
12
---
9
8
--
17
16
---
16
16
--
0
But you will recollect, without the necessity of writing it down,
that 13 contains 4 three times with a remainder 1; this 1 you will
place before 2, the next figure of the dividend, and you know that 12
contains 4 3 times exactly, and so on. It will be more convenient to
write down the quotient thus:
4)132976
-------
33244
While on this part of the subject, we may mention, that the shortest
way to multiply by 5 is to annex a cipher and divide by 2, which is
equivalent to taking the half of 10 times, or 5 times. To divide by
5, multiply by 2 and strike off the last figure, which leaves the
quotient; half the last figure is the remainder. To multiply by 25,
annex two ciphers and divide by 4. To divide by 25, multiply by 4 and
strike off the last two figures, which leaves the quotient; one fourth
of the last two figures, taken as one number, is the remainder. To
multiply a number by 9, annex a cipher, and subtract the number, which
is equivalent to taking the number ten times, and then subtracting it
once. To multiply by 99, annex two ciphers and subtract the number, &c.
In order that a number may be divisible by 2 without remainder, its
units’ figure must be an even number.[11] That it may be divisible by
4, its last two figures must be divisible by 4. Take the example 1236:
this is composed of 12 hundreds and 36, the first part of which, being
hundreds, is divisible by 4, and gives 12 twenty-fives; it depends then
upon 36, the last two figures, whether 1236 is divisible by 4 or not.
A number is divisible by 8 if the last three figures are divisible by
8; for every digit, except the last three, is a number of thousands,
and 1000 is divisible by 8; whether therefore the whole shall be
divisible by 8 or not depends on the last three figures: thus, 127946
is not divisible by 8, since 946 is not so. A number is divisible by 3
or 9 only when the sum of its digits is divisible by 3 or 9. Take for
example 1234; this is
[11] Among the even figures we include 0.
1 thousand, or 999 and 1
2 hundred, or twice 99 and 2
3 tens, or three times 9 and 3
and 4 or 4
Now 9, 99, 999, &c. are all obviously divisible by 9 and by 3, and so
will be any number made by the repetition of all or any of them any
number of times. It therefore depends on 1 + 2 + 3 + 4, or the sum of
the digits, whether 1234 shall be divisible by 9 or 3, or not. From
the above we gather, that a number is divisible by 6 when it is even,
and when the sum of its digits is divisible by 3. Lastly, a number is
divisible by 5 only when the last figure is 0 or 5.
82. Where the divisor is unity followed by ciphers, the rule becomes
extremely simple, as you will see by the following examples:
100)33429(334
300
----
342
300
----
429
400
---
29
This is, then, the rule: Cut off as many figures from the right hand of
the dividend as there are ciphers. These figures will be the remainder,
and the rest of the dividend will be the quotient.
10)2717316
--------
271731 and rem. 6.
Or we may prove these results thus: from (20), 2717316 is 271731 tens
and 6; of which the first contains 10 271731 times, and the second not
at all; the quotient is therefore 271731, and the remainder 6 (72).
Again (20), 33429 is 334 hundreds and 29; of which the first contains
100 334 times, and the second not at all; the quotient is therefore
334, and the remainder 29.
83. The following examples will shew how the rule may be shortened when
there are ciphers in the divisor. With each example is placed another
containing the same process, all unnecessary figures being removed; and
from the comparison of the two, the rule at the end of this article is
derived.
I. 1782000)6424700000(3605 1782)6424700(3605
5346000 5346
-------- ----
10787000 10787
10692000 10692
---------- -------
9500000 9500
8910000 8910
------- -------
590000 590000
II. 12300000)42176189300(3428 123)421761(3428
36900000 369
--------- ----
52761893 527
49200000 492
--------- ----
35618930 356
24600000 246
--------- ----
110189300 1101
98400000 984
-------- ----------
11789300 11789300
The rule, then, is: Strike out as many _figures_[12] from the right of
the dividend as there are _ciphers_ at the right of the divisor. Strike
out all the ciphers from the divisor, and divide in the usual way; but
at the end of the process place on the right of the remainder all those
figures which were struck out of the dividend.
[12] Including both ciphers and others.
84. EXERCISES.
Dividend. | Divisor. |Quotient.|Remainder.
9694 | 47 | 206 | 12
175618 | 3136 | 56 | 2
23796484 | 130000 | 183 | 6484
14002564 | 1871 | 7484 | 0
310314420 | 7878 | 39390 | 0
3939040647 | 6889 | 571787 | 4
22876792454961 | 43046721 | 531441 | 0
Shew that
100 × 100 × 100 - 43 × 43 × 43
I. ------------------------------ = 100 × 100 + 100 × 43 + 43 × 43.
100 - 43
100 × 100 × 100 + 43 × 43 × 43
II. ------------------------------ = 100 × 100 - 100 × 43 + 43 × 43.
100 + 43
76 × 76 + 2 × 76 × 52 + 52 × 52
III. -------------------------------- = 76 + 52.
76 + 52
12 × 12 × 12 × 12 - 1
IV. 1 + 12 + 12 × 12 + 12 × 12 × 12 = ----------------------.
12 - 1
What is the nearest number to 1376429 which can be divided by 36300
without remainder?--_Answer_, 1379400.
If 36 oxen can eat 216 acres of grass in one year, and if a sheep eat
half as much as an ox, how long will it take 49 oxen and 136 sheep
together to eat 17550 acres?--_Answer_, 25 years.
85. Take any two numbers, one of which divides the other without
remainder; for example, 32 and 4. Multiply both these numbers by any
other number; for example, 6. The products will be 192 and 24. Now,
192 contains 24 just as often as 32 contains 4. Suppose 6 baskets,
each containing 32 pebbles, the whole number of which will be 192.
Take 4 from one basket, time after time, until that basket is empty.
It is plain that if, instead of taking 4 from that basket, I take 4
from each, the whole 6 will be emptied together: that is, 6 times 32
contains 6 times 4 just as often as 32 contains 4. The same reasoning
applies to other numbers, and therefore _we do not alter the quotient
if we multiply the dividend and divisor by the same number_.
86. Again, suppose that 200 is to be divided by 50. Divide both the
dividend and divisor by the same number; for example, 5. Then, 200 is 5
times 40, and 50 is 5 times 10. But by (85), 40 divided by 10 gives the
same quotient as 5 times 40 divided by 5 times 10, and therefore _the
quotient of two numbers is not altered by dividing both the dividend
and divisor by the same number_.
87. From (55), if a number be multiplied successively by two others, it
is multiplied by their product. Thus, 27, first multiplied by 5, and
the product multiplied by 3, is the same as 27 multiplied by 5 times
3, or 15. Also, if a number be divided by any number, and the quotient
be divided by another, it is the same as if the first number had been
divided by the product of the other two. For example, divide 60 by 4,
which gives 15, and the quotient by 3, which gives 5. It is plain, that
if each of the four fifteens of which 60 is composed be divided into
three equal parts, there are twelve equal parts in all; or, a division
by 4, and then by 3, is equivalent to a division by 4 × 3, or 12.
88. The following rules will be better understood by stating them in
an example. If 32 be multiplied by 24 and divided by 6, the result is
the same as if 32 had been multiplied by the quotient of 24 divided
by 6, that is, by 4; for the sixth part of 24 being 4, the sixth part
of any number repeated 24 times is that number repeated 4 times; or,
multiplying by 24 and dividing by 6 is equivalent to multiplying by 4.
89. Again, if 48 be multiplied by 4, and that product be divided by
24, it is the same thing as if 48 were divided at once by the quotient
of 24 divided by 4, that is, by 6. For, every unit which is repeated 6
times in 48 is repeated 4 times as often, or 24 times, in 4 times 48,
or the quotient of 48 and 6 is the same as the quotient of 48 × 4 and 6
× 4.
90. The results of the last five articles may be algebraically
expressed thus:
_ma_ _a_
---- = ---- (85)
_mb_ _b_
If _n_ divide _a_ and _b_ without remainder,
_a_
----
_n_ _a_
------ = ---- (86)
_b_ _b_
----
_n_
_a_
----
_b_ _a_
------ = ---- (87)
_c_ _bc_
_ab_ _b_
------ = _a_ × ---- (88)
_c_ _c_
_ac_ _a_
----- = ------ (89)
_b_ _b_
----
_c_
It must be recollected, however, that these have only been proved in
the case where all the divisions are without remainder.
91. When one number divides another without leaving any remainder,
or is contained an exact number of times in it, it is said to be a
_measure_ of that number, or to _measure_ it. Thus, 4 is a measure of
136, or measures 136; but it does not measure 137. The reason for
using the word measure is this: Suppose you have a rod 4 feet long,
with nothing marked upon it, with which you want to measure some
length; for example, the length of a street. If that street should
happen to be 136 feet in length, you will be able to _measure_ it with
the rod, because, since 136 contains 4 34 times, you will find that the
street is exactly 34 times the length of the rod. But if the street
should happen to be 137 feet long, you cannot measure it with the rod;
for when you have measured 34 of the rods, you will find a remainder,
whose length you cannot tell without some shorter measure. Hence 4 is
said to measure 136, but not to measure 137. A measure, then, is a
divisor which leaves no remainder.
92. When one number is a measure of two others, it is called a _common
measure_ of the two. Thus, 15 is a common measure of 180 and 75. Two
numbers may have several common measures. For example, 360 and 168
have the common measures 2, 3, 4, 6, 24, and several others. Now, this
question maybe asked: Of all the common measures of 360 and 168, which
is the greatest? The answer to this question is derived from a rule of
arithmetic, called the rule for finding the GREATEST COMMON MEASURE,
which we proceed to consider.
93. If one quantity measure two others, it measures their sum and
difference. Thus, 7 measures 21 and 56. It therefore measures 56 + 21
and 56-21, or 77 and 35. This is only another way of saying what was
said in (74).
94. If one number measure a second, it measures every number which the
second measures. Thus, 5 measures 15, and 15 measures 30, 45, 60, 75,
&c.; all which numbers are measured by 5. It is plain that if
15 contains 5 3 times,
30, or 15 + 15 contains 5 3 + 3 times, or 6 times,
45, or 15 + 15 + 15 contains 5 3 + 3 + 3 or 9 times;
and so on.
95. Every number which measures both the dividend and divisor measures
the remainder also. To shew this, divide 360 by 112. The quotient is
3, and the remainder 24, that is (72) 360 is three times 112 and 24,
or 360 = 112 × 3 + 24. From this it follows, that 24 is the difference
between 360 and 3 times 112, or 24 = 360-112 × 3. Take any number which
measures both 360 and 112; for example, 4. Then
4 measures 360,
4 measures 112, and therefore (94) measures 112 × 3,
or 112 + 112 + 112.
Therefore (93) it measures 360-112 × 3, which is the remainder 24. The
same reasoning may be applied to all other measures of 360 and 112; and
the result is, that every quantity which measures both the dividend and
divisor also measures the remainder. Hence, every _common measure_ of
a dividend and divisor is also a _common measure_ of the divisor and
remainder.
96. Every common measure of the divisor and remainder is also a
common measure of the dividend and divisor. Take the same example,
and recollect that 360 = 112 × 3 + 24. Take any common measure of the
remainder 24 and the divisor 112; for example, 8. Then
8 measures 24;
and 8 measures 112, and therefore (94) measures 112 × 3.
Therefore (93) 8 measures 112 × 3 + 24, or measures the dividend 360.
Then every common measure of the remainder and divisor is also a common
measure of the divisor and dividend, or there is no common measure of
the remainder and divisor which is not also a common measure of the
divisor and dividend.
97. I. It is proved in (95) that the remainder and divisor have all the
common measures which are in the dividend and divisor.
II. It is proved in (96) that they have no others.
It therefore follows, that the greatest of the common measures of the
first two is the greatest of those of the second two, which shews how
to find the greatest common measure of any two numbers,[13] as follows:
98. Take the preceding example, and let it be required to find the g.
c. m. of 360 and 112, and observe that
360 divided by 112 gives the remainder 24,
112 divided by 24 gives the remainder 16,
24 divided by 16 gives the remainder 8,
16 divided by 8 gives no remainder.
[13] For shortness, I abbreviate the words _greatest common measure_
into their initial letters, g. c. m.
Now, since 8 divides 16 without remainder, and since it also divides
itself without remainder, 8 is the g. c. m. of 8 and 16, because it is
impossible to divide 8 by any number greater than 8; so that, even if
16 had a greater measure than 8, it could not be _common_ to 16 and 8.
Therefore 8 is g. c. m. of 16 and 8,
(97) g. c. m. of 16 and 8 is g. c. m. of 24 and 16,
g. c. m. of 24 and 16 is g. c. m. of 112 and 24,
g. c. m. of 112 and 24 is g. c. m. of 360 and 112,
Therefore 8 is g. c. m. of 360 and 112.
The process carried on may be written down in either of the following
ways:
112)360(3
336
---
24)112(4 112 | 360 3
96 96 | 336 4
--- ----+-------
16)24(1 16 | 24 1
16 16 | 16 2
-- ----+-------
8)16(2 0 | 8
16
--
0
The rule for finding the greatest common measure of two numbers is,
I. Divide the greater of the two by the less.
II. Make the remainder a divisor, and the divisor a dividend, and find
another remainder.
III. Proceed in this way until there is no remainder, and the last
divisor is the greatest common measure required.
99. You may perhaps ask how the rule is to shew when the two numbers
have no common measure. The fact is, that there are, strictly speaking,
no such numbers, because all numbers are measured by 1; that is,
contain an exact number of units, and therefore 1 is a common measure
of every two numbers. If they have no other common measure, the last
divisor will be 1, as in the following example, where the greatest
common measure of 87 and 25 is found.
25)87(3
75
--
12)25(2
24
--
1)12(12
12
--
0
EXERCISES.
Numbers. g. c. m.
6197 9521 1
58363 2602 1
5547 147008443 1849
6281 326041 571
28915 31495 5
1509 300309 3
What are 36 × 36 + 2 × 36 × 72 + 72 × 72
and 36 × 36 × 36 + 72 × 72 × 72;
and what is their greatest common measure?--_Answer_, 11664.
100. If two numbers be divisible by a third, and if the quotients be
again divisible by a fourth, that third is not the greatest common
measure. For example, 360 and 504 are both divisible by 4. The
quotients are 90 and 126. Now 90 and 126 are both divisible by 9,
the quotients of which division are 10 and 14. By (87), dividing a
number by 4, and then dividing the quotient by 9, is the same thing
as dividing the number itself by 4 × 9, or by 36. Then, since 36 is
a common measure of 360 and 504, and is greater than 4, 4 is not the
greatest common measure. Again, since 10 and 14 are both divisible by
2, 36 is not the greatest common measure. It therefore follows, that
when two numbers are divided by their greatest common measure, the
quotients have no common measure except 1 (99). Otherwise, the number
which was called the greatest common measure in the last sentence is
not so in reality.
101. To find the greatest common measure of three numbers, find the g.
c. m. of the first and second, and of this and the third. For since
all common divisors of the first and second are contained in their g.
c. m., and no others, whatever is common to the first, second, and
third, is common also to the third and the g. c. m. of the first and
second, and no others. Similarly, to find the g. c. m. of four numbers,
find the g. c. m. of the first, second, and third, and of that and the
fourth.
102. When a first number contains a second, or is divisible by it
without remainder, the first is called a multiple of the second. The
words _multiple_ and _measure_ are thus connected: Since 4 is a
measure of 24, 24 is a multiple of 4. The number 96 is a multiple of
8, 12, 24, 48, and several others. It is therefore called a _common
multiple_ of 8, 12, 24. 48, &c. The product of any two numbers is
evidently a common multiple of both. Thus, 36 × 8, or 288, is a common
multiple of 36 and 8. But there are common multiples of 36 and 8 less
than 288; and because it is convenient, when a common multiple of two
quantities is wanted, to use the least of them, I now shew how to find
the least common multiple of two numbers.
103. Take, for example, 36 and 8. Find their greatest common measure,
which is 4, and observe that 36 is 9 × 4, and 8 is 2 × 4. The quotients
of 36 and 8, when divided by their greatest common measure, are
therefore 9 and 2. Multiply these quotients together, and multiply the
product by the greatest common measure, 4, which gives 9 × 2 × 4, or
72. This is a multiple of 8, or of 4 × 2 by (55); and also of 36 or of
4 × 9. It is also the least common multiple; but this cannot be proved
to you, because the demonstration cannot be thoroughly understood
without more practice in the use of letters to stand for numbers. But
you may satisfy yourself that it is the least in this case, and that
the same process will give the least common multiple in any other case
which you may take. It is not even necessary that you should know it is
the least. Whenever a common multiple is to be used, any one will do as
well as the least. It is only to avoid large numbers that the least is
used in preference to any other.
When the greatest common measure is 1, the least common multiple of the
two numbers is their product.
The rule then is: To find the least common multiple of two numbers,
find their greatest common measure, and multiply one of the numbers by
the quotient which the other gives when divided by the greatest common
measure. To find the least common multiple of three numbers, find the
least common multiple of the first two, and find the least common
multiple of that multiple and the third, and so on.
EXERCISES.
Numbers proposed. | Least common multiple.
14, 21 | 42
16, 5, 24 | 240
1, 2, 3, 4, 5, 6, 7, 8, 9, 10 | 2520
6, 8, 11, 16, 20 | 2640
876, 864 | 63072
868, 854 | 52948
A convenient mode of finding the least common multiple of several
numbers is as follows, when the common measures are easily visible:
Pick out a number of common measures of two or more, which have
themselves no divisors greater than unity. Write them as divisors,
and divide every number which will divide by one or more of them.
Bring down the quotients, and also the numbers which will not divide
by any of them. Repeat the process with the results, and so on until
the numbers brought down have no two of them any common measure except
unity. Then, for the least common multiple, multiply all the divisors
by all the numbers last brought down. For instance, let it be required
to find the least common multiple of all the numbers from 11 to 21.
2, 2, 3, 5, 7)11 12 13 14 15 16 17 18 19 20 21
---------------------------------
11 1 13 1 1 4 17 3 19 1 1
There are now no common measures left in the row, and the least common
multiple required is the product of 2, 2, 3, 5, 7, 11, 13, 4, 17, 3,
and 19; or 232792560.
SECTION V.
FRACTIONS.
104. Suppose it required to divide 49 yards into five equal parts, or,
as it is called, to find the fifth part of 49 yards. If we divide 45 by
5, the quotient is 9, and the remainder is 4; that is (72), 49 is made
up of 5 times 9 and 4. Let the line A B represent 49 yards:
A----------------------------------------B
C-------------- I --
D-------------- K --
E-------------- L --
F-------------- M --
G-------------- N --
I K L M N
H +-+-+-+-+-+
| | | | | |
Take 5 lines, C, D, E, F, and G, each 9 yards in length, and the line
H, 4 yards in length. Then, since 49 is 5 nines and 4, C, D, E, F, G,
and H, are together equal to A B. Divide H, which is 4 yards, into five
equal parts, I, K, L, M, and N, and place one of these parts opposite
to each of the lines, C, D, E, F, and G. It follows that the ten lines,
C, D, E, F, G, I, K, L, M, N, are together equal to A B, or 49 yards.
Now D and K together are of the same length as C and I together, and
so are E and L, F and M, and G and N. Therefore, C and I together,
repeated 5 times, will be 49 yards; that is, C and I together make up
the fifth part of 49 yards.
105. C is a certain number of yards, viz. 9; but I is a new sort of
quantity, to which hitherto we have never come. It is not an exact
number of yards, for it arises from dividing 4 yards into 5 parts, and
taking one of those parts. It is the fifth part of 4 yards, and is
called a FRACTION of a yard. It is written thus, ⁴/₅(23), and is what
we must add to 9 yards in order to make up the fifth part of 49 yards.
The same reasoning would apply to dividing 49 bushels of corn, or 49
acres of land, into 5 equal parts. We should find for the fifth part
of the first, 9 bushels and the fifth part of 4 bushels; and for the
second, 9 acres and the fifth part of 4 acres.
We say, then, once for all, that the fifth part of 49 is 9 and ⁴/₅, or
9 + ⁴/₅; which is usually written (9⁴/₅), or if we use signs, 49/5 =
(9⁴/₅).
EXERCISES.
What is the seventeenth part of 1237?--_Answer_, (72-¹³/₁₇).
10032 663819 22773399
What are -----, ------, and -------- ?
1974 23710 2424
162 23649 2343
_Answer_, (5 ----), (27 -----), (9394 ----).
1974 23710 2424
106. By the term fraction is understood a part of any number, or the
sum of any of the equal parts into which a number is divided. Thus,
⁴⁹/₅, ⁴/₅, ²⁰/₇, are fractions. The term fraction even includes whole
numbers:[14] for example, 17 is ¹⁷/₁, ³⁴/₂, ⁵¹/₃, &c.
[14] Numbers which contain an exact number of units, such as 5, 7,
100, &c., are called _whole numbers_ or _integers_, when we wish to
distinguish them from fractions.
The upper number is called the _numerator_, the lower number is
called the _denominator_, and both of these are called _terms_ of the
fraction. As long as the numerator is less than the denominator, the
fraction is less than a unit: thus, ⁶/₁₇ is less than a unit; for 6
divided into 6 parts gives 1 for each part, and must give less when
divided into 17 parts. Similarly, the fraction is equal to a unit when
the numerator and denominator are equal, and greater than a unit when
the numerator is greater than the denominator.
107. By ⅔ is meant the third part of 2. This is the same as twice the
third part of 1.
To prove this, let A B be two yards, and divide each of the yards A C
and C B into three equal parts.
|--|--|--|--|--|--|
A D E C F G B
Then, because A E, E F, and F B, are all equal to one another, A E is
the third part of 2. It is therefore ⅔. But A E is twice A D, and A
D is the third part of one yard, or ⅓; therefore ⅔ is twice ⅓; that
is, in order to get the length ⅔, it makes no difference whether we
divide _two_ yards at once into three parts, and take _one_ of them,
or whether we divide _one_ yard into three parts, and take _two_ of
them. By the same reasoning, ⅝ may be found either by dividing 5 into
8 parts, and taking one of them, or by dividing 1 into 8 parts, and
taking five of them. In future, of these two meanings I shall use that
which is most convenient at the time, as it is proved that they are
the same thing. This principle is the same as the following: The third
part of any number may be obtained by adding together the thirds of all
the units of which it consists. Thus, the third part of 2, or of two
units, is made by taking one-third out of each of the units, that is,
⅔ = ⅓ × 2.
This meaning appears ambiguous when the numerator is greater than the
denominator: thus, ¹⁵/₇ would mean that 1 is to be divided into 7
parts, and 15 of them are to be taken. We should here let as many units
be each divided into 7 parts as will give more than 15 of those parts,
and take 15 of them.
108. The value of a fraction is not altered by multiplying the
numerator and denominator by the same quantity. Take the fraction ¾,
multiply its numerator and denominator by 5, and it becomes ¹⁵/₂₀,
which is the same thing as ¾; that is, one-twentieth part of 15 yards
is the same thing as one-fourth of 3 yards: or, if our second meaning
of the word fraction be used, you get the same length by dividing a
yard into 20 parts and taking 15 of them, as you get by dividing it
into 4 parts and taking 3 of them. To prove this,
[Illustration]
let A B represent a yard; divide it into 4 equal parts, A C, C D, D E,
and E B, and divide each of these parts into 5 equal parts. Then A E is
¾. But the second division cuts the line into 20 equal parts, of which
A E contains 15. It is therefore ¹⁵/₂₀. Therefore, ¹⁵/₂₀ and ¾ are the
same thing.
Again, since ¾ is made from ¹⁵/₂₀ by dividing both the numerator
and denominator by 5, the value of a fraction is not altered by
dividing both its numerator and denominator by the same quantity. This
principle, which is of so much importance in every part of arithmetic,
is often used in common language, as when we say that 14 out of 21 is 2
out of 3, &c.
109. Though the two fractions ¾ and ¹⁵/₂₀ are the same in value,
and either of them may be used for the other without error, yet the
first is more convenient than the second, not only because you have a
clearer idea of the fourth of three yards than of the twentieth part
of fifteen yards, but because the numbers in the first being smaller,
are more convenient for multiplication and division. It is therefore
useful, when a fraction is given, to find out whether its numerator
and denominator have any common divisors or common measures. In (98)
was given a rule for finding the greatest common measure of any two
numbers; and it was shewn that when the two numbers are divided by
their greatest common measure, the quotients have no common measure
except 1. Find the greatest common measure of the terms of the
fraction, and divide them by that number. The fraction is then said to
be _reduced to its lowest terms_, and is in the state in which the best
notion can be formed of its magnitude.
EXERCISES.
With each fraction is written the same reduced to its lowest terms.
2794 22 × 127 22
---- = ---------- = ----
2921 23 × 127 23
2788 17 × 164 17
---- = ---------- = ----
4920 30 × 164 30
93208 764 × 122 764
----- = ---------- = ---
13786 113 × 122 113
888800 22 × 40400 22
-------- = ------------ = -----
40359600 999 × 40400 999
95469 121 × 789 121
------ = ----------- = ---
359784 456 × 789 456
110. When the terms of the fraction given are already in factors,[15]
any one factor in the numerator may be divided by a number, provided
some one factor in the denominator is divided by the same. This follows
from (88) and (108). In the following examples the figures altered by
division are accented.
[15] A factor of a number is a number which divides it without
remainder: thus, 4, 6, 8, are factors of 24, and 6 × 4, 8 × 3, 2 × 2 ×
2 × 3, are several ways of decomposing 24 into factors.
12 × 11 × 10 3′ × 11 × 10 1′ × 11 × 5′
------------ = ------------- = ------------- = 55.
2 × 3 × 4 2 × 3 × 1′ 1′ × 1′ × 1′
18 × 15 × 13 2′ × 3′ × 1′ 1′ × 1′ × 1′
------------ = ------------- = ------------- = ¹/₁₆.
20 × 54 × 52 4′ × 6′ × 4′ 2′ × 2′ × 4′
27 × 28 3′ × 4′ 3′ × 2′
------- = --------- = -------- = ⁶/₅.
9 × 70 1′ × 10′ 1′ × 5′
111. As we can, by (108), multiply the numerator and denominator of a
fraction by any number, without altering its value, we can now readily
reduce two fractions to two others, which shall have the same value as
the first two, and which shall have the same denominator. Take, for
example, ⅔ and ⁴/₇; multiply both terms of ⅔ by 7, and both terms of
⁴/₇ by 3. It then appears that
2 × 7
⅔ is ------- or ¹⁴/₂₁
3 × 7
4 × 3
⁴/₇ is ------- or ¹²/₂₁.
7 × 3
Here are then two fractions ¹⁴/₂₁ and ¹²/₂₁, equal to ⅔ and ⁴/₇, and
having the same denominator, 21; in this case, ⅔ and ⁴/₇ are said to be
_reduced to a common denominator_.
It is required to reduce ⅒, ⅚, and ⁷/₉ to a common denominator.
Multiply both terms of the first by the product of 6 and 9; of the
second by the product of 10 and 9; and of the third by the product of
10 and 6. Then it appears (108) that
1 × 6 × 9
⅒ is ----------- or ⁵⁴/₅₄₀
10 × 6 × 9
5 × 10 × 9
⅚ is ------------ or ⁴⁵⁰/₅₄₀
6 × 10 × 9
7 × 10 × 6
⁷/₉ is ------------ or ⁴²⁰/₅₄₀.
9 × 10 × 6
On looking at these last fractions, we see that all the numerators and
the common denominator are divisible by 6, and (108) this division will
not alter their values. On dividing the numerators and denominators of
⁵⁴/₅₄₀, ⁴⁵⁰/₅₄₀, and ⁴²⁰/₅₄₀ by 6, the resulting fractions are, ⁹/₉₀,
⁷⁵/₉₀, and ⁷⁰/₉₀. These are fractions with a common denominator, and
which are the same as ⅒, ⅚, and ⁷/₉; and therefore these are a more
simple answer to the question than the first fractions. Observe also
that 540 is one common multiple of 10, 6, and 9, namely, 10 × 6 × 9,
but that 90 is _the least_ common multiple of 10, 6, and 9 (103). The
following process, therefore, is better. To reduce the fractions ⅒,
⅚, and ⁷/₉, to others having the same value and a common denominator,
begin by finding the least common multiple of 10, 6, and 9, by the rule
in (103), which is 90. Observe that 10, 6, and 9 are contained in 90 9,
15, and 10 times. Multiply both terms of the first by 9, of the second
by 15, and of the third by 10, and the fractions thus produced are
⁹/₉₀, ⁷⁵/₉₀, and ⁷⁰/₉₀, the same as before.
If one of the numbers be a whole number, it may be reduced to a
fraction having the common denominator of the rest, by (106).
EXERCISES.
Fractions proposed reduced to a common denominator.
2 1 1 | 20 6 5
--- --- --- | ---- --- ---
3 5 6 | 30 30 30
|
|
1 2 3 12 3 | 28 24 18 48 63
--- --- --- ---- --- | --- --- --- --- ---
3 7 14 21 4 | 84 84 84 84 84
|
4 5 6 | 3000 400 50 6
3 --- ---- ---- | ---- ----- ----- ----
10 100 1000 | 1000 1000 1000 1000
|
33 281 | 22341 106499
---- ---- | -------- ------
379 677 | 256583 256583
112. By reducing two fractions to a common denominator, we are able
to compare them; that is, to tell which is the greater and which the
less of the two. For example, take ½ and ⁷/₁₅. These fractions reduced,
without alteration of their value, to a common denominator, are ¹⁵/₃₀
and ¹⁴/₃₁. Of these the first must be the greater, because (107) it may
be obtained by dividing 1 into 30 equal parts and taking 15 of them,
but the second is made by taking 14 of those parts.
It is evident that of two fractions which have the same denominator,
the greater has the greater numerator; and also that of two fractions
which have the same numerator, the greater has the less denominator.
Thus, ⁸/₇ is greater than ⁸/⁹, since the first is a 7th, and the
last only a 9th part of 8. Also, any numerator may be made to belong
to as small a fraction as we please, by sufficiently increasing the
denominator. Thus, ¹⁰/₁₀₀ is ¹/₁₀, ¹⁰/₁₀₀₀ is ¹/₁₀₀, and ¹⁰/₁₀₀₀₀₀₀ is
¹/₁₀₀₀₀₀₀ (108).
We can now also increase and diminish the first fraction by the second.
For the first fraction is made up of 15 of the 30 equal parts into
which 1 is divided. The second fraction is 14 of those parts. The sum
of the two, therefore, must be 15 + 14, or 29 of those parts; that is,
½ + ⁷/₁₅ is ²⁹/₃₀. The difference of the two must be 15-14, or 1 of
those parts; that is, ½-⁷/₁₅ = ¹/₃₀.
113. From the last two articles the following rules are obtained:
I. To compare, to add, or to subtract fractions, first reduce them to
a common denominator. When this has been done, that is the greatest of
the fractions which has the greatest numerator.
Their sum has the sum of the numerators for its numerator, and the
common denominator for its denominator.
Their difference has the difference of the numerators for its
numerator, and the common denominator for its denominator.
EXERCISES.
1 1 1 1 53 44 153 18329
--- + --- + --- - --- = ---- ---- - ----- = -------
2 3 4 5 60 3 427 1282
8 3 4 1834 1 12 253
1 + ---- + ---- + ---- = ---- 2 - --- + ---- = ----
10 100 1000 1000 7 13 91
1 8 94 3 163 97 93066
--- + --- + ---- = --- --- - ---- = -------
2 16 188 2 521 881 459001
114. Suppose it required to add a whole number to a fraction, for
example, 6 to ⁴/₉. By (106) 6 is ⁵⁴/₉, and ⁵⁴/₉ + ⁴/₉ is ⁵⁸/⁹; that is,
6 + ⁴/⁹, or as it is usually written, (6⁴/₉), is ⁵⁸/₉. The rule in this
case is: Multiply the whole number by the denominator of the fraction,
and to the product add the numerator of the fraction; the sum will be
the numerator of the result, and the denominator of the fraction will
be its denominator. Thus, (3¼) = ¹³/₄, (22⁵/₉) = ²⁰³/₉, (74²/₅₅) =
⁴⁰⁷²/₅₅. This rule is the opposite of that in (105).
115. From the last rule it appears that
907 17230907 225 667225
1723 ------ is --------, 667 ----- is ------,
10000 10000 1000 1000
99 2300099
and 23 ------ is -------.
10000 10000
Hence, when a whole number is to be added to a fraction whose
denominator is 1 followed by _ciphers_, the number of which is not less
than the number of _figures_ in the numerator, the rule is: Write the
whole number first, and then the numerator of the fraction, with as
many ciphers between them as the number of _ciphers_ in the denominator
exceeds the number of _figures_ in the numerator. This is the numerator
of the result, and the denominator of the fraction is its denominator.
If the number of ciphers in the denominator be equal to the number of
figures in the numerator, write no ciphers between the whole number and
the numerator.
EXERCISES.
Reduce the following mixed quantities to fractions:
23707 6 299 2210
1 ------, 2457 ---, 1207 --------, and 233 -----.
100000 10 10000000 10000
116. Suppose it required to multiply ⅔ by 4. This by (48) is taking ⅔
four times; that is, finding ⅔ + ⅔ + ⅔ + ⅔. This by (112) is ⁸/₃; so
that to multiply a fraction by a whole number the rule is: Multiply the
numerator by the whole number, and let the denominator remain.
117. If the denominator of the fraction be divisible by the whole
number, the rule may be stated thus: Divide the denominator of the
fraction by the whole number, and let the numerator remain. For
example, multiply ⁷/₃₆ by 6. This (116) is ⁴²/₃₆, which, since the
numerator and denominator are now divisible by 6, is (108) the same as
⁷/₆. It is plain that ⁷/₆ is made from ⁷/₃₆ in the manner stated in the
rule.
118. Multiplication has been defined to be the taking as many of one
number as there are units in another. Thus, to multiply 12 by 7 is to
take as many twelves as there are units in 7, or to take 12 as many
times as you must take 1 in order to make 7. Thus, what is done with 1
in order to make 7, is done with 12 to make 7 times 12. For example,
7 is 1 + 1 + 1 + 1 + 1 + 1 + 1
7 times 12 is 12 + 12 + 12 + 12 + 12 + 12 + 12.
When the same thing is done with two fractions, the result is still
called their product, and the process is still called multiplication.
There is this difference, that whereas a whole number is made by adding
1 to itself a number of times, a fraction is made by dividing 1 into
a number of equal parts, and adding _one of these parts_ to itself a
number of times. This being the meaning of the word multiplication,
as applied to fractions, what is ¾ multiplied by ⅞? Whatever is done
with 1 in order to make ⅞ must now be done with ¾; but to make ⅞, 1 is
divided into 8 parts, and 7 of them are taken. Therefore, to make ¾ ×
⅞, ¾ must be divided into 8 parts, and 7 of them must be taken. Now ¾
is, by (108), the same thing as ²⁴/₃₂. Since ²⁴/₃₂ is made by dividing
1 into 32 parts, and taking 24 of them, or, which is the same thing,
taking 3 of them 8 times, if ²⁴/₃₂ be divided into 8 equal parts, each
of them is ³/₃₂; and if 7 of these parts be taken, the result is ²¹/₃₂
(116): therefore ¾ multiplied by ⅞ is ²¹/₃₂; and the same reasoning
may be applied to any other fractions. But ²¹/₃₂ is made from ¾ and ⅞
by multiplying the two numerators together for the numerator, and the
two denominators for the denominator; which furnishes a rule for the
multiplication of fractions.
119. If this product ²¹/₃₂ is to be multiplied by a third fraction, for
example, by ⁵/₉, the result is, by the same rule, ¹⁰⁵/₂₈₈; and so on.
The general rule for multiplying any number of fractions together is
therefore:
Multiply all the numerators together for the numerator of the product,
and all the denominators together for its denominator.
120. Suppose it required to multiply together ¹⁵/₁₆ and ⁸/₁₀. The
product may be written thus:
15 × 8 120
-------, and is ----,
16 × 10 160
which reduced to its lowest terms (109) is ¾. This result might have
been obtained directly, by observing that 15 and 10 are both measured
by 5, and 8 and 16 are both measured by 8, and that the fraction may be
written thus:
3 × 5 × 8
-------------.
2 × 8 × 2 × 5
Divide both its numerator and denominator by 5 × 8 (108) and (87),
and the result is at once ¾; therefore, before proceeding to multiply
any number of fractions together, if there be any numerator and any
denominator, whether belonging to the same fraction or not, which have
a common measure, divide them both by that common measure, and use the
quotients instead of the dividends.
A whole number may be considered as a fraction whose denominator is 1;
thus, 16 is ¹⁶/₁ (106); and the same rule will apply when one or more
of the quantities are whole numbers.
EXERCISES
136 268 36448 18224
---- × --- = ------- = -------
7470 919 6864930 3432465
1 2 3 4 1 2 17 2
--- × --- × --- × --- = --- ---- × ---- = ----
2 3 4 5 5 17 45 45
2 13 241 6266 13 601 7813
--- × ---- × ---- = ------ ---- × ---- = -----
59 7 19 7847 461 11 5071
Fraction proposed. Square. Cube.
