,transcriber's ,notes3 ,underscores 80 before and after a word or phrase indicate italics in the original text4 ,equal signs 80 before and after a word or phrase indicate bold in the original text4 ,carat symbol 80 designates a superscript4 ,small capitals have been converted to ,,solid capitals4 ,illustrations have been moved so they do not break up paragraphs4 ,old or antiquated spellings have been preserved4 ,typographical and punctuation errors have been silently corrected4 ,,elements ,,of ,,arithmetic4 ,,by ,,augustus ,,de ,,morgan1 ,,of ,,trinity ,,college1 ,,cambridge2 ,,fellow ,,of ,,the ,,royal ,,astronomical ,,society1 ,,and ,,of ,,the ,,cambridge ,,philosophical ,,society2 ,,professor ,,of ,,mathematics ,,in ,,university ,,college1 ,,london4 8,hominis studiosi est intelligere1 quas utilitates proprie afferat arithmetica his1 qui solidam et perfectam doctrinam in cteris philosophi partibus explicant4 ,quod enim vulgo dicunt1 principium esse dimidium totius1 id vel maxime in philosophi partibus conspicitur40--,,melancthon4 8,ce n'est point par la routine qu'on e'instruit1 c'est par sa propre rflexion2 et il est essentiel de contracter l'habitude de se rendre raison de ce qu'on fait3 cette habitude s'acquiert plus facilement qu'on ne pense2 et une fois acquise1 elle ne se perd plus40--,,condillac4 ,,seventeenth ,,thousand4 ,,london3 ,,walton ,,and ,,maberly1 ,,upper ,,gower ,,street1 ,,and ,,ivy ,,lane1 ,,paternoster ,,row4 ,m4,,dccc4,,lviii4 ,,london3 ,,printed ,,by ,j4 ,,wertheimer ,,and ,,co41 ,,circus-,,place1 ,,finsbury-,,circus4 ,,preface4 ,the preceding editions of this work were published in #ahcj1 #ahcb1 #ahce1 and #ahdj4 ,this fifth edition differs from the three preceding1 as to the body of the work1 in nothing which need prevent the four1 or any two of them1 from being used together in a class4 ,but it is considerably augmented by the addition of eleven new ,appendixes1#a relating to matters on which it is most desirable that the advanced student should possess information4 ,the first ,appendix1 on ,computation1 and the sixth1 on ,decimal ,money1 should be read and practised by every student with as much attention as any part of the work4 ,the mastery of the rules for instantaneous conversion of the usual fractions of a pound sterling into decimal fractions1 gives the possessor the greater part of the advantage which he would derive from the introduction of a decimal coinage4 ,at the time when this work was first published1 the importance of establishing arithmetic in the young mind upon reason and demonstration1 was not admitted by many4 ,the case is now altered3 schools exist in which rational arithmetic is taught1 and mere rules are made to do no more than their proper duty4 ,there is no necessity to advocate a change which is actually in progress1 as the works which are published every day sufficiently shew4 ,and my principal reason for alluding to the subject here1 is merely to warn those who want nothing but routine1 that this is not the book for their purpose4 ,a4 ,,de ,,morgan4 ,london1 ,may #a1 #ahdf4 #a ,some separate copies of these ,appendixes are printed1 for those who may desire to add them to the former editions4 ,,table ,,of ,,contents4 ,,book ,i4 ,,section ,,page ,i4 ,numeration #a ,,ii4 ,addition and ,subtraction #ad ,,iii4 ,multiplication #bd ,,iv4 ,division #cd ,v4 ,fractions #ea ,,vi4 ,decimal ,fractions #fe ,,vii4 ,square ,root #hi ,,viii4 ,proportion #ajj ,,ix4 ,permutations and ,combinations #aah ,,book ,,ii4 ,i4 ,weights and ,measures1 c4 #abd ,,ii4 ,rule of ,three #add ,,iii4 ,interest1 c4 #aej ,,appendix4 ,i4 ,on the mode of computing #afa ,,ii4 ,on verification by casting out nines and elevens #aff ,,iii4 ,on scales of notation #afh ,,iv4 ,on the definition of fractions #aga ,v4 ,on characteristics #agd ,,vi4 ,on decimal money #agf ,,vii4 ,on the main principle of book-keeping #ahj ,,viii4 ,on the reduction of fractions to others of nearly equal value #aij ,,ix4 ,on some general properties of numbers #aic ,x4 ,on combinations #bja ,,xi4 ,on ,horner's method of solving equations #baj ,,xii4 ,rules for the application of arithmetic to geometry #bag ,,elements ,,of ,,arithmetic4 ,,book ,i4 ,,principles ,,of ,,arithmetic4 ,,section ,i4 ,,numeration4 #a4 ,imagine a multitude of objects of the same kind assembled together2 for example1 a company of horsemen4 ,one of the first things that must strike a spectator1 although unused to counting1 is1 that to each man there is a horse4 ,now1 though men and horses are things perfectly unlike1 yet1 because there is one of the first kind to every one of the second1 one man to every horse1 a new notion will be formed in the mind of the observer1 which we express in words by saying that there is the same number of men as of horses4 ,a savage1 who had no other way of counting1 might remember this number by taking a pebble for each man4 ,out of a method as rude as this has sprung our system of calculation1 by the steps which are pointed out in the following articles4 ,suppose that there are two companies of horsemen1 and a person wishes to know in which of them is the greater number1 and also to be able to recollect how many there are in each4 #b4 ,suppose that while the first company passes by1 he drops a pebble into a basket for each man whom he sees4 ,there is no connexion between the pebbles and the horsemen but this1 that for every horseman there is a pebble2 that is1 in common language1 the number of pebbles and of horsemen is the same4 ,suppose that while the second company passes1 he drops a pebble for each man into a second basket3 he will then have two baskets of pebbles1 by which he will be able to convey to any other person a notion of how many horsemen there were in each company4 ,when he wishes to know which company was the larger1 or contained most horsemen1 he will take a pebble out of each basket1 and put them aside4 ,he will go on doing this as often as he can1 that is1 until one of the baskets is emptied4 ,then1 if he also find the other basket empty1 he says that both companies contained the same number of horsemen2 if the second basket still contain some pebbles1 he can tell by them how many more were in the second than in the first4 #c4 ,in this way a savage could keep an account of any numbers in which he was interested4 ,he could thus register his children1 his cattle1 or the number of summers and winters which he had seen1 by means of pebbles1 or any other small objects which could be got in large numbers4 ,something of this sort is the practice of savage nations at this day1 and it has in some places lasted even after the invention of better methods of reckoning4 ,at ,rome1 in the time of the republic1 the prtor1 one of the magistrates1 used to go every year in great pomp1 and drive a nail into the door of the temple of ,jupiter2 a way of remembering the number of years which the city had been built1 which probably took its rise before the introduction of writing4 #d4 ,in process of time1 names would be given to those collections of pebbles which are met with most frequently4 ,but as long as small numbers only were required1 the most convenient way of reckoning them would be by means of the fingers4 ,any person could make with his two hands the little calculations which would be necessary for his purposes1 and would name all the different collections of the fingers4 ,he would thus get words in his own language answering to one1 two1 three1 four1 five1 six1 seven1 eight1 nine1 and ten4 ,as his wants increased1 he would find it necessary to give names to larger numbers2 but here he would be stopped by the immense quantity of words which he must have1 in order to express all the numbers which he would be obliged to make use of4 ,he must1 then1 after giving a separate name to a few of the first numbers1 manage to express all other numbers by means of those names4 #e4 ,i now shew how this has been done in our own language4 ,the ,english names of numbers have been formed from the ,saxon3 and in the following table each number after ten is written down in one column1 while another shews its connexion with those which have preceded it4 ,one eleven ten and one#b 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 c4 c4 c4 c4 seventy seven tens thirty three tens eighty eight tens c4 c4 ninety nine tens forty four tens a hundred ten tens c4 c4 a hundred and one ten tens and one c4 c4 a thousand ten hundreds ten thousand a hundred thousand a million ten hundred thousand or one thousand thousand ten millions a hundred millions c4 #b ,it has been supposed that eleven and twelve are derived from the ,saxon for one left and two left 7meaning1 after ten is removed72 but there seems better reason to think that leven is a word meaning ten1 and connected with decem4 #f4 ,words1 written down in ordinary language1 would very soon be too long for such continual repetition as takes place in calculation4 ,short signs would then be substituted for words2 but it would be impossible to have a distinct sign for every number3 so that when some few signs had been chosen1 it would be convenient to invent others for the rest out of those already made4 ,the signs which we use areas follow3 #j #a #b #c #d #e #f #g #h #i 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 numbers4 #g4 ,suppose a man first to hold up one finger1 then two1 and so on1 until he has held up every finger1 and suppose a number of men to do the same thing4 ,it is plain that we may thus distinguish one number from another1 by causing two different sets of persons to hold up each a certain number of fingers1 and that we may do this in many different ways4 ,for example1 the number fifteen might be indicated either by fifteen men each holding up one finger1 or by four men each holding up two fingers and a fifth holding up seven1 and so on4 ,the question is1 of all these contrivances for expressing the number1 which is the most convenient8 ,in the choice which is made for this purpose consists what is called the method of numeration4 #h4 ,i have used the foregoing explanation because it is very probable that our system of numeration1 and almost every other which is used in the world1 sprung from the practice of reckoning on the fingers1 which children usually follow when first they begin to count4 ,the method which ,i have described is the rudest possible2 but1 by a little alteration1 a system may be formed which will enable us to express enormous numbers with great ease4 #i4 ,suppose that you are going to count some large number1 for example1 to measure a number of yards of cloth4 ,opposite to yourself suppose a man to be placed1 who keeps his eye upon you1 and holds up a finger for every yard which he sees you measure4 ,when ten yards have been measured he will have held up ten fingers1 and will not be able to count any further unless he begin again1 holding up one finger at the eleventh yard1 two at the twelfth1 and so on4 ,but to know how many have been counted1 you must know1 not only how many fingers he holds up1 but also how many times he has begun again4 ,you may keep this in view by placing another man on the right of the former1 who directs his eye towards his companion1 and holds up one finger the moment he perceives him ready to begin again1 that is1 as soon as ten yards have been measured4 ,each finger of the first man stands only