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If you are not located in the United States, you -will have to check the laws of the country where you are located before -using this eBook. - -Title: Elements of arithmetic - -Author: Augustus De Morgan - -Release Date: August 1, 2022 [eBook #68662] - -Language: English - -Produced by: Richard Tonsing and the Online Distributed Proofreading - Team at https://www.pgdp.net (This file was produced from - images generously made available by The Internet Archive) - -*** START OF THE PROJECT GUTENBERG EBOOK ELEMENTS OF ARITHMETIC *** - - - - - -Transcriber’s Notes: - - Underscores “_” before and after a word or phrase indicate _italics_ - in the original text. - Equal signs “=” before and after a word or phrase indicate =bold= - in the original text. - Carat symbol “^” designates a superscript. - Small capitals have been converted to SOLID capitals. - Illustrations have been moved so they do not break up paragraphs. - Old or antiquated spellings have been preserved. - Typographical and punctuation errors have been silently corrected. - - - - - 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. - - - - - =WALTON AND MABERLY’S= - CATALOGUE OF EDUCATIONAL WORKS, AND WORKS - IN SCIENCE AND GENERAL LITERATURE. - - - =ENGLISH=. - - _Dr. R. G. Latham. 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