701 491401 344472101
----- ------- ---------
158 24964 3944312
140 19600 2744000
----- ------ --------
141 19881 2803221
355 126025 44738875
----- ------- ---------
113 12769 1442897
From 100 acres of ground, two-thirds of them are taken away; 50 acres
are then added to the result, and ⁵/₇ of the whole is taken; what
number of acres does this produce?--_Answer_, (59¹¹/₂₁).
121. In dividing one whole number by another, for example, 108 by 9,
this question is asked,--Can we, by the addition of any number of
nines, produce 108? and if so, how many nines will be sufficient for
that purpose?
Suppose we take two fractions, for example, ⅔ and ⅘, and ask, Can we,
by dividing ⅘ into some number of equal parts, and adding a number of
these parts together, produce ⅔? if so, into _how many parts_ must we
divide ⅘, and _how many of them_ must we add together? The solution of
this question is still called the division of ⅔ by ⅘; and the fraction
whose denominator is the number of parts into which ⅘ is divided, and
whose numerator is the number of them which is taken, is called the
quotient. The solution of this question is as follows: Reduce both
these fractions to a common denominator (111), which does not alter
their value (108); they then become ¹⁰/₁₅ and ¹²/₁₅. The question
now is, to divide ¹²/₁₅ into a number of parts, and to produce ¹⁰/₁₅
by taking a number of these parts. Since ¹²/₁₅ is made by dividing 1
into 15 parts and taking 12 of them, if we divide ¹²/₁₅ into 12 equal
parts, each of these parts is ¹/₁₅; if we take 10 of these parts, the
result is ¹⁰/₁₅. Therefore, in order to produce ¹⁰/₁₅ or ⅔ (108), we
must divide ¹²/₁₅ or ⅘ into 12 parts, and take 10 of them; that is, the
quotient is ¹⁰/₁₂. If we call ⅔ the dividend, and ⅘ the divisor, as
before, the quotient in this case is derived from the following rule,
which the same reasoning will shew to apply to other cases:
The numerator of the quotient is the numerator of the dividend
multiplied by the denominator of the divisor. The denominator of the
quotient is the denominator of the dividend multiplied by the numerator
of the divisor. This rule is the reverse of multiplication, as will be
seen by comparing what is required in both cases. In multiplying ⅘ by
¹⁰/₁₂, I ask, if out of ⅘ be taken 10 parts out of 12, how much _of a
unit_ is taken, and the answer is ⁴⁰/⁶⁰, or ⅔. Again, in dividing ⅔ by
⅘, I ask what part of ⅘ is ⅔, the answer to which is ¹⁰/₁₂.
122. By taking the following instance, we shall see that this rule can
be sometimes simplified. Divide ¹⁶/₃₃ by ²⁸/₁₅. Observe that 16 is 4 ×
4, and 28 is 4 × 7; 33 is 3 × 11, and 15 is 3 × 5; therefore the two
fractions are
4 × 4 4 × 7
------ and -----,
3 × 11 3 × 5
and their quotient, according to the rule, is
4 × 4 × 3 × 5
--------------,
3 × 11 × 4 × 7
in which 4 × 3 is found both in the numerator and denominator. The
fraction is therefore (108) the same as
4 × 5 20
------, or ----.
11 × 7 77
The rule of the last article, therefore, admits of this modification:
If the two numerators or the two denominators have a common measure,
divide by that common measure, and use the quotients instead of the
dividends.
123. In dividing a fraction by a whole number, for example, ⅔ by 15,
consider 15 as the fraction ¹⁵/₁. The rule gives ²/⁴⁵ as the quotient.
Therefore, to divide a fraction by a whole number, multiply the
denominator by that whole number.
EXERCISES.
Dividend. Divisor. Quotient.
41 63 41
---- ---- -----
33 11 189
467 907 47167
---- ---- -------
151 101 136957
7813 601 13
----- ---- ----
5071 11 461
¹/₅ × ¹/₅ × ¹/₅ - ²/₁₇× ²/₁₇ × ²/₁₇
What are -----------------------------------,
¹/₅ - ²/₁₇
and ⁸/₁₁ × ⁸/₁₁ - ³/₁₁ × ³/₁₁
----------------------- ?
⁸/₁₁ - ³/₁₁
559
_Answer_, ----, and 1.
7225
A can reap a field in 12 days, B in 6, and C in 4 days; in what time
can they all do it together?[16]--_Answer_, 2 days.
[16] The method of solving this and the following question may be shewn
thus: If the number of days in which each could reap the field is
given, the part which each could do in a day by himself can be found,
and thence the part which all could do together; this being known, the
number of days which it would take all to do the whole can be found.
In what time would a cistern be filled by cocks which would separately
fill it in 12, 11, 10, and 9 hours?--_Answer_, (2⁴⁵⁴/₇₆₃) hours.
124. The principal results of this section may be exhibited
algebraically as follows; let _a_, _b_, _c_, &c. stand for any whole
numbers. Then
_a_ 1 _a_ _ma_
(107) ---- = ---- × _a_ (108) ---- = ----
_b_ _b_ _b_ _mb_
_a_ _c_ _ad_ _bc_
(111) --- and --- are the same as ---- and ----
_b_ _d_ _bd_ _bd_
_a_ _b_ _a_ + _b_
(112) --- + --- = ---------
_c_ _c_ _c_
_a_ _b_ _a_ - _b_
--- - --- = ---------
_c_ _c_ _c_
_a_ _c_ _ad_ + _bc_
(113) --- + --- = -----------
_b_ _d_ _bd_
_a_ _c_ _ad_ - _bc_
--- - --- = -----------
_b_ _d_ _bd_
_a_ _c_ _ac_
(118) --- × --- = ----
_b_ _d_ _bd_
_a_ _c_ _a_/_b_ _ad_
(121) --- divᵈ. by --- or --------- = ----
_b_ _d_ _c_/_d_ _bc_
125. These results are true even when the letters themselves represent
fractions. For example, take the fraction
_a_/_b_
-------,
_c_/_d_
whose numerator and denominator are fractional, and multiply its
numerator and denominator by the fraction
_e_ _ae_/_bf_
---, which gives ----------,
_f_ _ce_/_df_
_aedf_
which (121) is -------,
_bfce_
which, dividing the numerator and denominator by _ef_ (108), is
_ad_
----.
_bc_
But the original fraction itself is
_ad_ _a_/_b_ (_a_/_b_) × (_e_/_f_)
----; hence ------- = ---------------------
_bc_ _c_/_d_ (_c_/_d_) × (_e_/_f_)
which corresponds to the second formula[17] in (124). In a similar
manner it may be shewn, that the other formulæ of the same article
are true when the letters there used either represent fractions, or
are removed and fractions introduced in their place. All formulæ
established throughout this work are equally true when fractions are
substituted for whole numbers. For example (54), (_m_ + _n_)_a_ = _ma_
+ _na_. Let _m_, _n_, and _a_ be respectively the fractions
_p_ _r_ _b_
---, ---, and ---.
_q_ _s_ _c_
Then _m_ + _n_ is
_p_ _r_ _ps_ + _qr_
--- + ---, or -----------
_q_ _s_ _qs_
and (_m_ + _n_)_a_ is
_ps_ + _qr_ _b_ (_ps_ + _qr_)_b_
----------- × ---, or ----------------
_qs_ _c_ _qsc_
_psb_ + _qrb_
or -------------.
_qsc_
_psb_ _qrb_ _pb_ _rb_
But this (112) is ----- + -----, which is ---- + ----,
_qsc_ _qsc_ _qc_ _sc_
_psb_ _pb_ _qrb_ _rb_
since ----- = ----, and ----- = ---- (103).
_qsc_ _qc_ _qsc_ _sc_
_pb_ _p_ _b_ _rb_ _r_ _b_
But ---- = --- × ---, and ---- = --- × ---.
_qc_ _q_ _c_ _sc_ _s_ _c_
Therefore (_m_ + _n_)_a_, or
(_p_ _r_ )_b_ _p_ _b_ _r_ _b_
(--- + --- )--- = --- × --- + --- × ---.
(_q_ _s_ )_c_ _q_ _c_ _s_ _c_
In a similar manner the same may be proved of any other formula.
[17] A formula is a name given to any algebraical expression which is
commonly used.
The following examples may be useful:
_a_ _c_ _e_ _g_
--- × --- + --- × ---
_b_ _d_ _f_ _h_ _acfh_ + _bdeg_
--------------------- = ---------------
_a_ _e_ _c_ _g_ _aedh_ + _bcfg_
--- × --- + --- × ---
_b_ _f_ _d_ _h_
1 _b_
---------- = ---------
1 _ab_ + 1
_a_ + ---
_b_
1 1 _bc_ + 1
--------------- = -------------- = -----------------
1 _c_ _abc_ + _a_ + _c_
_a_ + --------- _a_ + --------
1 _bc_ + 1
_b_ + ---
_c_
1 1 57
Thus, ------------ = --------- = ---
1 8 350
6 + ------- 6 + ---
1 57
7 + ---
8
The rules that have been proved to hold good for all numbers may be
applied when the numbers are represented by letters.
SECTION VI.
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).
SECTION VII.
ON THE EXTRACTION OF THE SQUARE ROOT.
156. We have already remarked (66), that a number multiplied by itself
produces what is called the _square_ of that number. Thus, 169, or 13 ×
13, is the square of 13. Conversely, 13 is called the _square root_ of
169, and 5 is the square root of 25; and any number is the square root
of another, which when multiplied by itself will produce that other.
The square root is signified by the sign
_
√ or √ ;
_______
thus, √25 means the square root of 25, or 5; √(16 + 9)
means the square root of 16 + 9, and is 5, and must not be confounded
with √16 + √9, which is 4 + 3, or 7.
157. The following equations are evident from the definition:
___ ___
√_a_ × √_a_ = _a_
____
√_aa_ = _a_
___ ___ ___
√_ab_ × √_ab_ = _ab_
___ ___ ___ ___ ___ ___ ___ ___
(√_a_ × √_b_) × (√_a_ × √_b_) = √_a_ × √_a_ × √_b_ × √_b_ = _ab_
___ ___ ____
whence √_a_ × √_b_ = √_ab_
158. It does not follow that a number has a square root because it
has a square; thus, though 5 can be multiplied by itself, there is
no number which multiplied by itself will produce 5. It is proved in
algebra, that no fraction[22] multiplied by itself can produce a whole
number, which may be found true in any number of instances; therefore
5 has neither a whole nor a fractional square root; that is, it has
no square root at all. Nevertheless, there are methods of finding
fractions whose squares shall be as _near_ to 5 as we please, though
not exactly equal to it. One of these methods gives ¹⁵¹²⁷/₆₇₆₅, whose
square, viz.
15127 15127 228826129
----- × ----- or ---------,
6765 6765 45765225
differs from 5 by only ⁴/₄₅₇₆₅₂₂₅, which is less than ·0000001: hence
we are enabled to use √5 in arithmetical and algebraical reasoning: but
when we come to the practice of any problem, we must substitute for
√5 one of the fractions whose square is nearly 5, and on the degree
of accuracy we want, depends what fraction is to be used. For some
purposes, ¹²³/₅₅ may be sufficient, as its square only differs from 5
by ⁴/₃₀₂₅; for others, the fraction first given might be necessary,
or one whose square is even nearer to 5. We proceed to shew how to
find the square root of a number, when it has one, and from thence how
to find fractions whose squares shall be as near as we please to the
number, when it has not. We premise, what is sufficiently evident, that
of two numbers, the greater has the greater square; and that if one
number lie between two others, its square lies between the squares of
those others.
[22] Meaning, of course, a really fractional number, such as ⅞ or
¹⁵/₁₁, not one which, though fractional in form, is whole in reality,
such as ¹⁰/₅ or ²⁷/₃.
159. Let _x_ be a number consisting of any number of parts, for
example, four, viz. _a_, _b_, _c_, and _d_; that is, let
_x_ = _a_ + _b_ + _c_ + _d_
The square of this number, found as in (68), will be
_aa_ + 2_a_(_b_ + _c_ + _d_)
+ _bb_ + 2_b_(_c_ + _d_)
+ _cc_ + 2_cd_
+ _dd_
The rule there found for squaring a number consisting of parts was:
Square each part, and multiply all that come after by twice that part,
the sum of all the results so obtained will be the square of the whole
number. In the expression above obtained, instead of multiplying 2_a_
by _each_ of the succeeding parts, _b_, _c_, and _d_, and adding the
results, we multiplied 2_a_ by the _sum of all_ the succeeding parts,
which (52) is the same thing; and as the parts, however disposed,
make up the number, we may reverse their order, putting the last
first, &c.; and the rule for squaring will be: Square each part, and
multiply all that come before by twice that part. Hence a reverse rule
for extracting the square root presents itself with more than usual
simplicity. It is: To extract the square root of a number N, choose
a number A, and see if N will bear the subtraction of the square of
A; if so, take the remainder, choose a second number B, and see if
the remainder will bear the subtraction of the square of B, and twice
B multiplied by the preceding part A: if it will, there is a second
remainder. Choose a third number C, and see if the second remainder
will bear the subtraction of the square of C, and twice C multiplied by
A + B: go on in this way either until there is no remainder, or else
until the remainder will not bear the subtraction arising from any new
part, even though that part were the least number, which is 1. In the
first case, the square root is the sum of A, B, C, &c.; in the second,
there is no square root.
160. For example, I wish to know if 2025 has a square root. I choose 20
as the first part, and find that 400, the square of 20, subtracted from
2025, gives 1625, the first remainder. I again choose 20, whose square,
together with twice itself, multiplied by the preceding part, is 20
× 20 + 2 × 20 × 20, or 1200; which subtracted from 1625, the first
remainder, gives 425, the second remainder. I choose 7 for the third
part, which appears to be too great, since 7 × 7, increased by 2 × 7
multiplied by the sum of the preceding parts 20 + 20, gives 609, which
is more than 425. I therefore choose 5, which closes the process, since
5 × 5, together with 2 × 5 multiplied by 20 + 20, gives exactly 425.
The square root of 2025 is therefore 20 + 20 + 5, or 45, which will be
found, by trial, to be correct; since 45 × 45 = 2025. Again, I ask if
13340 has, or has not, a square root. Let 100 be the first part, whose
square is 10000, and the first remainder is 3340. Let 10 be the second
part. Here 10 × 10 + 2 × 10 × 100 is 2100, and the second remainder, or
3340-2100, is 1240. Let 5 be the third part; then 5 × 5 + 2 × 5 × (100
+ 10) is 1125, which, subtracted from 1240, leaves 115. There is, then,
no square root; for a single additional unit will give a subtraction
of 1 × 1 + 2 × 1 × (100 + 10 + 5), or 231, which is greater than 115.
But if the number proposed had been less by 115, each of the remainders
would have been 115 less, and the last remainder would have been
nothing. Therefore 13340-115, or 13225, has the square root 100 + 10 +
5, or 115; and the answer is, that 13340 has no square root, and that
13225 is the next number below it which has one, namely, 115.
161. It only remains to put the rule in such a shape as will guide us
to those parts which it is most convenient to choose. It is evident
(57) that any number which terminates with ciphers, as 4000, has double
the number of ciphers in its square. Thus, 4000 × 4000 = 16000000;
therefore, any square number,[23] as 49, with an even number of ciphers
annexed, as 490000, is a square number. The root[24] of 490000 is 700.
This being premised, take any number, for example, 76176; setting out
from the right hand towards the left, cut off two figures; then two
more, and so on, until one or two figures only are left: thus, 7,61,76.
This number is greater than 7,00,00, of which the first figure is not
a square number, the nearest square below it being 4. Hence, 4,00,00
is the nearest square number below 7,00,00, which has four ciphers,
and its square root is 200. Let this be the first part chosen: its
square subtracted from 76176 leaves 36176, the first remainder; and
it is evident that we have obtained the highest number of the highest
denomination which is to be found in the square root of 76176; for
300 is too great, its square, 9,00,00, being greater than 76176: and
any denomination higher than hundreds has a square still greater. It
remains, then, to choose a second part, as in the examples of (160),
with the remainder 36176. This part cannot be as great as 100, by what
has just been said; its highest denomination is therefore a number of
tens. Let N stand for a number of tens, which is one of the simple
numbers 1, 2, 3, &c.; that is, let the new part be 10N, whose square
is 10N × 10N, or 100NN, and whose double multiplied by the former part
is 20N × 200, or 4000N; the two together are 4000N + 100NN. Now, N
must be so taken that this may not be greater than 36176: still more
4000N must not be greater than 36176. We may therefore try, for N, the
number of times which 36176 contains 4000, or that which 36 contains
4. The remark in (80) applies here. Let us try 9 tens or 90. Then, 2 ×
90 × 200 + 90 × 90, or 44100, is to be subtracted, which is too great,
since the whole remainder is 36176. We then try 8 tens or 80, which
gives 2 × 80 × 200 + 80 × 80, or 38400, which is likewise too great. On
trying 7 tens, or 70, we find 2 × 70 × 200 + 70 × 70, or 32900, which
subtracted from 36176 gives 3276, the second remainder. The rest of
the square root can only be units. As before, let N be this number of
units. Then, the sum of the preceding parts being 200 + 70, or 270,
the number to be subtracted is 270 × 2N + NN, or 540N + NN. Hence, as
before, 540N must be less than 3276, or N must not be greater than the
number of times which 3276 contains 540, or (80) which 327 contains
54. We therefore try if 6 will do, which gives 2 × 6 × 270 + 6 × 6, or
3276, to be subtracted. This being exactly the second remainder, the
third remainder is nothing, and the process is finished. The square
root required is therefore 200 + 70 + 6, or 276.
[23] By square number I mean, a number which has a square root. Thus,
25 is a square number, but 26 is not.
[24] The term ‘root’ is frequently used as an abbreviation of square
root.
The process of forming the numbers to be subtracted may be shortened
thus. Let A be the sum of the parts already found, and N a new part:
there must then be subtracted 2AN + NN, or (54) 2A + N multiplied by
N. The rule, therefore, for forming it is: Double the sum of all the
preceding parts, add the new part, and multiply the result by the new
part.
162. The process of the last article is as follows:
7,61,76(200 7,61,76(276
4 00 00 70 4
------- 6 ---
400)3,61,76 47)361
70)3 29 00 329
------- -----
400) 32 76 546)3276
140) 32 76 3276
6) ----- ----
0 0
In the first of these, the numbers are written at length, as we found
them; in the second, as in (79), unnecessary ciphers are struck off,
and the periods 61, 76, are not brought down, until, by the continuance
of the process, they cease to have ciphers under them. The following
is another example, to which the reasoning of the last article may be
applied.
34,86,78,44,01(50000 34,86,78,44,01(59049
25 00 00 00 00 9000 25
-------------- 40 ----
100000) 9 86 78 44 01 9 109) 986
9000) 9 81 00 00 00 981
------------- -------
100000) 5 78 44 01 11804) 57844
18000) 4 72 16 00 47216
40) ----------- --------
100000) 1 06 28 01 118089)1062801
18000) 1 06 28 01 1062801
80) ---------- -------
9) 0 0
163. The rule is as follows: To extract the square root of a number;--
I. Beginning from the right hand, cut off periods of two figures each,
until not more than two are left.
II. Find the root of the nearest square number next below the number
in the first period. This root is the first figure of the required
root; subtract its square from the first period, which gives the first
remainder.
III. Annex the second period to the right of the remainder, which gives
the first dividend.
IV. Double the first figure of the root; see how often this is
contained in the number made by cutting one figure from the right of
the first dividend, attending to IX., if necessary; use the quotient as
the second figure of the root; annex it to the right of the double of
the first figure, and call this the first divisor.
V. Multiply the first divisor by the second figure of the root; if the
product be greater than the first dividend, use a lower number for the
second figure of the root, and for the last figure of the divisor,
until the multiplication just mentioned gives the product less than the
first dividend; subtract this from the first dividend, which gives the
second remainder.
VI. Annex the third period to the second remainder, which gives the
second dividend.
VII. Double the first two figures of the root;[25] see how often the
result is contained in the number made by cutting one figure from the
right of the second dividend; use the quotient as the third figure of
the root; annex it to the right of the double of the first two figures,
and call this the second divisor.
[25] Or, more simply, add the second figure of the root to the first
divisor.
VIII. Get a new remainder, as in V., and repeat the process until all
the periods are exhausted; if there be then no remainder, the square
root is found; if there be a remainder, the proposed number has no
square root, and the number found as its square root is the square root
of the proposed number diminished by the remainder.
IX. When it happens that the double of the figures of the root is not
contained at all in all the dividend except the last figure, or when,
being contained once, 1 is found to give more than the dividend, put a
cipher in the square root and in the divisor, and bring down the next
period; should the same thing still happen, put another cipher in the
root and divisor, and bring down another period; and so on.
EXERCISES.
Numbers proposed. | Square roots.
73441 | 271
2992900 | 1730
6414247921 | 80089
903687890625 | 950625
42420747482776576 | 205962976
13422659310152401 | 115856201
164. Since the square of a fraction is obtained by squaring the
numerator and the denominator, the square root of a fraction is found
by taking the square root of both. Thus, the square root of ²⁵/₆₄ is ⅝,
since 5 × 5 is 25, and 8 × 8 is 64. If the numerator or denominator,
or both, be not square numbers, it does not therefore follow that the
fraction has no square root; for it may happen that multiplication
or division by the same number may convert both the numerator and
denominator into square numbers (108). Thus, ²⁷/₄₈, which appears at
first to have no square root, has one in reality, since it is the same
as ⁹/₁₆, whose square root is ¾.
165. We now proceed from (158), where it was stated that any number or
fraction being given, a second may be found, whose square is as near to
the first as we please. Thus, though we cannot solve the problem, “Find
a fraction whose square is 2,” we can solve the following, “Find a
fraction whose square shall not differ from 2 by so much as ·00000001.”
Instead of this last, a still smaller fraction may be substituted;
in fact, any one however small: and in this process we are said to
approximate to the square root of 2. This can be done to any extent,
as follows: Suppose we wish to find the square root of 2 within ¹/₅₇
of the truth; by which I mean, to find a fraction _a_/_b_ whose square
is less than 2, but such that the square of _a_/_b_ + ¹/₅₇ is greater
than 2. Multiply the numerator and denominator of ²/₁ by the square
of 57, or 3249, which gives ⁶⁴⁹⁸/₃₂₄₉. On attempting to extract the
square root of the numerator, I find (163) that there is a remainder
98, and that the square number next below 6498 is 6400, whose root is
80. Hence, the square of 80 is less than 6498, while that of 81 is
greater. The square root of the denominator is of course 57. Hence,
the square of ⁸⁰/⁵⁷ is less than ⁶⁴⁹⁸/₃₂₄₉, or 2, while that of ⁸¹/₅₇
is greater, and these two fractions only differ by ¹/₅₇; which was
required to be done.
166. In practice, it is usual to find the square root true to a certain
number of places of decimals. Thus, 1·4142 is the square root of 2 true
to four places of decimals, since the square of 1·4142, or 1·99996164,
is less than 2, while an increase of only 1 in the fourth decimal
place, giving 1·4143, gives the square 2·00024449, which is greater
than 2. To take a more general case: Suppose it required to find the
square root of 1·637 true to four places of decimals. The fraction is
¹⁶³⁷/₁₀₀₀, whose square root is to be found within ·0001, or ¹/₁₀₀₀₀.
Annex ciphers to the numerator and denominator, until the denominator
becomes the square of ¹/₁₀₀₀₀, which gives ¹⁶³⁷⁰⁰⁰⁰⁰/₁₀₀₀₀₀₀₀₀,
extract the square root of the numerator, as in (163), which shews
that the square number nearest to it is 163700000-13564, whose root is
12794. Hence, ¹²⁷⁹⁴/₁₀₀₀₀, or 1·2794, gives a square less than 1·637,
while 1·2795 gives a square greater. In fact, these two squares are
1·63686436 and 1·63712025.
167. The rule, then, for extracting the square root of a number or
decimal to any number of places is: Annex ciphers until there are twice
as many places following the units’ place as there are to be decimal
places in the root; extract the nearest square root of this number,
and mark off the given number of decimals. Or, more simply: Divide the
number into periods, so that the units’ figure shall be the last of
a period; proceed in the usual way; and if, when decimals follow the
units’ place, there is one figure on the right, in a period by itself,
annex a cipher in bringing down that period, and afterwards let each
new period consist of two ciphers. Place the decimal point after that
figure in forming which the period containing the units was used.
168. For example, what is the square root of (1⅜) to five places of
decimals? This is (145) 1·375, and the process is the first example
over leaf. The second example is the extraction of the root of ·081
to seven places, the first period being 08, from which the cipher is
omitted as useless.
1,37,5(1·17260
1
---
21) 37
21
----
227)1650
1589
------
2342) 6100
4684
------
23446)141600
140676
--------
23452) 92400
8,1(·2846049
4
---
48)410
384
------
564) 2600
2256
------
5686) 34400
34116
---------
569204) 2840000
2276816
---------
569208) 56318400
·000002413672221(·001553599
1
---
25) 141
125
-----
305) 1636
1525
------
3103) 11172
9309
------
31065) 186322
155325
--------
310709) 3099710
2796381
---------
30332900
169. When more than half the decimals required have been found, the
others may be simply found by dividing the dividend by the divisor, as
in (155). The extraction of the square root of 12 to ten places, which
will be found in the next page, is an example. It must, however, be
observed in this process, as in all others where decimals are obtained
by approximation, that the last place cannot always be depended upon:
on which account it is advisable to carry the process so far, that
one or even two more decimals shall be obtained than are absolutely
required to be correct.
A
12(3·46410161513
9
---
64) 300
256
-----
686) 4400
4116
------
6924) 28400
27696
------
69281) 70400
69281
----------
6928201) 11190000
6928201
-------+--
69282026) 4261799|00
4156921|56
-------+----
692820321) 104877|4400
69282|0321
-----+------
6928203225) 35595|407900
34641|016125
-----+--------
69282032301) 954|39177500
692|82032301
---+----------
692820323023) 261|5714519900
207|8460969069
---+----------
53|7253550831
B
692820323026) 537253550831(77545870549
484974226118
------------
52279324713
48497422611
-----------
3781902102
3464101615
----------
317800487
277128129
---------
40672358
34641016
--------
6031342
5542562
-------
488780
484974
------
3806
3464
----
342
277
---
65
62
--
3
If from any remainder we cut off the ciphers, and all figures which
would come under or on the right of these ciphers, by a vertical line,
we find on the left of that line a contracted division, such as those
in (155). Thus, after having found the root as far as 3·464101, we
have the remainder 4261799, and the divisor 6928202. The figures on
the left of the line are nothing more than the contracted division of
this remainder by the divisor, with this difference, however, that we
have to begin by striking a figure off the divisor, instead of using
the whole divisor once, and then striking off the first figure. By this
alone we might have doubled our number of decimal places, and got the
additional figures 615137, the last 7 being obtained by carrying the
contracted division one step further with the remainder 53. We have,
then, this rule: When half the number of decimal places have been
obtained, instead of annexing two ciphers to the remainder, strike off
a figure from what would be the divisor if the process were continued
at length, and divide the remainder by this contracted divisor, as in
(155).
As an example, let us double the number of decimal places already
obtained, which are contained in 3·46410161513. The remainder is
537253550831, the divisor 692820323026, and the process is as in (B).
Hence the square root of 12 is,
3·4641016151377545870549;
which is true to the last figure, and a little too great; but the
substitution of 8 instead of 9 on the right hand would make it too
small.
EXERCISES.
Numbers. | Square roots.
·001728 | ·0415692194
64·34 | 8·02122185
8074 | 89·8554394
10 | 3·16227766
1·57 | 1·2529964086141667788495
SECTION VIII.
ON THE PROPORTION OF NUMBERS.
170. When two numbers are named in any problem, it is usually
necessary, in some way or other, to compare the two; that is, by
considering the two together, to establish some connexion between
them, which may be useful in future operations. The first method
which suggests itself, and the most simple, is to observe which is
the greater, and by how much it differs from the other. The connexion
thus established between two numbers may also hold good of two other
numbers; for example, 8 differs from 19 by 11, and 100 differs from
111 by the same number. In this point of view, 8 stands to 19 in the
same situation in which 100 stands to 111, the first of both couples
differing in the same degree from the second. The four numbers thus
noticed, viz.:
8, 19, 100, 111,
are said to be in _arithmetical[26] proportion_. When four numbers are
thus placed, the first and last are called the _extremes_, and the
second and third the _means_. It is obvious that 111 + 8 = 100 + 19,
that is, the sum of the extremes is equal to the sum of the means.
And this is not accidental, arising from the particular numbers we
have taken, but must be the case in every arithmetical proportion; for
in 111 + 8, by (35), any diminution of 111 will not affect the sum,
provided a corresponding increase be given to 8; and, by the definition
just given, one mean is as much less than 111 as the other is greater
than 8.
[26] This is a very incorrect name, since the term ‘arithmetical’
applies equally to every notion in this book. It is necessary, however,
that the pupil should use words in the sense in which they will be used
in his succeeding studies.
171. A set or series of numbers is said to be in _continued_
arithmetical proportion, or in arithmetical _progression_, when the
difference between every two succeeding terms of the series is the
same. This is the case in the following series:
1, 2, 3, 4, 5, &c.
3, 6, 9, 12, 15, &c.
(1½), 2, (2½), 3, (3½), &c.
The difference between two succeeding terms is called the common
difference. In the three series just given, the common differences are,
1, 3, and ½.
172. If a certain number of terms of any arithmetical series be taken,
the sum of the first and last terms is the same as that of any other
two terms, provided one is as distant from the beginning of the series
as the other is from the end. For example, let there be 7 terms, and
let them be,
_a_ _b_ _c_ _d_ _e_ _f_ _g_.
Then, since, by the nature of the series, _b_ is as much above _a_ as
_f_ is below _g_ (170), _a_ + _g_ = _b_ + _f_. Again, since _c_ is as
much above _b_ as _e_ is below _f_ (170), _b_ + _f_ = _c_ + _e_. But
_a_ + _g_ = _b_ + _f_; therefore _a_ + _g_ = _c_ + _e_, and so on.
Again, twice the middle term, or the term equally distant from the
beginning and the end (which exists only when the number of terms is
odd), is equal to the sum of the first and last terms; for since _c_
is as much below _d_ as _e_ is above it, we have _c_ + _e_ = _d_ + _d_
= 2_d_. But _c_ + _e_ = _a_ + _g_; therefore, _a_ + _g_ = 2_d_. This
will give a short rule for finding the sum of any number of terms of
an arithmetical series. Let there be 7, viz. those just given. Since
_a_ + _g_, _b_ + _f_, and _c_ + _e_, are the same, their sum is three
times (_a_ + _g_), which with _d_, the middle term, or half _a_ + _g_,
is three times and a half (_a_ + _g_), or the sum of the first and
last terms multiplied by (3½), or ⁷/₂, or half the number of terms. If
there had been an even number of terms, for example, six, viz. _a_,
_b_, _c_, _d_, _e_, and _f_, we know now that _a_ + _f_, _b_ + _e_, and
_c_ + _d_, are the same, whence the sum is three times (_a_ + _f_), or
the sum of the first and last terms multiplied by half the number of
terms, as before. The rule, then, is: To sum any number of terms of an
arithmetical progression, multiply the sum of the first and last terms
by half the number of terms. For example, what are 99 terms of the
series 1, 2, 3, &c.? The 99th term is 99, and the sum is
99 100 × 99
(99 + 1)---, or --------, or 4950.
2 2
The sum of 50 terms of the series
1 2 4 5 ( 1 50 ) 50
---, ---, 1, ---, ---, 2, &c. is (--- + ---)---,
3 3 3 3 ( 3 3 ) 2
or 17 × 25, or 425.
173. The first term being given, and also the common difference and
number of terms, the last term may be found by adding to the first term
the common difference multiplied by one less than the number of terms.
For it is evident that the second term differs from the first by the
common difference, the _third_ term by _twice_, the _fourth_ term by
_three_ times the common difference; and so on. Or, the passage from
the first to the _n_th term is made by _n_-1 steps, at each of which
the common difference is added.
EXERCISES.
_Given._ | _To find._
Series. |No. of terms.| Last term. | Sum.
4, (6½), 9, &c. | 33 | 84 | 1452
1, 3, 5, &c. | 28 | 55 | 784
2, 20, 38, &c. | 100,000 | 1799984 | 89999300000
174. The sum being given, the number of terms, and the first term,
we can thence find the common difference. Suppose, for example, the
first term of a series to be one, the number of terms 100, and the sum
10,000. Since 10,000 was made by multiplying the sum of the first and
last terms by ¹⁰⁰/₂, if we divide by this, we shall recover the sum
of the first and last terms. Now, ¹⁰,⁰⁰⁰/₁ divided by ¹⁰⁰/₂ is (122)
200, and the first term being 1, the last term is 199. We have then to
pass from 1 to 199, or through 198, by 99 equal steps. Each step is,
therefore, ¹⁹⁸/⁹⁹, or 2, which is the common difference; or the series
is 1, 3, 5, &c., up to 199.
_Given._ | _To find._
Sum. |No. of terms.|First term.|Last term.|Common diff.
1809025 | 1345 | 1 | 2689 | 2
44 | 10 | 3 | ²⁹/₅ | ¹⁴/₄₅
7075600 | 1330 | 4 | 10636 | 8
175. We now return to (170), in which we compared two numbers together
by their difference. This, however, is not the method of comparison
which we employ in common life, as any single familiar instance will
shew. For example, we say of A, who has 10 thousand pounds, that he is
much richer than B, who has only 3 thousand; but we do not say that
C, who has 107 thousand pounds, is much richer than D, who has 100
thousand, though the difference of fortune is the same in both cases,
viz. 7 thousand pounds. In comparing numbers we take into our reckoning
not only the differences, but the numbers themselves. Thus, if B and D
both received 7 thousand pounds, B would receive 233 pounds and a third
for every 100 pounds which he had before, while D for every 100 pounds
would receive only 7 pounds. And though, in the view taken in (170), 3
is as near to 10 as 100 is to 107, yet, in the light in which we now
regard them, 3 is not so near to 10 as 100 is to 107, for 3 differs
from 10 by more than twice itself, while 100 does not differ from 107
by so much as one-fifth of itself. This is expressed in mathematical
language by saying, that the _ratio_ or _proportion_ of 10 to 3 is
greater than the _ratio_ or _proportion_ of 107 to 100. We proceed to
define these terms more accurately.
176. When we use the term _part_ of a number or fraction in the
remainder of this section, we mean, one of the various sets of _equal_
parts into which it may be divided, either the half, the third, the
fourth, &c.: the term multiple has been already explained (102). By
the term _multiple-part_ of a number we mean, the abbreviation of the
words _multiple of a part_. Thus, 1, 2, 3, 4, and 6, are parts of 12;
½ is also a part of 12, being contained in it 24 times; 12, 24, 36,
&c., are multiples of 12; and 8, 9, ⁵/₂, &c. are multiple parts of 12,
being multiples of some of its parts. And when multiple parts generally
are spoken of, the parts themselves are supposed to be included, on
the same principle that 12 is counted among the multiples of 12, the
multiplier being 1. The multiples themselves are also included in this
term; for 24 is also 48 halves, and is therefore among the multiple
parts of 12. Each part is also in various ways a multiple-part; for
one-fourth is two-eighths, and three-twelfths, &c.
177. Every number or fraction is a multiple-part of every other number
or fraction. If, for example, we ask what part 12 is of 7, we see
that on dividing 7 into 7 parts, and repeating one of these parts 12
times, we obtain 12; or, on dividing 7 into 14 parts, each of which
is one-half, and repeating one of these parts 24 times, we obtain 24
halves, or 12. Hence, 12 is ¹²/₇, or ²⁴/₁₄, or ³⁶/₂₁ of 7; and so on.
Generally, when _a_ and _b_ are two whole numbers, _a_/_b_ expresses
the multiple-part which _a_ is of _b_, and _b_/_a_ that which _b_ is
of _a_. Again, suppose it required to determine what multiple-part
(2⅐) is of (3⅕), or ¹⁵/₇ of ¹⁶/₅. These fractions, reduced to a common
denominator, are ⁷⁵/₃₅ and ¹¹²/₃₅, of which the second, divided into
112 parts, gives ¹/₃₅, which repeated 75 times gives ⁷⁵/₃₅, the first.
Hence, the multiple-part which the first is of the second is ⁷⁵/₁₁₂,
which being obtained by the rule given in (121), shews that _a_/_b_, or
_a_ divided by _b_, according to the notion of division there given,
expresses the multiple-part which _a_ is of _b_ in every case.
178. When the first of four numbers is the same multiple-part of the
second which the third is of the fourth, the four are said to be
_geometrically[27] proportional_, or simply _proportional_. This is
a word in common use; and it remains to shew that our mathematical
definition of it, just given, is, in fact, the common notion attached
to it. For example, suppose a picture is copied on a smaller scale,
so that a line of two inches long in the original is represented by a
line of one inch and a half in the copy; we say that the copy is not
correct unless all the parts of the original are reduced in the same
proportion, namely, that of 2 to (1½). Since, on dividing two inches
into 4 parts, and taking 3 of them, we get (1½), the same must be done
with all the lines in the original, that is, the length of any line in
the copy must be three parts out of four of its length in the original.