for one yard1 but each finger of the second stands for as many as all the fingers of the first together1 that is1 for ten4 ,in this way a hundred may be counted1 because the first may now reckon his ten fingers once for each finger of the second man1 that is1 ten times in all1 and ten tens is one hundred 7#e74c ,now place a third man at the right of the second1 who shall hold up a finger whenever he perceives the second ready to begin again4 ,one finger of the third man counts as many as all the ten fingers of the second1 that is1 counts one hundred4 ,in this way we may proceed until the third has all his fingers extended1 which will signify that ten hundred or one thousand have been counted 7#e74 ,a fourth man would enable us to count as far as ten thousand1 a fifth as far as one hundred thousand1 a sixth as far as a million1 and so on4 #c ,the references are to the preceding articles4 #aj4 ,each new person placed himself towards your left in the rank opposite to you4 ,now rule columns as in the next page1 and to the right of them all place in words the number which you wish to represent2 in the first column on the right1 place the number of fingers which the first man will be holding up when that number of yards has been measured4 ,in the next column1 place the fingers which the second man will then be holding up2 and so on4 #gth4#fth4#eth4#dth4#crd4#bnd4#ast4 ,i4 #e #g fifty-seven4 ,,ii4 #a #j #d one hundred and four4 ,,iii4 #a #a #j one hundred and ten4 ,,iv4 #b #c #d #h two thousand three hundred and forty-eight4 ,v4 #a #e #i #j #f fifteen thousand nine hundred and six4 ,,vi4 #a #h #g #j #j #d one hundred and eighty-seven thousand and four4 ,,vii4 #c #f #i #g #b #h #e three million1 six hundred and ninety-seven thousand1 two hundred and eighty-five4 #aa4 ,in ,i4 the number fifty-seven is expressed4 ,this means 7#e7 five tens and seven4 ,the first has therefore counted all his fingers five times1 and has counted seven fingers more4 ,this is shewn by five fingers of the second man being held up1 and seven of the first4 ,in ,,ii4 the number one hundred and four is represented4 ,this number is 7#e7 ten tens and four4 ,the second person has therefore just reckoned all his fingers once1 which is denoted by the third person holding up one finger2 but he has not yet begun again1 because he does not hold up a finger until the first has counted ten1 of which ten only four are completed4 ,when all the last-mentioned ten have been counted1 he then holds up one finger1 and the first being ready to begin again1 has no fingers extended1 and the number obtained is eleven tens1 or ten tens and one ten1 or one hundred and ten4 ,this is the case in ,,iii4 ,you will now find no difficulty with the other numbers in the table4 #ab4 ,in all these numbers a figure in the first column stands for only as many yards as are written under that figure in 7#f74 ,a figure in the second column stands1 not for as many yards1 but for as many tens of yards2 a figure in the third column stands for as many hundreds of yards2 in the fourth column for as many thousands of yards2 and so on3 that is1 if we suppose a figure to move from any column to the one on its left1 it stands for ten times as many yards as before4 ,recollect this1 and you may cease to draw the lines between the columns1 because each figure will be sufficiently well known by the place in which it is2 that is1 by the number of figures which come upon the right hand of it4 #ac4 ,it is important to recollect that this way of writing numbers1 which has become so familiar as to seem the natural method1 is not more natural than any other4 ,for example1 we might agree to signify one ten by the figure of one with an accent1 thus1 #a2 twenty or two tens by #b2 and so on3 one hundred or ten tens by #a2 two hundred by #b2 one thousand by #a2 and so on3 putting ,roman figures for accents when they become too many to write with convenience4 ,the fourth number in the table would then be written #b #c #d #h1 which might also be expressed by #h #d #c #b1 #d #h #c #b2 or the order of the figures might be changed in any way1 because their meaning depends upon the accents which are attached to them1 and not upon the place in which they stand4 ,hence1 a cipher would never be necessary2 for #ajd would be distinguished from #ad by writing for the first #a #d1 and for the second #a #d4 ,the common method is preferred1 not because it is more exact than this1 but because it is more simple4 #ad4 ,the distinction between our method of numeration and that of the ancients1 is in the meaning of each figure depending partly upon the place in which it stands4 ,thus1 in #ddddd each four stands for four of something2 but in the first column on the right it signifies only four of the pebbles which are counted2 in the second1 it means four collections of ten pebbles each2 in the third1 four of one hundred each2 and so on4 #ae4 ,the things measured in 7#aa7 were yards of cloth4 ,in this case one yard of cloth is called the unit4 ,the first figure on the right is said to be in the units' place1 because it only stands for so many units as are in the number that is written under it in 7#f74 ,the second figure is said to be in the tens' place1 because it stands for a number of tens of units4 ,the third1 fourth1 and fifth figures are in the places of the hundreds1 thousands1 and tens of thousands1 for a similar reason4 #af4 ,if the quantity measured had been acres of land1 an acre of land would have been called the unit1 for the unit is one of the things which are measured4 ,quantities are of two sorts2 those which contain an exact number of units1 as #dg yards1 and those which do not1 as #dg yards and a half4 ,of these1 for the present1 we only consider the first4 #ag4 ,in most parts of arithmetic1 all quantities must have the same unit4 ,you cannot say that #b yards and #c feet make #e yards or #e feet1 because #b and #c make #e2 yet you may say that #b yards and #c yards make #e yards1 and that #b feet and #c feet make #e feet4 ,it would be absurd to try to measure a quantity of one kind with a unit which is a quantity of another kind2 for example1 to attempt to tell how many yards there are in a gallon1 or how many bushels of corn there are in a barrel of wine4 #ah4 ,all things which are true of some numbers of one unit are true of the same numbers of any other unit4 ,thus1 #ae pebbles and #g pebbles together make #bb pebbles2 #ae acres and #g acres together make #bb acres1 and so on4 ,from this we come to say that #ae and #g make #bb1 meaning that #ae things of the same kind1 and #g more of the same kind as the first1 together make #bb of that kind1 whether the kind mentioned be pebbles1 horsemen1 acres of land1 or any other4 ,for these it is but necessary to say1 once for all1 that #ae and #g make #bb4 ,therefore1 in future1 on this part of the subject ,i shall cease to talk of any particular units1 such as pebbles or acres1 and speak of numbers only4 ,a number1 considered without intending to allude to any particular things1 is called an abstract number3 and it then merely signifies repetitions of a unit1 or the number of times a unit is repeated4 #ai4 ,i will now repeat the principal things which have been mentioned in this chapter4 ,i4 ,ten signs are used1 one to stand for nothing1 the rest for the first nine numbers4 ,they are #j1 #a1 #b1 #c1 #d1 #e1 #f1 #g1 #h1 #i4 ,the first of these is called a cipher4 ,,ii4 ,higher numbers have not signs for themselves1 but are signified by placing the signs already mentioned by the side of each other1 and agreeing that the first figure on the right hand shall keep the value which it has when it stands alone2 that the second on the right hand shall mean ten times as many as it does when it stands alone2 that the third figure shall mean one hundred times as many as it does when it stands alone2 the fourth1 one thousand times as many2 and so on4 ,,iii4 ,the right hand figure is said to be in the units' place1 the next to that in the tens' place1 the third in the hundreds' place1 and so on4 ,,iv4 ,when a number is itself an exact number of tens1 hundreds1 or thousands1 c41 as many ciphers must be placed on the right of it as will bring the number into the place which is intended for it4 ,the following are examples3 ,fifty1 or five tens1 #ej3 seven hundred1 #gjj4 ,five hundred and twenty-eight thousand1 #ebhjjj4 ,if it were not for the ciphers1 these numbers would be mistaken for #e1 #g1 and #ebh4 ,v4 ,a cipher in the middle of a number becomes necessary when any one of the denominations1 units1 tens1 c4 is wanting4 ,thus1 twenty thousand and six is #bjjjf1 two hundred and six is #bjf4 ,ciphers might be placed at the beginning of a number1 but they would have no meaning4 ,thus #jbf is the same as #bf1 since the cipher merely shews that there are no hundreds1 which is evident from the number itself4 #bj4 ,if we take out of a number1 as #afghe1 any of those figures which come together1 as #fg1 and ask1 what does this sixty-seven mean8 of what is it sixty-seven8 the answer is1 sixty-seven of the same collections as the #g1 when it was in the number2 that is1 #fg hundreds4 ,for the #f is #f thousands1 or #f ten hundreds1 or sixty hundreds2 which1 with the #g1 or #g hundreds1 is #fg hundreds3 similarly1 the #fgh is #fgh tens4 ,this number may then be expressed either as #a ten thousand #f thousands #g hundreds #h tens and #e2 or #af thousands #gh tens and #e2 or #a ten thousand #fgh tens and #e2 or #afg hundreds #h tens and #e2 or #afgh tens and #e1 and so on4 #ba4 ,,exercises4 ,i4 ,write down the signs for--four hundred and seventy-six2 two thousand and ninety-seven2 sixty-four thousand three hundred and fifty2 two millions seven hundred and four2 five hundred and seventy-eight millions of millions4 ,,ii4 ,write at full length #ec1 #ahje1 #ahcj1 #ffgjg1 #ahjiagcbd1 #ffgacgba1 #ijigfcij1 #bejjjjjj4 ,,iii4 ,what alteration takes place in a number made up entirely of nines1 such as #iiiii1 by adding one to it8 ,,iv4 ,shew that a number which has five figures in it must be greater than one which has four1 though the first have none but small figures in it1 and the second none but large ones4 ,for example1 that #ajaaa is greater than #ihgi4 #bb4 ,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 placed4 ,the same advantage arises from counting in a similar way all the articles which are used in every-day life4 ,for example1 we count money by dividing it into pounds1 shillings1 and pence1 of which a shilling is #ab pence1 and a pound #bj shillings1 or #bdj pence4 ,we write a number of pounds1 shillings1 and pence in three columns1 generally placing points between the columns4 ,thus1 #bfc pence would not be written as #bfc1 but as #a4 #a4 #aa1 where shews that the #a in the first column is a pound4 ,here is a system of numeration in which a number in the second column on the right means #ab times as much as the same number in the first2 and one in the third column is twenty times as great as the same in the second1 or #bdj times as great as the same in the first4 ,in each of the tables of measures which you will hereafter meet with1 you will see a separate system of numeration1 but the methods of calculation for all will be the same4 #bc4 ,in order to make the language of arithmetic shorter1 some other signs are used4 ,they are as follow3 ,i4 #ae #ch means