Again, interest being at 5 per cent, that is, £5 being given for the
use of £100, a similar proportion of every other sum would be given;
the interest of £70, for example, would be just such a part of £70 as
£5 is of £100.
[27] The same remark may be made here as was made in the note on the
term ‘arithmetical proportion,’ page 101. The word ‘geometrical’ is,
generally speaking, dropped, except when we wish to distinguish between
this kind of proportion and that which has been called arithmetical.
Since, then, the part which _a_ is of _b_ is expressed by the fraction
_a_/_b_, or any other fraction which is equivalent to it, and that
which _c_ is of _d_ by _c_/_d_, it follows, that when _a_, _b_, _c_,
and _d_, are proportional, _a_/_b_ = _c_/_d_. This equation will be
the foundation of all our reasoning on proportional quantities; and
in considering proportionals, it is necessary to observe not only the
quantities themselves, but also the order in which they come. Thus,
_a_, _b_, _c_, and _d_, being proportionals, that is, _a_ being the
same multiple-part of _b_ which _c_ is of _d_, it does not follow that
_a_, _d_, _b_, and _c_ are proportionals, that is, that _a_ is the
same multiple-part of _d_ which _b_ is of _c_. It is plain that _a_ is
greater than, equal to, or less than _b_, according as _c_ is greater
than, equal to, or less than _d_.
179. Four numbers, _a_, _b_, _c_, and _d_, being proportional in the
order written, _a_ and _d_ are called the _extremes_, and _b_ and _c_
the _means_, of the proportion. For convenience, we will call the two
extremes, or the two means, _similar_ terms, and an extreme and a mean,
_dissimilar_ terms. Thus, _a_ and _d_ are similar, and so are _b_ and
_c_; while _a_ and _b_, _a_ and _c_, _d_ and _b_, _d_ and _c_, are
dissimilar. It is customary to express the proportion by placing dots
between the numbers, thus:
_a_ : _b_ ∷ _c_ : _d_
180. Equal numbers will still remain equal when they have been
increased, diminished, multiplied, or divided, by equal quantities.
This amounts to saying that if
_a_ = _b_ and _p_ = _q_,
_a_ + _p_ = _b_ + _q_,
_a_ - _p_ = _b_ - _q_,
_ap_ = _bq_,
_a_ _b_
and --- = ---.
_p_ _q_
It is also evident, that _a_ + _p_-_p_, _a_ -_p_ + _p_, _ap_/_p_, and
_a_/_p_ × _p_, are all equal to _a_.
181. The product of the extremes is equal to the product of the means.
Let _a_/_b_ = _c_/_d_, and multiply these equal numbers by the product
_bd_. Then,
_a_ _abd_
--- × _bd_ = ----- (116) = _ad_,
_b_ _b_
_c_ _cbd_
and --- × _bd_ = ----- = _cb_: hence (180), _ad_ = _bc_.
_d_ _d_
Thus, 6, 8, 21, and 28, are proportional, since
6 3 3 × 7 21
--- = --- = ------ = --- (180);
8 4 4 × 7 28
and it appears that 6 × 28 = 8 × 21, since both products are 168.
182. If the product of two numbers be equal to the product of two
others, these numbers are proportional in any order whatever, provided
the numbers in the same product are so placed as to be similar terms;
that is, if _ab_ = _pq_, we have the following proportions:--
_a_ : _p_ ∷ _q_ : _b_
_a_ : _q_ ∷ _p_ : _b_
_b_ : _p_ ∷ _q_ : _a_
_b_ : _q_ ∷ _p_ : _a_
_p_ : _a_ ∷ _b_ : _q_
_p_ : _b_ ∷ _a_ : _q_
_q_ : _a_ ∷ _b_ : _p_
_q_ : _b_ ∷ _a_ : _p_
To prove any one of these, divide both _ab_ and _pq_ by the product of
its second and fourth terms; for example, to shew the truth of _a_: _q_
∷ _p_: _b_, divide both _ab_ and _pq_ by _bq_. Then,
_ab_ _a_ _pq_ _p_
---- = ---, and ---- = ---; hence (180),
_bq_ _q_ _bq_ _b_
_a_ _p_
--- = ---, or _a_ : _q_ ∷ _p_ : _b_.
_q_ _b_
The pupil should not fail to prove every one of the eight cases, and to
verify them by some simple examples, such as 1 × 6 = 2 × 3, which gives
1: 2 ∷ 3: 6, 3: 1 ∷ 6: 2, &c.
183. Hence, if four numbers be proportional, they are also proportional
in any other order, provided it be such that similar terms still remain
similar. For since, when
_a_ _c_
--- = ---,
_b_ _d_
it follows (181) that _ad_ = _bc_, all the proportions which follow
from _ad_ = _bc_, by the last article, follow also from
_a_ _c_
--- = ---,
_b_ _d_
184. From (114) it follows that
_a_ _b_ + _a_
1 + --- = ---------,
_b_ _b_
_a_
and if --- be less than 1,
_b_
_a_ _b_ - _a_
1 - --- = ---------,
_b_ _b_
_a_
while if --- be greater than 1,
_b_
_a_ _a_ - _b_
--- - 1 = ---------.
_b_ _b_
_a_ + _b_ _a_ - _b_
Also (122), if --------- be divided by ---------
_b_ _b_
_a_ + _b_
the result is ---------.
_a_ - _b_
Hence, _a_, _b_, _c_, and _d_, being proportionals, we may obtain other
proportions, thus:
_a_ _c_
Let --- = ---
_b_ _d_
_a_ _c_
Then (114) 1 + --- = 1 + ---
_b_ _d_
_a_ + _b_ _c_ + _d_
or --------- = ---------
_b_ _d_
or _a_ + _b_: _b_ ∷ _c_ + _d_: _d_
That is, the sum of the first and second is to the second as the sum of
the third and fourth is to the fourth. For brevity, we shall not state
in words any more of these proportions, since the pupil will easily
supply what is wanting.
Resuming the proportion _a_: _b_ ∷ _c_: _d_
_a_ _c_
or --- = ---
_b_ _d_
_a_ _c_ _a_
1 - --- = 1 - ---, if --- be less than 1,
_b_ _d_ _b_
_b_ - _a_ _d_ - _c_
or --------- = ---------
_b_ _d_
that is, _b_-_a_: _b_ ∷ _d_-_c_: _d_ or, _a_-_b_: _b_ ∷ _c_-_d_: _d_,
_a_
if --- be greater than 1.
_b_
_a_ + _b_ _c_ + _d_
Again, since --------- = ---------
_b_ _d_
_a_ - _b_ _c_ - _d_ _a_
and --------- = --------- (--- being greater than 1)
_b_ _d_ _b_
_a_ + _b_ _c_ + _d_
dividing the first by the second we have --------- = ----------,
_a_ - _b_ _c_ - _d_
or _a_ + _b_ : _a_ - _b_ ∷ _c_ + _d_ : _c_ - _d_
and also _a_ + _b_ : _b_ - _a_ ∷ _c_ + _d_ : _d_ - _c_,
_a_
if --- be less than 1.
_b_
185. Many other proportions might be obtained in the same manner. We
will, however, content ourselves with writing down a few which can be
obtained by combining the preceding articles.
_a_ + _b_ : _a_ ∷ _c_ + _d_ : _c_
_a_ : _a_ - _b_ ∷ _c_ : _c_ - _d_
_a_ + _c_ : _a_ - _c_ ∷ _b_ + _d_ : _b_ - _d_.
In these and all others it must be observed, that when such expressions
as _a_-_b_ and _c_-_d_ occur, it is supposed that _a_ is greater than
_b_, and _c_ greater than _d_.
186. If four numbers be proportional, and any two dissimilar terms be
both multiplied, or both divided by the same quantity, the results are
proportional. Thus, if _a_: _b_ ∷ _c_: _d_, and _m_ and _n_ be any two
numbers, we have also the following:
_ma_ : _b_ ∷ _mc_ : _d_
_a_ : _mb_ ∷ _c_ : _md_
_a_ _c_
--- : _mb_ ∷ --- : _md_
_n_ _n_
_ma_ : _nb_ ∷ _mc_ : _nd_
_a_ _b_ _c_ _d_
--- : --- ∷ --- : ---
_m_ _m_ _m_ _m_
_a_ _b_ _c_ _d_
--- : --- ∷ --- : ---
_m_ _m_ _n_ _n_
and various others. To prove any one of these, recollect that nothing
more is necessary to make four numbers proportional except that the
product of the extremes should be equal to that of the means. Take the
third of those just given; the product of its extremes is
_a_ _mad_
--- × _md_, or -----,
_n_ _n_
_c_ _mbc_
while that of the means is _mb_ × ---, or -----.
_n_ _n_
But since _a_ : _b_ ∷ _c_ : _d_, by (181) _ad_ = _bc_,
_mad_ _mbc_
whence, by (180), _mad_ = _mbc_, and ----- = -----.
_n_ _n_
_a_ _c_
Hence, ---, _mb_, ---, and _md_, are proportionals.
_n_ _n_
187. If the terms of one proportion be multiplied by the terms of a
second, the products are proportional; that is, if _a_: _b_ ∷ _c_:
_d_, and _p_: _q_ ∷ _r_: _s_, it follows that _ap_: _bq_ ∷ _cr_: _ds_.
For, since _ad_ = _bc_, and _ps_ = _qr_, by (180) _adps_ = _bcqr_, or
_ap_ × _ds_ = _bq_ × _cr_, whence (182) _ap_: _bq_ ∷ _cr_: _ds_.
188. If four numbers be proportional, any similar powers of these
numbers are also proportional; that is, if
_a_ : _b_ ∷ _c_ : _d_
Then _aa_ : _bb_ ∷ _cc_ : _dd_
_aaa_ : _bbb_ ∷ _ccc_ : _ddd_
&c. &c.
For, if we write the proportion twice, thus,
_a_ : _b_ ∷ _c_ : _d_
_a_ : _b_ ∷ _c_ : _d_
by (187) _aa_ : _bb_ ∷ _cc_ : _dd_
But _a_ : _b_ ∷ _c_ : _d_
Whence (187) _aaa_ : _bbb_ ∷ _ccc_ : _ddd_; and so on.
189. An expression is said to be homogeneous with respect to any two or
more letters, for instance, _a_, _b_, and _c_, when every term of it
contains the same number of letters, counting _a_, _b_, and _c_ only.
Thus, _maab_ + _nabc_ + _rccc_ is homogeneous with respect to _a_, _b_,
and _c_; and of the third degree, since in each term there is either
_a_, _b_, and _c_, or one of these repeated alone, or with another, so
as to make three in all. Thus, 8_aaabc_, 12_abccc_, _maaaaa_, _naabbc_,
are all homogeneous, and of the fifth degree, with respect to _a_, _b_,
and _c_ only; and any expression made by adding or subtracting these
from one another, will be homogeneous and of the fifth degree. Again
_ma_ + _mnb_ is homogeneous with respect to _a_ and _b_, and of the
first degree; but it is not homogeneous with respect to _m_ and _n_,
though it is so with respect to _a_ and _n_. This being premised, we
proceed to a theorem,[28] which will contain all the results of (184),
(185), and (188).
[28] A theorem is a general mathematical fact: thus, that every number
is divisible by four when its last two figures are divisible by four,
is a theorem; that in every proportion the product of the extremes is
equal to the product of the means, is another.
190. If any four numbers be proportional, and if from the first two,
_a_ and _b_, any two homogeneous expressions of the same degree be
formed; and if from the last two, two other expressions be formed, in
precisely the same manner, the four results will be proportional. For
example, if _a_: _b_ ∷ _c_: _d_, and if 2_aaa_ + 3_aab_ and _bbb_ +
_abb_ be chosen, which are both homogeneous with respect to _a_ and
_b_, and both of the third degree; and if the corresponding expressions
2_ccc_ + 3_ccd_ and _ddd_ + _cdd_ be formed, which are made from _c_
and _d_ precisely in the same manner as the two former ones from _a_
and _b_, then will
2_aaa_ + 3_aab_ : _bbb_ + _abb_ ∷ 2_ccc_ + 3_ccd_ : _ddd_ + _cdd_
_a_
To prove this, let --- be called _x_.
_b_
_a_ _a_ _c_
Then, since --- = _x_, and --- = ---,
_b_ _b_ _d_
_c_
it follows that --- = _x_.
_d_
But since _a_ divided by _b_ gives _x_, _x_ multiplied by _b_ will give
_a_, or _a_ = _bx_. For a similar reason, _c_ = _dx_. Put _bx_ and _dx_
instead of _a_ and _c_ in the four expressions just given, recollecting
that when quantities are multiplied together, the result is the same
in whatever order the multiplications are made; that, for example,
_bxbxbx_ is the same as _bbbxxx_.
Hence, 2_aaa_ + 3_aab_ = 2_bxbxbx_ + 3_bxbxb_
= 2_bbbxxx_ + 3_bbbxx_
which is _bbb_ multiplied by 2_xxx_ + 3_xx_
or _bbb_ (2_xxx_ + 3_xx_)[29]
Similarly, 2_ccc_ + 3_ccd_ = _ddd_ (2_xxx_ + 3_xx_)
Also, _bbb_ + _abb_ = _bbb_ + _bxbb_
= _bbb_ multiplied by 1 + _x_
or _bbb_(1 + _x_)
Similarly, _ddd_ + _cdd_ = _ddd_ (1 + _x_)
Now, _bbb_ : _bbb_ ∷ _ddd_ : _ddd_
[29] If _bx_ be substituted for _a_ in any expression which is
homogeneous with respect to _a_ and _b_, the pupil may easily see
that _b_ must occur in every term as often as there are units in the
degree of the expression: thus, _aa_ + _ab_ becomes _bxbx_ + _bxb_ or
_bb_(_xx_ + _x_); _aaa_ + _bbb_ becomes _bxbxbx_ + _bbb_ or _bbb_(_xxx_
+ 1); and so on.
Whence (186), _bbb_(2_xxx_ + 3_xx_): _bbb_(1 + _x_) ∷ _ddd_(2_xxx_ +
3_xx_): _ddd_(1 + _x_), which, when instead of these expressions their
equals just found are substituted, becomes 2_aaa_ + 3_aab_: _bbb_ +
_abb_ ∷ 2_ccc_ + 3_ccd_: _ddd_ + _cdd_.
The same reasoning may be applied to any other case, and the pupil may
in this way prove the following theorems:
If _a_ : _b_ ∷ _c_ : _d_
2_a_ + 3_b_ : _b_ ∷ 2_c_ + 3_d_ : _d_
_aa_ + _bb_ : _aa_ - _bb_ ∷ _cc_ + _dd_ : _cc_ - _dd_
_mab_ : 2_aa_ + _bb_ ∷ _mcd_ : 2_cc_ + _dd_
191. If the two means of a proportion be the same, that is, if _a_ :
_b_ ∷ _b_: _c_, the three numbers, _a_, _b_, and _c_, are said to be in
_continued_ proportion, or in _geometrical progression_. The same terms
are applied to a series of numbers, of which any three that follow one
another are in continued proportion, such as
1 2 4 8 16 32 64 &c.
2 2 2 2 2 2
2 --- --- --- --- ---- ---- &c.
3 9 27 81 243 729
Which are in continued proportion, since
2 2 2
1 : 2 ∷ 2 : 4 2 : --- ∷ --- : ---
3 3 9
2 2 2 2
2 : 4 ∷ 4 : 8 --- : --- ∷ --- : ---
3 9 9 27
&c. &c.
192. Let _a_, _b_, _c_, _d_, _e_ be in continued proportion; we have
then
_a_ _b_
_a_ : _b_ ∷ _b_ : _c_ or --- = --- or _ac_ = _bb_
_b_ _c_
_b_ _c_
_b_ : _c_ ∷ _c_ : _d_ --- = --- _bd_ = _cc_
_c_ _d_
_c_ _d_
_c_ : _d_ ∷ _d_ : _e_ --- = --- _ce_ = _dd_
_d_ _e_
Each term is formed from the preceding, by multiplying it by the same
number. Thus,
_b_ _c_
_b_ = --- × _a_ (180); _c_ = ---× _b_;
_a_ _b_
_a_ _b_ _b_ _c_ _b_
and since --- = ---, --- = --- or _c_ = --- × _b_.
_b_ _c_ _a_ _b_ _a_
_d_ _d_ _c_ _b_
Again, _d_ = --- × _c_, but --- = ---, which is = ---;
_c_ _c_ _b_ _a_
_b_
therefore, _d_ = --- × _c_, and so on.
_c_
_b_
If, then, ---
_a_
(which is called the _common ratio_ of the series) be denoted by _r_,
we have
_b_ = _ar_ _c_ = _br_ = _arr_ _d_ = _cr_ = _arrr_
and so on; whence the series
_a_ _b_ _c_ _d_ &c.
is _a_ _ar_ _arr_ _arrr_ &c.
Hence _a_ : _c_ ∷ _a_ : _arr_
(186) ∷ _aa_ : _aarr_
∷ _aa_ : _bb_
because, _b_ being _ar_, _bb_ is _arar_ or _aarr_. Again,
_a_ : _d_ ∷ _a_ : _arrr_
(186) ∷ _aaa_ : _aaarrr_
∷ _aaa_ : _bbb_
Also _a_ : _e_ ∷ _aaaa_ : _bbbb_, and so on;
that is, the first bears to the _n_ᵗʰ term from the first the same
proportion as the _n_ᵗʰ power of the first to the _n_ᵗʰ power of the
second.
193. A short rule may be found for adding together any number of terms
of a continued proportion. Let it be first required to add together the
terms 1, _r_, _rr_, &c. where _r_ is greater than unity. It is evident
that we do not alter any expression by adding or subtracting any
numbers, provided we afterwards subtract or add the same. For example,
_p_ = _p_-_q_ + _q_-_r_ + _r_- _s_ + _s_
Let us take four terms of the series, 1, _r_, _rr_, &c. or,
1 + _r_ + _rr_ + _rrr_
It is plain that
_rrrr_-1 = _rrrr_-_rrr_ + _rrr_-_rr_ + _rr_-_r_ + _r_-1
Now (54), _rr_-_r_ = _r_(_r_-1), _rrr_ -_rr_ = _rr_(_r_-1),
_rrrr_-_rrr_ = _rrr_(_r_-1), and the above equation becomes _rrrr_ -1 =
_rrr_(_r_-1) + _rr_ (_r_-1) + _r_ (_r_-1) + _r_-1; which is (54) _rrr_
+ _rr_ + _r_ + 1 taken _r_-1 times. Hence, _rrrr_-1 divided by _r_-1
will give 1 + _r_ + _rr_ + _rrr_, the sum of the terms required. In
this way may be proved the following series of equations:
_rr_ - 1
1 + _r_ = --------
_r_ - 1
_rrr_ - 1
1 + _r_ + _rr_ = ---------
_r_ - 1
_rrrr_ - 1
1 + _r_ + _rr_ + _rrr_ = ----------
_r_ - 1
_rrrrr_ - 1
1 + _r_ + _rr_ + _rrr_ + _rrrr_ = -----------
_r_ - 1
If _r_ be less than unity, in order to find 1 + _r_ + _rr_ + _rrr_,
observe that
1 - _rrrr_ = 1 - _r_ + _r_ - _rr_ + _rr_ - _rrr_ + _rrr_ - _rrrr_
= 1 - _r_ + _r_(1 - _r_) + _rr_(1 - _r_) + _rrr_(1 - _r_);
whence, by similar reasoning, 1 + _r_ + _rr_ + _rrr_ is found by
dividing 1-_rrrr_ by 1-_r_; and equations similar to these just given
may be found, which are,
1 - _rr_
1 + _r_ = --------
1 - _r_
1 - _rrr_
1 + _r_ + _rr_ = ---------
1 - _r_
1 - _rrrr_
1 + _r_ + _rr_ + _rrr_ = ----------
1 - _r_
1 - _rrrrr_
1 + _r_ + _rr_ + _rrr_ + _rrrr_ = -----------
1 - _r_
The rule is: To find the sum of n terms of the series, 1 + _r_ + _rr_
+ &c., divide the difference between 1 and the (_n_ + 1)ᵗʰ term by the
difference between 1 and _r_.
194. This may be applied to finding the sum of any number of terms of
a continued proportion. Let _a_, _b_, _c_, &c. be the terms of which
it is required to sum four, that is, to find _a_ + _b_ + _c_ + _d_, or
(192) _a_ + _ar_ + _arr_ + _arrr_, or (54) a(1 + _r_ + _rr_ + _rrr_),
which (193) is
_rrrr_ - 1 1 - _rrrr_
---------- × _a_, or ---------- × _a_,
_r_ - 1 1 - _r_
according as _r_ is greater or less than unity. The first fraction is
_arrrr_ - _a_ _e_ - _a_
-------------, or (192) ---------.
_r_ - 1 _r_ - 1
_a_ - _e_
Similarly, the second is ---------.
1 - _r_
The rule, therefore, is: To sum _n_ terms of a continued proportion,
divide the difference of the (_n_ + 1)ᵗʰ and first terms by the
difference between unity and the common measure. For example, the
sum of 10 terms of the series 1 + 3 + 9 + 27 + &c. is required. The
eleventh term is 59049, and ⁽⁵⁹⁰⁴⁹ ⁻ ¹⁾/₍₃₋₁₎ is 29524. Again, the sum
of 18 terms of the series 2 + 1 + ½ + ½ + &c. of which the nineteenth
term is ¹/₁₃₁₀₇₂, is
1
2 - ------
131072 131070
----------- = 3 ------.
1 - ½ 131072
EXAMPLES.
9 terms of 1 + 4 + 16 + &c. are 87381
6 12 847422675
10 ...... 3 + --- + ---- + &c. ... ---------
7 49 201768035
1 1 1 1048575
20 ...... --- + --- + --- + &c. ... -------
2 4 8 1048576
195. The powers of a number or fraction greater than unity increase;
for since 2½ is greater than 1, 2½ × 2½ is 2½ taken more than once,
that is, is greater than 2½, and so on. This increase goes on without
limit; that is, there is no quantity so great but that some power of
2½ is greater. To prove this, observe that every power of 2½ is made
by multiplying the preceding power by 2½, or by 1 + 1½, that is, by
adding to the former power that power itself and its half. There will,
therefore, be more added to the 10th power to form the 11th, than was
added to the 9th power to form the 10th. But it is evident that if any
given quantity, however small, be continually added to 2½, the result
will come in time to exceed any other quantity that was also given,
however great; much more, then, will it do so if the quantity added to
2½ be increased at each step, which is the case when the successive
powers of 2½ are formed. It is evident, also, that the powers of 1
never increase, being always 1; thus, 1 × 1 = 1, &c. Also, if _a_ be
greater than _m_ times _b_, the square of _a_ is greater than _mm_
times the square of _b_. Thus, if _a_ = 2_b_ + _c_, where _a_ is
greater than 2_b_, the square of _a_, or _aa_, which is (68) 4_bb_ +
4_bc_ + _cc_ is greater than 4_bb_, and so on.
196. The powers of a fraction less than unity continually decrease;
thus, the square of ⅖, or ⅖ × ⅖, is less than ⅖, being only two-fifths
of it. This decrease continues without limit; that is, there is no
quantity so small but that some power of ⅖ is less. For if
5 2 1 1 1
--- = _x_, --- = ---, and the powers of ⅖ are ----, -----,
2 5 _x_ _xx_ _xxx_
and so on. Since _x_ is greater than 1 (195), some power of _x_ may be
found which shall be greater than a given quantity. Let this be called
_m_; then 1/_m_ is the corresponding power of ⅖; and a fraction whose
denominator can be made as great as we please, can itself be made as
small as we please (112).
197. We have, then, in the series
1 _r_ _rr_ _rrr_ _rrrr_ &c.
I. A series of increasing terms, if _r_ be greater than 1. II. Of
terms having the same value, if _r_ be equal to 1. III. A series of
decreasing terms, if _r_ be less than 1. In the first two cases, the sum
1 + _r_ + _rr_ + _rrr_ + &c.
may evidently be made as great as we please, by sufficiently increasing
the number of terms. But in the third this may or may not be the case;
for though something is added at each step, yet, as that augmentation
diminishes at every step, we may not certainly say that we can, by any
number of such augmentations, make the result as great as we please. To
shew the contrary in a simple instance, consider the series,
1 + ½ + ¼ + ⅛ + ¹/₁₆ + &c.
Carry this series to what extent we may, it will always be necessary to
add the last term in order to make as much as 2. Thus,
(1 + ½ + ¼) + ¼ = 1 + ½ + ½ = 1 + 1 = 2
(1 + ½ + ¼ + ⅛) + ⅛ = 2.
(1 + ½ + ¼ + ⅛ + ¹/₁₆) + ¹/₁₆ = 2, &c.
But in the series, every term is only the half of the preceding;
consequently no number of terms, however great, can be made as great as
2 by adding one more. The sum, therefore, of 1, ½, ¼, ⅛ &c. continually
approaches to 2, diminishing its distance from 2 at every step, but
never reaching it. Hence, 2 is celled the _limit_ of 1 + ½ + ¼ + &c. We
are not, therefore, to conclude that _every_ series of decreasing terms
has a limit. The contrary may be shewn in the very simple series, 1 + ½
+ ⅓ + ¼ + &c. which may be written thus:
1 + ½ + (⅓ + ¼) + (⅕ + ... up to ⅛) + (⅑ + ... up to ¹/₁₆)
+ (¹/₁₇ + ... up to ¹/₃₂) + &c.
We have thus divided all the series, except the first two terms, into
lots, each containing half as many terms as there are units in the
denominator of its last term. Thus, the fourth lot contains 16 or ³²/₂2
terms. Each of these lots may be shewn to be greater than ½. Take the
third, for example, consisting of ⅑, ¹/₁₀, ¹/₁₁, ¹/₁₂, ¹/₁₃, ¹/₁₄,
¹/₁₅, and ¹/₁₆. All except ¹/₁₆, the last, are greater than ¹/₁₆;
consequently, by substituting ¹/₁₆ for each of them, the amount of the
whole lot would be lessened; and as it would then become ⁸/₁₆, or ½,
the lot itself is greater than ½. Now, if to 1 + ½, ½ be continually
added, the result will in time exceed any given number. Still more will
this be the case if, instead of ½, the several lots written above be
added one after the other. But it is thus that the series 1 + ½ + ⅓,
&c. is composed, which proves what was said, that this series has no
limit.
198. The series 1 + _r_ + _rr_ + _rrr_ + &c. always has a limit when
_r_ is less than 1. To prove this, let the term succeeding that at
which we stop be _a_, whence (194) the sum is
1 - _a_ 1 _a_
-------, or (112) ------- - ------.
1 - _r_ 1 - _r_ 1 - _r_
The terms decrease without limit (196), whence we may take a term so
far distant from the beginning, that _a_, and therefore
_a_
-------,
1 - _r_
shall be as small as we please. But it is evident that in this case
1 _a_
------- - ------- though always less than
1 - _r_ 1 - _r_
1 1
-------- may be brought as near to -------
1 - _r_ 1 - _r_
as we please; that is, the series 1 + _r_ + _rr_ + &c. continually
approaches to the limit
1
--------.
1 - _r_
Thus 1 + ½ + ¼ + ⅛ + &c. where _r_ = ½, continually approaches to
1
----- or 2, as was shewn in the last article.
1 - ½
EXERCISES.
2 2
The limit of 2 + --- + --- + &c.
3 9
1 1
or 2(1 + --- + --- + &c.) is 3
3 9
9 81
... 1 + --- + ---- + &c. ... 10
10 100
15 45
... 5 + ---- + ---- + &c. ... 8¾
7 49
199. When the fraction _a_/_b_ is not equal to _c_/_d_, but greater,
_a_ is said to have to _b_ a greater ratio than _c_ has to _d_; and
when _a_/_b_ is less than _c_/_d_, _a_ is said to have to _b_ a less
ratio than _c_ has to _d_. We propose the following questions as
exercises, since they follow very simply from this definition.
I. If _a_ be greater than _b_, and _c_ less than or equal to _d_, _a_
will have a greater ratio to _b_ than _c_ has to _d_.
II. If _a_ be less than _b_, and _c_ greater than or equal to _d_, _a_
has a less ratio to _b_ than _c_ has to _d_.
III. If _a_ be to _b_ as _c_ is to _d_, and if _a_ have a greater ratio
to _b_ than _c_ has to _x_, _d_ is less than _x_; and if _a_ have a
less ratio to _b_ than _c_ to _x_, _d_ is greater than _x_.
IV. _a_ has to _b_ a greater ratio than _ax_ to _bx_ + _y_, and a less
ratio than _ax_ to _bx_- _y_.
200. If _a_ have to _b_ a greater ratio than _c_ has to _d_, _a_ + _c_
has to _b_ + _d_ a less ratio than _a_ has to _b_, but a greater ratio
than _c_ has to _d_; or, in other words, if _a_/_b_ be the greater of
the two fractions _a_/_b_ and _c_/_d_,
_a_ + _c_
---------
_b_ + _d_
will be greater than _c_/_d_, but less than _a_/_b_. To shew this,
observe that (_mx_ + _ny_)/(_m_ + _n_) must lie between _x_ and _y_,
if _x_ and _y_ be unequal: for if _x_ be the less of the two, it is
certainly greater than
_mx_ + _nx_
----------- or than _x_;
_m_ + _n_
and if _y_ be the greater of the two, it is certainly less than
_my_ + _ny_
-----------, or than _y_.
_m_ + _n_
It therefore lies between _x_ and _y_. Now let _a_/_b_ be _x_, and let
_c_/_d_ be _y_: then _a_ = _bx_, _c_ = _dy_. Now
_bx_ + _dy_
-----------
_b_ + _d_
is something between _x_ and _y_, as was just proved; therefore
_a_ + _c_
---------
_b_ + _d_
is something between _a_/_b_ and _c_/_d_. Again, since _a_/_b_ and
_c_/_d_ are respectively equal to _ap_/_bp_ and _cq_/_dq_, and since,
as has just been proved,
_ap_ + _cq_
-----------
_bp_ + _dq_
lies between the two last, it also lies between the two first; that is,
if _p_ and _q_ be any numbers or fractions whatsoever,
_ap_ + _cq_
-----------
_bp_ + _dq_
lies between _a_/_b_ and _c_/_d_.
201. By the last article we may often form some notion of the value of
an expression too complicated to be easily calculated. Thus,
1 + _x_ 1 _x_ 1
-------- lies between --- and ----, or 1 and ---;
1 + _xx_ 1 _xx_ _x_
_ax_ + _by_ _ax_ _by_
-------------- lies between ----- and ------,
_axx_ + _bbyy_ _axx_ _bbyy_
that is, between 1/_x_ and 1/_by_. And it has been shewn that (_a_ +
_b_)/2 lies between _a_ and _b_, the denominator being considered as 1
+ 1.
202. It may also be proved that a fraction such as
_a_ + _b_ + _c_ + _d_
---------------------
_p_ + _q_ + _r_ + _s_
_a_ _b_ _c_ _d_
always lies among ---, ---, ---, and ---,
_p_ _q_ _r_ _s_
that is, is less than the greatest of them, and greater than the
least. Let these fractions be arranged in order of magnitude; that is,
let _a_/_p_ be greater than _b_/_q_, _b_/_q_ be greater than _c_/_r_,
and _c_/_r_ greater than _d_/_s_. Then by (200)
is and
less greater
_a_ + _b_ _a_ than than _b_ _c_
--------- --- --- and ---
_p_ + _q_ _p_ _q_ _r_
_a_ + _b_ + _c_ _a_ + _b_ _a_ _c_ _d_
--------------- --------- and --- --- and ---
_p_ + _q_ + _r_ _p_ + _q_ _p_ _r_ _s_
_a_ + _b_ + _c_ + _d_ _a_ + _b_ + _c_ _a_ _d_
--------------------- --------------- and --- ---
_p_ + _q_ + _r_ + _s_ _p_ + _q_ + _r_ _p_ _s_
whence the proposition is evident.
203. It is usual to signify “_a_ is greater than _b_” by _a_ > _b_ and
“_a_ is less than _b_” by _a_ < _b_; the opening of V being turned
towards the greater quantity. The pupil is recommended to make himself
familiar with these signs.
SECTION IX.
ON PERMUTATIONS AND COMBINATIONS.
204. If a number of counters, distinguished by different letters, be
placed on the table, and any number of them, say four, be taken away,
the question is, to determine in how many different ways this can be
done. Each way of doing it gives what is called a _combination_ of
four, but which might with more propriety be called a _selection_
of four. Two combinations or selections are called different, which
differ in any way whatever; thus, _abcd_ and _abce_ are different,
_d_ being in one and _e_ in the other, the remaining parts being the
same. Let there be six counters, _a_, _b_, _c_, _d_, _e_, and _f_;
the combinations of three which can be made out of them are twenty in
number, as follow:
_abc_ _ace_ _bcd_ _bef_
_abd_ _acf_ _bce_ _cde_
_abe_ _ade_ _bcf_ _cdf_
_abf_ _adf_ _bde_ _cef_
_acd_ _aef_ _bdf_ _def_
The combinations of four are fifteen in number, namely,
_abcd_ _abde_ _acde_ _adef_ _bcef_
_abce_ _abdf_ _acdf_ _bcde_ _bdcf_
_abcf_ _abef_ _acef_ _bcdf_ _cdef_
and so on.
205. Each of these combinations may be written in several different
orders; thus, _abcd_ may be disposed in any of the following ways:
_abcd_ _acbd_ _acdb_ _abdc_ _adbc_ _adcb_
_bacd_ _cabd_ _cadb_ _badc_ _dabc_ _dacb_
_bcad_ _cbad_ _cdab_ _bdac_ _dbac_ _dcab_
_bcda_ _cbda_ _cdba_ _bdca_ _dbca_ _dcba_
of which no two are entirely in the same order. Each of these is
said to be a distinct _permutation_ of _abcd_. Considered as a
_combination_, they are all the same, as each contains _a_, _b_, _c_,
and _d_.
206. We now proceed to find how many _permutations_, each containing
one given number, can be made from the counters in another given
number, six, for example. If we knew how to find all the permutations
containing four counters, we might make those which contain five
thus: Take any one which contains four, for example, _abcf_ in which
_d_ and _e_ are omitted; write _d_ and _e_ successively at the end,
which gives _abcfd_, _abcfe_, and repeat the same process with every
other permutation of four; thus, _dabc_ gives _dabce_ and _dabcf_.
No permutation of five can escape us if we proceed in this manner,
provided only we know those of four; for any given permutation of five,
as _dbfea_, will arise in the course of the process from _dbfe_, which,
according to our rule, furnishes _dbfea_. Neither will any permutation
be repeated twice, for _dbfea_, if the rule be followed, can only
arise from the permutation _dbfe_. If we begin in this way to find the
permutations of two out of the six,
_a_ _b_ _c_ _d_ _e_ _f_
each of these gives five; thus,
_a_ gives _ab_ _ac_ _ad_ _ae_ _af_
_b_ ... _ba_ _bc_ _bd_ _be_ _bf_
and the whole number is 6 × 5, or 30.
Again, _ab_ gives _abc_ _abd_ _abe_ _abf_
_ac_ ... _acb_ _acd_ _ace_ _acf_
and here are 30, or 6 × 5 permutations of 2, each of which gives 4
permutations of 3; the whole number of the last is therefore 6 × 5 × 4,
or 120.
Again, _abc_ gives _abcd_ _abce_ _abcf_
_abd_ ... _abdc_ _abde_ _abdf_
and here are 120, or 6 × 5 × 4, permutations of three, each of which
gives 3 permutations of four; the whole number of the last is therefore
6 × 5 × 4 × 3, or 360.
In the same way, the number of permutations of 5 is 6 × 5 × 4 × 3 ×
2, and the number of permutations of six, or the number of different
ways in which the whole six can be arranged, is 6 × 5 × 4 × 3 × 2
× 1. The last two results are the same, which must be; for since a
permutation of five only omits one, it can only furnish one permutation
of six. If instead of six we choose any other number, _x_, the number
of permutations of two will be _x_(_x_-1), that of three will be
_x_(_x_-1)(_x_-2), that of four _x_(_x_ -1)(_x_-2)(_x_-3), the rule
being: Multiply the whole number of counters by the next less number,
and the result by the next less, and so on, until as many numbers
have been multiplied together as there are to be counters in each
permutation: the product will be the whole number of permutations of
the sort required. Thus, out of 12 counters, permutations of four may
be made to the number of 12 × 11 × 10 × 9, or 11880.
EXERCISES.
207. In how many different ways can eight persons be arranged on eight
seats?
_Answer_, 40320.
In how many ways can eight persons be seated at a round table, so that
all shall not have the same neighbours in any two arrangements?[30]
_Answer_, 5040.
[30] The difference between this problem and the last is left to the
ingenuity of the pupil.
If the hundredth part of a farthing be given for every different
arrangement which can be made of fifteen persons, to how much will the
whole amount?
_Answer_, £13621608.
Out of seventeen consonants and five vowels, how many words can be
made, having two consonants and one vowel in each?
_Answer_, 4080.