that #ch is to be added to #ae1 and is the same thing as #ec4 ,this is the sum of #ae and #ch1 and is read fifteen plus thirty-eight 7plus is the ,latin for more74 ,,ii4 #fd-ab means that #ab is to be taken away from #fd1 and is the same thing as #eb4 ,this is the difference of #fd and #ab1 and is read sixty-four minus twelve 7minus is the ,latin for less74 ,,iii4 #i #h means that #h is to be taken #i times1 and is the same thing as #gb4 ,this is the product of #i and #h1 and is read nine into eight4 ,,iv4 #ajh_/f means that #ajh is to be divided by #f1 or that you must find out how many sixes there are in #ajh2 and is the same thing as #ah4 ,this is the quotient of #ajh and #f2 and is read a hundred and eight by six4 ,v4 ,when two numbers1 or collections of numbers1 with the foregoing signs1 are the same1 the sign is put between them4 ,thus1 that #g and #e make #ab1 is written in this way1 #g #e #ab4 ,this is called an equation1 and is read1 seven plus five equals twelve4 ,it is plain that we may construct as many equations as we please4 ,thus3 #ab #g #i - #c #ab #a2 --- - #a #c #b #aa1 #b and so on4 #bd4 ,it often becomes necessary to speak of something which is true not of any one number only1 but of all numbers4 ,for example1 take #aj and #g2 their sum#d is #ag1 their difference is #c4 ,if this sum and difference be added together1 we get #bj1 which is twice the greater of the two numbers first chosen4 ,if from #ag we take #c1 we get #ad1 which is twice the less of the two numbers4 ,the same thing will be found to hold good of any two numbers1 which gives this general proposition1--,if the sum and difference of two numbers be added together1 the result is twice the greater of the two2 if the difference be taken from the sum1 the result is twice the lesser of the two4 ,if1 then1 we take any numbers1 and call them the first number and the second number1 and let the first number be the greater2 we have #d ,any little computations which occur in the rest of this section may be made on the fingers1 or with counters4 7#ast ,no4 #b;d ,no47 7#ast ,no4 - #b;d ,no47 twice #ast ,no4 7#ast ,no4 #b;d ,no47 - 7#ast ,no4 - #b;d ,no47 twice #b;d ,no4 ,the brackets here enclose the things which must be first done1 before the signs which join the brackets are made use of4 ,thus1 #h-7b #a7 7#a #a7 signifies that #b #a must be taken #a #a times1 and the product must be subtracted from #h4 ,in the same manner1 any result made from two or more numbers1 which is true whatever numbers are taken1 may be represented by using first ,no41 second ,no41 c41 to stand for them1 and by the signs in 7#bc74 ,but this may be much shortened2 for as first ,no41 second ,no41 c41 may mean any numbers1 the letters a and b may be used instead of these words2 and it must now be recollected that a and b stand for two numbers1 provided only that a is greater than b4 ,let twice a be represented by #b;a1 and twice b by #b;b4 ,the equations then become 7a b7 7a - b7 #b;a1 and 7a b7 - 7a - b7 #b;b4 ,this may be explained still further1 as follows3 #be4 ,suppose a number of sealed packets1 marked a1 b1 c1 d1 c41 on the outside1 each of which contains a distinct but unknown number of counters4 ,as long as we do not know how many counters each contains1 we can make the letter which belongs to each stand for its number1 so as to talk of the number a1 instead of the number in the packet marked a4 ,and because we do not know the numbers1 it does not therefore follow that we know nothing whatever about them2 for there are some connexions which exist between all numbers1 which we call general properties of numbers4 ,for example1 take any number1 multiply it by itself1 and subtract one from the result2 and then subtract one from the number itself4 ,the first of these will always contain the second exactly as many times as the original number increased by one4 ,take the number #f2 this multiplied by itself is #cf1 which diminished by one is #ce2 again1 #f diminished by #a is #e2 and #ce contains #e1 #g times1 that is1 #f #a times4 ,this will be found to be true of any number1 and1 when proved1 may be said to be true of the number contained in the packet marked a1 or of the number a4 ,if we represent a multiplied by itself by aa1#e we have1 by 7#bc7 aa - #a ------------- a #a4 a - #a #e ,this should be 7#bc7 a a1 but the sign is unnecessary here4 ,it is used with numbers1 as in #b #g1 to prevent confounding this1 which is #ad1 with #bg4 #bf4 ,when1 therefore1 we wish to talk of a number without specifying any one in particular1 we use a letter to represent it4 ,thus3 ,suppose we wish to reason upon what will follow from dividing a number into three parts1 without considering what the number is1 or what are the parts into which it is divided4 ,let a stand for the number1 and b1 c1 and d1 for the parts into which it is divided4 ,then1 by our supposition1 a b c d4 ,on this we can reason1 and produce results which do not belong to any particular number1 but are true of all4 ,thus1 if one part be taken away from the number1 the other two will remain1 or a - b c d4 ,if each part be doubled1 the whole number will be doubled1 or #b;a #b;b #b;c #b;d4 ,if we diminish one of the parts1 as d1 by a number x1 we diminish the whole number just as much1 or a - x b c 7d - x74 #bg4 ,,exercises4 ,what is a #b;b - c1 where a #ab1 b #ah1 c #g8--;,answer1 #da4 aa - bb ,what is ----------- 1 a - b where a #f and b #b8--;,ans4 #h4 ,what is the difference between 7a b77c d7 and a bc d1 for the following values of a1 b1 c1 and d8 a b c d ,ans4 #a #b #c #d #aj #b #ab #g #a #be #a #a #a #a #a ,,section ,,ii4 ,,addition ,,and ,,subtraction4 #bh4 ,there is no process in arithmetic which does not consist entirely in the increase or diminution of numbers4 ,there is then nothing which might not be done with collections of pebbles4 ,probably1 at first1 either these or the fingers were used4 ,our word calculation is derived from the ,latin word calculus1 which means a pebble4 ,shorter ways of counting have been invented1 by which many calculations1 which would require long and tedious reckoning if pebbles were used1 are made at once with very little trouble4 ,the four great methods are1 ,addition1 ,subtraction1 ,multiplication1 and ,division2 of which1 the last two are only ways of doing several of the first and second at once4 #bi4 ,when one number is increased by others1 the number which is as large as all the numbers together is called their sum4 ,the process of finding the sum of two or more numbers is called ,,addition1 and1 as was said before1 is denoted by placing a cross 77 between the numbers which are to be added together4 ,suppose it required to find the sum of #ahcd and #bgii4 ,in order to add these numbers1 take them to pieces1 dividing each into its units1 tens1 hundreds1 and thousands3 #ahcd is #a thous4 #h hund4 #c tens and #d2 #bgii is #b thous4 #g hund4 #i tens and #i4 ,each number is thus broken up into four parts4 ,if to each part of the first you add the part of the second which is under it1 and then put together what you get from these additions1 you will have added #ahcd and #bgii4 ,in the first number are #d units1 and in the second #i3 these will1 when the numbers are added together1 contribute #ac units to the sum4 ,again1 the #c tens in the first and the #i tens in the second will contribute #ab tens to the sum4 ,the #h hundreds in the first and the #g hundreds in the second will add #ae hundreds to the sum2 and the thousand in the first with the #b thousands in the second will contribute #c thousands to the sum2 therefore the sum required is #c thousands1 #ae hundreds1 #ab tens1 and #ac units4 ,to simplify this result1 you must recollect that-- #ac units are #a ten and #c units4 #ab tens are #a hund4 and #b tens4 #ae hund4 are #a thous4 and #e hund4 #c thous4 are #c thous4 ,now collect the numbers on the right hand side together1 as was done before1 and this will give1 as the sum of #ahcd and #bgii1 #d thousands1 #f hundreds1 #c tens1 and #c units1 which 7#ai7 is written #dfcc4 #cj4 ,the former process1 written with the signs of 7#bc7 is as follows3 #ahcd #a #ajjj #h #ajj #c #aj #d #bgii #b #ajjj #g #ajj #i #aj #i ,therefore1 #ahcd #bgii #c #ajjj #ae #ajj #ab #aj #ac ,but #ac #a #aj #c #ab #aj #a #ajj #b #aj #ae #ajj #a #ajjj #e #ajj #c #ajjj #c #ajjj ,therefore1 #ahcd #bgii #d #ajjj #f #ajj #c #aj #c #dfcc4 #ca4 ,the same process is to be followed in all cases1 but not at the same length4 ,in order to be able to go through it1 you must know how to add together the simple numbers4 ,this can only be done by memory2 and to help the memory you should make the following table three or four times for yourself3 ---------------------------------------- #a #b #c #d #e #f #g #h #i ---------------------------------------- #a #b #c #d #e #f #g #h #i #aj ---------------------------------------- #b #c #d #e #f #g #h #i #aj #aa ---------------------------------------- #c #d #e #f #g #h #i #aj #aa #ab ---------------------------------------- #d #e #f #g #h #i #aj #aa #ab #ac ---------------------------------------- #e #f #g #h #i #aj #aa #ab #ac #ad ---------------------------------------- #f #g #h #i #aj #aa #ab #ac #ad #ae ---------------------------------------- #g #h #i #aj #aa #ab #ac #ad #ae #af ---------------------------------------- #h #i #aj #aa #ab #ac #ad #ae #af #ag ---------------------------------------- #i #aj #aa #ab #ac #ad #ae #af #ag #ah ---------------------------------------- ,the use of this table is as follows3 ,suppose you want to find the sum of #h and #g4 ,look in the left-hand column for either of them1 #h1 for example2 and look in the top column for #g4 ,on the same line as #h1 and underneath #g1 you find #ae1 their sum4 #cb4 ,when this table has been thoroughly committed to memory1 so that you can tell at once the sum of any two numbers1 neither of which exceeds #i1 you should exercise yourself in adding and subtracting two numbers1 one of which is greater than #i and the other less4 ,you should write down a great number of such sentences as the following1 which will exercise you at the same time in addition1 and in the use of the signs mentioned in 7#bc74 #ab #f #ah #bb #f #bh #ai #h #bg #ed #i #fc #ef #g #fc #bb #h #cj #ajj - #i #ia #bg - #h #ai #dd - #f #ch1 c4 #cc4 ,when the last two articles have been thoroughly studied1 you will be able to find the sum of any numbers by the following process1#f which is the same as that in 7#bi74 #f ,in this and all other processes1 the student is strongly recommended to look at and follow the first ,appendix4 ,,rule ,i4 ,place the numbers under one another1 units under units1 tens under tens1 and so on4 ,,ii4 ,add together the units of all1 and part the whole number thus obtained into units and tens4 ,thus1 if #he be the number1 part it into #h tens and #e units2 if #acf be the number1 part it into #ac tens and #f units 7#bj74 ,,iii4 ,write down the units of this number under the units of the rest1 and keep in memory the number of tens4 ,,iv4 ,add together all the numbers in the column of tens1 remembering to take in 7or carry1 as it is called7 the tens which you were told to recollect in ,,iii41 and divide this number of tens into tens and hundreds4 ,thus1 if #cce tens be the number obtained1 part this into #cc hundreds and #e tens4 ,v4 ,place the number of tens under the tens1 and remember