208. If two or more of the counters have the same letter upon them, the
number of distinct permutations is less than that given by the last
rule. Let there be _a_, _a_, _a_, _b_, _c_, _d_, and, for a moment,
let us distinguish between the three as thus, _a_, _a′_, _a″_. Then,
_abca′a″d_, and _a″bcaa′d_ are reckoned as distinct permutations in
the rule, whereas they would not have been so, had it not been for the
accents. To compute the number of distinct permutations, let us make
one with _b_, _c_, and _d_, leaving places for the _a_s, thus, ( ) _bc_
( ) ( ) _d_. If the _a_s had been distinguished as _a_, _a′_, _a″_,
we might have made 3 × 2 × 1 distinct permutations, by filling up the
vacant places in the above, all which six are the same when the _a_s
are not distinguished. Hence, to deduce the number of permutations of
_a_, _a_, _a_, _b_, _c_, _d_, from that of _aa′a″bcd_, we must divide
the latter by 3 × 2 × 1, or 6, which gives
6 × 5 × 4 × 3 × 2 × 1
--------------------- or 120.
3 × 2 × 1
Similarly, the number of permutations of _aaaabbbcc_ is
9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
---------------------------------
4 × 3 × 2 × 1 × 3 × 2 × 1 × 2 × 1.
EXERCISE.
How many variations can be made of the order of the letters in the word
antitrinitarian?
_Answer_, 126126000.
209. From the number of permutations we can easily deduce the number of
combinations. But, in order to form these combinations independently,
we will shew a method similar to that in (206). If we know the
combinations of two which can be made out of _a_, _b_, _c_, _d_, _e_,
we can find the combinations of three, by writing successively at the
end of each combination of two, the letters which come after the last
contained in it. Thus, _ab_ gives _abc_, _abd_, _abe_; _ad_ gives _ade_
only. No combination of three can escape us if we proceed in this
manner, provided only we know the combinations of two; for any given
combination of three, as _acd_, will arise in the course of the process
from _ac_, which, according to our rule, furnishes _acd_. Neither will
any combination be repeated twice, for _acd_, if the rule be followed,
can only arise from _ac_, since neither _ad_ nor _cd_ furnishes it. If
we begin in this way to find the combinations of the five,
_a_ _b_ _c_ _d_ _e_
_a_ gives _ab_ _ac_ _ad_ _ae_
_b_ ···· _bc_ _bd_ _be_
_c_ ···· _cd_ _ce_
_d_ ···· _de_
Of these, _ab_ gives _abc_ _abd_ _abe_
_ac_ ···· _acd_ _ace_
_ad_ ···· _ade_
_bc_ ···· _bcd_ _bce_
_bd_ ···· _bde_
_cd_ ···· _cde_
_ae_ _be_ _ce_ and _de_ give none.
Of these, _abc_ gives _abcd_ _abce_
_abd_ ···· _abde_
_acd_ ···· _acde_
_bcd_ ···· _bcde_
Those which contain _e_ give none, as before.
Of the last, _abcd_ gives _abcde_, and the others none, which is
evidently true, since only one selection of five can be made out of
five things.
210. The rule for calculating the number of combinations is derived
directly from that for the number of permutations. Take 7 counters;
then, since the number of permutations of two is 7 × 6, and since two
permutations, _ba_ and _ab_, are in any combination _ab_, the number of
combinations is half that of the permutations, or (7 × 6)/2. Since the
number of permutations of three is 7 × 6 × 5, and as each combination
_abc_ has 3 × 2 × 1 permutations, the number of combinations of three is
7 × 6 × 5
----------.
1 × 2 × 3
Also, since any combination of four, _abcd_, contains 4 × 3 × 2 × 1
permutations, the number of combinations of four is
7 × 6 × 5 × 4
-------------,
1 × 2 × 3 × 4
and so on. The rule is: To find the number of combinations, each
containing _n_ counters, divide the corresponding number of
permutations by the product of 1, 2, 3, &c. up to _n_. If _x_ be the
whole number, the number of combinations of two is
_x_(_x_ - 1)
-------------;
1 × 2
that of three is
_x_(_x_ - 1)(_x_ - 2)
---------------------;
1 × 2 × 3
that of four is
_x_(_x_ - 1)(_x_ - 2)(_x_ - 3)
------------------------------; and so on.
1 × 2 × 3 × 4
211. The rule may in half the cases be simplified, as follows. Out of
ten counters, for every distinct selection of seven which is taken, a
distinct combination of 3 is left. Hence, the number of combinations
of seven is as many as that of three. We may, therefore, find the
combinations of three instead of those of seven; and we must moreover
expect, and may even assert, that the two formulæ for finding these two
numbers of combinations are the same in result, though different in
form. And so it proves; for the number of combinations of seven out of
ten is
10 × 9 × 8 × 7 × 6 × 5 × 4
--------------------------,
1 × 2 × 3 × 4 × 5 × 6 × 7
in which the product 7 × 6 × 5 × 4 occurs in both terms, and therefore
may be removed from both (108), leaving
10 × 9 × 8
----------,
1 × 2 × 3
which is the number of combinations of three out of ten. The same may
be shewn in other cases.
EXERCISES.
How many combinations of four can be made out of twelve things?
_Answer_, 495.
What number { 6 } { 8 } { 28
of combinations { 4 } out of { 11 } _Answer_, { 330
can be made of { 26 } { 28 } { 378
{ 6 } { 15 } { 5005
How many combinations can be made of 13 out of 52; or how many
different hands may a person hold at the game of whist?
_Answer_, 635013559600.
BOOK II.
COMMERCIAL ARITHMETIC.
SECTION I.
WEIGHTS, MEASURES, &C.
212. In making the calculations which are necessary in commercial
affairs, no more processes are required than those which have been
explained in the preceding book. But there is still one thing
wanted--not to insure the accuracy of our calculations, but to enable
us to compare and judge of their results. We have hitherto made use of
a single unit (15), and have treated of other quantities which are made
up of a number of units, in Sections II., III., and IV., and of those
which contain parts of that unit in Sections V. and VI. Thus, if we are
talking of distances, and take a mile as the unit, any other length may
be represented,[31] either by a certain number of miles, or a certain
number of parts of a mile, and (1 meaning one mile) may be expressed
either by a whole number or a fraction. But we can easily see that in
many cases inconveniences would arise. Suppose, for example, I say,
that the length of one room is ¹/₁₈₀ of a mile, and of another ¹/₁₇₄
of a mile, what idea can we form as to how much the second is longer
than the first? It is necessary to have some smaller measure; and if
we divide a mile into 1760 equal parts, and call each of these parts a
yard, we shall find that the length of the first room is 9 yards and
⁷/₉ of a yard, and that of the second 10 yards and ¹⁰/₈₇ of a yard.
From this we form a much better notion of these different lengths,
but still not a very perfect one, on account of the fractions ⁷/₉ and
¹⁰/₈₇. To get a clearer idea of these, suppose the yard to be divided
into three equal parts, and each of these parts to be called a foot;
then ⁷/₉ of a yard contains 2⅓ feet, and ¹⁰/₈₇ of a yard contains ³⁰/₈₇
of a foot, or a little more than ⅓ of a foot. Therefore the length of
the first room is now 9 yards, 2 feet, and ⅓ of a foot; that of the
second is 10 yards and a little more than ⅓ of a foot. We see, then,
the convenience of having large measures for large quantities, and
smaller measures for small ones; but this is done for convenience only,
for it is _possible_ to perform calculations upon any sort of quantity,
with one measure alone, as certainly as with more than one; and not
only possible, but more convenient, as far as the mere calculation is
concerned.
[31] It is not true, that if we choose any quantity as a unit, _any_
other quantity of the same kind can be exactly represented either by
a certain number of units, or of parts of a unit. To understand how
this is proved, the pupil would require more knowledge than he can be
supposed to have; but we can shew him that, for any thing he knows
to the contrary, there may be quantities which are neither units nor
parts of the unit. Take a mathematical line of one foot in length,
divide it into ten parts, each of those parts into ten parts, and so
on continually. If a point A be taken at hazard in the line, it does
not appear self-evident that if the decimal division be continued
ever so far, one of the points of division must at last fall exactly
on A: neither would the same appear necessarily true if the division
were made into sevenths, or elevenths, or in any other way. There may
then possibly be a part of a foot which is no exact numerical fraction
whatever of the foot; and this, in a higher branch of mathematics, is
found to be the case times without number. What is meant in the words
on which this note is written, is, that any part of a foot can be
represented as nearly as we please by a numerical fraction of it; and
this is sufficient for practical purposes.
The measures which are used in this country are not those which would
have been chosen had they been made all at one time, and by a people
well acquainted with arithmetic and natural philosophy. We proceed
to shew how the results of the latter science are made useful in
our system of measures. Whether the circumstances introduced are
sufficiently well known to render the following methods exact enough
for the recovery of _astronomical_ standards, may be matter of opinion;
but no doubt can be entertained of their being amply correct for
commercial purposes.
It is evidently desirable that weights and measures should always
continue the same, and that posterity should be able to replace any
one of them when the original measure is lost. It is true that a yard,
which is now exact, is kept by the public authorities; but if this were
burnt by accident,[32] how are those who shall live 500 years hence to
know what was the length which their ancestors called a yard? To ensure
them this knowledge, the measure must be derived from something which
cannot be altered by man, either from design or accident. We find such
a quantity in the time of the daily revolution of the earth, and also
in the length of the year, both of which, as is shewn in astronomy,
will remain the same, at least for an enormous number of centuries,
unless some great and totally unknown change take place in the solar
system. So long as astronomy is cultivated, it is impossible to suppose
that either of these will be lost, and it is known that the latter is
365·24224 mean solar days, or about 365¼ of the average interval which
elapses between noon and noon, that is, between the times when the sun
is highest in the heavens. Our year is made to consist of 365 days,
and the odd quarter is allowed for by adding one day to every fourth
year, which gives what we call leap-year. This is the same as adding ¼
of a day to each year, and is rather too much, since the excess of the
year above 365 days is not ·25 but ·24224 of a day. The difference is
·00776 of a day, which is the quantity by which our average year is too
long. This amounts to a day in about 128 years, or to about 3 days in 4
centuries. The error is corrected by allowing only one out of four of
the years which close the centuries to be leap-years. Thus, A.D. 1800
and 1900 are not leap-years, but 2000 is so.
[32] Since this was first written, the accident has happened. The
_standard yard_ was so injured as to be rendered useless by the fire at
the Houses of Parliament.
213. The day is therefore the first measure obtained, and is divided
into 24 parts or hours, each of which is divided into 60 parts or
minutes, and each of these again into 60 parts or seconds. One second,
marked thus, 1″,[33] is therefore the 86400ᵗʰ part of a day, and the
following is the
MEASURE OF TIME.[34]
60 _seconds_ are 1 _minute_ 1 m.
60 _minutes_ ” 1 _hour_ 1 h.
24 _hours_ ” 1 _day_ 1 d.
7 _days_ ” 1 _week_ 1 wk.
365 _days_ ” 1 _year_ 1 yr.
214. The _second_ having been obtained, a pendulum can be constructed
which shall, when put in motion, perform one vibration in exactly
one second, in the latitude of Greenwich.[35] If we were inventing
measures, it would be convenient to call the length of this pendulum a
yard, and make it the standard of all our measures of length. But as
there is a yard already established, it will do equally well to tell
the length of the pendulum in yards. It was found by commissioners
appointed for the purpose, that this pendulum in London was 39·1393
inches, or about one yard, three inches, and ⁵/₃₆ of an inch. The
following is the division of the yard.
MEASURES OF LENGTH.
The lowest measure is a barleycorn.[36]
3 _barleycorns_ are 1 _inch_ 1 in.
12 _inches_ 1 _foot_ 1 ft.
3 _feet_ 1 _yard_ 1 yd.
5½ _yards_ 1 _pole_ 1 po.
40 _poles_ or 220 _yards_ 1 _furlong_ 1 fur.
8 _furlongs_ or 1760 _yards_ 1 _mile_ 1 mi.
Also 6 _feet_ 1 _fathom_ 1 fth.
69⅓ _miles_ 1 _degree_ 1 deg. or 1°.
[33] The minute and second are often marked thus, 1′, 1″: but this
notation is now almost entirely appropriated to the minute and second
of _angular_ measure.
[34] The measures in italics are those which it is most necessary that
the student should learn by heart.
[35] The lengths of the pendulums which will vibrate in one second are
slightly different in different latitudes. Greenwich is chosen as the
station of the Royal Observatory. We may add, that much doubt is now
entertained as to the system of standards derived from nature being
capable of that extreme accuracy which was once attributed to it.
[36] The inch is said to have been originally obtained by putting
together three grains of barley.
A geographical mile is ¹/₆₀th of a degree, and three such miles are one
nautical league.
In the measurement of cloth or linen the following are also used:
2¼ inches are 1 nail 1 nl.
4 nails 1 quarter (of a yard) 1 qr.
3 quarters 1 Flemish ell 1 Fl. e.
5 quarters 1 English ell 1 E. e.
6 quarters 1 French ell 1 Fr. e.
215. MEASURES OF SURFACE, OR SUPERFICIES.
All surfaces are measured by square inches, square feet, &c.; the
square inch being a square whose side is an inch in length, and so on.
The following measures may be deduced from the last, as will afterwards
appear.
144 square inches are 1 square foot 1 sq. ft.
9 square feet 1 square yard 1 sq. yd.
30¼ square yards 1 square pole 1 sq. p.
40 square poles 1 rood 1 rd.
4 roods 1 acre 1 ac.
Thus, the acre contains 4840 square yards, which is ten times a square
of 22 yards in length and breadth. This 22 yards is the length which
land-surveyors’ chains are made to have, and the chain is divided into
100 links, each ·22 of a yard or 7·92 inches. An acre is then 10 square
chains. It may also be noticed that a square whose side is 69⁴/₇ yards
is nearly an acre, not exceeding it by ⅕ of a square foot.
216. MEASURES OF SOLIDITY OR CAPACITY.[37]
Cubes are solids having the figure of dice. A cubic inch is a cube each
of whose sides is an inch, and so on.
1728 cubic inches are 1 cubic foot 1 c. ft.
27 cubic feet 1 cubic yard 1 c. yd.
[37] ‘Capacity’ is a term which cannot be better explained than by its
use. When one measure holds more than another, it is said to be more
capacious, or to have a greater capacity.
This measure is not much used, except in purely mathematical questions.
In the measurements of different commodities various measures were
used, which are now reduced, by act of parliament, to one. This is
commonly called the imperial measure, and is as follows:
MEASURE OF LIQUIDS AND OF ALL DRY GOODS.
4 _gills_ are 1 _pint_ 1 pt.
2 _pints_ 1 _quart_ 1 qt.
4 _quarts_ 1 _gallon_ 1 gall.
2 _gallons_ 1 _peck_[38] 1 pk.
4 _pecks_ 1 _bushel_ 1 bu.
8 _bushels_ 1 _quarter_ 1 qr.
5 _quarters_ 1 _load_ 1 ld.
The gallon in this measure is about 277·274 cubic inches; that is, very
nearly 277¼ cubic inches.[39]
217. The smallest weight in use is the grain, which is thus determined.
A vessel whose interior is a cubic inch, when filled with water,[40]
has its weight increased by 252·458 grains. Of the grains so
determined, 7000 are a pound _averdupois_, and 5760 a pound _troy_.
The first pound is always used, except in weighing precious metals and
stones, and also medicines. It is divided as follows:
[38] This measure, and those which follow, are used for dry goods only.
[39] Since the publication of the third edition, the _heaped_ measure,
which was part of the new system, has been abolished. The following
paragraph from the third edition will serve for reference to it:
“The other imperial measure is applied to goods which it is customary
to sell by _heaped measure_, and is as follows:
2 gallons 1 peck
4 pecks 1 bushel
3 bushels 1 sack
12 sacks 1 chaldron.
The gallon and bushel in this measure hold the same when only just
filled, as in the last. The bushel, however, heaped up as directed by
the act of parliament, is a little more than one-fourth greater than
before.”
[40] Pure water, cleared from foreign substances by distillation, at a
temperature of 62° Fahr.
AVERDUPOIS WEIGHT.
27¹¹/₃₂ _grains_ are 1 _dram_ 1 dr.
6 _drams_, or _drachms_ 1 _ounce_[41] 1 oz.
16 _ounces_ 1 _pound_ 1 lb.
28 _pounds_ 1 _quarter_ 1 qr.
4 _quarters_ 1 _hundred-weight_ 1 cwt.
20 _hundred-weight_ 1 _ton_ 1 ton.
[41] It is more common to divide the ounce into four quarters than into
sixteen drams.
The pound averdupois contains 7000 grains. A cubic foot of water weighs
62·3210606 pounds averdupois, or 997·1369691 ounces.
For the precious metals and for medicines, the pound troy, containing
5760 grains, is used, but is differently divided in the two cases. The
measures are as follow:
TROY WEIGHT.
24 _grains_ are 1 _pennyweight_ 1 dwt.
20 _pennyweights_ 1 _ounce_ 1 oz.
12 _ounces_ 1 _pound_ 1 lb.
The pound troy contains 5760 grains. A cubic foot of water weighs
75·7374 pounds troy, or 908·8488 ounces.
APOTHECARIES’ WEIGHT.
20 _grains_ are 1 _scruple_ ℈
3 _scruples_ 1 _dram_ ʒ
8 _drams_ 1 _ounce_ ℥
12 _ounces_ 1 _pound_ lb
218. The standard coins of copper, silver, and gold, are,--the penny,
which is 10⅔ drams of copper; the shilling, which weighs 3 pennyweights
15 grains, of which 3 parts out of 40 are alloy, and the rest pure
silver; and the sovereign, weighing 5 pennyweights and 3¼ grains, of
which 1 part out of 12 is copper, and the rest pure gold.
MEASURES OF MONEY.
The lowest coin is a farthing, which is marked thus, ¼, being one
fourth of a penny.
2 _farthings_ are 1 _halfpenny_ ½_d_.
2 _halfpence_ 1 _penny_ 1_d_.
12 _pence_ 1 _shilling_ 1_s_.
20 _shillings_ 1 _pound_[42] or _sovereign_ £1
21 _shillings_ 1 _guinea_.[43]
219. When any quantity is made up of several others, expressed in
different units, such as £1. 14. 6, or 2cwt. 1qr. 3lbs., it is called a
_compound quantity_. From these tables it is evident that any compound
quantity of any substance can be measured in several different ways.
For example, the sum of money which we call five pounds four shillings
is also 104 shillings, or 1248 pence, or 4992 farthings. It is easy to
reduce any quantity from one of these measurements to another; and the
following examples will be sufficient to shew how to apply the same
process, usually called REDUCTION, to all sorts of quantities.
I. How many farthings are there in £18. 12. 6¾?[44]
[42] The English pound is generally called a _pound sterling_, which
distinguishes it from the weight called a pound, and also from foreign
coins.
[43] The coin called a guinea is now no longer in use, but the name is
still given, from custom, to 21 shillings. The pound, which was not a
coin, but a note promising to pay 20 shillings to the bearer, is also
disused for the present, and the sovereign supplies its place; but the
name pound is still given to 20 shillings.
[44] Farthings are never written but as parts of a penny. Thus, three
farthings being 3/4 of a penny, is written 3/4, or ¾. One halfpenny may
be written either as 2/4 or ½; the latter is most common.
Since there are 20 shillings in a pound, there are, in £18, 18 × 20, or
360 shillings; therefore, £18. 12 is 360 + 12, or 372 shillings. Since
there are 12 pence in a shilling, in 372 shillings there are 372 × 12,
or 4464 pence; and, therefore, in £18. 12. 6 there are 4464 + 6, or
4470 pence.
Since there are 4 farthings in a penny, in 4470 pence there are 4470 ×
4, or 17880 farthings; and, therefore, in £18. 12. 6¾ there are 17880
+ 3, or 17883 farthings. The whole of this process may be written as
follows:
£18 . 12 . 6¾
20
--------
360 + 12 = 372
12
-----
4464 + 6 = 4470
4
-----
17880 + 3 = 17883
II. In 17883 farthings, how many pounds, shillings, pence, and
farthings are there?
Since 17883, divided by 4, gives the quotient 4470, and the remainder
3, 17883 farthings are 4470 pence and 3 farthings (218).
Since 4470, divided by 12, gives the quotient 372, and the remainder 6,
4470 pence is 372 shillings and 6 pence.
Since 372, divided by 20, gives the quotient 18, and the remainder 12,
372 shillings is 18 pounds and 12 shillings.
Therefore, 17883 farthings is 4470¾_d_., which is 372s. 6¾_d_., which
is £18. 12. 6¾.
The process may be written as follows:
4)17883
-----
12)4470 ... 3
----
20)372 ... 6
£18 . 12 . 6¾
EXERCISES.
A has £100. 4. 11½, and B has 64392 farthings. If A receive 1492
farthings, and B £1. 2. 3½, which will then have the most, and by how
much?--_Answer_, A will have £33. 12. 3 more than B.
In the following table the quantities written opposite to each other
are the same: each line furnishes two exercises.
£15 . 18 . 9½ | 15302 farthings.
115ˡᵇˢ 1ᵒᶻ 8ᵈᵚᵗ | 663072 grains.
3ˡᵇˢ 14ᵒᶻ 9ᵈʳ | 1001 drams.
3ᵐ 149 yds 2ᶠᵗ 9 in | 195477 inches.
19ᵇᵘ 2ᵖᵏˢ 1 gall 2 qᵗˢ | 1260 pints.
16 ʰ 23ᵐ 47ˢ | 59027 seconds.
220. The same may be done where the number first expressed is
fractional. For example, how many shillings and pence are there in ⁴/₁₅
of a pound? Now, ⁴/₁₅ of a pound is ⁴/₁₅ of 20 shillings; ⁴/₁₅ of 20 is
4 × 20 4 × 4 16
------, or ----- (110), or ---,
15 3 3
or (105) 5⅓ of a shilling. Again, ⅓ of a shilling is ⅓ of 12 pence, or
4 pence. Therefore, £⁴/₁₅ = 5_s._ 4_d._
Also, ·23 of a day is ·23 × 24 in hours, or 5ʰ·52; and ·52 of an hour
is ·52 × 60 in minutes, or 3ᵐ·2; and ·2 of a minute is ·2 × 60 in
seconds, or 12ˢ; whence ·23 of a day is 5ʰ 31ᵐ 12ˢ.
Again, suppose it required to find what part of a pound 6_s_. 8_d_. is.
Since 6_s._ 8_d._ is 80 pence, and since the whole pound contains 20
× 12 or 240 pence, 6_s._ 8_d._ is made by dividing the pound into 240
parts, and taking 80 of them. It is therefore £⁸⁰/₂₄₀ (107), but ⁸⁰/₂₄₀
= ⅓ (108); therefore, 6_s._ 8_d._ = £⅓.
EXERCISES.
⅖ of a day is 9ʰ 36ᵐ
·12841 of a day 3ʰ 4ᵐ 54ᔆ·624[45]
·257 of a cwt. 28ˡᵇˢ 12ᵒᶻ 8ᵈʳ·704
£·14936 2ˢ 11ᵈ 3ᶠ·3856
[45] When a decimal follows a whole number, the decimal is always of
the same unit as the whole number. Thus, 5ᔆ·5 is five _seconds_ and
five-tenths of a _second_. Thus, 0ᔆ·5 means five-tenths of a second;
0ʰ·3, three-tenths of an hour.
221, 222. I have thought it best to refer the mode of converting
shillings, pence, and farthings into decimals of a pound to the
Appendix (See Appendix _On Decimal Money_). I should strongly recommend
the reader to make himself perfectly familiar with the modes given in
that Appendix. To prevent the subsequent sections from being altered in
their numbering, I have numbered this paragraph as above.
223. The rule of addition[46] of two compound quantities of the same
sort will be evident from the following example. Suppose it required to
add £192. 14. 2½ to £64. 13. 11¾. The sum of these two is the whole of
that which arises from adding their several parts. Now
¾_d._ + ½_d._ = ⁵/₄_d._ = £0 . 0 . 1¼ (219)
11_d._ + 2_d._ = 13_d._ = 0 . 1 . 1
13_s._ + 14_s._ = 27_s._ = 1 . 7 . 0
£64 + £192 = 256 . 0 . 0
-----------
The sum of all of which is £257. 8 . 2¼
This may be done at once, and written as follows:
£192 . 14 . 2½
64 . 13 . 11¾
----------------
£257 . 8 . 2¼
[46] Before reading this article and the next, articles (29) and (42)
should be read again carefully.
Begin by adding together the farthings, and reduce the result to pence
and farthings. Set down the last only, carry the first to the line
of pence, and add the pence in both lines to it. Reduce the sum to
shillings and pence; set down the last only, and carry the first to the
line of shillings, and so on. The same method must be followed when the
quantities are of any other sort; and if the tables be kept in memory,
the process will be easy.
224. SUBTRACTION is performed on the same principle as in (40), namely,
that the difference of two quantities is not altered by adding the same
quantity to both. Suppose it required to subtract £19 . 13. 10¾ from
£24. 5. 7½. Write these quantities under one another thus:
£24. 5. 7½
19. 13. 10¾
Since ¾ cannot be taken from ½ or ²/₄, add 1_d._ to both quantities,
which will not alter their difference; or, which is the same thing,
add 4 farthings to the first, and 1_d._ to the second. The pence and
farthings in the two lines then stand thus: 7⁶/₄_d._ and 11¾_d._ Now
subtract ¾ from ⁶/₄, and the difference is ¾ which must be written
under the farthings. Again, since 11_d._ cannot be subtracted from
7_d._, add 1_s._ to both quantities by adding 12_d._ to the first, and
1_s._ to the second. The pence in the first line are then 19, and in
the second 11, and the difference is 8, which write under the pence.
Since the shillings in the lower line were increased by 1, there are
now 14_s._ in the lower, and 5_s._ in the upper one. Add 20_s._ to the
upper and £1 to the lower line, and the subtraction of the shillings in
the second from those in the first leaves 11_s._ Again, there are now
£20 in the lower, and £24 in the upper line, the difference of which is
£4; therefore the whole difference of the two sums is £4. 11. 8¾. If we
write down the two sums with all the additions which have been made,
the process will stand thus:
£24 . 25 . 19⁶/₄
20 . 14 . 11¾
------------------
Difference £4 . 11 . 8¾
225. The same method may be applied to any of the quantities in the
tables. The following is another example:
From 7 cwt. 2 qrs. 21 lbs. 14 oz.
Subtract 2 cwt. 3 qrs. 27 lbs. 12 oz.
After alterations have been made similar to those in the last article,
the question becomes:
From 7 cwt. 6 qrs. 49 lbs. 14 oz.
Subtract 3 cwt. 4 qrs. 27 lbs. 12 oz.
----------------------------
The difference is 4 cwt. 2 qrs. 22 lbs. 2 oz.
In this example, and almost every other, the process may be a little
shortened in the following way. Here we do not subtract 27 lbs. from 21
lbs., which is impossible, but we increase 21 lbs. by 1 qr. or 28 lbs.
and then subtract 27 lbs. from the sum. It would be shorter, and lead
to the same result, first to subtract 27 lbs. from 1 qr. or 28 lbs. and
add the difference to 21 lbs.
226. EXERCISES.
A man has the following sums to receive: £193. 14. 11¼, £22. 0. 6¾,
£6473. 0. 0, and £49. 14. 4½; and the following debts to pay: £200 .
19. 6¼, £305. 16. 11, £22, and £19. 6. 0½. How much will remain after
paying the debts?
_Answer_, £6190. 7. 4¾.
There are four towns, in the order A, B, C, and D. If a man can go from
A to B in 5ʰ 20ᵐ 33ˢ, from B to C in 6ʰ 49ᵐ 2ˢ and from A to D in 19ʰ
0ᵐ 17ˢ, how long will he be in going from B to D, and from C to D?
_Answer_, 13ʰ 39ᵐ 44ˢ, and 6ʰ 50ᵐ 42ˢ.
227. In order to perform the process of MULTIPLICATION, it must be
recollected that, as in (52), if a quantity be divided into several
parts, and each of these parts be multiplied by a number, and the
products be added, the result is the same as would arise from
multiplying the whole quantity by that number.
It is required to multiply £7. 13. 6¼ by 13. The first quantity is made
up of 7 pounds, 13 shillings, 6 pence, and 1 farthing. And
1 farth. × 13 is 13 farth. or £0 . 0 . 3¼ (219)
6 pence × 13 is 78 pence, or 0 . 6 . 6
13 shill. × 13 is 169 shill. or 8 . 9 . 0
7 pounds × 13 is 91 pounds, or 91 . 0 . 0
--------------
The sum of all these is £99 . 15 . 9¼
which is therefore £7. 13. 6¼ × 13.
This process is usually written as follows:
£7 . 13 . 6¼
13
-------------
£99 . 15 . 9¼
228. DIVISION is performed upon the same principle as in (74), viz.
that if a quantity be divided into any number of parts, and each part
be divided by any number, the different quotients added together will
make up the quotient of the whole quantity divided by that number.
Suppose it required to divide £99. 15. 9¼ by 13. Since 99 divided by 13
gives the quotient 7, and the remainder 8, the quantity is made up of
£13 × 7, or £91, and £8. 15. 9¼. The quotient of the first, 13 being
the divisor, is £7: it remains to find that of the second. Since £8 is
160_s._, £8. 15. 9¼ is 175_s._ 9¼_d._, and 175 divided by 13 gives the
quotient 13, and the remainder 6; that is, 175_s._ 9¼_d._ is made up of
169_s._ and 6_s._ 9¼_d._, the quotient of the first of which is 13_s._,
and it remains to find that of the second. Since 6_s._ is 72_d._,
6_s._ 9¼_d._ is 81¼_d._, and 81 divided by 13 gives the quotient 6
and remainder 3; that is, 81¼_d._ is 78_d._ and 3¼_d._, of the first
of which the quotient is 6_d._ Again, since 3_d._ is ¹²/₄, or 12
farthings, 3¼_d._ is 13 farthings, the quotient of which is 1 farthing,
or ¼, without remainder. We have then divided £99. 15. 9¼ into four
parts, each of which is divisible by 13, viz. £91, 169_s._, 78_d._, and
13 farthings; so that the thirteenth part of this quantity is £7. 13.
6¼. The whole process may be written down as follows; and the same sort
of process may be applied to the exercises which follow:
£ _s._ _d._ £ _s._ _d._
13)99 15 9¼( 7 13 6¼
91
--
8
20
---
160 + 15 = 175
13
---
45
39
--
6
12
--
72 + 9 = 81
78
--
3
4
--
12 + 1 = 13
13
--
0
Here, each of the numbers 99, 175, 81, and 13, is divided by 13 in the
usual way, though the divisor is only written before the first of them.
EXERCISES.
2 cwt. 1 qr. 21 lbs. 7 oz. × 53 = 129 cwt. 1 qr. 16 lbs. 3 oz.
2ᵈ 4ʰ 3ᵐ 27ˢ × 109 = 236ᵈ 10ʰ 16ᵐ 3ˢ
£27 . 10 . 8 × 569 = £15666 . 9 . 4
£7 . 4 . 8 × 123 = £889 . 14
£166 × ₈/₃₃ = £40 . 4 . 10⁶/₃₃
£187 . 6 . 7 × ³/₁₀₀ = £5 . 12 . 4¾ ²/₂₅
4_s._ 6½_d._ × 1121 = £254 . 11 . 2½
4_s._ 4_d._ × 4260 = 6_s._ 6_d._ × 2840
229. Suppose it required to find how many times 1s. 4¼_d._ is contained
in £3. 19. 10¾. The way to do this is to find the number of farthings
in each. By 219, in the first there are 65, and in the second 3835
farthings. Now, 3835 contains 65 59 times; and therefore the second
quantity is 59 times as great as the first. In the case, however, of
pounds, shillings, and pence, it would be best to use decimals of a
pound, which will give a sufficiently exact answer. Thus 1s. 4¼_d._ is
£·067, and £3. 19. 10¾ is £3·994, and 3·994 divided by ·067 is 3994 by
67, or 59⁴¹/₆₇. This is an extreme case, for the smaller the divisor,
the greater the effect of an error in a given place of decimals.
EXERCISES.
How many times does 6 cwt. 2 qrs. contain 1 qr. 14 lbs. 1 oz.? and 1ᵈ
2ʰ 0ᵐ 47ˢ contain 3ᵐ 46ˢ?
_Answer_, 17·30758 and 414·367257.
If 2 cwt. 3 qrs. 1 lb. cost £150. 13. 10, how much does 1 lb. cost?
_Answer_, 9_s._ 9_d._ ¹³/₃₀₉.
A grocer mixes 2 cwt. 15 lbs. of sugar at 11_d._ per pound with 14
cwt. 3 lbs. at 5_d._ per pound. At how much per pound must he sell the
mixture so as not to lose by mixing them?
_Answer_, 5_d._ ¾ ¹⁵³/₉₀₅.
230. There is a convenient method of multiplication called PRACTICE.
Suppose I ask, How much do 153 tons cost if each ton cost £2. 15. 7½?
It is plain that if this sum be multiplied by 153, the product is the
price of the whole. But this is also evident, that, if I buy 153 tons
at £2. 15. 7½ each ton, payment may be made by first putting down £2
for each ton, then 10s. for each, then 5_s._, then 6_d._, and then
1½_d._ These sums together make up £2. 15. 7½, and the reason for this
separation of £2. 15 . 7½ into different parts will be soon apparent.
The process may be carried on as follows:
1. 153 tons, at £2 each ton, will cost £306 0 0
2. Since 10s. is £½, 153 tons, at 10_s._ each,
will cost £15³/₂, which is 76 10 0
3. Since 5_s._ is ½ of 10_s._, 153 tons, at 5_s._,
will cost half as much as the same number at
10_s._ each, that is, ½ of £76 . 10, which is 38 5 0
4. Since 6_d._ is ⅒ of 5_s._, 153 tons, at 6_d._
each, will cost ⅒ of what the same number
costs at 5_s._each, that is, ⅒ of £38 . 5,
which is 3 16 6
5. Since 1½ or 3 halfpence is ¼ of 6_d._ or 12
halfpence, 153 tons, at 1½_d._ each, will cost
¼ of what the same number costs at 6_d._ each,
that is, ¼ of£3 . 16 . 6, which is 0 19 1½
-----------
The sum of all these quantities is 425 10 7½
which is, therefore, £2 . 15 . 7½ × 153.
The whole process may be written down as follows:
or what
153 tons
would
| £153 0 0 cost at £1 per ton.
| ----------- -----------
£2 is 2 × £1 | 306 0 0 2 0 0
10_s._ is ½ of £1 | 76 10 0 0 10 0
5_s._ is ½ of 10_s._ | 38 5 0 0 5 0
6_d._ is ⅒ of 5_s._ | 3 16 6 0 0 6
1½_d._ is ¼ of 6_d._ | 0 19 1½ 0 0 1½
| ------------ ----------
Sum | £425 10 7½ £2 15 7½
ANOTHER EXAMPLE.
What do 1735 lbs. cost at 9_s._ 10¾_d._ per lb.? The price 9_s._
10¾_d_. is made up of 5_s._, 4_s._, 10_d._, ½_d._, and ¼_d._; of which
5_s._ is ¼ of £1, 4_s._ is ⅕ of £1, 10_d._ is ⅙ of 5_s._, ½_d._ is ¹/₂₀
of 10_d._, and ¼_d._ is ½ of ½_d._ Follow the same method as in the
last example, which gives the following:
or what
1735 tons
would
| £1735 0 0 cost at £1 per lb.
| ------------ -----------
5_s._ is ¼ of £1 | 433 15 0 0 5 0
|
4_s._ is ⅕ of £1 | 347 0 0 0 4 0
|
10_d._ is ⅙ of 5_s._ | 72 5 10 0 0 10
|
½_d._ is ¹/₂₀ of 10_d._ | 3 12 3½ 0 0 0½
|
¼_d._ is ½ of ½_d._ | 1 16 1¾ 0 0 0¼
| ------------- ------------
by addition ... | £858 9 3¼ £0 9 10¾
In all cases, the price must first be divided into a number of parts,
each of which is a simple fraction[47] of some one which goes before.
No rule can be given for doing this, but practice will enable the
student immediately to find out the best method for each case. When
that is done, he must find how much the whole quantity would cost if
each of these parts were the price, and then add the results together.
[47] Any fraction of a unit, whose numerator is unity, is generally
called an _aliquot part_ of that unit. Thus, 2_s._ and 10_s._ are both
aliquot parts of a pound, being £⅒ and £½.
EXERCISES.
What is the cost of
243 cwt. at £14 . 18 . 8¼ per cwt.?--_Answer_, £3629 . 1 . 0¾.
169 bushels at £2 . 1 . 3¼ per bushel?--_Answer_, £348 . 14 . 9¼.
273 qrs. at 19_s._ 2_d._ per quarter?--_Answer_, £261 . 12. 6.
2627 sacks at 7_s._ 8½_d._ per sack?--_Answer_, £1012 . 9 . 9½.
231. Throughout this section it must be observed, that the rules can be
applied to cases where the quantities given are expressed in common or
decimal fractions, instead of the measures in the tables. The following
are examples:
What is the price of 272·3479 cwt. at £2. 1. 3½ per cwt.?
_Answer_, £562·2849, or £562. 5. 8¼.
66½lbs. at 1_s._ 4½_d._ per lb. cost £4. 11. 5¼.
How many pounds, shillings, and pence, will 279·301 acres let for if
each acre lets for £3·1076?--_Answer_, £867·9558, or £867. 19. 1¼.
What does ¼ of ³/₁₃ of 17 bush. cost at ⅙ of ⅔ of £17. 14 per bushel?
_Answer_, £2·3146, or £2. 6. 3½.
What is the cost of 19lbs. 8oz. 12dwt. 8gr. at £4. 4. 6 per
ounce?--_Answer_, £999. 14. 1¼ ⅙.
232. It is often required to find to how much a certain sum per day
will amount in a year. This may be shortly done, since it happens that
the number of days in a year is 240 + 120 + 5; so that a penny per day
is a pound, half a pound, and 5 pence per year. Hence the following
rule: To find how much any sum per day amounts to in a year, turn it
into pence and fractions of a penny; to this add the half of itself,
and let the pence be pounds, and each farthing five shillings; then
add five times the daily sum, and the total is the yearly amount. For
example, what does 12_s._ 3¾_d._ amount to in a year? This is 147¾_d._,
and its half is 73⅞_d._, which added to 147¾_d._ gives 221⅝_d._, which
turned into pounds is £221. 12. 6. Also, 12_s._ 3¾_d._ × 5 is £3. 1.
6¾, which added to the former sum gives £224. 14. 0¾ for the yearly
amount. In the same way the yearly amount of 2_s._ 3½_d._ is £41. 16.
5½; that of 6¾_d._ is £10. 5. 3¾; and that of 11_d._ is £16. 14. 7.
233. An inverse rule may be formed, sufficiently correct for every
purpose, in the following way: If the year consisted of 360 days, or
³/₂ of 240, the subtraction of one-third from any sum per year would
give the proportion which belongs to 240 days; and every pound so
obtained would be one penny per day. But as the year is not 360, but
365 days, if we divide each day’s share into 365 parts, and take 5
away, the whole of the subtracted sum, or 360 × 5 such parts, will
give 360 parts for each of the 5 days which we neglected at first.
But 360 such parts are left behind for each of the 360 first days;
therefore, this additional process divides the whole annual amount
equally among the 365 days. Now, 5 parts out of 365 is one out of
73, or the 73d part of the first result must be subtracted from it
to produce the true result. Unless the daily sum be very large, the
72d part will do equally well, which, as 72 farthings are 18 pence,
is equivalent to subtracting at the rate of one farthing for 18_d._,
or ½_d._ for 3_s._, or 10_d._ for £3. The rule, then, is as follows:
To find how much per day will produce a given sum per year, turn
the shillings, &c. in the given sum into decimals of a pound (221);
subtract one-third; consider the result as pence; and diminish it by
one farthing for every eighteen pence, or ten pence for every £3. For
example, how much per day will give £224. 14. 0¾ per year? This is
224·703, and its third is 74·901, which subtracted from 224·703, gives
149·802, which, if they be pence, amounts to 12_s._ 5·802_d._, in which
1_s._ 6_d._ is contained 8 times. Subtract 8 farthings, or 2_d._, and
we have 12_s._ 3·802_d._, which differs from the truth only about ¹/₂₀
of a farthing. In the same way, £100 per year is 5_s._ 5¾_d._ per day.
234. The following connexion between the measures of length and the
measures of surface is the foundation of the application of arithmetic
to geometry.
[Illustration]
Suppose an oblong figure, A, B, C, D, as here drawn (which is called
a _rectangle_ in geometry), with the side A B 6 inches, and the side
A C 4 inches. Divide A B and C D (which are equal) each into 6 inches
by the points _a, b, c, l, m_, &c.; and A C and B D (which are also
equal) into 4 inches by the points _f, g, h, x, y_, and _z_. Join _a_
and l, _b_ and _m_, &c., and _f_ and _x_, &c. Then, the figure A B C D
is divided into a number of squares; for a square is a rectangle whose
sides are equal, and therefore A _a f_ E is square, since A _a_ is of
the same length as A _f_, both being 1 inch. There are also four rows
of these squares, with six squares in each row; that is, there are 6
× 4, or 24 squares altogether. Each of these squares has its sides 1
inch in length, and is what was called in (215) _a square inch_. By the
same reasoning, if one side had contained 6 _yards_, and the other 4
_yards_, the surface would have contained 6 × 4 _square yards_; and so
on.
[Illustration]
235. Let us now suppose that the sides of A B C D, instead of being
a whole number of inches, contain some inches and a fraction. For
example, let A B be 3½ inches, or (114) ⁷/₂ of an inch, and let A C
contain 2½ inches, or ⁹/₄ of an inch. Draw A E twice as long as A B,
and A F four times as long as A C, and complete the rectangle A E F G.
The rest of the figure needs no description. Then, since A E is twice
A B, or twice ⁷/₂ inches, it is 7 inches. And since A F is four times
A C, or four times ⁹/₄ inches, it is 9 inches. Therefore, the whole
rectangle A E F G contains, by (234), 7 × 9 or 63 square inches. But
the rectangle A E F G contains 8 rectangles, all of the same figure
as A B C D; and therefore A B C D is one-eighth part of A E F G, and
contains ⁶³/₈ square inches. But ⁶³/₈ is made by multiplying ⁹/₄ and
⁷/₂ together (118). From this and the last article it appears, that,
whether the sides of a rectangle be a whole or a fractional number of
inches, the number of square inches in its surface is the product of
the numbers of inches in its sides. The square itself is a rectangle
whose sides are all equal, and therefore the number of square inches
which a square contains is found by multiplying the number of inches in
its side by itself. For example, a square whose side is 13 inches in
length contains 13 × 13 or 169 square inches.
236. EXERCISES.
What is the content, in square feet and inches, of a room whose sides
are 42 ft. 5 inch. and 31 ft. 9 inch.? and supposing the piece from
which its carpet is taken to be three quarters of a yard in breadth,
what length of it must be cut off?--_Answer_, The content is 1346
square feet 105 square inches, and the length of carpet required is 598
feet 6⁵/₉ inches.
The sides of a rectangular field are 253 yards and a quarter of a mile;
how many acres does it contain?--_Answer_, 23.
What is the difference between 18 _square miles_, and a square of 18
miles long, or 18 _miles square_?--_Answer_, 306 square miles.
237. It is by this rule that the measure in (215) is deduced from
that in (214); for it is evident that twelve inches being a foot, the
square foot is 12 × 12 or 144 square inches, and so on. In a similar
way it may be shewn that the content in cubic inches of a cube, or
parallelepiped,[48] may be found by multiplying together the number of
inches in those three sides which meet in a point. Thus, a cube of 6
inches contains 6 × 6 × 6, or 216 cubic inches; a chest whose sides are
6, 8, and 5 feet, contains 6 × 8 × 5, or 240 cubic feet. By this rule
the measure in (216) was deduced from that in (214).
[48] A parallelepiped, or more properly, a _rectangular_
parallelepiped, is a figure of the form of a brick; its sides, however,
may be of any length; thus, the figure of a plank has the same name. A
cube is a parallelepiped with equal sides, such as is a die.
SECTION II.
RULE OF THREE.
238. Suppose it required to find what 156 yards will cost, if 22 yards
cost 17_s._ 4_d._ This quantity, reduced to pence, is 208_d._; and if
22 yards cost 208_d._, each yard costs ²⁰⁸/₂₂_d_. But 156 yards cost
156 times the price of one yard, and therefore cost
208 208 × 156
---- × 156 pence, or --------- pence (117).
22 22
Again, if 25½ French francs be 20 shillings sterling, how many francs
are in £20. 15? Since 25½ francs are 20 shillings, twice the number of
francs must be twice the number of shillings; that is, 51 francs are
40 shillings, and one shilling is the fortieth part of 51 francs, or
⁵¹/₄₀ francs. But £20 15_s._ contain 415 shillings (219); and since 1
shilling is ⁵¹/₄₀ francs, 415 shillings is
51 × 415
⁵¹/₄₀ × 415 francs, or (117) -------- francs.
40
239. Such questions as the last two belong to the most extensive rule
in Commercial Arithmetic, which is called the RULE OF THREE, because in
it three quantities are given, and a fourth is required to be found.
From both the preceding examples the following rule may be deduced,
which the same reasoning will shew to apply to all similar cases.
It must be observed, that in these questions there are two quantities
which are of the same sort, and a third of another sort, of which last
the answer must be. Thus, in the first question there are 22 and 156
yards and 208 pence, and the thing required to be found is a number
of pence. In the second question there are 20 and 415 shillings and
25½ francs, and what is to be found is a number of francs. Write the
three quantities in a line, putting that one last which is the only one
of its kind, and that one first which is connected with the last in
the question.[49] Put the third quantity in the middle. In the first
question the quantities will be placed thus:
22 yds. 156 yds. 17_s._ 4_d._
In the second question they will be placed thus:
20_s._ £20 15_s._ 25½ francs.
[49] This generally comes in the same member of the sentence. In some
cases the ingenuity of the student must be employed in detecting it.
The reasoning of (238) is the best guide. The following may be very
often applied. If it be evident that the answer must be less than the
given quantity of its kind, multiply that given quantity by the less of
the other two; if greater, by the greater. Thus, in the first question,
156 yards must cost more than 22; multiply, therefore, by 156.
Reduce the first and second quantities, if necessary, to quantities of
the same denomination. Thus, in the second question, £20 15_s._ must
be reduced to shillings (219). The third quantity may also be reduced
to any other denomination, if convenient; or the first and third may
be multiplied by any quantity we please, as was done in the second
question; and, on looking at the answer in (238), and at (108), it
will be seen that no change is made by that multiplication. Multiply
the second and third quantities together, and divide by the first. The
result is a quantity of the same sort as the third in the line, and is
the answer required. Thus, to the first question the answer is (238)
208 × 156 17_s._ 4_d_. × 156
----------pence, or, which is the same thing, -------------------.
22 22
240. The whole process in the first question is as follows:[50]
yds. yds. _s._ _d._
22 : 156 ∷ 17 . 4
12
---
208 pence.
156
----
1248
1040
208
-----
22)32448(1474¾_d._ and ¹⁴/₂₂, or ⁷/₁₁ of a farthing,
22 or (219) £6 . 2 . 10¾-⁷/₁₁.
---
104
88
----
164
154
----
108
88
--
20
(228) 4
--
80
66
--
14
[50] It is usual to place points, in the manner here shewn, between the
quantities. Those who have read Section VIII. will see that the Rule
of Three is no more than the process for finding the fourth term of a
proportion from the other three.
The question might have been solved without reducing 17_s._ 4_d._ to
pence, thus:
yds. yds. _s._ _d._
22 : 156 ∷ 17 . 4
156 (227)
----------
22)£135 . 4 . 0(£6 . 2 . 10¾-⁷/₁₁ (228)
132
---
3 × 20 + 4 = 64
44
--
20 × 12 = 240
220
---
20 × 4 = 80
66
--
14
The student must learn by practice which is the most convenient method
for any particular case, as no rule can be given.
241. It may happen that the three given quantities are all of one
denomination; nevertheless it will be found that two of them are of
one, and the third of another sort. For example: What must an income
of £400 pay towards an income-tax of 4_s._ 6_d._ in the pound? Here
the three given quantities are, £400, 4_s._ 6_d._, and £1, which are
all of the same species, viz. money. Nevertheless, the first and third
are income; the second is a tax, and the answer is also a tax; and
therefore, by (152), the quantities must be placed thus:
£1 : £400 ∷ 4_s._ 6_d._
242. The following exercises either depend directly upon this rule,
or can be shewn to do so by a little consideration. There are many
questions of the sort, which will require some exercise of ingenuity
before the method of applying the rule can be found.
EXERCISES.
If 15 cwt. 2 qrs. cost £198. 15. 4, what does 1 qr. 22 lbs. cost?
_Answer_, £5 . 14 . 5 ¾ ¹⁸⁵/₂₁₇.
If a horse go 14 m. 3 fur. 27 yds. in 3ʰ 26ᵐ 12ˢ, how long will he be
in going 23 miles?
_Answer_, 5ʰ 29ᵐ 34ˢ(²⁴⁶²/₂₅₃₂₇).
Two persons, A and B, are bankrupts, and owe exactly the same sum; A
can pay 15_s._ 4½_d._ in the pound, and B only 7_s._ (6¾)_d._ At the
same time A has in his possession £1304. 17 more than B; what do the
debts of each amount to?
_Answer_, £3340 . 8 . 3 ¾ ⁹/₂₅.
For every (12½) acres which one country contains, a second contains
(56¼). The second country contains 17,300 square miles. How much does
the first contain? Again, for every 3 people in the first, there are 5
in the second; and there are in the first 27 people on every 20 acres.
How many are there in each country?--_Answer_, The number of square
miles in the first is 3844⁴/₉, and its population 3,321,600; and the
population of the second is 5,536,000.
If (42½) yds. of cloth, 18 in. wide, cost £59. 14. 2, how much will
(118¼) yds. cost, if the width be 1 yd.?
_Answer_, £332. 5. (2⁴/₁₇).
If £9. 3. 6 last six weeks, how long will £100 last?
_Answer_, (65¹⁴⁵/₃₆₇) weeks.
How much sugar, worth (9¾d). a pound, must be given for 2 cwt. of tea,
worth 10_d._ an ounce?
_Answer_, 32 cwt. 3 qrs. 7 lbs. ³⁵/₃₉.
243. Suppose the following question asked: How long will it take 15 men
to do that which 45 men can finish in 10 days? It is evident that one
man would take 45 × 10, or 450 days, to do the same thing, and that 15
men would do it in one-fifteenth part of the time which it employs one
man, that is, in (450 ÷ 15) or 30 days. By this and similar reasoning
the following questions can be solved.
EXERCISES.
If 15 oxen eat an acre of grass in 12 days, how long will it take 26
oxen to eat 14 acres? _Answer_, (96¹²/₁₃) days.
If 22 masons build a wall 5 feet high in 6 days, how long will it take
43 masons to build 10 feet? _Answer_, (6⁶/₄₃) days.
244. The questions in the preceding article form part of a more general
class of questions, whose solution is called the DOUBLE RULE OF THREE,
but which might, with more correctness, be called the Rule of _Five_,
since five quantities are given, and a sixth is to be found. The
following is an example: If 5 men can make 30 yards of cloth in 3 days,
how long will it take 4 men to make 68 yards? The first thing to be
done is to find out, from the first part of the question, the time it
will take one man to make one yard. Now, since one man, in 3 days, will
do the fifth part of what 5 men can do, he will in 3 days make ³⁰/₅,
or 6 yards. He will, therefore, make one yard in ³/₆6 or (3 × 5)/30 of
a day. From this we are to find how long it will take 4 men to make 68
yards. Since one man makes a yard in
3 × 5 3 × 5
----- of a day, he will make 68 yards in ----- × 68 days,
30 30
3 × 5 × 68
or (116) in ---------- days; and 4 men will do this in one-fourth
30
3 × 5 × 68
of the time, that is (123), in ---------- days, or in 8½ days.
30 × 4
Again, suppose the question to be: If 5 men can make 30 yards in 3
days, how much can 6 men do in 12 days? Here we must first find the
quantity one man can do in one day, which appears, on reasoning similar
to that in the last example, to be 30/(3 × 5) yards. Hence, 6 men, in
one day, will make
6 × 30 12 × 6 × 30
------ yards, and in 12 days will make ----------- or 144 yards.
5 × 3 5 × 3
From these examples the following rule may be drawn. Write the given
quantities in two lines, keeping quantities of the same sort under one
another, and those which are connected with each other, in the same
line. In the two examples above given, the quantities must be written
thus:
[Illustration]
SECOND EXAMPLE.
[Illustration]
Draw a curve through the middle of each line, and the extremities of
the other. There will be three quantities on one curve and two on the
other. Divide the product of the three by the product of the two, and
the quotient is the answer to the question.
If necessary, the quantities in each line must be reduced to more
simple denominations (219), as was done in the common Rule of Three
(238).
EXERCISES.
If 6 horses can, in 2 days, plough 17 acres, how many acres will 93
horses plough in 4½ days?
_Answer_, 592⅞.
If 20 men, in 3¼ days, can dig 7 rectangular fields, the sides of each
of which are 40 and 50 yards, how long will 37 men be in digging 53
fields, the sides of each of which are 90 and 125½ yards?
2451
_Answer_, 75----- days.
20720
If the carriage of 60 cwt. through 20 miles cost £14 10_s._, what
weight ought to be carried 30 miles for £5. 8. 9?
_Answer_, 15 cwt.
If £100 gain £5 in a year, how much will £850 gain in 3 years and 8
months?
_Answer_, £155. 16. 8.
SECTION III.
INTEREST, ETC.
245. In the questions contained in this Section, almost the only
process which will be employed is the taking a fractional part of a
sum of money, which has been done before in several cases. Suppose it
required to take 7 parts out of 40 from £16, that is, to divide £16
into 40 equal parts, and take 7 of them. Each of these parts is
16 16 16 × 7
£----, and 7 of them make ---- × 7, or ------ pounds (116).
40 40 40
The process may be written as below:
£16
7
-----
40)112(£2 . 16_s._
80
--
32
20
---
640
40
---
240
240
---
0
Suppose it required to take 13 parts out of a hundred from £56. 13. 7½.
56 . 13 . 7½
13
----------------
100) 736 . 17 . 1½ ( £7 . 7 . 4 ¼ ¹/₄₁
700
---
36 × 20 + 17 = 737
700
---
37 × 12 + 1 = 445
400
---
45 × 4 × 2 = 182
100
---
82
Let it be required to take 2½ parts out of a hundred from £3 12_s._ The
result, by the same rule is
£3 12_s._ × 2½ 5
-------------------, or 123 £3 12_s._ × ---;
100 200
so that taking 2½ out of a hundred is the same as taking 5 parts out of
200.
EXERCISES.
Take 7⅓ parts out of 53 from £1 10_s._
129
_Answer_, 4_s._ 1---_d._
159
Take 5 parts out of 100 from £107 13_s._ 4¾_d._
_Answer_, £5. 7. 8 and ³/₂₀ of a farthing.
£56 3_s._ 2_d._ is equally divided among 32 persons. How much does the
share of 23 of them exceed that of the rest?
_Answer_, £24. 11. 4½ ½.
246. It is usual, in mercantile business, to mention the fraction which
one sum is of another, by saying how many parts out of a hundred must
be taken from the second in order to make the first. Thus, instead of
saying that £16 12_s._ is the half of £33 4_s._, it is said that the
first is 50 per cent of the second. Thus, £5 is 2½ per cent of £200;
because, if £200 be divided into 100 parts, 2½ of those parts are £5.
Also, £13 is 150 per cent of £8. 13. 4, since the first is the second
and half the second. Suppose it asked, How much per cent is 23 parts
out of 56 of any sum? The question amounts to this: If he who has £56
gets £100 for them, how much will he who has 23 receive? This, by 238,
is 23 × ¹⁰⁰/₅₆ or ²³⁰⁰/₅₆ or 41¹/₁₄. Hence, 23 out of 56 is 41¹/₁₄ per
cent.
Similarly 16 parts out of 18 is 16 × ¹⁰⁰/₁₈, or 88⁸/₉ per cent, and 2
parts out of 5 is 2 × ¹⁰⁰/₅, or 40 per cent.
From which the method of reducing other fractions to the rate per cent
is evident.
Suppose it asked, How much per cent is £6. 12. 2 of £12. 3? Since the
first contains 1586_d._, and the second 2916_d._, the first is 1586
out of 2916 parts of the second; that is, by the last rule, it is
¹⁵⁸⁶⁰⁰/₂₉₁₆, or 54¹¹³⁶/₂₉₁₆, or £54. 7. 9½ per cent, very nearly. The
more expeditious way of doing this is to reduce the shillings, &c.
to decimals of a pound. Three decimal places will give the rate per
cent to the nearest shilling, which is near enough for all practical
purposes. For instance, in the last example, which is to find how much
£6·608 is of £12·15, 6·608 × 100 is 660·8, which divided by 12·15 gives
£54·38, or £54. 7. Greater correctness may be had, if necessary, as in
the Appendix.
EXERCISES.
How much per cent is 198¼ out of 233 parts?--_Ans._ £85. 1. 8¾.
Goods which are bought for £193. 12, are sold for £216. 13. 4; how much
per cent has been gained by them?
_Answer_, A little less than £11. 18. 6.
A sells goods for B to the amount of £230. 12, and is allowed a
commission[51] of 3 per cent; what does that amount to?
_Answer_, £6 . 18. 4¼ ⁷/₂₅.
[51] Commission is what is allowed by one merchant to another for
buying or selling goods for him, and is usually a per-centage on the
whole sum employed. Brokerage is an allowance similar to commission,
under a different name, principally used in the buying and selling of
stock in the funds.
Insurance is a per-centage paid to those who engage to make good to the
payers any loss they may sustain by accidents from fire, or storms,
according to the agreement, up to a certain amount which is named,
and is a per-centage upon this amount. Tare, tret, and cloff, are
allowances made in selling goods by wholesale, for the weight of the
boxes or barrels which contain them, waste, &c.; and are usually either
the price of a certain number of pounds of the goods for each box or
barrel, or a certain allowance on each cwt.
A stockbroker buys £1700 stock, brokerage being at £⅛ per cent; what
does he receive?--_Answer_, £2. 2. 6.
A ship whose value is £15,423 is insured at 19⅔ per cent; what does the
insurance amount to?--_Answer_, £3033. 3. 9½ ²/₅.
247. In reckoning how much a bankrupt is able to pay his creditors, as
also to how much a tax or rate amounts, it is usual to find how many
shillings in the pound is paid. Thus, if a person who owes £100 can
only pay £50, he is said to pay 10_s._ in the pound. The rule is easily
derived from the same reasoning as in 246. For example, £50 out of £82
is
50 50×20
£---- out of £1, or ----- shillings,
82 82
or 12_s._ 2½ ¹⁵/₄₁ in the pound.
248. INTEREST is money paid for the use of other money, and is always
a per-centage upon the sum lent. It may be paid either yearly,
half-yearly, or quarterly; but when it is said that £100 is lent at 4
per cent, it must be understood to mean 4 per cent per annum; that is,
that 4 pounds are paid every year for the use of £100.
The sum lent is called the _principal_, and the interest upon it is
of two kinds. If the borrower pay the interest as soon as, from the
agreement, it becomes due, it is evident that he has to pay the same
sum every year; and that the whole of the interest which he has to pay
in any number of years is one year’s interest multiplied by the number
of years. But if he do not pay the interest at once, but keeps it in
his hands until he returns the principal, he will then have more of
his creditor’s money in his hands every year, and if it were so agreed
will have to pay interest upon each year’s interest for the time during
which he keeps it after it becomes due. In the first case, the interest
is called _simple_, and in the second _compound_. The interest and
principal together are called the _amount_.
249. What is the simple interest of £1049. 16. 6 for 6 years and
one-third, at 4½ per cent? This interest must be 6⅓ times the interest
of the same sum for one year, which (245) is found by multiplying the
sum by 4½, and dividing by 100. The process is as follows:
(230) (_a_) |£1049 . 16 . 6
+--------------
_a_ × 4 | 4199 . 6 . 0
_a_ × ½ | 524 . 18 . 3
+--------------
(82) 100) 47,24 . 4 . 3(£47 . 4 . 10¹¹/₁₀₀
20
----
(228) 4,84[52]
12
------
10,11[53]
(_b_) £47 . 4 . 10¹¹/₁₀₀ Int. for one yr.
+------------------
_b_ × 6 | 283 . 9 . 0⁶⁶/₁₀₀
_b_ × ⅓ | 15 . 14 . 11³⁷/₁₀₀
+---------------------
£299 . 4 . 0³/₁₀₀ Int. for 6⅓ yrs.
[52] Here the 4_s._ from the dividend is taken in.
[53] Here the 3_d._ from the dividend is taken in.
EXERCISES.
What is the interest of £105. 6. 2 for 19 years and 7 weeks at 3 per
cent?
_Answer_, £60. 9, very nearly.
What is the difference between the interest of £50. 19 for 7 years at 3
per cent, and for 8 years at 2½ per cent? _Answer_, 10_s._ (2½)_d._
What is the interest of £157. 17. 6 for one year at 5 per cent?
_Answer_, £7. 17. 10½.
Shew that the interest of any sum for 9 years at 4 per cent is the same
as that of the same sum for 4 years at 9 per cent.
250. In order to find the interest of any sum at compound interest,
it is necessary to find the amount of the principal and interest at
the end of every year; because in this case (248) it is the amount of
both principal and interest at the end of the first year, upon which
interest accumulates during the second year. Suppose, for example, it
is required to find the interest, for 3 years, on £100, at 5 per cent,
compound interest. The following is the process:
£100 First principal.
5 First year’s interest.
---
105 Amount at the end of the first year.
(249) 5 . 5 Interest for the second year on £105.
--------
110 . 5 Amount at the end of two years.
5 . 10 . 3 Interest due for the third year.
------------
115 . 15 . 3 Amount at the end of three years.
100 . 0 . 0 First principal.
------------
15 . 15 . 3 Interest gained in the three years.
When the number of years is great, and the sum considerable, this
process is very troublesome; on which account tables[54] are
constructed to shew the amount of one pound, for different numbers of
years, at different rates of interest. To make use of these tables in
the present example, look into the column headed “5 per cent;” and
opposite to the number 3, in the column headed “Number of years,” is
found 1·157625; meaning that £1 will become £1·157625 in 3 years. Now,
£100 must become 100 times as great; and 1·157625 × 100 is 115·7625
(141); but (221) £·7625 is 15_s._ 3_d._; therefore the whole amount of
£100 is £115. 15. 3, as before.
[54] Sufficient tables for all common purposes are contained in
the article on Interest in the Penny Cyclopædia; and ample ones in
the Treatise on Annuities and Reversions, in the Library of Useful
Knowledge.
251. Suppose that a sum of money has lain at simple interest 4 years,
at 5 per cent, and has, with its interest, amounted to £350; it is
required to find what the sum was at first. Whatever the sum was, if we
suppose it divided into 100 parts, 5 of those parts were added every
year for 4 years, as interest; that is, 20 of those parts have been
added to the first sum to make £350. If, therefore, £350 be divided
into 120 parts, 100 of those parts are the principal which we want to
find, and 20 parts are interest upon it; that is, the principal is
£(350 × 100)/150, or £291. 13. 4.
252. Suppose that A was engaged to pay B £350 at the end of four years
from this time, and that it is agreed between them that the debt shall
be paid immediately; suppose, also, that money can be employed at 5 per
cent, simple interest; it is plain that A ought not to pay the whole
sum, £350, because, if he did, he would lose 4 years’ interest of the
money, and B would gain it. It is fair, therefore, that he should only
pay to B as much as will, _with interest_, amount in four years to
£350, that is (251), £291. 13. 4. Therefore, £58. 6. 8 must be struck
off the debt in consideration of its being paid before the time. This
is called DISCOUNT;[55] and £291. 13. 4 is called the _present value_
of £350 due four years hence, discount being at 5 per cent. The rule
for finding the present value of a sum of money (251) is: Multiply the
sum by 100, and divide the product by 100 increased by the product of
the rate per cent and number of years. If the time that the debt has
yet to run be expressed in years and months, or months only, the months
must be reduced to the equivalent fraction of a year.
[55] This rule is obsolete in business. When a bill, for instance, of
£100 having a year to run, is _discounted_ (as people now say) at 5 per
cent, this means that 5 per cent of £100, or £5, is struck off.
EXERCISES.
What is the discount on a bill of £138. 14. 4, due 2 years hence,
discount being at 4½ per cent?
_Answer_, £11. 9. 1.
What is the present value of £1031. 17, due 6 months hence, interest
being at 3 per cent?
_Answer_, £1016. 12.
253. If we multiply by _a_ + _b_, or by _a_-_b_, when we should
multiply by _a_, the result is wrong by the fraction
_b_ _b_
--- + _b_, or ---------,
_a_ _a_ - _b_
of itself: being too great in the first case, and too small in the
second. Again, if we divide by _a_ + _b_, where we should have divided
by _a_, the result is too small by the fraction _b_/_a_ of itself;
while, if we divide by _a_-_b_ instead of _a_, the result is too great
by the same fraction of itself. Thus, if we divide by 20 instead of
17, the result is ³/₁₇ of itself too small; and if we divide by 360
instead of 365, the result is too great by ⁵/₃₆₅, or ¹/₇₃ of itself.
If, then, we wish to find the interest of a sum of money for a portion
of a year, and have not the assistance of tables, it will be found
convenient to suppose the year to contain only 360 days, in which case
its 73d part (the 72d part will generally do) must be subtracted from
the result, to make the alteration of 360 into 365. The number 360 has
so large a number of divisors, that the rule of Practice (230) may
always be readily applied. Thus, it is required to find the portion
which belongs to 274 days, the yearly interest being £18. 9. 10, or
18·491.
274 18·491
------
180 is ½ of 360 9·246
---
94
90 is ½ of 180 4·623
--
4 is ¹/₉₀ of 360 ·205
------
9)14·074
------
8)1·564
-----
·196
13·878 = £13 . 17 . 7 _Answer._
But if the nearest farthing be wanted, the best way is to take 2-tenths
of the number of days as a multiplier, and 73 as a divisor; since _m_ ÷
365 is 2_m_ ÷ 730, or (²/₁₀)_m_ ÷ 73. Thus, in the preceding instance,
we multiply by 54·8 and divide by 73; and 54·8 × 18·491 = 1013·3068,
which divided by 73 gives 13·881, very nearly agreeing with the former,
and giving £13. 17. 7½, which is certainly within a farthing of the
truth.
254. Suppose it required to divide £100 among three persons in such a
way that their shares may be as 6, 5, and 9; that is, so that for every
£6 which the first has, the second may have £5, and the third £9. It is
plain that if we divide the £100 into 6 + 5 + 9, or 20 parts, the first
must have 6 of those parts, the second 5, and the third 9. Therefore
(245) their shares are respectively,
100 × 6 100 × 5 100 × 9
£-------, £------- and £-------, or £30, £25, and £45.
20 20 20
EXERCISES.
Divide £394. 12 among four persons, so that their shares may be as 1,
6, 7, and 18.--_Answer_, £12. 6. 7½; £73. 19. 9; £86. 6. 4½; £221. 19.
3.
Divide £20 among 6 persons, so that the share of each may be as much
as those of all who come before put together.--_Answer_, The first two
have 12_s._ 6_d._; the third £1. 5; the fourth £2. 10; the fifth £5;
and the sixth £10.
255. When two or more persons employ their money together, and gain
or lose a certain sum, it is evidently not fair that the gain or loss
should be equally divided among them all, unless each contributed the
same sum. Suppose, for example, A contributes twice as much as B, and
they gain £15, A ought to gain twice as much as B; that is, if the
whole gain be divided into 3 parts, A ought to have two of them and B
one, or A should gain £10 and B £5. Suppose that A, B, and C engage in
an adventure, in which A embarks £250, B £130, and C £45. They gain
£1000. How much of it ought each to have? Each one ought to gain as
much for £1 as the others. Now, since there are 250 + 130 + 45, or 425
pounds embarked, which gain £1000, for each pound there is a gain of
£¹⁰⁰⁰/₄₂₄. Therefore A should gain 1000 × ²⁵⁰/₄₂₅ pounds, B should gain
1000 × ¹³⁰/₄₂₅ pounds, and C 1000 × ⁴⁵/₄₂₅ pounds. On these principles,
by the process in (245), the following questions may be answered.
A ship is to be insured, in which A has ventured £1928, and B £4963.
The expense of insurance is £474. 10. 2. How much ought each to pay of
it?
_Answer_, A must pay £132. 15. (2½).
A loss of £149 is to be made good by three persons, A, B, and C. Had
there been a gain, A would have gained 4 times as much as B, and C as
much as A and B together. How much of the loss must each bear?
_Answer_, A pays £59. 12, B £14. 18, and C £74. 10.
256. It may happen that several individuals employ several sums of
money together for different times. In such a case, unless there be
a special agreement to the contrary, it is right that the more time
a sum is employed, the more profit should be made upon it. If, for
example, A and B employ the same sum for the same purpose, but A’s
money is employed twice as long as B’s, A ought to gain twice as much
as B. The principle is, that one pound employed for one month, or one
year, ought to give the same return to each. Suppose, for example, that
A employs £3 for 6 months, B £4 for 7 months, and C £12 for 2 months,
and the gain is £100; how much ought each to have of it? Now, since
A employs £3 for six months, he must gain 6 times as much as if he
employed it one month only; that is, as much as if he employed £6 × 3,
or £18, for one month; also, B gains as much as if he had employed £4 ×
7 for one month; and C as if he had employed £12 × 2 for one month. If,
then, we divide £100 into 6 × 3 + 4 × 7 + 12 × 2, or 70 parts, A must
have 6 × 3, or 18, B must have 4 × 7, or 28, and C 12 × 2, or 24 of
those parts. The shares of the three are, therefore,
6 × 3 × 100 4 × 7 × 100
£----------------------, £----------------------,
6 × 3 + 4 × 7 + 12 × 2 6 × 3 + 4 × 7 + 12 × 2
12 × 2 × 100
and £----------------------.
6 × 3 + 4 × 7 + 12 × 2
EXERCISES.
A, B, and C embark in an undertaking; A placing £3. 6 for 2 years, B
£100 for 1 year, and C £12 for 1½ years. They gain £4276. 7 How much
must each receive of the gain?
_Answer_, A £226. 10. 4; B £3432. 1. 3; C £617. 15. 5.
A, B, and C rent a house together for 2 years, at £150 per annum. A
remains in it the whole time, B 16 months, and C 4½ months, during the
occupancy of B. How much must each pay of the rent?[56]
_Answer_, A should pay £190. 12. 6; B £90. 12. 6; C £18. 15.
[56] This question does not at first appear to fall under the rule. A
little thought will serve to shew that what probably will be the first
idea of the proper method of solution is erroneous.
257. These are the principal rules employed in the application of
arithmetic to commerce. There are others, which, as no one who
understands the principles here laid down can fail to see, are
virtually contained in those which have been given. Such is what is
commonly called the Rule of Exchange, for such questions as the
following: If 20 shillings be worth 25½ francs, in France, what is £160
worth? This may evidently be done by the Rule of Three. The rules here
given are those which are most useful in common life; and the student
who understands them need not fear that any ordinary question will be
above his reach. But no student must imagine that from this or any
other book of arithmetic he will learn precisely the modes of operation
which are best adapted to the wants of the particular kind of business
in which his future life may be passed. There is no such thing as a set
of rules which are at once most convenient for a butcher and a banker’s
clerk, a grocer and an actuary, a farmer and a bill-broker; but a
person with a good knowledge of the _principles_ laid down in this
work, will be able to examine and meet his own future wants, or, at
worst, to catch with readiness the manner in which those who have gone
before him have done so for themselves.
APPENDIX TO THE FIFTH EDITION OF
DE MORGAN’S ELEMENTS OF ARITHMETIC.
I. ON THE MODE OF COMPUTING.
The rules in the preceding work are given in the usual form, and the
examples are worked in the usual manner. But if the student really wish
to become a ready computer, he should strictly follow the methods laid
down in this Appendix; and he may depend upon it that he will thereby
save himself trouble in the end, as well as acquire habits of quick and
accurate calculation.
I. In numeration learn to connect each primary decimal number, 10,
100, 1000, &c. not with the place in which the unit falls, but with
the number of ciphers following. Call ten a _one-cipher_ number, a
hundred a _two-cipher_ number, a million a _six-cipher_ number, and so
on. If _five_ figures be cut off from a number, those that are left
are hundred-thousands; for 100,000 is a _five-cipher_ number. Learn
to connect tens, hundreds, thousands, tens of thousands, hundreds of
thousands, millions, &c. with 1, 2, 3, 4, 5, 6, &c. in the mind. What
is a _seventeen-cipher_ number? For every 6 in seventeen say _million_,
for the remaining 5 say _hundred-thousand_: the answer is a hundred
thousand millions of millions. If twelve places be cut off from the
right of a number, what does the remaining number stand for?--_Answer_,
As many millions of millions as there are units in it when standing by
itself.
II. After learning to count forwards and backwards with rapidity, as
in 1, 2, 3, 4, &c. or 30, 29, 28, 27, &c., learn to count forwards or
backwards by twos, threes, &c. up to nines at least, beginning from
any number. Thus, beginning from four and proceeding by sevens, we
have 4, 11, 18, 25, 32, &c., along which series you must learn to go
as easily as along the series 1, 2, 3, 4, &c.; that is, as quick as
you can pronounce the words. The act of addition must be made in the
mind without assistance: you must not permit yourself to say, 4 and 7
are 11, 11 and 7 are 18, &c.; but only 4, 11, 18, &c. And it would be
desirable, though not so necessary, that you should go back as readily
as forward; by sevens for instance, from sixty, as in 60, 53, 46, 39,
&c.
III. Seeing a number and another both of one figure, learn to catch
instantly the number you must add to the smaller to get the greater.