the number of hundreds4 ,,vi4 ,proceed in this way through every column1 and at the last column1 instead of separating the number you obtain into two parts1 write it all down before the rest4 ,,example4--,what is #ahje #cf #aigbg #c #adgd #bjjh #ahje #cf #aigbg #c #adgd #bjjh ----- #bejec ,the addition of the units' line1 or #h #d #c #g #f #e1 gives #cc1 that is1 #c tens and #c units4 ,put #c in the units' place1 and add together the line of tens1 taking in at the beginning the #c tens which were created by the addition of the units' line4 ,that is1 find #c #j #g #b #c #j1 which gives #ae for the number of tens2 that is1 #a hundred and #e tens4 ,add the line of hundreds together1 taking care to add the #a hundred which arose in the addition of the line of tens2 that is1 find #a #j #d #g #h1 which gives exactly #bj hundreds1 or #b thousands and no hundreds4 ,put a cipher in the hundreds' place 7because1 if you do not1 the next figure will be taken for hundreds instead of thousands71 and add the figures in the thousands' line together1 remembering the #b thousands which arose from the hundreds' line2 that is1 find #b #b #a #i #a1 which gives #ae thousands1 or #a ten thousand and #e thousand4 ,write #e under the line of thousands1 and collect the figures in the line of tens of thousands1 remembering the ten thousand which arose out of the thousands' line2 that is1 find #a #a1 or #b ten thousands4 ,write #b under the ten thousands' line1 and the operation is completed4 #cd4 ,as an exercise in addition1 you may satisfy yourself that what ,i now say of the following square is correct4 ,the numbers in every row1 whether reckoned upright1 or from right to left1 or from corner to corner1 when added together give the number #bdaef4 ---------------------------------------- ---- #bjafdbabafefchebabifcdib #icfcacb #egfbggb #baf ---------------------------------------- ---- #bebbjebdbdhafibchhhaccbcebh #igbcafh #fabbdab ---------------------------------------- ---- #bddh #bhhbjhhdbhdagbhcibdacfhcefdajjhbhjh #fdh ---------------------------------------- ---- #fhdbdhd #cbdbabddcbjagfdcifjadjdcbjdajddbhdd ---------------------------------------- ---- #bhhj #gbjbebj #cfjbafjdcefahjjcfjjaddjcbdjajhj ---------------------------------------- ---- #aaafbiaf #gefbeef #cifbaifciifahcfcfcfadgfcbgf ---------------------------------------- ---- #ccabaaebbieb #gibbeib #cfbbcbdjcbahgbcfgbaeab ---------------------------------------- ---- #aedhccdhaahhbihh #dcbbfbh #gbbbfhdjfhaijhcgjh ---------------------------------------- ---- #cgddaehdcchd #hbhcjbd #dfhbffd #ajhbcjddajdaidd ---------------------------------------- ---- #aihjcghjabbdcdbj #hfdcjfj #ejdbgjj #addbcdjdadj ---------------------------------------- ---- #dagfafbjchafabfjcdef #ijjcjif #edjbgcf #ahjbcgf ---------------------------------------- ---- #ce4 ,if two numbers must be added together1 it will not alter the sum if you take away a part of one1 provided you put on as much to the other4 ,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 one1 and putting them into the other4 ,thus1 #ae #g is the same as #ab #aj1 since #ab is #c less than #ae1 and #aj is three more than #g4 ,this was the principle upon which the whole of the process in 7#bi7 was conducted4 #cf4 ,let a and b stand for two numbers1 as in 7#bd74 ,it is impossible to tell what their sum will be until the numbers themselves are known4 ,in the mean while a b stands for this sum4 ,to say1 in algebraical language1 that the sum of a and b is not altered by adding c to a1 provided we take away c from b1 we have the following equation3 7a c7 7b - c7 a b2 which may be written without brackets1 thus1 a c b - c a b4 ,for the meaning of these two equations will appear to be the same1 on consideration4 #cg4 ,if a be taken twice1 three times1 c41 the results are represented in algebra by #b;a1 #c;a1 #d;a1 c4 ,the sum of any two of this series may be expressed in a shorter form than by writing the sign between them2 for though we do not know what number a stands for1 we know that1 be it what it may1 #b;a #b;a #d;a1 #c;a #b;a #e;a1 #d;a #i;a #ac;a2 and generally1 if a taken m times be added to a taken n times1 the result is a taken m n times1 or ma na 7m n7a4 #ch4 ,the use of the brackets must here be noticed4 ,they mean1 that the expression contained inside them must be used exactly as a single letter would be used in the same place4 ,thus1 pa signifies that a is taken p times1 and 7m n7a1 that a is taken m n times4 ,it is1 therefore1 a different thing from m na1 which means that a1 after being taken n times1 is added to m4 ,thus 7#c #d7 #b is #g #b or #ad2 while #c #d #b is #c #h1 or #aa4 #ci4 ,when one number is taken away from another1 the number which is left is called the difference or remainder4 ,the process of finding the difference is called ,,subtraction4 ,the number which is to be taken away must be of course the lesser of the two4 #dj4 ,the process of subtraction depends upon these two principles4 ,i4 ,the difference of two numbers is not altered by adding a number to the first1 if you add the same number to the second2 or by subtracting a number from the first1 if you subtract the same number from the second4 ,conceive two baskets with pebbles in them1 in the first of which are #ajj pebbles more than in the second4 ,if ,i put #ej more pebbles into each of them1 there are still only #ajj more in the first than in the second1 and the same if ,i take #ej from each4 ,therefore1 in finding the difference of two numbers1 if it should be convenient1 ,i may add any number ,i please to both of them1 because1 though ,i alter the numbers themselves by so doing1 ,i do not alter their difference4 ,,ii4 ,since #f exceeds #d by #b1 and #c exceeds #b by #a1 and #ab exceeds #e by #g1 #f1 #c1 and #ab together1 or #ba1 exceed #d1 #b1 and #e together1 or #aa1 by #b1 #a1 and #g together1 or #aj3 the same thing may be said of any other numbers4 #da4 ,if a1 b1 and c be three numbers1 of which a is greater than b 7#dj71 ,i4 leads to the following1 7a c7 - 7b c7 a - b4 ,again1 if c be less than a and b1 7a - c7 - 7b - c7 a - b4 ,the brackets cannot be here removed as in 7#cf74 ,that is1 p- 7q-r7 is not the same thing as p-q- r4 ,for1 in the first1 the difference of q and r is subtracted from p2 but in the second1 first q and then r are subtracted from p1 which is the same as subtracting as much as q and r together1 or q r4 ,therefore p-q-r is p-7q r74 ,in order to shew how to remove the brackets from p -7q-r7 without altering the value of the result1 let us take the simple instance #ab-7h-e74 ,if we subtract #h from #ab1 or form #ab-h1 we subtract too much2 because it is not #h which is to be taken away1 but as much of #h as is left after diminishing it by #e4 ,in forming #ab-h we have therefore subtracted #e too much4 ,this must be set right by adding #e to the result1 which gives #ab-h #e for the value of #ab-7h-e74 ,the same reasoning applies to every case1 and we have therefore1 p - 7q r7 p - q - r4 p - 7q - r7 p - q r4 ,by the same kind of reasoning1 a - 7b c - d - e7 a - b - c d e4 #b;a #c;b - 7a - #b;b7 #b;a #c;b - a #b;b a #e;b4 #dx y - 7#agx - #iy7 #dx y - #agx #iy #ajy - #acx4 #db4 ,i want to find the difference of the numbers #eggfb and #cdfca4 ,take these to pieces as in 7#bi7 and #eggfb is #e ten-th4 #g th4 #g hund4 #f tens and #b units4 #cdfca is #c ten-th4 #d th4 #f hund4 #c tens and #a unit4 ,now #b units exceed #a unit by #a unit4 #f tens #c tens #c tens4 #g hundreds #f hundreds #a hundred4 #g thousands #d thousands #c thousands4 #e ten-thousands #c ten-thous4 #b ten-thous4 ,therefore1 by 7#dj1 ,principle ,,ii47 all the first column together exceeds all the second column by all the third column1 that is1 by #b ten-th4 #c th4 #a hund4 #c tens and #a unit1 which is #bcaca4 ,therefore the difference of #eggfb and #cdfca is #bcaca1 or #eggfb-cdfca #bcaca4 #dc4 ,suppose ,i want to find the difference between #fabgd and #cifbh4 ,write them at length1 and #fabgd is #f ten-th4 #a th4 #b hund4 #g tens and #d units4 #cifbh is #c ten-th4 #i th4 #f hund4 #b tens and #h units4 ,if we attempt to do the same as in the last article1 there is a difficulty immediately1 since #h1 being greater than #d1 cannot be taken from it4 ,but from 7#dj7 it appears that we shall not alter the difference of two numbers if we add the same number to both of them4 ,add ten to the first number1 that is1 let there be #ad units instead of four1 and add ten also to the second number1 but instead of adding ten to the number of units1 add one to the number of tens1 which is the same thing4 ,the numbers will then stand thus1 #f ten-thous4 #a thous4 #b hund4 #g tens and #ad units4#g #c ten-thous4 #i thous4 #f hund4 #c tens and #h units4 #g ,those numbers which have been altered are put in italics4 ,you now see that the units and tens in the lower can be subtracted from those in the upper line1 but that the hundreds cannot4 ,to remedy this1 add one thousand or #aj hundred to both numbers1 which will not alter their difference1 and remember to increase the hundreds in the upper line by #aj1 and the thousands in the lower line by #a1 which are the same things4 ,and since the thousands in the lower cannot be subtracted from the thousands in the upper line1 add #a ten thousand or #aj thousand to both numbers1 and increase the thousands in the upper line by #aj1 and the ten thousands in the lower line by #a1 which are the same things2 and at the close the numbers which we get will be1 #f ten-thous4 #aa thous4 #ab hund4 #g tens and #ad units4 #d ten-thous4 #aj thous4 #f hund4 #c tens and #h units4 ,these numbers are not1 it is true1 the same as those given at the beginning of this article1 but their difference is the same1 by 7#dj74 ,with the last-mentioned numbers proceed in the same way as in 7#db71 which will give1 as their difference1 #b ten-thous4 #a thous4 #f hund4 #d tens1 and #f units1 which is #bafdf4 #dd4 ,from this we deduce the following rules for subtraction3 ,i4 ,write the number which is to be subtracted 7which is1 of course1 the lesser of the two1 and is called the subtrahend7 under the other1 so that its units shall fall under the units of the other1 and so on4 ,,ii4 ,subtract each figure of the lower line from the one above it1 if that can be done4 ,where that cannot be done1 add ten to the upper figure1 and then subtract the lower figure2 but recollect in this case always to increase the next figure in the lower line by #a1 before you begin to subtract it from the upper one4 #de4 ,if there should not be as many figures in the lower line as in the upper one1 proceed as if there were as many ciphers at the beginning of the lower line as will make the number of figures equal4 ,you do not alter a number by placing ciphers at the beginning of it4 ,for example1 #jjhah is the same number as #hah1 for it means #j ten-thous4 #j thous4 #h hunds4 #a ten and #h units2 the first two signs are nothing1 and the rest is #h hundreds1 #a ten1 and #h units1 or #hah4 ,the second does not differ from the first1 except in its being said that there are no thousands and no tens of thousands in the number1 which may be known without their being mentioned at all4 ,you may ask1 perhaps1 why this does not apply to a cipher placed in the middle of a number1 or at the right of it1 as1 for example1 in #bhjjg and #cigjj8 ,but you must recollect1 that if it were not for the two ciphers in the first1 the #h would be taken for #h tens1 instead of #h thousands2 and if it were not for the ciphers in the second1 the #g would be taken for #g units1 instead of #g hundreds4 #df4 ,,example4 ,what is the difference between #cgjhbiafdjjcjagd and #cjhacfdibgfahh ---------------- ,difference #cfggdggiijgecihf ,,exercises4 ,i4 ,what is #ahccg #adibfcbjj - #fdgbijb8--;,answer #adbhjhfce4 ,what is #ajjj - #dfd #cbgi - #fdf8--;,ans4 #cafi4 ,,ii4 ,subtract #fd #gf #add - #ah from #cc - #b #ajjjcg4--;,ans4 #iihjb4 ,,iii4 ,what shorter rule might be made for subtraction when all the figures in the upper line are ciphers except the first8 for example1 in finding #ajjjjjjj - #bgcafcd4 ,,iv4 ,find #ahcfb #bdfi and #ahcfb-bdfi1 add the second result to the first1 and then subtract #ahcfb2 subtract the second from the first1 and then subtract #bdfi4--;,answer #ahcfb and #bdfi4 ,v4 ,there are four places on the same line in the order ,a1 ,b1 ,c1 and ,d4 ,from ,a to ,d it is #adfc miles2 from ,a to ,c it is #gbh miles2 and from ,b to ,d it is #acag miles4 ,how far is it from ,a to ,b1 from ,b to ,c1 and from ,c to ,d8--,answer4 ,from ,a to ,b #adf1 from ,b to ,c #ehb1 and from ,c to ,d #gce miles4 ,,vi4 ,in the following table subtract ,b from ,a1 and ,b from the remainder1 and so on until ,b can be no longer subtracted4 ,find how many times ,b can be subtracted from ,a1 and what is the last remainder4 ,no4 of ,a ,b times4 ,remainder4 #bcfjd #iiii #b #cfjf #bjiifa #cgagc #e #bdjif #gdgab #fgib #aa #j #dhjbdfi #fedcba #g #bbbbbb #ahhdigdg #cadaeib #f #aie #ihgfedcba #abcdefghi #h #i ,,section ,,iii4 ,,multiplication4 #dg4 ,i have said that all questions in arithmetic require nothing but addition and subtraction4 ,i do not mean by this that no rule should ever be used except those given in the last section1 but that all other rules only shew shorter ways of finding what might be found1 if we pleased1 by the methods there deduced4 ,even the last two rules themselves are only short and convenient ways of doing what may be done with a number of pebbles or counters4 #dh4 ,i want to know the sum of five seventeens1 or ,i ask the following question3 ,there are five heaps of pebbles1 and seventeen pebbles in each heap2 how many are there in all8 ,write five seventeens in a column1 and make the addition1 which gives #he4 ,in this case #he is called the product of #e and #ag1 and the process of finding the product is called ,,multiplication1 which gives nothing more than the addition of a number of the same quantities4 ,here #ag is called the multiplicand1 and #e is called the multiplier4 #ag #ag #ag #ag #ag ---- #he #di4 ,if no question harder than this were ever proposed1 there would be no occasion for a shorter way than the one here followed4 ,but if there were #acfg heaps of pebbles1 and #dbi in each heap1 the whole number is then #acfg times #dbi1 or #dbi multiplied by #acfg4 ,i should have to write #dbi #acfg times1 and then to make an addition of enormous length4 ,to avoid this1 a shorter rule is necessary1 which ,i now proceed to explain4 #ej4 ,the student must first make himself acquainted with the products of all numbers as far as #aj times #aj by means of the following table1#h which must be committed to memory4 ---------------------------------------- -------- #a #b #c #d #e #f #g #h #i #aj #aa #ab ---------------------------------------- -------- #b #d #f #h #aj #ab #ad #af #ah #bj #bb #bd ---------------------------------------- -------- #c #f #i #ab #ae #ah #ba #bd #bg #cj #cc #cf ---------------------------------------- -------- #d #h #ab #af #bj #bd #bh #cb #cf #dj #dd #dh ---------------------------------------- -------- #e #aj #ae #bj #be #cj #ce #dj #de #ej #ee #fj ---------------------------------------- -------- #f #ab #ah #bd #cj #cf #db #dh #ed #fj #ff #gb ---------------------------------------- -------- #g #ad #ba #bh #ce #db #di #ef #fc #gj #gg #hd ---------------------------------------- -------- #h #af #bd #cb #dj #dh #ef #fd #gb #hj #hh #if ---------------------------------------- -------- #i #ah #bg #cf #de #ed #fc #gb #ha #ij #ii #ajh ---------------------------------------- -------- #aj #bj #cj #dj #ej #fj #gj #hj #ij #ajj #aaj #abj ---------------------------------------- -------- #aa #bb #cc #dd #ee #ff #gg #hh #ii #aaj #aba #acb ---------------------------------------- -------- #ab #bd #cf #dh #fj #gb #hd #if #ajh #abj #acb #add ---------------------------------------- -------- #h ,as it is usual to learn the product of numbers up to #ab times #ab1 ,i have extended the table thus far4 ,in my opinion1 all pupils who shew a tolerable capacity should slowly commit the products to memory as far as #bj times #bj1 in the course of their progress through this work4 ,if from this table you wish to know what is #g times #f1 look in the first upright column on the left for either of them2 #f for example4 ,proceed to the right until you come into the column marked #g at the top4 ,you there find #db1 which is the product of #f and #g4 #ea4 ,you may find1 in this way1 either #f times #g1 or #g times #f1 and for both you find #db4 ,that is1 six sevens is the same number as seven sixes4 ,this may be shewn as follows3 ,place seven counters in a line1 and repeat that line in all six times4 ,the number of counters in the whole is #f times #g1 or six sevens1 if ,i reckon the rows from the top to the bottom2 but if ,i count the rows that stand side by side1 ,i find seven of them1 and six in each row1 the whole number of which is #g times #f1 or seven sixes4 ,and the whole number is #db1 whichever way ,i count4 ,the same method may be applied to any other two numbers4 ,if the signs of 7#bc7 were used1 it would be said that #g #f #f #g4 #eb4 ,to take any quantity a number of times1 it will be enough to take every one of its parts the same number of times4 ,thus1 a sack of corn will be increased fifty-fold1 if each bushel which it contains be replaced by #ej bushels4 ,a country will be doubled by doubling every acre of land1 or every county1 which it contains4 ,simple as this may appear1 it is necessary to state it1 because it is one of the principles on which the rule of multiplication depends4 #ec4 ,in order to multiply by any number1 you may multiply separately by any parts into which you choose to divide that number1 and add the results4 ,for example1 #d and #b make #f4 ,to multiply #g by #f first multiply #g by #d1 and then by #b1 and add the products4 ,this will give #db1 which is the product of #g and #f4 ,again1 since #eg is made up of #cb and #be1 #eg times #ej is made up of #cb times #ej and #be times #ej1 and so on4 ,if the signs were used1 these would be written thus3 #g #f #g #d #g #b4 #ej #eg #ej #cb #ej #be4 #ed4 ,the principles in the last two articles may be expressed thus3 ,if a be made up of the parts x1 y1 and x1 ma is made up of mx1 my1 and mz2 or1 if a x y z4 ma mx my mz1 or1 m7x y z7 mx my mz4 ,a similar result may be obtained if a1 instead of being made up of x1 y1 and z1 is made by combined additions and subtractions1 such as x y-z1 x- y z1 x-y-z1 c4 ,to take the first as an instance3 ,let a x y - z1 then ma mx my - mz4 ,for1 if a had been x y1 ma would have been mx my4 ,but since a is less than x y by z1 too much by z has been repeated every time that x y has been repeated2--that is1 mz too much has been taken2 consequently1 ma is not mx my1 but mx my-mz4 ,similar reasoning may be applied to other cases1 and the following results may be obtained3 m7a b c - d7 ma mb mc - md4 a7a - b7 aa - ab4 b7a - b7 ba - bb4 #c7b;a - #d;b7 #f;a - #ab;b4 #g;a7#g #b;b7 #di;a #ad;ab4 7aa a #a7;a aaa aa a4 7#c;ab - #b;c7#d;abc #ab;aabbc - #h;abcc4 #ee4 ,there is another way in which two numbers may be multiplied together4 ,since #h is #d times #b1 #g times #h may be made by multiplying #g and #d1 and then multiplying that product by #b4 ,to shew this1 place #g counters in a line1 and repeat that line in all #h times1 as in figures ,i4 and ,,ii4 ,i4 --------------- ,a --------------- --------------- ,b --------------- ,,ii4 --------------- --------------- --------------- --------------- --------------- --------------- --------------- --------------- ,the number of counters in all is #h times #g1 or #ef4 ,but 7as in fig4 ,i47 enclose each four rows in oblong figures1 such as ,a and ,b4 ,the number in each oblong is #d times #g1 or #bh1 and there are two of those oblongs2 so that in the whole the number of counters is twice #bh1 or #bh x #b1 or #g first multiplied by #d1 and that product multiplied by #b4 ,in figure ,,ii4 it is shewn that #g multiplied by #h is also #g first multiplied by #b1 and that product multiplied by #d4 ,the same method may be applied to other numbers4 ,thus1 since #hj is #h times #aj1 #bef times #hj is #bef multiplied by #h1 and that product multiplied by #aj4 ,if we use the signs1 the foregoing assertions are made thus3 #g #h #g #d #b #g #b #d4 #bef #hj #bef #h #aj #bef #aj #h4 ,,exercises4 ,shew that #b #c #d #e #b #d #c #e #e #d #b #c1 c4 ,shew that #ah #ajj #ah #eg #ah #dc4 #ef4 ,articles 7#ea7 and 7#ee7 may be expressed in the following way1 where by ab we mean a taken b times2 by abc1 a taken b times1 and the result taken c times4 ab ba4 abc acb bca bac1 c4 abc a 7bc7 b 7ca7 c 7ab74 ,if we would say that the same results are produced by multiplying by b1 c1 and d1 one after the other1 and by the product bcd at once1 we write the following3 a b c d a bcd4 ,the fact is1 that if any numbers are to be multiplied together1 the product of any two or more may be formed1 and substituted instead of those two or more2 thus1 the product abcdef may be formed by multiplying ab cde f abf de c abc def c4 #eg4 ,in order to multiply by #aj1 annex a cipher to the right hand of the multiplicand4 ,thus1 #aj times #bcef is #bcefj4 ,to shew this1 write #bcef at length which is #b thousands1 #c hundreds1 #e tens1 and #f units4 ,take each of these parts ten times1 which1 by 7#eb71 is the same as multiplying the whole number by #aj1 and it will then become #b tens of thou4 #c tens of hun4 #e tens of tens1 and #f tens1 which is #b ten-thou4 #c thous4 #e hun4 and #f tens4 ,this must be written #bcefj1 because #f is not to be #f units1 but #f tens4 ,therefore #bcef #aj #bcefj4 ,in the same way you may shew1 that in order to multiply by #ajj you must affix two ciphers to the right2 to multiply by #ajjj you must affix three ciphers1 and so on4 ,the rule will be best caught from the following table3 #ac #aj #acj #ac #ajj #acjj #ac #ajjj #acjjj #ac #ajjjj #acjjjj #adb #ajjj #adbjjj #bcgjj #aj #bcgjjj #cjdj #ajjj #cjdjjjj #ajjjj #ajjjjj #ajjjjjjjjj #eh4 ,i now shew how to multiply by one of the numbers1 #b1 #c1 #d1 #e1 #f1 #g1 #h1 or #i4 ,i do not include #a1 because multiplying by #a1 or taking the number once1 is what is meant by simply writing down the number4 ,i want to multiply #acfh by #h4 ,write the first number at full length1 which is #a thousand1 #c hundreds1 #f tens1 and #h units4 ,to multiply this by #h1 multiply each of these parts by #h 7#ej7 and 7#eb71 which will give #h thousands1 #bd hundreds1 #dh tens1 and #fd units4 ,now #fd units are written thus #fd #dh tens #dhj #bd hundreds #bdjj #h thousands #hjjj ,add these together1 which gives #ajidd as the product of #acfh and #h1 or #acfh #h #ajidd4 ,by working a few examples in this way you will