Seeing 3 and 8, learn by practice to think of 5 without the necessity
of saying 3 _from_ 8 _and there remains_ 5. And if the second number be
the less, as 8 and 3, learn also by practice how to pass _up_ from 8 to
the next number which ends with 3 (or 13), and to catch the necessary
augmentation, _five_, without the necessity of formally undertaking in
words to subtract 8 from 13. Take rows of numbers, such as
4 2 6 0 5 0 1 8 6 4
and practise this rule upon every figure and the next, not permitting
yourself in this simple case ever to name the higher one. Thus, say 4
and 8 (4 first, 2 second, 4 from the next number that ends with 2, or
12, leaves 8), 2 and 4, 6 and 4, 0 and 5, 5 and 5, 0 and 1, 1 and 7, 8
and 8, 6 and 8.
IV. Study the same exercise as the last one with two figures and one.
Thus, seeing 27 and 6, pass from 27 up to the next number that ends
with 6 (or 36), catch the 9 through which you have to pass, and allow
yourself to repeat as much as “27 and 9 are 36.” Thus, the row of
figures 17729638109 will give the following practice: 17 and 0 are 17;
77 and 5 are 82; 72 and 7 are 79; 29 and 7 are 36; 96 and 7 are 103; 63
and 5 are 68; 38 and 3 are 41; 81 and 9 are 90; 10 and 9 are 19.
V. In a number of two figures, practise writing down the units at the
moment that you are keeping the attention fixed upon the tens. In the
preceding exercise, for instance, write down the results, repeating the
tens with emphasis at the instant of writing down the units.
VI. Learn the multiplication table so well as to name the product the
instant the factors are seen; that is, until 8 and 7, or 7 and 8,
suggest 56 at once, without the necessity of saying “7 times 8 are 56.”
Thus looking along a row of numbers, as 39706548, learn to name the
products of every successive pair of digits as fast as you can repeat
them, namely, 27, 63, 0, 0, 30, 20, 32.
VII. Having thoroughly mastered the last exercise, learn further, on
seeing three numbers, to augment the product of the first and second
by the third without any repetition of words. Practise until 3, 8, 4,
for instance, suggest 3 times 8 and 4, or 28, without the necessity of
saying “3 times 8 are 24, and 4 is 28.” Thus, 179236408 will suggest
the following practice, 16, 65, 21, 12, 22, 24, 8.
VIII. Now, carry the last still further, as follows: Seeing four
figures, as 2, 7, 6, 9, catch up the product of the first and second,
increased by the third, as in the last, without a helping word; name
the result, and add the next figure, name the whole result, laying
emphasis upon the tens. Thus, 2, 7, 6, 9, must immediately suggest “20
and 9 are 29.” The row of figures 773698974 will give the instances 52
and 6 are 58; 27 and 9 are 36; 27 and 8 are 35; 62 and 9 are 71; 81 and
7 are 88; 79 and 4 are 83.
IX. Having four numbers, as 2, 4, 7, 9, vary the last exercise as
follows: Catch the product of the first and second, increased by the
third; but instead of adding the fourth, go up to the next number
that ends with the fourth, as in exercise IV. Thus, 2, 4, 7, 9, are
to suggest “15 and 4 are 19.” And the row of figures 1723968929 will
afford the instances 9 and 4 are 13; 17 and 2 are 19; 15 and 1 are 16;
33 and 5 are 38; 62 and 7 are 69; 57 and 5 are 62; 74 and 5 are 79.
X. Learn to find rapidly the number of times a digit is contained
in given units and tens, with the remainder. Thus, seeing 8 and 53,
arrive at and repeat “6 and 5 over.” Common short division is the best
practice. Thus, in dividing 236410792 by 7,
7)236410792
---------
33772970, remainder 2.
All that is repeated should be 3 and 2; 3 and 5; 7 and 5; 7 and 2; 2
and 6; 9 and 4; 7 and 0; 0 and 2.
In performing the several rules, proceed as follows:
ADDITION. Not one word more than repeating the numbers written in the
following process: the accented figure is the one to be written down;
the doubly accented figure is carried (and don’t _say_ “carry 3,” but
do it).
47963
1598
26316
54792
819
6686
------
138174
6, 15, 17, 23, 31, 3″ 4′; 11, 12, 21, 22, 31, 3″7′; 9, 17, 24, 27, 32,
4″1′; 10, 14, 20, 21, 2″8′; 7, 9, 1′3′.
In verifying additions, instead of the usual way of omitting one line,
adding without it, and then adding the line omitted, verify each column
by adding it both upwards and downwards.
SUBTRACTION. The following process is enough. The carriages, being
always of _one_, need not be mentioned.
From 79436258190
Take 58645962738
-----------
20790295452
8 and 2′, 4 and 5′, 7 and 4′, 3 and 5′, 6 and 9′, 10 and 2′, 6 and 0′,
4 and 9′, 7 and 7′, 9 and 0′, 5 and 2′. It is useless to stop and say,
8 and 2 make 10; for as soon as the 2 is obtained, there is no occasion
to remember what it came from.
MULTIPLICATION. The following, put into words, is all that need be
repeated in the multiplying part; the addition is then done as usual.
The unaccented figures are carried.
670383
9876
-------
4022298 18′, 49′, 22′, 2′, 42′, 4′0′,
4692681 21′, 58′, 26′, 2′, 49′, 4′6′,
5363064 24′, 66′, 30′, 3′, 56′, 5′3′,
6033447 27′, 74′, 34′, 3′, 63′, 6′0′.
----------
6620702508
Verify each line of the multiplication and the final result by casting
out the nines. (_Appendix_ II. p. 166.)
It would be almost as easy, for a person who has well practised the 8th
exercise, to add each line to the one before in the process, thus:
670383
9876
-------
4022298
50949108
587255508
6620702508
8; 21 and 9 are 30′; 59 and 2 are 61′; 27 and 2 are 29; 2 and 2 are 4′;
49 and 0 are 49′; 46 and 4 are 5′0′.
On the right is all the process of forming the second line, which
completes the multiplication by 76, as the third line completes that by
876, and the fourth line that by 9876.
DIVISION. Make each multiplication and the following subtraction in one
step, by help of the process in the 9th exercise, as follows:
27693)441972809662(15959730
165042
265778
165410
269459
202226
83756
6772
The number of words by which 26577 is obtained from 165402 (the
multiplier being 5) is as follows: 15 and 7′ are 2″2; 47 and 7′ are
5″4; 35 and 5′ are 4″0; 39 and 6′ are 4″5; 14 and 2′ are 16.
The processes for extracting the square root, and for the solution of
equations (_Appendix_ XI.), should be abbreviated in the same manner as
the division.[57]
[57] The teacher will find further remarks on this subject in the
_Companion to the Almanac_ for 1844, and in the _Supplement to the
Penny Cyclopædia_, article _Computation_.
APPENDIX II.
ON VERIFICATION BY CASTING OUT NINES AND ELEVENS.
The process of _casting out the nines_, as it is called, is one which
the young computer should learn and practise, as a check upon his
computations. It is not a complete check, since if one figure were
made too small, and another as much too great, it would not detect
this double error; but as it is very unlikely that such a double error
should take place, the check furnishes a strong presumption of accuracy.
The proposition upon which this method depends is the following: If _a,
b, c, d_ be four numbers, such that
_a_ = _bc_ + _d_,
and if _m_ be any other number whatsoever, and if _a, b, c, d_,
severally divided by _m_, give the remainders _p, q, r, s_, then
_p_ and _qr_ + _s_
give the same remainder when divided by _m_ (and perhaps are themselves
equal).
For instance, 334 = 17 × 19 + 11;
divide these four numbers by 7, the remainders are 5, 3, 5, and 4. And
5 and 5 × 3 + 4, or 5 and 19, both leave the remainder 5 when divided
by 7.
Any number, therefore, being used as a divisor, may be made a check
upon the correctness of an operation. To provide a check which may be
most fit for use, we must take a divisor the remainder to which is most
easily found. The most convenient divisors are 3, 9, and 11, of which 9
is far the most useful.
As to the numbers 3 and 9, the remainder is always the same as that
of the sum of the digits. For instance, required the remainder of
246120377 divided by 9. The sum of the digits is 2 + 4 + 6 + 1 + 2 + 0
+ 3 + 7 + 7, or 32, which gives the remainder 5. But the easiest way
of proceeding is by throwing out nines as fast as they arise in the
sum. Thus, repeat 2, 6 (2 + 4), 12 (6 + 6), say 3 (throwing out 9),
4, 6, 9 (throw this away), 7, 14, (or throwing out the 9) 5. This is
the remainder required, as would appear by dividing 246120377 by 9. A
proof may be given thus: It is obvious that each of the numbers, 1,
10, 100, 1000, &c. divided by 9, leaves a remainder 1, since they are
1, 9 + 1, 99 + 1, &c. Consequently, 2, 20, 200, &c. leave the remainder
2; 3, 30, 300, the remainder 3; and so on. If, then, we divide, say
1764 by 9 in parcels, 1000 will be one more than an exact number of
nines, 700 will be seven more, and 60 will be six more. So, then, from
1, 7, 6, 4, put together, and the nines taken out, comes the only
remainder which can come from 1764.
To apply this process to a multiplication: It is asserted, in page 32,
that
10004569 × 3163 = 31644451747.
In casting out the nines from the first, all that is necessary to
repeat is, one, five, ten, one, _seven_; in the second, three, four,
ten, one, _four_; in the third, three, four, ten, one, five, nine,
four, nine, eight, twelve, three, ten, _one_. The remainders then are,
7, 4, 1. Now, 7 × 4 is 28, which, casting out the nines, gives 1, the
same as the product.
Again, in page 43, it is asserted that
23796484 = 130000 × 183 + 6484.
Cast out the nines from 13000, 183, 6484, and we have 4, 3, and 4. Now,
4 × 3 + 4, with the nines cast out, gives 7; and so does 23796484.
To avoid having to remember the result of one side of the equation,
or to write it down, in order to confront it with the result of the
other side, proceed as follows: Having got the remainder of the more
complicated side, into which two or more numbers enter, subtract it
from 9, and carry the remainder into the simple side, in which there is
only one number. Then the remainder of that side ought to be 0. Thus,
having got 7 from the left-hand of the preceding, take 2, the rest
of 9, forget 7, and carry in 2 as a beginning to the left-hand side,
giving 2, 4, 7, 14, 5, 11, 2, 6, 14, 5, 9, 0.
Practice will enable the student to cast out nines with great rapidity.
This process of casting out the nines does not detect any errors
in which the remainder to 9 happens to be correct. If a process be
tedious, and some additional check be desirable, the method of casting
out _elevens_ may be followed after that of casting out the nines.
Observe that 10 + 1, 100-1, 1000 + 1, 10000-1, &c. are all divisible by
eleven. From this the following rule for the remainder of division by
11 may be deduced, and readily used by those who know the algebraical
process of subtraction. For those who have not got so far, it may be
doubted whether the rule can be made easier than the actual division by
11.
Subtract the first figure from the second, the result from the third,
the result from the fourth, and so on. The final result, or the rest
of 11 if the figure be negative, is the remainder required. Thus, to
divide 1642915 by 11, and find the remainder, we have 1 from 6, 5; 5
from 4,-1;-1 from 2, 3; 3 from 9, 6; 6 from 1,-5;-5 from 5, 10; and
10 is the remainder. But 164 gives-1, and 10 is the remainder; 164291
gives-5, and 6 is the remainder. With very little practice these
remainders may be read as rapidly as the number itself. Thus, for
127619833424 need only be repeated, 1, 6, 0, 1, 8, 0, 3, 0, 4,-2, 6,
and 6 is the remainder.
When a question has been tried both by nines and elevens, there can be
no error unless it be one which makes the result wrong by a number of
times 99 exactly.
APPENDIX III.
ON SCALES OF NOTATION.
We are so well accustomed to 10, 100, &c., as standing for ten, ten
tens, &c., that we are not apt to remember that there is no reason why
10 might not stand for five, 100 for five fives, &c., or for twelve,
twelve twelves, &c. Because we invent different columns of numbers, and
let units in the different columns stand for collections of the units
in the preceding columns, we are not therefore bound to allow of no
collections except in tens.
If 10 stood for 2, that is, if every column had its unit double of the
unit in the column on the right, what we now represent by 1, 2, 3,
4, 5, 6, &c., would be represented by 1, 10, 11, 100, 101, 110, 111,
1000, 1001, 1010, 1011, 1100, &c. This is the _binary_ scale. If we
take the _ternary_ scale, in which 10 stands for 3, we have 1, 2, 10,
11, 12, 20, 21, 22, 100, 101, 102, 110, &c. In the _quinary_ scale, in
which 10 is five, 234 stands for 2 twenty-fives, 3 fives, and 4, or
sixty-nine. If we take the _duodenary_ scale, in which 10 is twelve, we
must invent new symbols for ten and eleven, because 10 and 11 now stand
for twelve and thirteen; use the letters _t_ and _e_. Then 176 means 1
twelve-twelves, 7 twelves, and 6, or two hundred and thirty-four; and
1_te_ means two hundred and seventy-five.
The number which 10 stands for is called the _radix_ of the _scale of
notation_. To change a number from one scale into another, divide the
number, written as in the first scale, by the number which is to be the
radix of the new scale; repeat this division again and again, and the
remainders are the digits required. For example, what, in the quinary
scale, is that number which, in the decimal scale, is 17036?
5)17036
-----
5)3407 Remʳ. 1
----
5)681 2
---
5)136 1
---
5)27 1
--
5)5 2
-
5)1 0
-
0 1
_Answer_ 1021121
Quinary. Decimal.
_Verification_, 1000000 means 15625
20000 1250
1000 125
100 25
20 10
1 1
------ -----
1021121 17036
The reason of this rule is easy. Our process of division is nothing but
telling off 17036 into 3407 fives and 1 over; we then find 3407 fives
to be 681 fives of fives and 2 _fives_ over. Next we form 681 fives of
fives into 136 fives of fives of fives and 1 five of fives over; and so
on.
It is a useful exercise to multiply and divide numbers represented in
other scales of notation than the common or decimal one. The rules are
in all respects the same for all systems, _the number carried being
always the radix of the system_. Thus, in the quinary system we carry
fives instead of tens. I now give an example of multiplication and
division:
Quinary. Decimal.
42143 means 2798
1234 194
------ -----
324232 11192
232034 25182
134341 2798
42143
--------- ------
114332222 542812
Duodecimal. Decimal.
4_t_9)76_t_4_e_08(16687 705)22610744(32071
4_t_9 1460
----- 5074
2814 1394
2546 689
-----
28_te_
2546
------
3650
3320
-----
3308
2_t_33
------
495
Another way of turning a number from one scale into another is as
follows: Multiply the first digit by the _old_ radix _in the new
scale_, and add the next digit; multiply the result again by the old
radix in the new scale, and take in the next digit, and so on to the
end, always using the radix of the scale you want to leave, and the
notation of the scale you want to end in.
Thus, suppose it required to turn 16687 (duodecimal) into the decimal
scale, and 16432 (septenary) into the quaternary scale:
16687 16432
Duodecimals into Decimals. Septenaries into Quaternaries.
1 × 12 + 6 = 18 1 × 7 + 6 = 31
× 12 + 6 × 7 + 4
--- ----
222 1133
× 12 + 8 × 7 + 3
---- -----
2672 22130
× 12 + 7 × 7 + 2
------ -------
_Answer_ 32071 1021012
Owing to our division of a foot into 12 equal parts, the duodecimal
scale often becomes very convenient. Let the square foot be also
divided into 12 parts, each part is 12 square inches, and the 12th of
the 12th is one square inch. Suppose, now, that the two sides of an
oblong piece of ground are 176 feet 9 inches 7-12ths of an inch, and
65 feet 11 inches 5-12ths of an inch. Using the duodecimal scale, and
_duodecimal fractions_, these numbers are 128·97 and 55·_e_5. Their
product, the number of square feet required, is thus found:
128·97
55·_e_5
---------
617_ee_
116095
617_ee_
617_ee_
------------
68_e_8144_e_
_Answer_, 68_e_8·144_e_ (duod.) square feet, or 11660 square feet 16
square inches ⁴/₁₂ and ¹¹/₁₄₄ of a square inch.
It would, however, be exact enough to allow 2-hundredths of a foot
for every quarter of an inch, an additional hundredth for every 3
inches,[58] and 1-hundredth more if there be a 12th or 2-12ths above
the quarter of an inch. Thus, 9⁷/₁₂ inches should be ·76 + ·03 + ·01,
or ·80, and 11⁵/₁₂ would be ·95; and the preceding might then be found
decimally as 176·8 × 65·95 as 11659·96 square feet, near enough for
every practical purpose.
[58] And at discretion one hundredth more for a large fraction of three
inches.
APPENDIX IV.
ON THE DEFINITION OF FRACTIONS.
The definition of a fraction given in the text shews that ⁷/₉, for
instance, is the _ninth_ part of _seven_, which is shewn to be the
same thing as _seven-ninths_ of a unit. But there are various modes of
speech under which a fraction may be signified, all of which are more
or less in use.
1. In ⁷/₉ we have the 9th part of 7.
2. 7-9ths of a unit.
3. The fraction which 7 is of 9.
4. The times and parts of a time (in this case part of a time only)
which 7 contains 9.
5. The multiplier which turns _nines_ into _sevens_.
6. The _ratio_ of 7 to 9, or the _proportion_ of 7 to 9.
7. The multiplier which alters a number in the ratio of 9 to 7.
8. The 4th proportional to 9, 1, and 7.
The first two views are in the text. The third is deduced thus: If
we divide 9 into 9 equal parts, each is 1, and 7 of the parts are 7;
consequently the fraction which 7 is of 9 is ⁷/₉. The fourth view
follows immediately: For _a time_ is only a word used to express one
of the repetitions which take place in multiplication, and we allow
ourselves, by an easy extension of language, to speak of a portion
of a number as being that number taken a _part of a time_. The fifth
view is nothing more than a change of words: A number reduced to ⁷/₉
of its amount has every 9 converted into a 7, and any fraction of a
9 which may remain over into the corresponding fraction of 7. This
is completely proved when we prove the equation ⁷/₉ of _a_ = 7 times
_a_/9. The sixth, seventh, and eighth views are illustrated in the
chapter on proportion.
When the student comes to algebra, he will find that, in all the
applications of that science, fractions such as _a_/_b_ most frequently
require that _a_ and _b_ should be themselves supposed to be fractions.
It is, therefore, of importance that he should learn to accommodate his
views of a fraction to this more complicated case.
2½
Suppose we take -----.
4³/₅
We shall find that we have, in this case, a better idea of the views
from and after the third inclusive, than of the first and second, which
are certainly the most simple ways of conceiving ⁷/₉. We have no notion
of the (4³/₅)th part of 2½,
1 ( 3 )
nor of 2 ---(4---)ths
2 ( 5 )
of a unit; indeed, we coin a new species of adjective when we talk of
the (4³/₅)th part of anything. But we can readily imagine that 2½ is
some fraction of 4³/₅; that the first is _some_ part of a time the
second; that there must be _some_ multiplier which turns every 4³/₅ in
a number into 2½; and so on. Let us now see whether we can invent a
distinct mode of applying the first and second views to such a compound
fraction as the above.
We can easily imagine a fourth part of a length, and a fifth part,
meaning the lines of which 4 and 5 make up the length in question;
and there is also in existence a length of which four lengths and
two-fifths of a length make up the original length in question. For
instance, we might say that 6, 6, 2 is a division of 14 into 2⅓ equal
parts--2 equal parts, 6, 6, and a third of a part, 2. So we might agree
to say, that the (2⅓)th, or (2⅓)rd, or (2⅓)st (the reader may coin the
adjective as he pleases) part of 14 is 6. If we divide the line A B
into eleven equal parts in C, D, E, &c., we must then say that A C is
the 11th part,
[Illustration]
A D the (5½)th, A E the (3⅔)th, A F the (2¾)th, A G the (2⅕)th, A H
the (1⅚)th, A I the (1⁴/₇)th, A K the (1⅜)th, A L the (1²/₉)th, A M
the (1⅒)th, and A B itself the 1st part of A B. The reader may refuse
the language if he likes (though it is not so much in defiance of
etymology as talking of _multiplying_ by ½); but when A B is called 1,
he must either call A F 1/(2¾), or make one definition of one class
of fractions and another of another. Whatever abbreviations they may
choose, all persons will agree that _a_/_b_ is a direction to find such
a fraction as, repeated _b_ times, will give 1, and then to take that
fraction _a_ times.
So, to get 2½/4⅗, the simplest way is to divide the whole unit into 46
parts; 10 of these parts, repeated 4⅗ times, give the whole. The
[Illustration]
4⅗th is then ¹⁰/₄₆, and 2½ such parts is ²⁵/₄₆, or A C. The student
should try several examples of this mode of interpreting complex
fractions.
But what are we to say when the denominator itself is less than unity,
as in 3¼/⅖? Are we to have a (⅖)th part of a unit? and what is it?
Had there been a 5 in the denominator, we should have taken the part
of which 5 will make a unit. As there is ⅖ in the denominator, we must
take the part of which ⅖ will be a unit. That part is larger than a
unit; it is 2½ units; 2½ is that of which ⅖ is 1. The above fraction
then directs us to repeat 2½ units 3¼ times. By extending our word
‘multiplication’ to the taking of a part of a time, all multiplications
are also divisions, and all divisions multiplications, and all the
terms connected with either are subject to be applied to the results of
the other.
If 2⅓ yards cost 3½ shillings, how much does one yard cost? In such a
case as this, the student looks at a more simple question. If 5 yards
cost 10 shillings, he sees that each yard costs ¹⁰/₅, or 2 shillings,
and, concluding that the same process will give the true result when
the data are fractional, he forms 3½/2⅓, reduces it by rules to ³/₂
or 1½, and concludes that 1 yard costs 18 pence. The answer happens
to be correct; but he is not to suppose that this rule of copying for
fractions whatever is seen to be true of integers is one which requires
no demonstration. In the above question we want money which, repeated
2⅓ times, shall give 3½ shillings. If we divide the shilling into 14
equal parts, 6 of these parts repeated 2⅓ times give the shilling. To
get 3½ times as much by the same repetition, we must take 3½ of these 6
parts at each step, or 21 parts. Hence, ²¹/₁₄, or 1½, is the number of
shillings in the price.
APPENDIX V.
ON CHARACTERISTICS.
When the student comes to use logarithms, he will find what follows
very useful. In the mean while, I give it merely as furnishing a rapid
rule for finding the place of a decimal point in the quotient before
the division is commenced.
When a bar is written over a number, thus, 7︤ let the number be called
negative, and let it be thus used: Let it be augmented by additions of
its own species, and diminished by subtractions; thus, 7︤ and 2︤ give
9︤, and let 7︤ with 2︤ subtracted give 5︤. But let the _addition_
of a number without the bar _diminish_ the negative number, and the
_subtraction increase_ it. Thus, 7︤ and 4 are 3︤, 7︤ and 12 make 5, 7︤
with 8 subtracted is 1︦5. In fact, consider 1, 2, 3, &c., as if they
were gains, and 1︤, 2︤, 3︤, as if they were losses: let the addition
of a gain or the removal of a loss be equivalent things, and also the
removal of a gain and the addition of a loss. Thus, when we say that
4︤ diminished by 1︦1 gives 7, we say that a loss of 4 incurred at the
moment when a loss of 11 is removed, is, on the whole, equivalent to a
gain of 7; and saying that 4︤ diminished by 2 is 6︤, we say that a loss
of 4, accompanied by the removal of a gain of 2, is altogether a loss
of 6.
By the _characteristic_ of a number understand as follows: When there
are places before the decimal point, it is one less than the number
of such places. Thus, 3·214, 1·0083, 8 (which is 8·00 ...) 9·999, all
have 0 for their characteristics. But 17·32, 48, 93·116, all have 1;
126·03 and 126 have 2; 11937264·666 has 7. But when there are no places
before the decimal point, look at the first decimal place which is
significant, and make the characteristic negative accordingly. Thus,
·612, ·121, ·9004, in all of which significance begins in the first
decimal place, have the characteristic 1︤; but ·018 and ·099 have 2︤;
·00017 has 4︤; ·000000001 has 9︤.
To find the characteristic of a quotient, subtract the characteristic
of the divisor from that of the dividend, carrying one before
subtraction if the first significant figures of the divisor are greater
than those of the dividend. For instance, in dividing 146·08 by ·00279.
The characteristics are 2 and 3︤; and 2 with 3︤ removed would be 5. But
on looking, we see that the first significant figures of the divisor,
27, taken by themselves, and without reference to their local value,
mean a larger number than 14, the first two figures of the dividend.
Consequently, to 3︤ we carry 1 before subtracting, and it then becomes
2︤, which, taken from 2, gives 4. And this 4 is the characteristic of
the quotient, so that the quotient has 5 places before the decimal
point. Or, if _abcdef_ be the first figures of the quotient, the
decimal point must be thus placed, _abcde·f_. But if it had been to
divide ·00279 by 146·08, no carriage would have been required; and 3︤
diminished by 2 is 5︤; that is, the first significant figure of the
quotient is in the 5th place. The quotient, then, has ·0000 before
any significant figure. A few applications of this rule will make it
easy to do it in the head, and thus to assign the meaning of the first
figure of the quotient even before it is found.
APPENDIX VI.
ON DECIMAL MONEY.
Of all the simplifications of commercial arithmetic, none is comparable
to that of expressing shillings, pence, and farthings as decimals of a
pound. The rules are thereby put almost upon as good a footing as if
the country possessed the advantage of a real decimal coinage.
Any fraction of a pound sterling may be decimalised by rules which can
be made to give the result at once.
Two shillings is £·100 |
One shilling is £·050 |
Sixpence is £·025 |
One farthing is £·001 | 04⅙
Thus, every pair of shillings is a unit in the first decimal place;
an odd shilling is a 50 in the second and third places; a farthing is
so nearly the thousandth part of a pound, that to say one farthing is
·001, two farthings is ·002, &c., is so near the truth that it makes
no error in the first three decimals till we arrive at sixpence, and
then 24 farthings is exactly ·025 or 25 thousandths. But 25 farthings
is ·026, 26 farthings is ·027, &c. Hence the rule for the _first three
places_ is
_One in the first for every pair of shillings; 50 in the second
and third for the odd shilling, if any; and 1 for every farthing
additional, with 1 extra for sixpence._
Thus, 0_s._ 3½_d._ = £·014
0_s._ 7¾_d._ = £·032
1_s._ 2½_d._ = £·060
1_s._ 11¼_d._ = £·096
2_s._ 6_d._ = £·125
2_s._ 9½_d._ = £·139
3_s._ 2¾_d._ = £·161
13_s._ 10¾_d._ = £·694
In the fourth and fifth places, and those which follow, it is obvious
that we have no produce from any farthings except those above sixpence.
For at every sixpence, ·00004⅙ is converted into ·001, and this has
been already accounted for. Consequently, to fill up the _fourth and
fifth_ places,
_Take 4 for every farthing[59] above the last sixpence, and an
additional 1 for every six farthings, or three halfpence._
[59] The student should remember all the multiples of 4 up to 4 × 25,
or 100.
The remaining places arise altogether from ·00000⅙ for every farthing
above the last three halfpence; for at every three halfpence complete,
·00000⅙ is converted into ·00001, and has been already accounted for.
Consequently, to fill up _all the places after the fifth_,
_Let the number of farthings above the last three halfpence be a
numerator, 6 a denominator, and annex the figures of the corresponding
decimal fraction._
It may be easily remembered that
The figures of ¹/₆ are 166666...
” ²/₆ ... 333333...
” ³/₆ ... 5
” ⁴/₆ ... 666666...
” ⁵/₆ ... 833333...
0_s._ 3½_d._ = ·014|58|3333...
0_s._ 7¾_d._ = ·032|29|1666...
1_s._ 2½_d._ = ·060|41|6666...
1_s._ 11¼_d._ = ·096|87|5
2_s._ 6_d._ = ·125|00|0000...
2_s._ 9½_d._ = ·139|58|3333...
3_s._ 2¾_d._ = ·161|45|83333...
13_s._ 10¾_d._ = ·694|79|1666...
The following examples will shew the use of this rule, if the student
will also work them in the common way.
To turn pounds, &c., into farthings: Multiply the pounds by 960, or
by 1000-40, or by 1000(1-⁴/₁₀₀); that is, from 1000 times the pounds
subtract 4 per cent of itself. Thus, required the number of farthings
in £1663. 11. 9¾.
1663·590625 × 1000 = 1663590·625
4 per cent of this, 66543·625
-----------
No. of farthings required, 1597047
What is 47½ per cent of £166. 13. 10 and ·6148 of £2971. 16. 9?
166·691
--------
40 p. c. 66·6764
5 p. c. 8·3346
2½ p. c. 4·1673
--------
79·1783
£79.3.6¾
2971·837
---------
·6 1783·1022
·01 29·7184
·004 11·8873
·0008 2·3775
---------
1827·0854
£1827.1.8½
The inverse rule for turning the decimal of a pound into shillings,
pence, and farthings, is obviously as follows:
_A pair of shillings for every unit in the first place; an odd shilling
for 50 (if there be 50) in the second and third places; and a farthing
for every thousandth left, after abating 1 if the number of thousandths
so left exceed 24._
The direct rule (with three places) gives too little, the inverse rule
too much, except at the end of a sixpence, when both are accurate.
Thus, £·183 is rather less than 3_s._ 8_d._, and 6_s._ 4¾_d._ is rather
greater than £319; or when the two do not exactly agree, the _common
money is the greatest_. But £·125 and £·35 are exactly 2_s._ 6_d._ and
7_s._
Required the price of 17 cwt. 81 lb. 13½ oz. at £3.11.9¾ per cwt. true
to the hundredth of a farthing.
3·590625
17
---------
61·040625
lb. 56 ½ 1·795313
16 ⅐ ·512946
7 ⅛ ·224414
2 ⅛ ·064118
oz. 8 ¼ ·016029
4 ½ ·008015
1 ¼ ·002004
½ ½ ·001002
---------
£63·664466
£63.13.3½
Three men, A, B, C, severally invest £191.12.7¾, £61.14.8, and
£122.1.9½ in an adventure which yields £511.12.6½. How ought the
proceeds to be divided among them?
A, 191·63229
B, 61·73333
C, 122·08958 Produce of £1.
---------
375·45520)511·62708(1·362686
136·17188
23·53532
1·00801
25710
3183
180
1·362686 1·362686 1·362686
92·236191 33·33716 85·980221
--------- --------- ---------
1·362686 8·17612 1·362686
1·226417 13627 272537
13627 9538 27254
8176 409 1090
409 41 122
27 4 7
3 -------- 1
1 8·41231 ---------
--------- 1·663697
2·611346
261·1346 ... A’s share £261.2.8¼
84·1231 ... B’s ... 84.2.5¾
166·3697 ... C’s ... 166.7.4¾
-------- -------------
511·6274 £511.2.6¾
If ever the fraction of a farthing be wanted, remember that the
_coinage_-result is larger than the decimal of a pound, when we use
only three places. From 1000 times the decimal take 4 per cent, and we
get the exact number of farthings, and we need only look at the decimal
then left to set the preceding right. Thus, in
134·6 123·1 369·7
5·38 4·92 14·79
----- ------ ------
·22 ·18 ·91
we see that (if we use four decimals only) the pence of the above
results are nearly 8_d._ ·22 of a farthing, 5½_d._ ·18, and 4½_d._ ·91.
A man can pay £2376. 4. 4½, his debts being £3293. 11. 0¾. How much per
cent can he pay, and how much in the pound?
3293·553)2376·2180(·7214756
70·7309
4·8598
1·5662
2488
183
18
_Answer_, £72. 2.11½ per cent.
0.14.5¼ per pound.
APPENDIX VII.
ON THE MAIN PRINCIPLE OF BOOK-KEEPING.
A brief notice of the principle on which accounts are kept (when they
are _properly_ kept) may perhaps be useful to students who are learning
book-keeping, as the treatises on that subject frequently give too
little in the way of explanation.
Any person who is engaged in business must desire to know accurately,
whenever an investigation of the state of his affairs is made.
1, What he had at the commencement of the account, or immediately after
the last investigation was made; 2, What he has gained and lost in the
interval in all the several branches of his business; 3, What he is now
worth. From the first two of these things he obviously knows the third.
In the interval between two investigations, he may at any one time
desire to know how any one account stands.
An _account_ is a recital of all that has happened, in reference to any
class of dealings, since the last investigation. It can only consist
of receipts and expenditures, and so it is said to have two sides, a
_debtor_ and a _creditor_ side.
All accounts are kept in _money_. If goods be bought, they are
estimated by the money paid for them. If a debtor give a bill of
exchange, being a promise to pay a certain sum at a certain time, it is
put down as worth that sum of money. All the tools, furniture, horses,
&c. used in the business are rated at their value in money. All the
actual coin, bank-notes, &c., which are in or come in, being the only
money in the books which really is money, is called _cash_.
The accounts are kept as if every different sort of account belonged
to a separate person, and had an interest of its own, which every
transaction either promotes or injures. If the student find that it
helps him, he may imagine a clerk to every account: one to take charge
of, and regulate, the actual cash; another for the bills which the
house is to receive when due; another for those which it is to pay when
due; another for the cloth (if the concern deal in cloth); another
for the sugar (if it deal in sugar); one for every person who has an
account with the house; one for the profits and losses; and so on.
All these clerks (or accounts) belonging to one merchant, must account
to him in the end--must either produce all they have taken in charge,
or relieve themselves by shewing to whom it went. For all that they
have received, for every responsibility they have undertaken to _the
concern itself_, they are bound, or are _debtors_; for everything
which has passed out of charge, or about which they are relieved from
answering _to the concern_, they are unbound, or are _creditors_.
These words must be taken in a very wide sense by any one to whom
book-keeping is not to be a mystery. Thus, whenever any account assumes
responsibility to any parties _out of the concern_, it must be creditor
in the books, and debtor whenever it discharges any other parties of
their responsibility. But whenever an account removes responsibility
from any other account in the same books it is debtor, and creditor
whenever it imposes the same.
To whom are all these parties, or accounts, bound, and from whom are
they released? Undoubtedly the merchant himself, or, more properly,
the _balance-clerk_, presently mentioned. But it is customary to say
that the accounts are debtors _to_ each other, and creditors _by_
each other. Thus, cash _debtor_ to bills receivable, means that the
cash account (or the clerk who keeps it) is bound to answer for a sum
which was paid on a bill of exchange due to the house. At full length
it would be: “Mr. C (who keeps the cash-box) has received, and is
answerable for, this sum which has been paid in by Mr. A, when he paid
his bill of exchange.” On the other hand, the corresponding entry in
the account of bills receivable runs--bills receivable, _creditor_
by cash. At full length: “Mr. B (who keeps the bills receivable) is
freed from all responsibility for Mr. A’s bill, which he once held, by
handing over to Mr. C, the cash-clerk, the money with which Mr. A took
it up.” Bills receivable creditor _by_ cash is intelligible, but cash
debtor _to_ bills receivable is a misnomer. The cash account is debtor
_to the merchant by_ the sum received for the bill, and it should be
cash debtor _by_ bill receivable. The fiction of debts, not one of
which is ever paid to the party _to_ whom it is said to be owing,
though of no consequence in practice, is a stumbling-block to the
learner; but he must keep the phrase, and remember its true meaning.
The account which is made _debtor_, or bound, is said to be _debited_;
that which is made _creditor_, or released, is said to be _credited_.
All who receive must be _debited_; all who give must be _credited_.
No cancel is ever made. If cash received be afterwards repaid, the
sum paid is not struck off the receipts (or debtor-side of the cash
account), but a discharge, or credit, is written on the expenditure (or
credit) side.
The book in which the accounts are kept is called a _ledger_. It
has double columns, or else the debtor-side is on one page, and the
creditor side on the opposite, of each account. The debtor-side is
always the left. Other books are used, but they are only to help in
keeping the ledger correct. Thus there may be a _waste-book_, in which
all transactions are entered as they occur, in common language; a
_journal_, in which the transactions described in the waste-book are
entered at stated periods, in the language of the ledger. The items
entered in the journal have references to the pages of the ledger
to which they are carried, and the items in the ledger have also
references to the pages of the journal from which they come; and by
this mode of reference it is easy to make a great deal of abbreviation
in the ledger. Thus, when it happens, in making up the journal to a
certain date, that several different sums were paid or received at or
near the same time, the totals may be entered in the ledger, and the
cash account may be made debtor to, or creditor by, sundry accounts,
or sundries; the sundry accounts being severally credited or debited
for their shares of the whole. The only book that need be explained is
the ledger. All the other books, and the manner in which they are kept,
important as they may be, have nothing to do with the main principle
of the method. Let us, then, suppose that all the items are entered
at once in the ledger as they arise. It has appeared that every item
is entered twice. If A pay on account of B, there is an entry, “A,
creditor by B;” and another, “B, debtor to A.” This is what is called
_double-entry_; and the consequence of it is, that the sum of all the
debtor items in the whole book is equal to the sum of all the creditor
items. For what is the first set but the second with the items in a
different order? If it were convenient, one entry of each sum might
be made a double-entry. The multiplication table is called a table of
_double-entry_, because 42, for instance, though it occurs only once,
appears in two different aspects, namely, as 6 times 7 and as 7 times
6. Suppose, for example, that there are five accounts, A, B, C, D, E,
and that each account has one transaction of its own with every other
account; and let the debits be in the _columns_, the credits in the
_rows_, as follows:
-------------+------+------+------+------+------+
Debtor | A | B | C | D | E |
-------------+------+------+------+------+------+
A, Creditor | | 23 | 19 | 32 | 4 |
+------+------+------+------+------+
B, Creditor | 17 | | 6 | 11 | 25 |
+------+------+------+------+------+
C, Creditor | 9 | 41 | | 10 | 2 |
+------+------+------+------+------+
D, Creditor | 14 | 28 | 16 | | 3 |
+------+------+------+------+------+
E, Creditor | 15 | 4 | 60 | 1 | |
+------+------+------+------+------+
Here the 16 is supposed to appear in D’s account as D creditor by C,
and in C’s account as C debtor to D. And to say that the sum of debtor
items is the same as that of creditor items, is merely to say that the
preceding numbers give the same sum, whether the rows or the columns be
first added up.