see for following rule4 #ei4 ,i4 ,multiply the first figure of the multiplicand by the multiplier1 write down the units' figure1 and reserve the tens4 ,,ii4 ,do the same with the second figure of the multiplicand1 and add to the product the number of tens from the first2 put down the units' figure of this1 and reserve the tens4 ,,iii4 ,proceed in this way till you come to the last figure1 and then write down the whole number obtained from that figure4 ,,iv4 ,if there be a cipher in the multiplicand1 treat it as if it were a number1 observing that #j #a #j1 #j #b #j1 c4 #fj4 ,in a similar way a number can be multiplied by a figure which is accompanied by ciphers1 as1 for example1 #hjjj4 ,for #hjjj is #h #ajjj1 and therefore 7#ee7 you must first multiply by #h and then by #ajjj1 which last operation 7#eg7 is done by placing #c ciphers on the right4 ,hence the rule in this case is1 multiply by the simple number1 and place the number of ciphers which follow it at the right of the product4 ,,example4 ,multiply #afgidbchjjhgb by #fjjjj ------------------ #ajjgfedbhjebcbjjjj #fa4 ,,exercises4 ,what is #ajjgcfj #g8 ,answer1 #gjeaebj4 #abcdefghi #i #aj and #abc #i #d8--;,ans4 #aaaaaaaaaa and #aaaa4 ,what is #acf #c #abi #d #adg #h #bg #cjjj8--;,ans4 #hcajj4 ,an army is made up of #cc regiments of infantry1 each containing #hjj men2 #ad of cavalry1 each containing #fjj men2 and #b of artillery1 each containing #cjj men4 ,the enemy has #f more regiments of infantry1 each containing #ajj more men2 #c more regiments of cavalry1 each containing #ajj men less2 and #d corps of artillery of the same magnitude as those of the first3 two regiments of cavalry and one of infantry desert from the former to the latter4 ,how many men has the second army more than the first8--,answer1 #acdjj4 #fb4 ,suppose it is required to multiply #bcgjg by #defg4 ,since #defg is made up of #djjj1 #ejj1 #fj1 and #g1 by 7#ec7 we must multiply #bcgjg by each of these1 and add the products4 ,now 7#eh7 #bcgjg #g is #afeidi 7#fj7 #bcgjg #fj is #adbbdbj #bcgjg #ejj is #aahecejj #bcgjg #djjj is #idhbhjjj --------- ,the sum of these is #ajhbfihfi which is the product required4 ,it will do as well if1 instead of writing the ciphers at the end of each line1 we keep the other figures in their places without them4 ,if we take away the ciphers1 the second line is one place to the left of the first1 the third one place to the left of the second1 and so on4 ,write the multiplier and the multiplicand over these lines1 and the process will stand thus3 #bcgjg #defg ------ #afeidi #adbbdb #aahece #idhbh --------- #ajhbfihfi #fc4 ,there is one more case to be noticed2 that is1 where there is a cipher in the middle of the multiplier4 ,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 arises4 ,suppose it required to multiply #cfe by #ajajja4 ,the multiplier is made up of #ajjjjj1 #ajjj and #a4 ,proceed as before1 and #cfe #a is #cfe 7#eg7 #cfe #ajjj is #cfejjj #cfe #ajjjjj is #cfejjjjj -------- ,the sum of which is #cfhfecfe and the whole process with the ciphers struck off is3 #cfe #ajajja ------ #cfe #cfe #cfe -------- #cfhfecfe #fd4 ,the following is the rule in all cases3 ,i4 ,place the multiplier under the multiplicand1 so that the units of one may be under those of the other4 ,,ii4 ,multiply the whole multiplicand by each figure of the multiplier 7#ei71 and place the unit of each line in the column under the figure of the multiplier from which it came4 ,,iii4 ,add together the lines obtained by ,,ii4 column by column4 #fe4 ,when the multiplier or multiplicand1 or both1 have ciphers on the right hand1 multiply the two together without the ciphers1 and then place on the right of the product all the ciphers that are on the right both of the multiplier and multiplicand4 ,for example1 what is #cbjj #cjjj8 ,first1 #cbjj is #cb #ajj1 or one hundred times as great as #cb4 ,again1 #cb #acjjj is #cb #ac1 with three ciphers affixed1 that is #daf1 with three ciphers affixed1 or #dafjjj4 ,but the product required must be #ajj times as great as this1 or must have two ciphers affixed4 ,it is therefore #dafjjjjj1 having as many ciphers as are in both multiplier and multiplicand4 #ff4 ,when any number is multiplied by itself any number of times1 the result is called a power of that number4 ,thus3 #f is called the first power of #f #f #f second power of #f #f #f #f third power of #f #f #f #f #f fourth power of #f c4 c4 ,the second and third powers are usually called the square and cube1 which are incorrect names1 derived from certain connexions of the second and third power with the square and cube in geometry4 ,as exercises in multiplication1 the following powers are to be found4 ,number proposed4 ,square4 ,cube4 #igb #iddghd #iahccjjdh #ajjh #ajafjfd #ajbdaibeab #cadb #ihgbafd #cajahccibhh #cafc #ajjjdefi #cafdddeagdg #eeee #cjhehjbe #agadafcbhhge #fghi #dfjijeba #cabijhedgjfi ,the fifth power of #cf is #fjdffagf fourth #ej #fbejjjj fourth #ajh #acfjdhhif fourth #bgg #ehhgccidda #fg4 ,it is required to multiply a b by c d1 that is1 to take a b as many times as there are units in c d4 ,by 7#ec7 a b must be taken c times1 and d times1 or the product required is 7a b7c 7a b7d4 ,but 7#eb7 7a b7c is ac bc1 and 7a b7d is ad bd2 whence the product required is ac bc ad bd2 or1 7a b77c d7 ac bc ad bd4 ,by similar reasoning 7a - b77c d7 is 7a - b7c 7a - b7d2 or1 7a - b77c d7 ac - bc ad - bd4 ,to multiply a-b by c-d1 first take a-b c times1 which gives ac-bc4 ,this is not correct2 for in taking it c times instead of c-d times1 we have taken it d times too many2 or have made a result which is 7a-b7d too great4 ,the real result is therefore ac-bc-7a -b7d4 ,but 7a-b7d is ad- bd1 and therefore 7a - b77c - d7 ac - bc - ad - bd ac - bc - ad bd 7#da7 ,from these three examples may be collected the following rule for the multiplication of algebraic quantities3 ,multiply each term of the multiplicand by each term of the multiplier2 when the two terms have both or both-before them1 put before their product2 when one has and the other-1 put-before their product4 ,in using the first terms1 which have no sign1 apply the rule as if they had the sign 4 #fh4 ,for example1 7a b77a b7 gives aa ab ab bb4 ,but ab ab is #b;ab2 hence the square of a b is aa #b;ab bb4 ,again 7a- b77a-b7 gives aa-ab-ab bb4 ,but two subtractions of ab are equivalent to subtracting #b;ab2 hence the square of a- b is aa-#b;ab bb4 ,again1 7a b77a-b7 gives aa ab-ab -bb4 ,but the addition and subtraction of ab makes no change2 hence the product of a b and a- b is aa-bb4 ,again1 the square of a b c d or 7a b c d77a b c d7 will be found to be aa #b;ab #b;ac #b;ad bb #b;bc #b;bd cc #b;cd dd2 or the rule for squaring such a quantity is3 ,square the first term1 and multiply all that come after by twice that term2 do the same with the second1 and so on to the end4 ,,section ,,iv4 ,,division4 #fi4 ,suppose ,i ask whether #aef can be divided into a number of parts each of which is #ac1 or how many thirteens #aef contains2 ,i propose a question1 the solution of which is called ,,division4 ,in this case1 #aef is called the dividend1 #ac the divisor1 and the number of parts required is the quotient2 and when ,i find the quotient1 ,i am said to divide #aef by #ac4 #gj4 ,the simplest method of doing this is to subtract #ac from #aef1 and then to subtract #ac from the remainder1 and so on2 or1 in common language1 to tell off #aef by thirteens4 ,a similar process has already occurred in the exercises on subtraction1 ,art4 7#df74 ,do this1 and mark one for every subtraction that is made1 to remind you that each subtraction takes #ac once from #aef1 which operations will stand as follows3 #aef #ac #a ------ #adc #ac #a ------ #acj #ac #a ------ #aag #ac #a ------ #ajd #ac #a ------ #ia #ac #a ------ #gh #ac #a ------ #fe #ac #a ------ #eb #ac #a ------ #ci #ac #a ------ #bf #ac #a ------ #ac #ac #a ------ #j ,begin by subtracting #ac from #aef1 which leaves #adc4 ,subtract #ac from #adc1 which leaves #acj2 and so on4 ,at last #ac only remains1 from which when #ac is subtracted1 there remains nothing4 ,upon counting the number of times which you have subtracted #ac1 you find that this number is #ab2 or #aef contains twelve thirteens1 or contains #ac twelve times4 ,this method is the most simple possible1 and might be done with pebbles4 ,of these you would first count #aef4 ,you would then take #ac from the heap1 and put them into one heap by themselves4 ,you would then take another #ac from the heap1 and place them in another heap by themselves2 and so on until there were none left4 ,you would then count the number of heaps1 which you would find to be #ab4 #ga4 ,division is the opposite of multiplication4 ,in multiplication you have a number of heaps1 with the same number of pebbles in each1 and you want to know how many pebbles there are in all4 ,in division you know how many there are in all1 and how many there are to be in each heap1 and you want to know how many heaps there are4 #gb4 ,in the last example a number was taken which contains an exact number of thirteens4 ,but this does not happen with every number4 ,take1 for example1 #aei4 ,follow the process of 7#gj71 and it will appear that after having subtracted #ac twelve times1 there remains #c1 from which #ac cannot be subtracted4 ,we may say then that #aei contains twelve thirteens and #c over2 or that #aei1 when divided by #ac1 gives a quotient #ab1 and a remainder #c4 ,if we use signs1 #aei #ac #ab #c4 ,,exercises4 #adf #bd #f #b1 or #adf contains six twenty-fours and #b over4 #adf #f #bd #b1 or #adf contains twenty-four sixes and #b over4 #cjj #db #g #f1 or #cjj contains seven forty-twos and #f over4 #cifbd #gbgg #e #cbci4 #gc4 ,if a contain b q times with a remainder r1 a must be greater than bq by r2 that is1 a bq r4 ,if there be no remainder1 a bq4 ,here a is the dividend1 b the divisor1 q the quotient1 and r the remainder4 ,in order to say that a contains b q times1 we write1 a_/b q1 or a 3 b q1 which in old books is often found written thus3 a b q4 #gd4 ,if ,i divide #aef into several parts1 and find how often #ac is contained in each of them1 it is plain that #aef contains #ac as often as all its parts together4 ,for example1 #aef is made up of #ia1 #ci1 and #bf4 ,of these #ia contains #ac #g times1 #ci contains #ac #c times1 #bf contains #ac #b times2 therefore #ia #ci #bf contains #ac #g #c #b times1 or #ab times4 ,again1 #aef is made up of #ajj1 #ej1 and #f4 ,now #ajj contains #ac #g times and #i over1 #ej contains #ac #c times and #aa over1 #f contains #ac #j times#i and #f over4 #i ,to speak always in the same way1 instead of saying that #f does not contain #ac1 ,i say that it contains it #j times and #f over1 which is merely saying that #f is #f more than nothing4 ,therefore #ajj #ej #f contains #ac #g #c #j times and #i #aa #f over2 or #aef contains #ac #aj times and #bf over4 ,but #bf is itself #b thirteens2 therefore #aef contains #aj thirteens and #b thirteens1 or #ab thirteens4 #ge4 ,the result of the last article is expressed by saying1 that if a b c d1 then a_/m b_/m c_/m d_/m #gf4 ,in the first example ,i did not take away #ac more than once at a time1 in order that the method might be as simple as possible4 ,but if ,i know what is twice #ac1 #c times #ac1 c41 ,i can take away as many thirteens at a time as ,i please1 if ,i take care to mark at each step how many ,i take away4 ,for example1 take away #ac ten times at once from #aef1 that is1 take away #acj1 and afterwards take away #ac twice1 or take away #bf1 and the process is as follows3 #aef #acj #aj times #ac4 --- #bf #bf #b times #ac4 --- #j ,therefore #aef contains #ac #aj #b1 or #ab times4 ,again1 to divide #cjif by #ah4 #cjif #ahjj #ajj times #ah4 ---- #abif #ijj #ej times #ah4 ---- #cif #cfj #bj times #ah4 ---- #cf #cf #b times #ah4 ---- #j ,therefore #cjif contains #ah #ajj #ej #bj #b1 or #agb times4 #gg4 ,you will now understand the following sentences1 and be able to make similar assertions of other numbers4 #dej is #ge #f2 it therefore contains any number1 as #e1 #f times as often as #ge contains it4 #ace contains #c more than #bf times2 therefore1 ,twice #ace 0 #c 0 #eb or twice #bf times4 #aj times #ace 0 #c 0 #bfj or #aj times #bf #ej times #ace 0 #c 0 #acjj or #ej times #bf #dgb contains #ah more than #ba times2 therefore1 #dgbj contains #ah more than #baj times1 #dgbjj contains #ah more than #bajj times1 #dgbjjj contains #ah more than #bajjj times1 #cb contains #ab more than #b times1 and less than #c times4 #cbj 0 #ab 0 #bj #cj #cbjj 0 #ab 0 #bjj #cjj #cbjjj 0 #ab 0 #bjjj #cjjj c4 c4 c4 #gh4 ,the foregoing articles contain the principles of division4 ,the question now is1 to apply them in the shortest and most convenient way4 ,suppose it required to divide #djfh by #ah1 or to find #djfh_/ah 7#bc74 ,if we divide #djfh into any number of parts1 we may1 by the process followed in 7#gd71 find how many times #ah is contained in each of these parts1 and from thence how many times it is contained in the whole4 ,now1 what separation of #djfh into parts will be most convenient8 ,observe that #d1 the first figure of #djfh1 does not contain #ah2 but that #dj1 the first and second figures together1 does contain #ah more than twice1 but less than three times4#aj ,but #djfh 7#bj7 is made up of #dj hundreds1 and #fh2 of which1 #dj hundreds 7#gg7 contains #ah more than #bjj times1 and less than #cjj times4 ,therefore1 #djfh also contains more than #bjj times #ah1 since it must contain #ah more times than #djjj does4 ,it also contains #ah less than #cjj times1 because #cjj times #ah is #edjj1 a greater number than #djfh4 ,subtract #ah #bjj times from #djfh2 that is1 subtract #cfjj1 and there remains #dfh4 ,therefore1 #djfh contains #ah #bjj times1 and as many more times as #dfh contains #ah4 #aj ,if you have any doubt as to this expression1 recollect that it means 8contains more than two eighteens1 but not so much as three40 ,it remains1 then1 to find how many times #dfh contains #ah4 ,proceed exactly as before4 ,observe that #df contains #ah more than twice1 and less than #c times2 therefore1 #dfj contains it more than #bj1 and less than #cj times 7#gg72 as does also #dfh4 ,subtract #ah #bj times from #dfh1 that is1 subtract #cfj2 the remainder is #ajh4 ,therefore1 #dfh contains #ah #bj times1 and as many more as #ajh contains it4 ,now1 #ajh is found to contain #ah #f times exactly2 therefore1 #dfh contains it #bj #f times1 and #djfh contains it #bjj #bj #f times1 or #bbf times4 ,if we write down the process that has been followed1 without any explanation1 putting the divisor1 dividend1 and quotient1 in a line separated by parentheses it will stand1 as in example7,a74 ,let it be required to divide #cfcbfeii by #acdb 7,b74 ,a4 ,b4 #ah7djfh7bjj #bj #f #acdb7cfcbfeii7bjjjj #gjjj #fj #i #cfjj #bfhdjjjj ---- -------- #dfh #idhfeii #cfj #icidjjj --- ------- #ajh #ibeii #ajh #hjebj --- ----- #j #abjgi #abjgh ----- #a ,as in the previous example1 #cfcbfeii is separated into #cfcbjjjj and #feii2 the first four figures #cfcb being separated from the rest1 because it takes four figures from the left of the dividend to make a number which is greater than the divisor4 ,again1 #cfcbjjjj is found to contain #acdb more than #bjjjj1 and less than #cjjjj times2 and #acdb #bjjjj is subtracted from the dividend1 after which the remainder is #idhfeii4 ,the same operation is repeated again and again1 and the result is found to be1 that there is a quotient #bjjjj #gjjj #fj #i1 or #bgjfi1 and a remainder #a4 ,before you proceed1 you should now repeat the foregoing article at length in the solution of the following questions4 ,what are #ajjichgd #ffggiibb #bgahbah -------- 1 -------- 1 ------- 8 #cbjg #aaddcc #acceb the quotients of which are #cadg1 #ehc1 #bjc2 and the remainders #adde1 #fedhc1 #ggfb4 #gi4 ,in the examples of the last article1 observe1 #ast1 that it is useless to write down the ciphers which are on the right of each subtrahend1 provided that without them you keep each of the other figures in its proper place3 #b;d1 that it is useless to put down the right hand figures of the dividend so long as they fall over ciphers1 because they do not begin to have any share in the making of the quotient until1 by continuing the process1 they cease to have ciphers under them3 #c;d1 that the quotient is only a number written at length1 instead of the usual way4 ,for example1 the first quotient is #bjj #bj #f1 or #bbf2 the second is #bjjjj #gjjj #fj #i1 or #bgjfi4 ,strike out1 therefore1 all the ciphers and the numbers which come above them1 except those in the first line1 and put the quotient in one line2 and the two examples of the last article will stand thus3 #ah7djfh7bbf #acdb7cfcbfeii7bgjfi #cf #bfhd --- ----- #df #idhf #cf #icid --- ----- #ajh #ibei #ajh #hjeb --- ----- #j #abjgi #abjgh ----- #a #hj4 ,hence the following rule is deduced3 ,i4 ,write the divisor and dividend in one line1 and place parentheses on each side of the dividend4 ,,ii4 ,take off from the left-hand of the dividend the least number of figures which make a number greater than the divisor2 find what number of times the divisor is contained in these1 and write this number as the first figure of the quotient4 ,,iii4 ,multiply the divisor by the last-mentioned figure1 and subtract the product from the number which was taken off at the left of the dividend4 ,,iv4 ,on the right of the remainder place the figure of the dividend which comes next after those already separated in ,,ii43 if the remainder thus increased be greater than the divisor1 find how many times the divisor is contained in it2 put this number at the right of the first figure of the quotient1 and repeat the process3 if not1 on the right place the next figure of the dividend1 and the next1 and so on until it is greater2 but remember to place a cipher in the quotient for every figure of the dividend which you are obliged to take1 except the first4 ,v4 ,proceed in this way until all the figures of the dividend are exhausted4 ,in judging how often one large number is contained in another1 a first and rough guess may be made by striking off the same number of figures from both1 and using the results instead of the numbers themselves4 ,thus1 #d1gcb is contained in #ad1cgi about the same number of times that #d is contained in #ad1 or about #c times4 ,the reason is1 that #d being contained in #ad as often as #djjj is in #adjjj1 and these last only differing from the proposed numbers by lower denominations1 viz4 hundreds1 c4 we may expect that there will not be much difference between the number of times which #adjjj contains #djjj1 and that which #adcgi contains #dgcb3 and it generally happens so4 ,but if the second figure of the divisor be #e1 or greater than #e1 it will be more accurate to increase the first figure of the divisor by #a1 before trying the method just explained4 ,nothing but practice can give facility in this sort of guess-work4 #ha4 ,this process may be made more simple when the divisor is not greater than #ab1 if you have sufficient knowledge of the multiplication table 7#ej74 ,for example1 ,i want to divide #acbigf by #d4 ,at full length the process stands thus3 #d7acbigf7ccbdd #ab --- #ab #ab --- #i #h -- #ag #af --- #af #af -- #j ,but you will recollect1 without the necessity of writing it down1 that #ac contains #d three times with a remainder #a2 this #a you will place before #b1 the next figure of the dividend1 and you know that #ab contains #d #c times exactly1 and so on4 ,it will be more convenient to write down the quotient thus3 #d7acbigf ------- #ccbdd ,while on this part of the subject1 we may mention1 that the shortest way to multiply by #e is to annex a cipher and divide by #b1 which is equivalent to taking the half of #aj times1 or #e times4 ,to divide by #e1 multiply by #b and strike off the last figure1 which leaves the quotient2 half the last figure is the remainder4 ,to multiply by #be1 annex two ciphers and divide by #d4 ,to divide by #be1 multiply by #d and strike off the last two figures1 which leaves the quotient2 one fourth of the last two figures1 taken as one number1 is the remainder4 ,to multiply a number by #i1 annex a cipher1 and subtract the number1 which is equivalent to taking the number ten times1 and then subtracting it once4 ,to multiply by #ii1 annex two ciphers and subtract the number1 c4 ,in order that a number may be divisible by #b without remainder1 its units' figure must be an even number4#aa ,that it may be divisible by #d1 its last two figures must be divisible by #d4 ,take the example #abcf3 this is composed of #ab hundreds and #cf1 the first part of which1 being hundreds1 is divisible by #d1 and gives #ab twenty-fives2 it depends then upon #cf1 the last two figures1 whether #abcf is divisible by #d or not4 ,a number is divisible by #h if the last three figures are divisible by #h2 for every digit1 except the last three1 is a number of thousands1 and #ajjj is divisible by #h2 whether therefore the whole shall be divisible by #h or not depends on the last three figures3 thus1 #abgidf is not divisible by #h1 since #idf is not so4 ,a number is divisible by #c or #i only when the sum of its digits is divisible by #c or #i4 ,take for example #abcd2 this is #aa ,among the even figures we include #j4 #a thousand1 or #iii and #a #b hundred1 or twice #ii and #b #c tens1 or three times #i and #c and #d or #d ,now #i1 #ii1 #iii1 c4 are all obviously divisible by #i and by #c1 and so will be any number made by the repetition of all or any of them any number of times4 ,it therefore depends on #a #b #c #d1 or the sum of the digits1 whether #abcd shall be divisible by #i or #c1 or not4 ,from the above we gather1 that a number is divisible by #f when it is even1 and when the sum of its digits is divisible by #c4 ,lastly1 a number is divisible by #e only when the last figure is #j or #e4 #hb4 ,where the divisor is unity followed by ciphers1 the rule becomes extremely simple1 as you will see by the following examples3 #ajj7ccdbi7ccd #cjj ---- #cdb #cjj ---- #dbi #djj --- #bi ,this is1 then1 the rule3 ,cut off as many figures from the right hand of the dividend as there are ciphers4 ,these figures will be the remainder1 and the rest of the dividend will be the quotient4 #aj7bgagcaf -------- #bgagca and rem4 #f4 ,or we may prove these results thus3 from 7#bj71 #bgagcaf is #bgagca tens and #f2 of which the first contains #aj #bgagca times1 and the second not at all2 the quotient is therefore #bgagca1 and the remainder #f 7#gb74 ,again 7#bj71 #ccdbi is #ccd hundreds and #bi2 of which the first contains #ajj #ccd times1 and the second not at all2 the quotient is therefore #ccd1 and the remainder #bi4 #hc4 ,the following