If it be desired to close the ledger when it stands as above, the
following is the way the accounts will stand: the lines in italics will
presently be explained.
A, Debtor. | A, Creditor.| B, Debtor. | B, Creditor.
To B 17 | By B 23 | To A 23 | By A 17
To C 9 | By C 19 | To C 41 | By C 6
To D 14 | By D 32 | To D 28 | By D 11
To E 15 | By E 4 | To E 4 | By E 25
To Balance 23 | | | By Balance 37
-- | -- | -- | --
78 | 78 | 96 | 96
--------------+------------------+---------------+---------------+
C, Debtor. | C, Creditor. | D, Debtor. | D, Creditor.
To A 19 | By A 9 | To A 32 | By A 14
To B 6 | By B 41 | To B 11 | By B 28
To D 16 | By D 10 | To C 10 | By C 16
To E 60 | By E 2 | To E 1 | By E 3
| By Balance 39 | To Balance 7 |
--- | --- | -- | --
101 | 101 | 61 | 61
--------------+------------------+-----------------+---------------
E, Debtor. | E, Creditor. | Balance, Debtor.| Balance, Cred.
To A 4 | By A 15 | To B 37 | By A 23
To B 25 | By B 4 | To C 39 | By D 7
To C 2 | By C 60 | | By E 46
| | -- | --
To D 3 | By D 1 | 76 | 76
To Balance 46 | | |
-- | -- | |
80 | 80 | |
In all the part of the above which is printed in Roman letters we see
nothing but the preceding table repeated. But when all the accounts
have been completed, and no more entries are left to be made, there
remains the last process, which is termed _balancing the ledger_.
To get an idea of this, suppose a new clerk, who goes round all the
accounts, collecting debts and credits, and taking them all upon
himself, that he alone may be entitled to claim the debts and to be
responsible for the assets of the concern. To this new clerk, whom I
will call the _balance-clerk_, every account gives up what it has,
whether the same be debt or credit. The cash-clerk gives up all the
cash; the clerks of the two kinds of bills give up all their documents,
whether bills receivable or entries of bills payable (remember that
any entry against which there is money set down in the books counts as
money when given up, that is, as money due or money owing); the clerks
of the several accounts of goods give up all their unsold remainders
at cost prices; the clerks of the several personal accounts give up
vouchers for the sums owing to or from the several parties; and so on.
But where more has been paid out than received, the balance-clerk
adjusts these accounts by giving instead of receiving; in fact, he so
acts as to make the debtor and creditor sides of the accounts he visits
equal in amount. For instance, the A account is indebted to the concern
55, while payments or discharges to the amount of 78 have been made by
it. The balance-clerk accordingly hands over 23 to that account, for
which it becomes debtor, while the balance enters itself as creditor to
the same amount. But in the B account there is 96 of receipt, and only
59 of payment or discharge. The balance-clerk then receives 37 from
this account, which is therefore credited by balance, while the balance
acknowledges as much of debt. The balance account must, of course,
exactly balance itself, if the accounts be all right; for of all the
equal and opposite entries of which the ledger consists, so far as
they do not balance one another, one goes into one side of the balance
account, and the other into the other. Thus the balance account becomes
a test of the accuracy of one part of the work: if its two sides do not
give the same sums, either there have been entries which have not had
their corresponding balancing entries correctly made, or else there has
been error in the additions.
But since the balance account must thus always give the _same sum_ on
both sides, and since _balance debtor_ implies what is favourable to
the concern, and _balance creditor_ what is unfavourable, does it not
appear as if this system could only be applied to cases in which there
is neither loss nor gain? This brings us to the two accounts in which
are entered all that the concern _began with_, and all that it _gains
or loses_--the _stock account_, and the _profit-and-loss account_.
In order to make all that there was to begin with a matter of double
entry, the opening of the ledger supposes the merchant himself to put
his several clerks in charge of their several departments. In the stock
account, _stock_, which here stands for the owner of the books, is made
creditor by all the property, and debtor by all the liabilities; while
the several accounts are made debtors for all they take from the stock,
and creditors by all the responsibilities they undertake. Suppose, for
instance, there are £500 in cash at the commencement of the ledger.
There will then appear that the merchant has handed over to the
cash-box £500, and in the stock account will appear, “Stock creditor
by cash, £500;” while in the cash account will appear, “Cash debtor to
stock, £500.” Suppose that at the beginning there is a debt outstanding
of £50 to Smith and Co., then there will appear in the stock account,
“Stock debtor to Smith and Co. £50,” and in Smith and Co.’s account
will appear “Smith and Co. creditors by stock, £50.” Thus there is
double entry for all that the concern begins with by this contrivance
of the stock account.
The account to which everything is placed for which an actual
equivalent is not seen in the books is the _profit-and-loss_ account.
This profit-and-loss account, or the clerk who keeps it, is made
answerable for every loss, and the supposed cause of every gain. This
account, then, becomes debtor for every loss, and creditor by every
gain. If goods be damaged to the amount of £20 by accident, and a loss
to that amount occur in their sale, say they cost £80 and sell for
£60 cash, it is clear that there is an entry “Cash debtor to goods
£60,” and “Goods creditor by cash £60.” Now, there is an entry of
£80 somewhere to the debit of the goods for cash laid out, or bills
given, for the whole of the goods. It would affect the accuracy of
the accounts to take no notice of this; for when the balance-clerk
comes to adjust this account, he would find he receives £20 less than
he might have reckoned upon, without any explanation of the reason;
and there would be a failure of the principle of double-entry. Since
it is convenient that the balance account of the goods should merely
represent the stock in hand at the close, the account of goods
therefore lays the responsibility of £20 upon the profit-and-loss
account, or there is the entry “Goods creditor by profit-and-loss,
£20,” and also “Profit-and-loss debtor to goods, £20.” Again, in all
payments which are not to bring in a specific return, such as house
and trade expenses, wages, &c. these several accounts are supposed to
adjust matters with the profit-and-loss account before the balance
begins. Thus, suppose the outgoings from the mere premises occupied
exceed anything those premises yield by £200, or the debits of the
house account exceed its credits by £200, the account should be
balanced by transferring the responsibility to the profit-and-loss
account, under the entries “House expenses creditor by profit-and-loss,
£200”, “Profit-and-loss debtor to house expenses, £200.” In this way
the profit-and-loss account steps in from time to time before the
balance account commences its operations, in order that that same
balance account may consist of _nothing but the necessary matters of
account for the next year’s ledger_.
This _transference of accounts_, or transfusion of one account into
another, requires attentive consideration. The receiving account
becomes creditor for the credits, and debtor for the debits, of the
transmitting account. The rule, therefore, is: Make the transmitting
account balance itself, and, on whichever side it is necessary to enter
a balancing sum, make the account debtor or creditor, as the case may
be, to the receiving account, and the latter creditor or debtor to the
former. Thus, suppose account A is to be transferred to account B, and
the latter is to arrange with the balance account. If the two stand as
in Roman letters, the processes in Italic letters will occur before the
final close.
A, Debtor. | A, Creditor. | B, Debtor. | B, Creditor.
| | |
To sundries £100|By sundries £500|To sundries £600|By sundries £400
To B . . . 400| |To Balance 200|By A . . . 400
----| ----| ----| ----
£500| £500| £800| £800
And the entry in the balance account will be, “Creditor by B, £200,”
shewing that, on these two accounts, the credits exceed the debits by
£200.
Still, before the balance account is made up, it is desirable that the
profit-and-loss account should be transferred to the stock account;
for the profit and loss of this year is of no moment as a part of
next year’s ledger, except in so far as it affects the stock at the
commencement of the latter. Let this be done, and the balance account
may then be made in the form required.
The stock account and the profit-and-loss account, the latter being
the only direct channel of alteration for the former, differ in a
peculiar manner[60] from the other preliminary accounts, and the
balance account is a species of umpire. They represent the merchant:
their interests are his interests; he is solvent upon the excess of
their credits over their debits, insolvent upon the excess of their
debits over their credits. It is exactly the reverse in all the other
accounts. If a malicious person were to get at the ledger, and put on
a cipher to the pounds in various items, with a view of making the
concern appear worse than it really is, he would make his alterations
on the _debtor_ sides of the stock and profit-and-loss accounts, and
on the _creditor_ sides of all the others. Accordingly, in the balance
account, the net stock, after the incorporation of the profit-and-loss
account, appears on the _creditor_ side (if not, it should be called
amount of _insolvency_, not _stock_), and the debts of the concern
appear on the same side. But on the debit side of the balance account
appear all the assets of the concern (for which the balance-clerk is
debtor to the clerks from whom he has taken them).
[60] The treatises on book-keeping have described this difference in as
peculiar a manner. They call these accounts the _fictitious accounts_.
Now they represent the merchant himself; their credits are gain to the
business, their debits losses or liabilities. If the terms real and
fictitious are to be used at all, they are the _real_ accounts, end all
the others are as _fictitious_ as the clerks whom we have supposed to
keep them.
The young student must endeavour to get the enlarged view of the words
debtor and creditor which is requisite, and must then learn by practice
(for nothing else will give it) facility in allotting the actual
entries in the waste-book to the proper sides of the proper accounts.
I do not here pretend to give more than such a view of the subject as
may assist him in studying a treatise on book-keeping, which he will
probably find to contain little more than examples.
APPENDIX VIII.
ON THE REDUCTION OF FRACTIONS TO OTHERS OF NEARLY EQUAL VALUE.
There is a useful method of finding fractions which shall be nearly
equal to a given fraction, and with which the computer ought to be
acquainted. Proceed as in the rule for finding the greatest common
measure of the numerator and denominator, and bring all the quotients
into a line. Then write down,
1 2nd Quot.
-------- ------------------------
1st Quot. 1st Quot. × 2d Quot. + 1
Then take the third quotient, multiply the numerator and denominator of
the second by it, and add to the products the preceding numerator and
denominator. Form a third fraction with the results for a numerator and
denominator. Then take the fourth quotient, and proceed with the third
and second fractions in the same way; and so on till the quotients are
exhausted. For example, let the fraction be ⁹¹³¹/₁₃₁₂₈.
9131)13128(1, 2
1137 3997(3, 1
551 586(1, 15
201 35(1, 2
26 9(1, 8
8 1
This is the process for finding the greatest common measure of 9131 and
13128 in its most compact form, and the quotients and fractions are:
1 2 3 1 1 15 1 2 1 8
1 2 7 9 16 249 265 779 1044 9131
--- --- --- --- ---- ----- ----- ----- ------ ------
1 3 10 13 23 358 381 1120 1501 13128
It will be seen that we have thus a set of fractions ending with the
original fraction itself, and formed by the above rule, as follows:
1 1
1st Fraction = -------- = ---
1st Quot. 1
2d Quot. 2
2d Fraction = ------------------------ = ---
1st Quot. × 2d Quot. + 1 3
2d Numʳ. × 3d Quot. + 1st Numʳ. 2 × 3 + 1 7
3d Fraction = ------------------------------- = --------- = ---
2d Denʳ. × 3d Quot. + 1st Denʳ. 3 × 3 + 1 10
3d Numʳ. × 4th Quot. + 2d Numʳ. 7 × 1 + 2 9
4th Fraction = ------------------------------- = --------- = ---;
3d Denʳ. × 4th Quot. + 2d Denʳ. 10 × 1 + 3 13
and so on. But we have done something more than merely reascend to the
original fraction by means of the quotients. The set of fractions,
¹/₁, ²/₃, ⁷/₁₀, ⁹/₁₃, &c. are continually approaching in value to the
original fraction, the first being too great, the second too small, the
third too great, and so on alternately, but each one being nearer to
the given fraction than any of those before it. Thus, ¹/₁ is too great,
and ²/₃ is too small; but ²/₃ is not so much too small as ¹/₁ is too
great. And again, ⁷/₁₀, though too great, is not so much too great as
²/₃ is too small.
Moreover, the difference of any of the fractions from the original
fraction is never greater than a fraction having unity for its
numerator and the product of the denominator and the next denominator
for its denominator. Thus, ¹/₁ does not err by so much as ¹/₃, nor ²/₃
by so much as ¹/₃₀, nor ⁷/₁₀ by so much as ¹/₁₃₀, nor ⁹/₁₃ by so much
as ¹/₂₉₉, &c.
Lastly, no fraction of a less numerator and denominator can come
so near to the given fraction as any one of the fractions in the
list. Thus, no fraction with a less numerator than 249, and a less
denominator than 358, can come so near to
9131 249
----- as ---.
13128 358
The reader may take any example for himself, and the test of the
accuracy of the process is the ultimate return to the fraction begun
with. Another test is as follows: The numerator of the difference of
any two consecutive approximating fractions ought to be unity. Thus,
in our instance, we have ¹⁶/₂₃ and ²⁴⁹/₃₅₈, which, with a common
denominator, 23 × 358, have 5728 and 5727 for their numerators.
As another example, let us examine this question: The length of the
year is 365·24224 days, which is called in common life 365¼ days. Take
the fraction ²⁴²²⁴/₁₀₀₀₀₀, and proceed as in the rule.
24224)100000(4, 7, 1, 4, 9, 2
2496 3104
64 608
0 32
1 7 8 39 359 757
--- --- --- ---- ---- ----
4 29 33 161 1482 3125
and ⁷⁵⁷/₃₁₂₅ is ·24224 in its lowest terms. Hence, it appears that the
excess of the year over 365 days amounts to about 1 day in 4 years,
which is not wrong by so much as 1 day in 116 years; more accurately,
to 7 days in 29 years, which is not wrong by so much as 1 day in 957
years; more accurately still, to 8 days in 33 years, which is not wrong
by so much as 1 day in 5313 years; and so on.
This method may be applied to finding fractions nearly equal to the
square roots of integers, in the following manner:
__
√43 = 6 + ...
6 | 1 5 4 5 5 4 5 1 6 6 |1 5 4, &c.
1 | 7 6 3 9 2 9 3 6 7 1 |7 6 3, &c.
--+----------------------+------
6 | 1 1 3 1 5 1 3 1 1 1 2|1 1 3, &c.
Set down the number whose square root is wanted, say 43. This square
root is 6 and a fraction. Set down the integer 6 in the first and third
row, and 1 in the second row always. Form the successive rows each from
the one before, in the following manner:
One row The next row has _b′_, _a′_, _c′_, formed in this order,
being thus,
_a_ _a′_ = excess of _b′c′_, already formed, over _a_.
_b_ _b′_ = quotient of 43 - _a_² divided by _b_.
_c_ _c′_ = integer in the quotient of 6 + _a_ divided by _b′_.
Thus the second row is formed from the first, as under:
6|1 = excess of 7 × 1 (both just found) over 6.
1|7 = 43 - 6 × 6 divided by 1.
--+--
6|1 = integer of 6 + 6 divided by 7 (just found).
The third row is formed from the second, thus:
1 5 = excess of 1 × 6 over 1.
7 6 = 43 - 1 × 1 divided by 7.
1 1 = integer of 6 + 1 divided by 6;
and so on. In process of time the second column, 1, 7, 1, occurs again,
after which the several columns are repeated in the same order. As a
final process, take the set in the lowest line (excluding the first,
6), namely, 1, 1, 3, 1, 5, 1, 3, &c. and use them by the rule given at
the beginning of this article, as follows:
1 1 3 1 5 1 3 1 1, &c.
1 1 4 5 29 34 131 165 296
--- --- --- --- ---- ---- ---- ---- ----
1 2 7 9 52 61 235 296 531
Hence, 6¹⁶⁵/₂₉₆ is very near the square root of 43, not erring by so
much as
1
---------.
296 × 531
If we try it, we shall find (⁶¹⁶⁵/₂₉₆) to be ¹⁹⁴¹/₂₉₆, the square of
which is ³⁷⁶⁷⁴⁸¹/₈₇₆₁₆, or 43⁷/₈₇₆₁₆.
This rule is of use when it is frequently wanted to use one square
root, and therefore desirable to ascertain whether any easy
approximation exists by means of a common fraction. For example, √2 is
often used.
_
√2 = 1 + ...
1|1 1
1|1 1
1|2 2 2 2 2 2
1 2 5 12 29 70
--- --- --- --- --- ---, &c.
2 5 12 29 70 169
Here it appears that
29 1 99 100 - 1
1---- does not err by --------; consequently, ---- or ------- is,
70 70 × 169 70 70
considering the ease of the operation, a fair approximation. In fact,
⁹⁹/₇₀ is 1·4142857 ... the truth being 1·4142135 ...
The following is an additional example:
__
√19 = 4 + ...
4 | 2 3 3 2 4 4 2
1 | 3 5 2 5 3 1 3
4 | 2 1 3 1 2 8 2 1 3 1 2, &c.
1 1 4 5 14
--- --- --- --- ---, &c.
2 3 11 14 39
APPENDIX IX.
ON SOME GENERAL PROPERTIES OF NUMBERS.
PROP. 1. If a fraction be reduced to its lowest terms, _so called_,[61]
that is, if neither, numerator nor denominator be divisible by any
integer greater than unity, then no fraction of a smaller numerator and
denominator can have the same value.
[61] This theorem shews that what is _called_ reducing a fraction to
its lowest terms (namely, dividing numerator and denominator by their
greatest common measure), is correctly so called.
Let _a_/_b_ be a fraction in which _a_ and _b_ have no common measure
greater than unity: and, if possible, let _c_/_d_ be a fraction of the
same value, _c_ being less than _a_, and _d_ less than _b_. Now, since
_a_ _c_ _a_ _b_
--- = ---, we have --- = ---;
_b_ _d_ _c_ _d_
let _m_ be the integer quotient of these last fractions (which must
exist, since _a_ > _c_, _b_ > _d_), and let _e_ and _f_ be the
remainders. Then
_a_ _mc_ + _e_ _c_ _mc_
--- or ---------- = --- = ----
_b_ _md_ + _f_ _d_ _md_
Hence,
_e_ _mc_
--- and ---- must be equal, for if not,
_f_ _md_
_mc_ + _e_ _mc_ _e_
---------- would lie between ---- and ---,
_md_ + _f_ _md_ _f_
instead of being equal to the former. Hence,
_a_ _e_
--- = ---;
_b_ _f_
so that if a fraction whose numerator and denominator have no common
measure greater than unity, be equal to a fraction of lower numerator
and denominator, it is equal to another in which the numerator and
denominator are still lower. If we proceed with
_a_ _e_
--- = --- in a similar manner, we find
_b_ _f_
_a_ _g_
--- = --- where _g_ < _e_, _h_ < _f_,
_b_ _h_
and so on. Now, if there be any process which perpetually diminishes
the terms of a fraction by one or more units at every step, it must at
last bring either the numerator or denominator, or both, to 0. Let
_a_ _v_
--- = ---
_b_ _w_
be one of the steps, and let _a_ = _kv_ + _x_, _b_ = _kw_ + _y_; so that
_kv_ + _x_ _v_
---------- = ---.
_kw_ + _y_ _w_
Now, if _x_ = 0 but not _y_, this is absurd, for it gives
_kv_ _kv_
---------- = ----.
_kw_ + _y_ _kw_
A similar absurdity follows if _y_ be 0, but not _x_; and if both _x_
and _y_ be = 0, then _a_ = _kv_, _b_ = _kw_, or _a_ and _b_ have a
common measure, _k_. Now _k_ must be greater than 1, for _v_ and _w_
are less than _c_ and _d_, which by hypothesis are less than _a_ and
_b_. Consequently _a_ and _b_ have a common measure _k_ greater than 1,
which by hypothesis they have not. If, then, _a_ and _b_ be integers
not divisible by any integer greater than 1, the fraction _a_/_b_ is
really _in its lowest terms_. Also _a_ and _b_ are said to be _prime to
one another_.
PROP. 2. If the product _ab_ be divisible by _c_, and if _c_ be prime
to _b_, it must divide _a_. Let
_ab_ _b_ _d_
---- = _d_, then --- = ---.
_c_ _c_ _c_
Now _b_/_c_ is in its lowest terms; therefore, by the last proposition,
_d_ and _a_ must have a common measure. Let the greatest common measure
be _k_, and let _a_ = _kl_, _d_ = _km_. Then
_b_ _km_ _m_ _m_
--- = ---- = ---, and ---
_c_ _kl_ _l_ _l_
is also in its lowest terms; but so is _b_/_c_; therefore we must have
_m_ = _b_, _l_ = _c_, for otherwise a fraction in its lowest terms
would be equal to another of lower terms. Therefore _a_ = _kc_, or _a_
is divisible by _c_. And from this it follows, that if a number be
prime to two others, it is prime to their product. Let _a_ be prime to
_b_ and _c_, then no measure of _a_ can measure either _b_ or _c_, and
no such measure can measure the product _bc_; for any measure of _bc_
which is prime to one must measure the other.
PROP. 3. If _a_ be prime to _b_, it is prime to all the powers of _b_.
Every measure[62] of _a_ is prime to _b_, and therefore does not divide
_b_. Hence, by the last, no measure of _a_ divides _b_²; hence, _a_ is
prime to _b_², and so is every measure of it; therefore, no measure of
_a_ divides _bb_², consequently _a_ is prime to _b_³, and so on.
Hence, if _a_ be prime to _b_, _a_ cannot divide without remainder
any power of _b_. This is the reason why no fraction can be made into
a decimal unless its denominator be measured by no prime[63] numbers
except 2 and 5. For if
_a_ _c_
--- = ---,
_b_ 10ⁿ
which last is the general form of a decimal fraction, let
_a_ 10ⁿ_a_
--- be in its lowest terms; then ------
_b_ _b_
is an integer, whence (Prop. 2) _b_ must divide 10ⁿ, and so must all
the divisors of _b_. If, then, among the divisors of _b_ there be any
prime numbers except 2 and 5, we have a prime number (which is of
course a number prime to 10) not dividing 10, but dividing one of its
powers, which is absurd.
[62] For that which measures a measure is itself a measure; so that if
a measure of _a_ could have a measure in common with _b_, _a_ itself
would have a common measure with _b_.
[63] A prime number is one which is prime to all numbers except its own
multiples, or has no divisors except 1 and itself.
PROP. 4. If _b_ be prime to _a_, all the multiples of _b_, as _b_,
2_b_, ... up to (_a_-1)_b_ must leave different remainders when divided
by _a_. For if, _m_ being greater than _n_, and both less than _a_,
we have _mb_ and _nb_ giving the same remainder, it follows that
_mb_-_nb_, or (_m_-_n_)_b_, is divisible by _a_; whence (Prop. 2), a
divides _m_-_n_, a number less than itself, which is absurd.
* * * * *
If a number be divided into its prime factors, or reduced to a product
of prime numbers only (as in 360 = 2 × 2 × 2 × 3 × 3 × 5), and if
_a_, _b_, _c_, &c. be the prime factors, and α, β, γ, &c. the number
of times they severally enter, so that the number is _a_{^α} × _b_ᵝ ×
_c_ᵞ × &c., then this can be done in only one way: For any prime number
_v_, not included in the above list, is prime to _a_, and therefore
to _a_{^α}, to _b_ and therefore to _b_ᵝ and therefore to _a_{^α} ×
_b_ᵝ Proceeding in this way, we prove that _v_ is prime to the complete
product above, or to the given number itself.
The number of divisors which the preceding number _a_{^α}_b_ᵝ_c_ᵞ
... can have, 0 and itself included, is (α + 1)(β+ 1)(γ + 1).... For
_a_{^α} as the divisors 1, _a_, _a_² ... _a_{^α} and no others, α + 1
in all. Similarly, _b_ᵝ has β+ 1 divisors, and so on. Now as all the
divisors are made by multiplying together one out of each set, their
number (page 202) is (α + 1)(β + 1)(γ+ 1)....
If a number, _n_, be divisible by certain prime numbers, say 3, 5, 7,
11, then the third part of all the numbers up to _n_ is divisible by 3,
the fifth part by 5, and so on. But more than this: when the multiples
of 3 are omitted, exactly the fifth part of _those which remain_ are
divisible by 5; for the fifth part of the whole are divisible by 5,
and the fifth part of those which are removed are divisible by 5,
therefore the fifth part of those which are left are divisible by 5.
Again, because the seventh part of the whole are divisible by 7, and
the seventh part of those which are divisible by 3, or by 5, or by 15,
it follows that when all those which are multiples of 3 or 5, or both,
are removed, the seventh part of those which remain are divisible by
7; and so on. Hence, the number of numbers not exceeding n, which are
not divisible by 3, 5, 7, or 11, is ¹⁰/₁₁ of ⁶/₇ of ⁴/₅ of ²/₃ of n.
Proceeding in this way, we find that the number of numbers which are
prime to _n_, that is, which are not divisible by any one of its prime
factors, _a_, _b_, _c_, ... is
_a_ - 1 _b_ - 1 _c_ - 1
_n_ ------- ------- ------- ...
_a_ _b_ _c_
or _a_{^α-1} - 1}_b_ᵝ⁻¹_c_ᵞ⁻¹ ... (_a_ - 1)(_b_ - 1)(_c_ - 1)....
Thus, 360 being 2³3²5, its number of divisors is 4 × 3 × 2, or 24, and
there are 2³3.1.2.4 or 96 numbers less than 360 which are prime to it.
PROP. 5. If _a_ be prime to _b_, then the terms of the series, _a_,
_a_², _a_³, ... severally divided by _b_, must all leave different
remainders, until 1 occurs as a remainder, after which the cycle of
remainders will be again repeated.
Let _a_ + _b_ give the remainder _r_ (not unity); then _a_² ÷ _b_ gives
the same remainder as _r__a_ + _b_, which (Prop. 4) cannot be _r_: let
it be _s_. Then _a_ˢ ÷ _b_ gives the same remainder as _s__a_ ÷ _b_,
which (Prop. 4) cannot be either _r_ or _s_, unless _s_ be 1: let it be
_t_. Then _a_ᵗ ÷ _b_ gives the same remainder as _ta_ ÷ _b_; if _t_ be
not 1, this cannot be either _r_, _s_, or _t_: let it be _u_. So we go
on getting different remainders, until 1 occurs as a remainder; after
which, at the next step, the remainder of _a_ ÷ _b_ is repeated. Now, 1
must come at last; for division by _b_ cannot give any remainders but
0, 1, 2, ... _b_- 1; and 0 never arrives (Prop. 3), so that as soon as
_b_-2 _different_ remainders have occurred, no one of which is unity,
the next, which must be different from all that precede, must be 1. If
not before, then at _a_ᵇ⁻¹ we must have a remainder 1; after which the
cycle will obviously be repeated.
Thus, 7, 7², 7³, 7⁴, &c. will, when divided by 5, be found to give the
remainders 2, 4, 3, 1, &c.
PROP. 6. The difference of two _m_th powers is always divisible without
remainder by the difference of the roots; or _a_ᵐ -_b_ᵐ is divisible by
_a_-_b_; for
_a_ᵐ - _b_ᵐ = _a_ᵐ - _a_ᵐ⁻¹_b_ + _a_ᵐ⁻¹_b_ - _b_ᵐ
= _a_ᵐ⁻¹(_a_ - _b_) + _b_(_a_ᵐ⁻¹ - _b_ᵐ⁻¹)
From which, if _aᵐ⁻¹_-_bᵐ⁻¹_ is divisible by _a_ -_b_, so is _a_ᵐ-_b_ᵐ.
But _a_-_b_ is divisible by _a_-_b_; so therefore is _a_²- _b_²; so
therefore is _a_³-_b_³; and so on.
Therefore, if _a_ and _b_, divided by _c_, leave the same remainder,
_a_² and _b_², _a_³ and _b_³, &c. severally divided by _c_, leave the
same remainders; for this means that _a_-_b_ is divisible by _c_. But
_a_ᵐ - _b_ᵐ is divisible by _a_-_b_, and therefore by every measure of
_a_-_b_, or by _c_; but _a_ᵐ-_b_ᵐ cannot be divisible by _c_, unless
_a_ᵐ and _b_ᵐ, severally divided by _c_, give the same remainder.
PROP. 7. If _b_ be a prime number, and _a_ be not divisible by _b_,
then _a_ᵇ and (_a_-1)ᵇ + 1 leave the same remainder when divided by
_b_. This proposition cannot be proved here, as it requires a little
more of algebra than the reader of this work possesses.[64]
[64] Expand (_a_-1)ᵇ by the binomial theorem; shew that _when b is a
prime number_ every coefficient which is not unity is divisible by _b_;
and the proposition follows.
PROP. 8. In the last case, _a_ᵇ⁻¹ divided by _b_ leaves a remainder
1. From the last, _a_ᵇ-_a_ leaves the same remainder as (_a_-1)ᵇ +
1-_a_ or (_a_-1)ᵇ- (_a_-1); that is, the remainder of _a_ᵇ-_a_ is
not altered if _a_ be reduced by a unit. By the same rule, it may be
reduced another unit, and so on, still without any alteration of the
remainder. At last it becomes 1ᵇ-1, or 0, the remainder of which is 0.
Accordingly, _a_ᵇ-_a_, which is _a_(_a_ᵇ⁻¹- 1), is divisible by _b_;
and since _b_ is prime to _a_, it must (Prop. 2) divide _a_ᵇ⁻¹-1; that
is, _a_ᵇ⁻¹, divided by _b_, leaves a remainder 1, if _b_ be a prime
number and _a_ be not divisible by _b_.
From the above it appears (Prop. 5 and 7), that if _a_ be prime to
_b_, the set 1, _a_, _a_², _a_³, &c. successively divided by _b_, give
a set of remainders beginning with 1, and in which 1 occurs again at
_a_ᵇ⁻¹, if not before, and at _a_ᵇ⁻¹ certainly (whether before or not),
if _b_ be a prime number. From the point at which 1 occurs, the cycle
of remainders recommences, and 1 is always the beginning of a cycle.
If, then, _a_ᵐ be the first power which gives 1 for remainder, _m_ must
either be _b_-1, or a measure of it, _when b is a prime number_.
But if we divide the terms of the series _m_, _ma_, _ma_², _ma_³, &c.
by _b_, _m_ being less than _b_, we have cycles of remainders beginning
with _m_. If 1, _r_, _s_, _t_, &c. be the first set of remainders, then
the second set is the set of remainders arising from _m_, _mr_, _ms_,
_mt_, &c. If 1 never occur in the first set before _a_ᵇ⁻¹ (except at
the beginning), then all the numbers under _b_-1 inclusive are found
among the set 1, _r_, _s_, _t_, &c.; and if _m_ be prime to _b_ (Prop.
4), all the same numbers are found, in a different order, among the
remainders of _m_, _mr_, &c. But should it happen that the set 1, _r_,
_s_, _t_, &c. is not complete, then _m_, _mr_, _ms_, &c. may give a
different set of remainders.
All these last theorems are constantly verified in the process for
reducing a fraction to a decimal fraction. If _m_ be prime to _b_, or
the fraction _m_/_b_ in its lowest terms, the process involves the
successive division of _m_, _m_ × 10, _m_ × 10², &c. by _b_. This
process can never come to an end unless some power of 10, say 10ⁿ, is
divisible by _b_; which cannot be, if _b_ contain any prime factors
except 2 and 5. In every other case the quotient repeats itself, the
repeating part sometimes commencing from the first figure, sometimes
from a later figure. Thus, ¹/₇ yields ·142857142857, &c., but ¹/₁₄
gives ·07(142857)(142857), &c., and ¹/₂₈ gives ·03(571428)(571428), &c.
In _m_/_b_, the quotient always repeats from the very beginning
whenever _b_ is a prime number and _m_ is less than _b_; and the number
of figures in the repeating part is then always _b_-1, or a measure of
it. That it must be so, appears from the above propositions.
Before proceeding farther, we write down the repeating part of a
quotient, with the remainders which are left after the several figures
are formed. Let the fraction be ¹/₁₇, we have
0₁₀5₁₅8₁₄8₄2₆3₉5₅2₁₆9₇4₂1₃1₁₃7₁₁6₈4₁₂7₁
This may be read thus: 10 by 17, quotient 0, remainder 10; 10² by 17,
quotient 05, remainder 15; 10³ by 17, quotient 058, remainder 14; and
so on. It thus appears that 10¹⁶ by 17 leaves a remainder 1, which is
according to the theorem.
If we multiply 0588, &c. by _any number under_ 17, the same cycle is
obtained with a different beginning. Thus, if we multiply by 13, we have
7647058823529411
beginning with what comes after remainder 13 in the first number. If
we multiply by 7, we have 4117, &c. The reason is obvious: ¹/₁₇ × 13,
or ¹³/₁₇, when turned into a decimal fraction, starts with the divisor
130, and we proceed just as we do in forming ¹/₁₇, when within four
figures of the close of the cycle.
It will also be seen, that in the last half of the cycle the quotient
figures are complements to 9 of those in the first half, and that
the remainders are complements to 17. Thus, in 0₁₀5₁₅8₁₄8₄, &c. and
9₇4₂1₃1₁₃, &c. we see 0 + 9 = 9, 5 + 4 = 9, 8 + 1 = 9, &c., and 10 + 7
= 17, 15 + 2 = 17, 14 + 3 = 17, &c. We may shew the necessity of this
as follows: If the remainder 1 never occur till we come to use _a_ᵇ⁻¹,
then, _b_ being prime, _b_-1 is even; let it be 2_k_. Accordingly,
_a_²ᵏ-1 is divisible by _b_; but this is the product of _a_ᵏ-1 and _a_ᵏ
+ 1, one of which must be divisible by _b_. It cannot be _a_ᵏ-1, for
then a power of _a_ preceding the (_b_-1)th would leave remainder 1,
which is not the case in our instance: it must then be _a_ᵏ + 1, so
that _a_ᵏ divided by _b_ leaves a remainder _b_-1; and the _k_th step
concludes the first half of the process. Accordingly, in our instance,
we see, _b_ being 17 and _a_ being 10, that remainder 16 occurs at
the 8th step of the process. At the next step, the remainder is that
yielded by 10(_b_-1), or 9_b_ + _b_-10, which gives the remainder
_b_-10. But the first remainder of all was 10, and 10 + (_b_-10) =
_b_. If ever this complemental character occur in any step, it must
continue, which we shew as follows: Let _r_ be a remainder, and _b_-_r_
a subsequent remainder, the sum being _b_. At the next step after the
first remainder, we divide 10_r_ by _b_, and, at the next step after
the second remainder, we divide 10_b_-10_r_ by _b_. Now, since the sum
of 10_r_ and 10_b_-10_r_ is divisible by _b_, the two remainders from
these new steps must be such as added together will give _b_, and so
on; and the _quotients_ added together must give 9, for the sum of the
remainders 10_r_ and 10_b_-10_r_ yields a quotient 10, of which the two
remainders give 1.
If ¹/₅₉ and ¹/₆₁ be taken, the repeating parts will be found to contain
58 and 60 figures. Of these we write down only the first halves, as the
reader may supply the rest by the complemental property just given.
01694915254237288135593220338, &c.
016393442622950819672131147540, &c.
Here, then, are two numbers, the first of which multiplied by any
number under 59, and the second by any number under 61, can have the
products formed by carrying certain of the figures from one end to the
other.
But, _b_ being still prime, it may happen that remainder 1 may occur
before _b_-1 figures are obtained; in which case, as shewn, the
number of figures must be a measure of _b_-1. For example, take ¹/₄₁.
The repeating quotient, written as above, has only 5 figures, and 5
measures 41-1.
0₁₀2₁₈4₁₆3₃₇9₁
Now, this period, it will be found, has its figures merely transposed,
if we multiply by 10, 18, 16, or 37. But if we multiply by any other
number under 41, we convert this period into the period of another
fraction whose denominator is 41. The following are 8 periods which may
be found.
0₁₀2₁₈4₁₆3₃₇9₁ | 1₉2₈1₃₉9₂₁5₅
0₂₀4₃₆8₃₂7₃₃8₂ | 1₁₉4₂₆6₁₄3₁₇4₆
0₃₀7₁₃3₇1₂₉7₃ | 2₂₈6₃₄8₁₂2₃₈9₁₁
0₄₀9₈₁7₂₃5₂₅6₄ | 3₂₇6₂₄5₃₅8₂₂5₁₅
To find _m_/41, look out for _m_ among the remainders, and take the
period in which it is, beginning after the remainder. Thus, ³⁴/₄₁ is
·8292682926, &c., and ¹⁵/₄₁ is ·3658536585, &c. These periods are
complemental, four and four, as 02439 and 97560, 07317 and 92682, &c.
And if the first number, 02439, be multiplied by any number under 41,
look for that number among the remainders, and the product is found in
the period of that remainder by beginning after the remainder. Thus,
02439 multiplied by 23 gives 56097, and by 6 gives 14634.
The reader may try to decipher for himself how it is that, with no more
figures than the following, we can extend the result of our division.
The fraction of which the period is to be found is ¹/₈₇.
87)100(01149425
130
430
820 01149425 × 25
370 28735625 × 25
220 718390625 × 25
460 17959765625 × 25
25 448994140625
0114942528735625
718390625
1795976 5625
448994
----------------------------+------
0114942528735632183908045977|011494
|
APPENDIX X.
ON COMBINATIONS.
There are some things connected with combinations which I place in an
appendix, because I intend to demonstrate them more briefly than the
matters in the text.
Suppose a number of boxes, say 4, in each of which there are counters,
say 5, 7, 3, and 11 severally. In how many ways can one counter be
taken out of each box, the order of going to the boxes not being
regarded. _Answer_, in 5 × 7 × 3 × 11 ways. For out of the first box we
may draw a counter in 5 different ways, and to each such drawing we may
annex a drawing from the second in 7 different ways--giving 5 × 7 ways
of making a drawing from the first two. To each of these we may annex
a drawing from the third box in 3 ways--giving 5 × 7 × 3 drawings from
the first three; and so on. The following statements may now be easily
demonstrated, and similar ones made as to other cases.
If the order of going to the boxes make a difference, and if _a_, _b_,
_c_, _d_ be the numbers of counters in the several boxes, there are
4 × 2 × 3 × 1 × _a_ × _b_ × _c_ × _d_ distinct ways. If we want to
draw, say 2 out of the first box, 3 out of the second, 1 out of the
third, and 3 out of the fourth, and if the order of the boxes be not
considered, the number of ways is
_a_ - 1 _b_ - 1 _b_ - 2 _d_ - 1 _d_ - 2
_a_------- × _b_------- -------- × _c_ × _d_------- -------
2 2 3 2 3
If the order of going to the boxes be considered, we must multiply the
preceding by 4 × 3 × 2 × 1. If the order of the drawings out of the
boxes makes a difference, but not the order of the boxes, then the
number of ways is
_a_(_a_-1)_b_(_b_-1)(_b_-2)_cd_(_d_-1)(_d_-2)
The nth power of _a_, or _a_ⁿ, represents the number of ways in which
_a_ counters _differently marked_ can be distributed in _n_ boxes,
order of placing them in each box not being considered. Suppose we want
to distribute 4 differently-marked counters among 7 boxes. The first
counter may go into either box, which gives 7 ways; the second counter
may go into either; and any of the first 7 allotments may be combined
with any one of the second 7, giving 7 × 7 distinct ways; the third
counter varies each of these in 7 different ways, giving 7 × 7 × 7 in
all; and so on. But if the counters be undistinguishable, the problem
is a very different thing.
Required the number of ways in which a number can be compounded of
other numbers, different orders counting as different ways. Thus, 1 +
3 + 1 and 1 + 1 + 3 are to be considered as distinct ways of making 5.
It will be obvious, on a little examination, that each number can be
composed in exactly twice as many ways as the preceding number. Take
8 for instance. If every possible way of making 7 be written down, 8
may be made either by increasing the last component by a unit, or by
annexing a unit at the end. Thus, 1 + 3 + 2 + 1 may yield 1 + 3 + 2
+ 2, or 1 + 3 + 2 + 1 + 1: and all the ways of making 8 will thus be
obtained; for any way of making 8, say _a_ + _b_ + _c_ + _d_, must
proceed from the following mode of making 7, _a_ + _b_ + _c_ + (_d_-1).
Now, (_d_-1) is either 0--that is, _d_ is unity and is struck out--or
(_d_-1) remains, a number 1 less than _d_. Hence it follows that the
number of ways of making _n_ is 2ⁿ⁻¹. For there is obviously 1 way of
making 1, 2 of making 2; then there must be, by our rule, 2² ways of
making 3, 2³ ways of making 4; and so on.
{ 1 + 1 + 1 { 1 + 1 + 1 + 1
{ 1 + 1 { { 1 + 1 + 2
{ { 1 + 2 { 1 + 2 + 1
1 { { 1 + 3
{ { 2 + 1 { 2 + 1 + 1
{ 2 { { 2 + 2
{ 3 { 3 + 1
{ 4
This table exhibits the ways of making 1, 2, 3, and 4. Hence it follows
(which I leave the reader to investigate) that there are twice as many
ways of forming _a_ + _b_ as there are of forming _a_ and then annexing
to it a formation of _b_; four times as many ways of forming _a_ + _b_
+ _c_ as there are of annexing to a formation of _a_ formations of _b_
and of _c_; and so on. Also, in summing numbers which make up _a_ +
_b_, there are ways in which _a_ is a rest, and ways in which it is
not, and as many of one as of the other.
Required the number of ways in which a number can be compounded of odd
numbers, different orders counting as different ways. If _a_ be the
number of ways in which _n_ can be so made, and _b_ the number of ways
in which _n_ + 1 can be made, then _a_ + _b_ must be the number of ways
in which _n_ + 2 can be made; for every way of making 12 out of odd
numbers is either a way of making 10 with the last number increased by
2, or a way of making 11 with a 1 annexed. Thus, 1 + 5 + 3 + 3 gives
12, formed from 1 + 5 + 3 + 1 giving 10. But 1 + 9 + 1 + 1 is formed
from 1 + 9 + 1 giving 11. Consequently, the number of ways of forming
12 is the sum of the number of ways of forming 10 and of forming 11.
Now, 1 can only be formed in 1 way, and 2 can only be formed in 1 way;
hence 3 can only be formed in 1 + 1 or 2 ways, 4 in only 1 + 2 or 3
ways. If we take the series 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, &c.
in which each number is the sum of the two preceding, then the _n_th
number of this set is the number of ways (orders counting) in which _n_
can be formed of odd numbers. Thus, 10 can be formed in 55 ways, 11 in
89 ways, &c.
Shew that the number of ways in which _mk_ can be made of numbers
divisible by _m_ (orders counting) is 2ᵏ⁻¹.
In the two series, 1 1 1 2 3 4 6 9 13 19 28, &c.
0 1 0 1 1 1 2 2 3 4 5, &c.,
the first has each new term after the third equal to the sum of the
last and last but two; the second has each new term after the third
equal to the sum of the last but one and last but two. Shew that the
_n_th number in the first is the number of ways in which _n_ can be
made up of numbers which, divided by 3, leave a remainder 1; and that
the _n_th number in the second is the number of ways in which _n_ can
be made up of numbers which, divided by 3, leave a remainder 2.
It is very easy to shew in how many ways a number can be made up of
a given number of numbers, if different orders count as different
ways. Suppose, for instance, we would know in how many ways 12 can
be thus made of 7 numbers. If we write down 12 units, there are 11
intervals between unit and unit. There is no way of making 12 out of 7
numbers which does not answer to distributing 6 partition-marks in the
intervals, 1 in each of 6, and collecting all the units which are not
separated by partition-marks. Thus, 1 + 1 + 3 + 2 + 1 + 2 + 2, which is
one way of making 12 out of 7 numbers, answers to
| | | | | |
1 | 1 | 111 | 11 | 1 | 11 | 11
| | | | | |
in which the partition-marks come in the 1st, 2d, 5th, 7th, 8th, and
10th of the 11 intervals. Consequently, to ask in how many ways 12 can
be made of 7 numbers, is to ask in how many ways 6 partition-marks can
be placed in 11 intervals; or, how many combinations or selections can
be made of 6 out of 11. The answer is,
11 × 10 × 9 × 8 × 7 × 6
-----------------------, or 462.
1 × 2 × 3 × 4 × 5 × 6
Let us denote by _m_ₙ the number of ways in which _m_ things can be
taken out of _n_ things, so that _m_ₙ is the abbreviation for
_n_ - 1 _n_ - 2 _n_ - _m_ + 1
_n_ × ------ × ------- ... as far as -------------
2 3 _m_
Then _m_ₙ also represents the number of ways in which _m_ + 1 numbers
can be put together to make _n_ + 1. What we proved above is, that 6₁₁
is the number of ways in which we can put together 7 numbers to make
12. There will now be no difficulty in proving the following:
2ⁿ = 1 + 1ₙ + 2ₙ + 3ₙ ... + _n_ₙ
In the preceding question, 0 did not enter into the list of numbers
used. Thus, 3 + 1 + 0 + 0 was not considered as one of the ways of
putting together four numbers to make 5. But let us now ask, what is
the number of ways of putting together 7 numbers to make 12, allowing 0
to be in the list of numbers. There can be no more (nor fewer) ways of
doing this than of putting 7 numbers together, among which 0 is _not_
included, to make 19. Take every way of making 12 (0 included), and put
on 1 to each number, and we get a way of making 19 (0 not included).
Take any way of making 19 (0 not included), and strike off 1 from
each number, and we have one of the ways of making 12 (0 included).
Accordingly, 6₁₈ is the number of ways of putting together 7 numbers (0
being allowed) to make 12. And (_m_- 1)ₙ₊ₘ₋₁ is the number of ways of
putting together _m_ numbers to make _n_, 0 being included.
This last amounts to the solution of the following: In how many ways
can _n_ counters (undistinguishable from each other) be distributed
into _m_ boxes? And the following will now be easily proved: The number
of ways of distributing _c_ undistinguishable counters into _b_ boxes is
(_b_ - 1)_{_b_ + _c_ - 1}, if any box or boxes may be left empty. But
if there must be 1 at least in each box, the number of ways is (_b_ -
1)_{_c_ - 1}; if there must be 2 at least in each box, it is (_b_ -
1)_{_c- b_-1}; if there must be 3 at least in each box, it is (_b_ -
1)_{_c_ - 2_b_ - 1}; and so on.
The number of ways in which _m odd_ numbers can be put together to make
_n_, is the same as the number of ways in which _m_ even numbers (0
included) can be put together to make _n_-_m_; and this is the number
of ways in which _m_ numbers (odd or even, 0 included) can be put
together to make ½(_n_-_m_). Accordingly, the number of ways in which m
odd numbers can be put together to make _n_ is the same as the number
of combinations of _m_-1 things out of ½(_n_-_m_) + _m_-1, or ½(_n_ +
_m_)-1. Unless _n_ and _m_ be both even or both odd, the problem is
evidently impossible.
There are curious and useful relations existing between numbers of
combinations, some of which may readily be exhibited, under the simple
expression of _m_ₙ to stand for the number of ways in which _m_ things
may be taken out of _n_. Suppose we have to take 5 out of 12: Let the
12 things be marked A, B, C, &c. and set apart one of them, A. Every
collection of 5 out of the 12 either does or does not include A. The
number of the latter sort must be 5₁₁; the number of the former sort
must be 4₁₁, since it is the number of ways in which the _other four_
can be chosen out of all but A. Consequently, 5₁₂ must be 5₁₁ + 4₁₁,
and thus we prove in every case,
_m_ₙ′ = _m_ₙ₋₁ + (_m_ - 1)ₙ₋₁
0ₙ and _n_ₙ both are 1; for there is but one way of taking _none_, and
but one way of taking _all_. And again _m_ₙ and (_n_-_m_)ₙ are the same
things. And if _m_ be greater than _n_, _m_ₙ is 0; for there are no
ways of doing it. We make one of our preceding results more symmetrical
if we write it thus,
2ⁿ = 0ₙ + 1ₙ + 2ₙ + ... + _n_ₙ
If we now write down the table of symbols in which the (_m_ + 1)th
0 1 2 3, &c.
+--------------------------------------
1 | 0₁ 1₁ 2₁ 3₁, &c.
2 | 0₂ 1₂ 2₂ 3₂, &c.
3 | 0₃ 1₃ 2₃ 3₃, &c.
&c. | &c. &c. &c. &c.
number of the _n_th row represents _m_ₙ, the number of combinations of
_m_ out of _n_, we see it proved above that the law of formation of
this table is as follows: Each number is to be the sum of the number
above it and the number preceding the number above it. Now, the first
row must be 1, 1, 0, 0, 0, &c. and the first column must be 1, 1, 1, 1,
&c. so that we have a table of the following kind, which may be carried
as far as we please:
0 1 2 3 4 5 6 7 8 9 10
+----------------------------------------------------
1 | 1 1 0 0 0 0 0 0 0 0 0
2 | 1 2 1 0 0 0 0 0 0 0 0
3 | 1 3 3 1 0 0 0 0 0 0 0
4 | 1 4 6 4 1 0 0 0 0 0 0
5 | 1 5 10 10 5 1 0 0 0 0 0
6 | 1 6 15 20 15 6 1 0 0 0 0
7 | 1 7 21 35 35 21 7 1 0 0 0
8 | 1 8 28 56 70 56 28 8 1 0 0
9 | 1 9 36 84 126 126 84 36 9 1 0
10 | 1 10 45 120 210 252 210 120 45 10 1
Thus, in the row 9, under the column headed 4, we see 126, which is 9
× 8 × 7 × 6 ÷ (1 × 2 × 3 × 4), the number of ways in which 4 can be
chosen out of 9, which we represent by 4-{9}.
If we add the several rows, we have 1 + 1 or 2, 1 + 2 + 1 or 2², next
1 + 3 + 3 + 1 or 2³, &c. which verify a theorem already announced; and
the law of formation shews us that the several columns are formed thus:
1 1 1 2 1 1 3 3 1
1 1 1 2 1 1 3 3 1
----- ------- ---------
1 2 1 1 3 3 1 1 4 6 4 1, &c.
so that the sum in each row must be double of the sum in the preceding.
But we can carry the consequences of this mode of formation further. If
we make the powers of 1 + _x_ by actual algebraical multiplication, we
see that the process makes the same oblique addition in the formation
of the numerical multipliers of the powers of _x_.
1 + _x_
1 + _x_
-------
1 + _x_
_x_ + _x_²
---------------
1 + 2_x_ + _x_²
1 + 2_x_ + _x_²
1 + _x_
---------------
1 + 2_x_ + _x_²
_x_ + 2_x_² + _x_³
-----------------------
1 + 3_x_ + 3_x_² + _x_³
Here are the second and third powers of 1 + _x_: the fourth, we can
tell beforehand from the table, must be 1 + 4_x_ + 6_x_² + 4_x_³ +
_x_⁴; and so on. Hence we have
(1 + _x_)ⁿ = 0ₙ + 1ₙ_x_ + 2ₙ_x_² + 3ₙ_x_³ + ... + _n_ₙ_x_ⁿ
which is usually written with the symbols 0ₙ, 1ₙ, &c. at length, thus,
_n_ - 1 _n_ - 1 _n_ - 2
(1 + _x_)ⁿ = 1 + _nx_ + _n_-------_x_² + _n_------- -------_x_³ + &c.
2 2 3
This is the simplest case of what in algebra is called the _binomial
theorem_. If instead of 1 + _x_ we use _x_ + _a_, we get
(_x_+_a_)ⁿ = _x_ⁿ+1ₙ_ax_ⁿ⁻¹+2ₙ_a_²_x_ⁿ⁻²+3ₙ_a_³_x_ⁿ⁻³+... +_n_ₙ_a_ⁿ
We can make the same table in another form. If we take a row of ciphers
beginning with unity, and setting down the first, add the next, and
then the next, and so on, and then repeat the process with one step
less, and then again with one step less, we have the following:
1 0 0 0 0 0 0
1 1 1 1 1 1 1
1 2 3 4 5 6
1 3 6 10 15
1 4 10 20
1 5 15
1 6
1
In the oblique columns we see 1 1, 1 2 1, 1 3 3 1, &c. the same as
in the original table, and formed by the same additions. If, before
making the additions, we had always multiplied by _a_, we should have
got the several components of the powers of 1 + _a_, thus,
1 0 0 0 0
1 _a_ _a_² _a_³ _a_⁴
1 2_a_ 3_a_² 4_a_³
1 3_a_ 6_a_²
1 4_a_
1
where the oblique columns 1 + _a_, 1 + 2_a_ + _a_², 1 + 3_a_ + 3_a_² +
_a_³, &c., give the several powers of 1 + _a_. If instead of beginning
with 1, 0, 0, &c. we had begun with _p_, 0, 0, &c. we should have got
_p_, _p_ × 4_a_, _p_ × 6_a_², &c. at the bottom of the several columns;
and if we had written at the top _x_⁴, _x_³, _x_², _x_, 1, we should
have had all the materials for forming _p_(_x_ + _a_)⁴ by multiplying
the terms at the top and bottom of each column together, and adding the
results.
Suppose we follow this mode of forming _p_(_x_ + _a_)³ + _q_(_x_ +
_a_)² + _r_(_x_ + _a_) + _s_.
_x_³ _x_² _x_ 1 _x_² _x_ 1 _x_ 1 1
_p_ 0 0 0 _q_ 0 0 _r_ 0 3
_p_ _pa_ _pa_² _pa_³ _q_ _qa_ _qa_² _r_ _ra_
_p_ 2_pa_ 3_pa_² _q_ 2_qa_ _r_
_p_ 3_pa_ _q_
_p_
_px_³ + 3_pax_² + 3_pa_²_x_ + _pa_³ + _qx_² + 2_qax_ + _qa_²
+ _rx_ + _ra_ + _s_
= _px_³ + (3_pa_ + _q_)_x_² + (3_pa_² + 2_qa_ + _r_)_x_ + _pa_³
+ _qa_² + _ra_ + _s_
Now, observe that all this might be done in one process, by entering
_q_, _r_, and _s_ under their proper powers of _x_ in the first
process, as follows
_x_³ _x_² _x_ 1
_p_ _q_ _r_ _s_
_p_ _pa_ + _q_ _pa_² + _qa_ + _r_ _pa_³ + _qa_² + _ra_ + _s_
_p_ 2_pa_ + _q_ 3_pa_² + 2_qa_ + _r_
_p_ 3_pa_ + _q_
_p_
This process[65] is the one used in Appendix XI., with the slight
alteration of varying the sign of the last letter, and making
subtractions instead of additions in the last column. As it stands, it
is the most convenient mode of writing _x_ + _a_ instead of _x_ in a
large class of algebraical expressions. For instance, what does 2_x_⁵ +
_x_⁴ + 3_x_² + 7_x_ + 9 become when _x_ + 5 is written instead of _x_?
The expression, made complete, is,
2_x_⁵ + 1_x_⁴ + 0_x_³ + 3_x_² + 7_x_ + 9
1 0 3 7 9
2 11 55 278 1397 6994
2 21 160 1078 6787
2 31 315 2653
2 41 520
2 51
_Answer_, 2_x_⁵ + 51_x_⁴ + 520_x_³ + 2653_x_² + 6787_x_ + 6994.
[65] The principle of this mode of demonstration of Horner’s method was
stated in Young’s Algebra (1823), being the earliest elementary work in
which that method was given.
APPENDIX XI.
ON HORNER’S METHOD OF SOLVING EQUATIONS.
The rule given in this chapter is inserted on account of its excellence
as an exercise in computation. The examples chosen will require but
little use of algebraical signs, that they may be understood by those
who know no more of algebra than is contained in the present work.
To solve an equation such as
2_x_⁴ + _x_² - 3_x_ = 416793,
or, as it is usually written,
2_x_⁴ + _x_² - 3_x_ - 416793 = 0,
we must first ascertain by trial not only the first figure of the root,
but also the denomination of it: if it be a 2, for instance, we must
know whether it be 2, or 20, or 200, &c., or ·2, or ·02, or ·002,
&c. This must be found by trial; and the shortest way of making the
trial is as follows: Write the expression in its complete form. In the
preceding case the form is not complete, and the complete form is
2_x_⁴ + 0_x_³ + 1_x_² - 3_x_ - 416793.
To find what this is when x is any number, for instance, 3000, the best
way is to take the first multiplier (2), multiply it by 3000, and take
in the next multiplier (0), multiply the result by 3000, and take in
the next multiplier (1), and so on to the end, as follows:
2 × 3000 + 0 = 6000; 6000 × 3000 + 1 = 18000001
18000001 × 3000 - 3 = 54000002997
54000002997 × 3000 - 416793 = 162000008574207
Now try the value of the above when _x_ = 30. We have then, for the
steps, 60 (2 × 30 + 0), 1801, 54027, and lastly,
1620810-416793,
or _x_ = 30 makes the first terms greater than 416793. Now try _x_ = 20
which gives 40, 801, 16017, and lastly,
320340-416793,
or _x_ = 20 makes the first terms less than 416793. Between 20 and
30, then, must be a value of _x_ which makes 2_x_⁴ + _x_²-3x equal to
416793. And this is the preliminary step of the process.
Having got thus far, write down the coefficients +2, 0, +1,-3, and
-416793, each with its proper algebraical sign, except the last, in
which let the sign be changed. This is the most convenient way when the
last sign is-. But if the last sign be +, it may be more convenient
to let it stand, and change all which come before. Thus, in solving
_x_³-12_x_ + 1 = 0, we might write
-1 0 +12 1
whereas in the instance before us, we write
+2 0 +1 -3 416793
Having done this, take the highest figure of the root, properly named,
which is 2 tens, or 20. Begin with the first column, multiply by 20,
and join it to the number in the next column; multiply that by 20,
and join it to the number in the next column; and so on. But when you
come to the last column, subtract the product which comes out of the
preceding column, or join it to the last column after changing its
sign. When this has been done, repeat the process with the numbers
which now stand in the columns, omitting the last, that is, the
subtracting step; then repeat it again, going only as far as the last
column but two, and so on, until the columns present a set of rows of
the following appearance:
_a_ _b_ _c_ _d_ _e_
_f_ _g_ _h_ _i_
_k_ _l_ _m_
_n_ _o_
_p_
to the formation of which the following is the key:
_f_ = 20_a_ + _b_,
_g_ = 20_f_ + _c_,
_h_ = 20_g_ + _d_,
_i_ = _e_ - 20_h_,
_k_ = 20_a_ + _f_,
_l_ = 20_k_ + _g_,
_m_ = 20_l_ + _h_,
_n_ = 20_a_ + _k_,
_o_ = 20_n_ + _l_,
_p_ = 20_a_ + _n_.
We call this _Horner’s Process_, from the name of its inventor. The
result is as follows:
2 0 1 -3 416793 (20
40 801 16017 96453
80 2401 64037
120 4801
160
We have now before us the row
2 160 4801 64037 96453
which furnishes our means of guessing at the next, or units’ figure of
the root.
Call the last column the _dividend_, the last but one the _divisor_,
and all that come before _antecedents_. See how often the dividend
contains the divisor; this gives the guess at the next figure. The
guess is a true one,[66] if, on applying Horner’s process, the divisor
result, augmented as it is by the antecedent processes, still go as
many times in the dividend. For example, in the case before us, 96453
contains 64037 once; let 1 be put on its trial. Horner’s process is
found to succeed, and we have for the second process,
2 160 4801 64037 96453
162 4963 69000 27453
164 5127 74127
166 5293
168
As soon as we come to the fractional portion of the root, the process
assumes a more[67] methodical form.
The equation being of the _fourth_ degree, annex _four_ ciphers to
the dividend, _three_ to the divisor, _two_ to the antecedent, and
_one_ to the previous antecedent, leaving the first column as it is;
then find the new figure by the dividend and divisor, as before,[68]
and apply Horner’s process. Annex ciphers to the results, as before,
and proceed in the same way. The annexing of the ciphers prevents our
having any thing to do with decimal points, and enables us to use the
quotient-figures without paying any attention to their _local_ values.
The following exhibits the whole process from the beginning, carried
as far as it is here intended to go before beginning the contraction,
which will give more figures, as in the rule for the square root. The
following, then, is the process as far as one decimal place:
[66] Various exceptions may arise when an equation has two nearly equal
roots. But I do not here introduce algebraical difficulties; and a
student might give himself a hundred examples, taken at hazard, without
much chance of lighting upon one which gives any difficulty.
[67] This form might be also applied to the integer portions; but
it is hardly needed in such instances as usually occur. See the
article _Involution and Evolution_ in the _Supplement_ to the _Penny
Cyclopædia_.
[68] After the second step, the trial will rarely fail to give the true
figure.
2 0 1 -3 416793(213
40 801 16017 96453
80 2401 64037 -----
120 4801 ----- 274530000
160 ---- 69000 47339778
--- 4963 74127000 ---------
162 5127 --------
164 529300 75730074
166 ------ 77348376
1680 534358
---- 539434
1686 544528
1692
1698
1704
----
If we now begin the contraction, it is good to know beforehand on
what number of additional root-figures we may reckon. We may be
pretty certain of having nearly as many as there are figures in the
divisor when we begin to contract--one less, or at least two less.
Thus, there being now eight figures in the divisor, we may conclude
that the contraction will give us at least six more figures. To begin
the contraction, let the dividend stand, cut off one figure from the
divisor, two from the column before that, three from the one before
that, and so on. Thus, our contraction begins with
| | | |
|0002 1|704 5445|28 7734837|6 47339778
| | | |
The first column is rendered quite useless here. Conduct the process
as before, using only the figures which are not cut off. But it will
be better to go as far as the first figure cut off, carrying from the
second figure cut off. We shall then have as follows:
| | |
1|704 5445|28 7734837|6 47339778(6
| 5455|5 7767570|6 734354
5465|7 7800364|8
5475|9 |
|
At the next contraction the column 1|704 becomes |001704, and is quite
useless. The next step, separately written (which is not, however,
necessary in working), is
| |
54|759 780036|48 734354(0
| |
Here the dividend 734354 does not contain the divisor 780036, and we,
therefore, write 0 as a root figure and make another contraction, or
begin with
| |
|54759 78003|648 734354(9
| 78008|5 32277
78013|4
|
At the next contraction the first column becomes |0054759, and is
quite useless, so that the remainder of the process is the contracted
division.
|
7801|34)32277(4137
| 1072
292
58
3
and the root required is 21·36094137.
I now write down the complete process for another equation, one root of
which lies between 3 and 4: it is
_x_³ - 10_x_ + 1 = 0
1 0 -10 -1(3·1110390520730990796
3 -1 2000
6 1700 209000
9 0 1791 19769000
9 1 188300 743369000000
9 2 189231 172311710273000
9 30 19016300 991247447681
9 31 19025631 39462875420
9 32 1903496300 0 0 1391491559
9 33 0 1903524299 0 9 58993123
9 33 1 1903552298 2 7 0 0 1886047
9 33 2 1903560698 0 5|9|1 172835
9 33 30 0 1903569097 8 5|6|3 1515
9 33 30 3 1903569144 5 2|2| 183
9 33 30 6 1903569191 1|8|8 12
9 33 30 90 1903569193 0|6| 1
9 33|30|99 1903569194|9|3|
9 33|31|08 | | |
|09|33|31|17
| | | |
The student need not repeat the rows of figures so far as they come
under one another: thus, it is not necessary to repeat 190356. But he
must use his own discretion as to how much it would be safe for him to
omit. I have set down the whole process here as a guide.
The following examples will serve for exercise:
1. 2_x_³ - 100_x_ - 7 = 0
_x_ = 7·10581133.
2. _x_⁴ + _x_³ + _x_² + _x_ = 6000
_x_ = 8·531437726.
3. _x_³ + 3_x_² - 4_x_ - 10 = 0
_x_ = 1·895694916504.
4. _x_³ + 100_x_² - 5_x_ - 2173 = 0
_x_ = 4·582246071058464.
_
5. ∛2 = 1·259921049894873164767210607278.[69]
6. _x_³ - 6_x_ = 100
_x_ = 5·071351748731.
7. _x_³ + 2_x_² + 3_x_ = 300
_x_ = 5·95525967122398.
8. _x_³ + _x_ = 1000
_x_ = 9·96666679.
9. 27000_x_³ + 27000_x_ = 26999999
_x_ = 9·9666666.....
10. _x_³ - 6_x_ = 100
_x_ = 5·0713517487.
11. _x_⁵ - 4_x_⁴ + 7_x_³ - 863 = 0
_x_ = 4·5195507.
12. _x_³ - 20_x_ + 8 = 0
_x_ = 4·66003769300087278.
13. _x_³ + _x_² + _x_ - 10 = 0
_x_ = 1·737370233.
14. _x_³ - 46_x_² - 36_x_ + 18 = 0
_x_ = 46·7616301847, or _x_ = ·3471623192.
15. _x_³ + 46_x_² - 36_x_ - 18 = 0
_x_ = 1·1087925037.
16. 8991_x_³ - 162838_x_² + 746271_x_ - 81000 = 0
_x_ = ·111222333444555....
17. 729_x_³ - 486_x_² + 99_x_ - 6 = 0
_x_ = ·1111..., or ·2222..., or ·3333....
18. 2_x_³ + 3_x_² - 4_x_ = 500
_x_ = 5·93481796231515279.
19. _x_³ + 2_x_² + _x_ - 150 = 0
_x_ = 4·6684090145541983253742991201705899.
20. _x_³ + _x_ = _x_² + 500
_x_ = 8·240963558144858526963.
21. _x_³ + 2_x_² + 3_x_ - 10000 = 0
_x_ = 20·852905526009.
22. _x_⁵ - 4_x_ - 2000 = 0
_x_ = 4·581400362.
23. 10_x_³ - 33_x_² - 11_x_ - 100 = 0
_x_ = 4·146797808584278785.
24. _x_⁴ + _x_³ + _x_² + _x_ = 127694
_x_ = 18·64482373095.
25. 10_x_³ + 11_x_² + 12_x_ = 100000
_x_ = 21·1655995554508805.
26. _x_³ + _x_ = 13
_x_ = 2·209753301208849.
27. _x_³ + _x_² - 4_x_ - 1600 = 0
_x_ = 11·482837157.
28. _x_³ - 2_x_ = 5
_x_ = 2·094551481542326591482386540579302963857306105628239.
29. _x_⁴ - 80_x_³ + 24_x_² - 6_x_ - 80379639 = 0
_x_ = 123.[70]
30. _x_³ - 242_x_² - 6315_x_ + 2577096 = 0
_x_ = 123.[71]
31. 2_x_⁴ - 3_x_³ + 6_x_ - 8 = 0
_x_ = 1·414213562373095048803.[72]
32. _x_⁴ - 19_x_³ + 132_x_² - 302_x_ + 200 = 0
_x_ = 1·02804, or 4, or 6·57653, or 7·39543[73].
33. 7_x_⁴ - 11_x_³ + 6_x_² + 5_x_ = 215
_x_ = 2·70648049385791.[74]
34. 7_x_⁵ + 6_x_⁴ + 5_x_³ + 4_x_² + 3_x_ = 11
_x_ = ·770768819622658522379296505.[75]
35. 4_x_⁶ + 7_x_⁵ + 9_x_⁴ + 6_x_³ + 5_x_² + 3_x_ = 792
_x_ = 2·0520421768796053652140434012812019734602755995
45541724214.[76]
36. 2187_x_⁴ - 2430_x_³ + 945_x_² - 150_x_ + 8 = 0
_x_ = ·1111...., or ·2222...., or ·3333...., or ·4444....
[69] The solution of _x_³ + 0_x_² + 0_x_-2 = 0.
[70] Taken from a paper on the subject, by Mr. Peter Gray, in the
_Mechanics’ Magazine_.
[71] Taken from a paper on the subject, by Mr. Peter Gray, in the
_Mechanics’ Magazine_.
[72] Taken from a paper on the subject, by Mr. Peter Gray, in the
_Mechanics’ Magazine_.
[73] Taken from the late Mr. Peter Nicholson’s Essay on Involution and
Evolution.
[74] Taken from the late Mr. Peter Nicholson’s Essay on Involution and
Evolution.
[75] Taken from the late Mr. Peter Nicholson’s Essay on Involution and
Evolution.
[76] Taken from the late Mr. Peter Nicholson’s Essay on Involution and
Evolution.
APPENDIX XII.
RULES FOR THE APPLICATION OF ARITHMETIC TO GEOMETRY.
The student should make himself familiar with the most common terms of
geometry, after which the following rules will present no difficulty.
In them all, it must be understood, that when we talk of multiplying
one line by another, we mean the repetition of one line as often as
there are units of a given kind, as feet or inches, in another. In any
other sense, it is absurd to talk of multiplying a quantity by another
quantity. All quantities of the same kind should be represented in
numbers of the same unit; thus, all the lines should be either feet
and decimals of a foot, or inches and decimals of an inch, &c. And in
whatever unit a length is represented, a surface is expressed in the
corresponding square units, and a solid in the corresponding cubic
units. This being understood, the rules apply to all sorts of units.
_To find the area of a rectangle._ Multiply together the units in
two sides which meet, or multiply together two sides which meet; the
product is the number of square units in the area. Thus, if 6 feet and
5 feet be the sides, the area is 6 × 5, or 30 square feet. Similarly,
the area of a square of 6 feet long is 6 × 6, or 36 square feet (234).
_To find the area of a parallelogram._ Multiply one side by the
perpendicular distance between it and the opposite side; the product is
the area required in square units.
_To find the area of a trapezium._[77] Multiply either of the two sides
which are not parallel by the perpendicular let fall upon it from the
middle point of the other.
[77] A four-sided figure, which has two sides parallel, and two sides
not parallel.
_To find the area of a triangle._ Multiply any side by the
perpendicular let fall upon it from the opposite vertex, and take half
the product. Or, halve the sum of the three sides, subtract the three
sides severally from this half sum, multiply the four results together,
and find the square root of the product. The result is the number of
square units in the area; and twice this, divided by either side, is
the perpendicular distance of that side from its opposite vertex.
_To find the radius of the internal circle which touches the three
sides of a triangle._ Divide the area, found in the last paragraph, by
half the sum of the sides.
_Given the two sides of a right-angled triangle, to find the
hypothenuse._ Add the squares of the sides, and extract the square root
of the sum.
_Given the hypothenuse and one of the sides, to find the other side._
Multiply the sum of the given lines by their difference, and extract
the square root of the product.
_To find the circumference of a circle from its radius, very
nearly._ Multiply twice the radius, or the diameter, by 3·1415927,
taking as many decimal places as may be thought necessary. For a
rough computation, multiply by 22 and divide by 7. For a very exact
computation, in which decimals shall be avoided, multiply by 355 and
divide by 113. See (131), last example.
_To find the arc of a circular sector, very nearly, knowing the radius
and the angle._ Turn the angle into seconds,[78] multiply by the
radius, and divide the product by 206265. The result will be the number
of units in the arc.
[78] The right angle is divided into 90 equal parts called _degrees_,
each degree into 60 equal parts called _minutes_, and each minute into
60 equal parts called _seconds_. Thus, 2° 15′ 40″ means 2 degrees, 15
minutes, and 40 seconds.
_To find the area of a circle from its radius, very nearly._ Multiply
the square of the radius by 3·1415927.
_To find the area of a sector, very nearly, knowing the radius and the
angle._ Turn the angle into seconds, multiply by the square of the
radius, and divide by 206265 × 2, or 412530.
_To find the solid content of a rectangular parallelopiped._ Multiply
together three sides which meet: the result is the number of cubic
units required. If the figure be not rectangular, multiply the area
of one of its planes by the perpendicular distance between it and its
opposite plane.
_To find the solid content of a pyramid._ Multiply the area of the base
by the perpendicular let fall from the vertex upon the base, and divide
by 3.
_To find the solid content of a prism._ Multiply the area of the base
by the perpendicular distance between the opposite bases.
_To find the surface of a sphere._ Multiply 4 times the square of the
radius by 3·1415927.
_To find the solid content of a sphere._ Multiply the cube of the
radius by 3·1415927 × ⁴/₃, or 4·18879.
_To find the surface of a right cone._ Take half the product of the
circumference of the base and slanting side. _To find the solid
content_, take one-third of the product of the base and the altitude.
_To find the surface of a right cylinder._ Multiply the circumference
of the base by the altitude. _To find the solid content_, multiply the
area of the base by the altitude.
The weight of a body may be found, when its solid content is known, if
the weight of one cubic inch or foot of the body be known. But it is
usual to form tables, not of the weights of a cubic unit of different
bodies, but of the proportion which these weights bear to some one
amongst them. The one chosen is usually distilled water, and the
proportion just mentioned is called the _specific gravity_. Thus, the
specific gravity of gold is 19·362, or a cubic foot of gold is 19·362
times as heavy as a cubic foot of distilled water. Suppose now the
weight of a sphere of gold is required, whose radius is 4 inches. The
content of this sphere is 4 × 4 × 4 × 4·1888, or 268·0832 cubic inches;
and since, by (217), each cubic inch of water weighs 252·458 grains,
each cubic inch of gold weighs 252·458 × 19·362, or 4888·091 grains; so
that 268·0832 cubic inches of gold weigh 268·0832 × 4888·091 grains, or
227½ pounds troy nearly. Tables of specific gravities may be found in
most works of chemistry and practical mechanics.
The cubic foot of water is 908·8488 troy ounces, 75·7374 troy pounds,
997·1369691 averdupois ounces, and 62·3210606 averdupois pounds. For
all rough purposes it will do to consider the cubic foot of water as
being 1000 common ounces, which reduces tables of specific gravities
to common terms in an obvious way. Thus, when we read of a substance
which has the specific gravity 4·1172, we may take it that a cubic foot
of the substance weighs 4117 ounces. For greater correctness, diminish
this result by 3 parts out of a thousand.
THE END.
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