examples will shew how the rule may be shortened when there are ciphers in the divisor4 ,with each example is placed another containing the same process1 all unnecessary figures being removed2 and from the comparison of the two1 the rule at the end of this article is derived4 ,i4 #aghbjjj7fdbdgjjjjj7cfje #aghb7fdbdgjj7cfje #ecdfjjj #ecdf -------- ---- #ajghgjjj #ajghg #ajfibjjj #ajfib ---------- ------- #iejjjjj #iejj #hiajjjj #hiaj ------- ------- #eijjjj #eijjjj ,,ii4 #abcjjjjj7dbagfahicjj7cdbh #abc7dbagfa7cdbh #cfijjjjj #cfi --------- ---- #ebgfahic #ebg #dibjjjjj #dib --------- ---- #cefahicj #cef #bdfjjjjj #bdf --------- ---- #aajahicjj #aaja #ihdjjjjj #ihd -------- ---------- #aaghicjj #aaghicjj ,the rule1 then1 is3 ,strike out as many figures#ab from the right of the dividend as there are ciphers at the right of the divisor4 ,strike out all the ciphers from the divisor1 and divide in the usual way2 but at the end of the process place on the right of the remainder all those figures which were struck out of the dividend4 #ab ,including both ciphers and others4 #hd4 ,,exercises4 ,dividend4 ,divisor4 ,quotient4,remainder4 #ifid #dg #bjf #ab #agefah #cacf #ef #b #bcgifdhd #acjjjj #ahc #fdhd #adjjbefd #ahga #gdhd #j #cajcaddbj #ghgh #cicij #j #cicijdjfdg #fhhi #egaghg #d #bbhgfgibdedifa #dcjdfgba #ecadda #j ,shew that #ajj #ajj #ajj - #dc #dc #dc ,i4 ------------------------------ #ajj #ajj #ajj #dc #dc #dc4 #ajj - #dc #ajj #ajj #ajj #dc #dc #dc ,,ii4 ------------------------------ #ajj #ajj - #ajj #dc #dc #dc4 #ajj #dc #gf #gf #b #gf #eb #eb #eb ,,iii4 -------------------------------- #gf #eb4 #gf #eb #ab #ab #ab #ab - #a ,,iv4 #a #ab #ab #ab #ab #ab #ab ----------------------4 #ab - #a ,what is the nearest number to #acgfdbi which can be divided by #cfcjj without remainder8--,answer1 #acgidjj4 ,if #cf oxen can eat #baf acres of grass in one year1 and if a sheep eat half as much as an ox1 how long will it take #di oxen and #acf sheep together to eat #ageej acres8--,answer1 #be years4 #he4 ,take any two numbers1 one of which divides the other without remainder2 for example1 #cb and #d4 ,multiply both these numbers by any other number2 for example1 #f4 ,the products will be #aib and #bd4 ,now1 #aib contains #bd just as often as #cb contains #d4 ,suppose #f baskets1 each containing #cb pebbles1 the whole number of which will be #aib4 ,take #d from one basket1 time after time1 until that basket is empty4 ,it is plain that if1 instead of taking #d from that basket1 ,i take #d from each1 the whole #f will be emptied together3 that is1 #f times #cb contains #f times #d just as often as #cb contains #d4 ,the same reasoning applies to other numbers1 and therefore we do not alter the quotient if we multiply the dividend and divisor by the same number4 #hf4 ,again1 suppose that #bjj is to be divided by #ej4 ,divide both the dividend and divisor by the same number2 for example1 #e4 ,then1 #bjj is #e times #dj1 and #ej is #e times #aj4 ,but by 7#he71 #dj divided by #aj gives the same quotient as #e times #dj divided by #e times #aj1 and therefore the quotient of two numbers is not altered by dividing both the dividend and divisor by the same number4 #hg4 ,from 7#ee71 if a number be multiplied successively by two others1 it is multiplied by their product4 ,thus1 #bg1 first multiplied by #e1 and the product multiplied by #c1 is the same as #bg multiplied by #e times #c1 or #ae4 ,also1 if a number be divided by any number1 and the quotient be divided by another1 it is the same as if the first number had been divided by the product of the other two4 ,for example1 divide #fj by #d1 which gives #ae1 and the quotient by #c1 which gives #e4 ,it is plain1 that if each of the four fifteens of which #fj is composed be divided into three equal parts1 there are twelve equal parts in all2 or1 a division by #d1 and then by #c1 is equivalent to a division by #d #c1 or #ab4 #hh4 ,the following rules will be better understood by stating them in an example4 ,if #cb be multiplied by #bd and divided by #f1 the result is the same as if #cb had been multiplied by the quotient of #bd divided by #f1 that is1 by #d2 for the sixth part of #bd being #d1 the sixth part of any number repeated #bd times is that number repeated #d times2 or1 multiplying by #bd and dividing by #f is equivalent to multiplying by #d4 #hi4 ,again1 if #dh be multiplied by #d1 and that product be divided by #bd1 it is the same thing as if #dh were divided at once by the quotient of #bd divided by #d1 that is1 by #f4 ,for1 every unit which is repeated #f times in #dh is repeated #d times as often1 or #bd times1 in #d times #dh1 or the quotient of #dh and #f is the same as the quotient of #dh #d and #f #d4 #ij4 ,the results of the last five articles may be algebraically expressed thus3 ma a ---- ---- 7#he7 mb b ,if n divide a and b without remainder1 a ---- n a ------ ---- 7#hf7 b b ---- n a ---- b a ------ ---- 7#hg7 c bc ab b ------ a ---- 7#hh7 c c ac a ----- ------ 7#hi7 b b ---- c ,it must be recollected1 however1 that these have only been proved in the case where all the divisions are without remainder4 #ia4 ,when one number divides another without leaving any remainder1 or is contained an exact number of times in it1 it is said to be a measure of that number1 or to measure it4 ,thus1 #d is a measure of #acf1 or measures #acf2 but it does not measure #acg4 ,the reason for using the word measure is this3 ,suppose you have a rod #d feet long1 with nothing marked upon it1 with which you want to measure some length2 for example1 the length of a street4 ,if that street should happen to be #acf feet in length1 you will be able to measure it with the rod1 because1 since #acf contains #d #cd times1 you will find that the street is exactly #cd times the length of the rod4 ,but if the street should happen to be #acg feet long1 you cannot measure it with the rod2 for when you have measured #cd of the rods1 you will find a remainder1 whose length you cannot tell without some shorter measure4 ,hence #d is said to measure #acf1 but not to measure #acg4 ,a measure1 then1 is a divisor which leaves no remainder4 #ib4 ,when one number is a measure of two others1 it is called a common measure of the two4 ,thus1 #ae is a common measure of #ahj and #ge4 ,two numbers may have several common measures4 ,for example1 #cfj and #afh have the common measures #b1 #c1 #d1 #f1 #bd1 and several others4 ,now1 this question maybe asked3 ,of all the common measures of #cfj and #afh1 which is the greatest8 ,the answer to this question is derived from a rule of arithmetic1 called the rule for finding the ,,greatest ,,common ,,measure1 which we proceed to consider4 #ic4 ,if one quantity measure two others1 it measures their sum and difference4 ,thus1 #g measures #ba and #ef4 ,it therefore measures #ef #ba and #ef-ba1 or #gg and #ce4 ,this is only another way of saying what was said in 7#gd74 #id4 ,if one number measure a second1 it measures every number which the second measures4 ,thus1 #e measures #ae1 and #ae measures #cj1 #de1 #fj1 #ge1 c42 all which numbers are measured by #e4 ,it is plain that if #ae contains #e #c times1 #cj1 or #ae #ae contains #e #c #c times1 or #f times1 #de1 or #ae #ae #ae contains #e #c #c #c or #i times2 and so on4 #ie4 ,every number which measures both the dividend and divisor measures the remainder also4 ,to shew this1 divide #cfj by #aab4 ,the quotient is #c1 and the remainder #bd1 that is 7#gb7 #cfj is three times #aab and #bd1 or #cfj #aab #c #bd4 ,from this it follows1 that #bd is the difference between #cfj and #c times #aab1 or #bd #cfj-aab #c4 ,take any number which measures both #cfj and #aab2 for example1 #d4 ,then #d measures #cfj1 #d measures #aab1 and therefore 7#id7 measures #aab #c1 or #aab #aab #aab4 ,therefore 7#ic7 it measures #cfj-aab #c1 which is the remainder #bd4 ,the same reasoning may be applied to all other measures of #cfj and #aab2 and the result is1 that every quantity which measures both the dividend and divisor also measures the remainder4 ,hence1 every common measure of a dividend and divisor is also a common measure of the divisor and remainder4 #if4 ,every common measure of the divisor and remainder is also a common measure of the dividend and divisor4 ,take the same example1 and recollect that #cfj #aab #c #bd4 ,take any common measure of the remainder #bd and the divisor #aab2 for example1 #h4 ,then #h measures #bd2 and #h measures #aab1 and therefore 7#id7 measures #aab #c4 ,therefore 7#ic7 #h measures #aab #c #bd1 or measures the dividend #cfj4 ,then every common measure of the remainder and divisor is also a common measure of the divisor and dividend1 or there is no common measure of the remainder and divisor which is not also a common measure of the divisor and dividend4 #ig4 ,i4 ,it is proved in 7#ie7 that the remainder and divisor have all the common measures which are in the dividend and divisor4 ,,ii4 ,it is proved in 7#if7 that they have no others4 ,it therefore follows1 that the greatest of the common measures of the first two is the greatest of those of the second two1 which shews how to find the greatest common measure of any two numbers1#ac as follows3 #ih4 ,take the preceding example1 and let it be required to find the g4 c4 m4 of #cfj and #aab1 and observe that #cfj divided by #aab gives the remainder #bd1 #aab divided by #bd gives the remainder #af1 #bd divided by #af gives the remainder #h1 #af divided by #h gives no remainder4 #ac ,for shortness1 ,i abbreviate the words greatest common measure into their initial letters1 g4 c4 m4 ,now1 since #h divides #af without remainder1 and since it also divides itself without remainder1 #h is the g4 c4 m4 of #h and #af1 because it is impossible to divide #h by any number greater than #h2 so that1 even if #af had a greater measure than #h1 it could not be common to #af and #h4 ,therefore #h is g4 c4 m4 of #af and #h1 7#ig7 g4 c4 m4 of #af and #h is g4 c4 m4 of #bd and #af1 g4 c4 m4 of #bd and #af is g4 c4 m4 of #aab and #bd1 g4 c4 m4 of #aab and #bd is g4 c4 m4 of #cfj and #aab1 ,therefore #h is g4 c4 m4 of #cfj and #aab4 ,the process carried on may be written down in either of the following ways3 #aab7cfj7c #ccf --- #bd7aab7d #aab #cfj #c #if #if #ccf #d --- ----------- #af7bd7a #af #bd #a #af #af #af #b -- ----------- #h7af7b #j #h #af -- #j ,the rule for finding the greatest common measure of two numbers is1 ,i4 ,divide the greater of the two by the less4 ,,ii4 ,make the remainder a divisor1 and the divisor a dividend1 and find another remainder4 ,,iii4 ,proceed in this way until there is no remainder1 and the last divisor is the greatest common measure required4 #ii4 ,you may perhaps ask how the rule is to shew when the two numbers have no common measure4 ,the fact is1 that there are1 strictly speaking1 no such numbers1 because all numbers are measured by #a2 that is1 contain an exact number of units1 and therefore #a is a common measure of every two numbers4 ,if they have no other common measure1 the last divisor will be #a1 as in the following example1 where the greatest common measure of #hg and #be is found4 #be7hg7c #ge -- #ab7be7b #bd -- #a7ab7ab #ab -- #j ,,exercises4 ,numbers4 g4 c4 m4 #faig #ieba #a #ehcfc #bfjb #a #eedg #adgjjhddc #ahdi #fbha #cbfjda #ega #bhiae #cadie #e #aeji #cjjcji #c ,what are #cf #cf #b #cf #gb #gb #gb and #cf #cf #cf #gb #gb #gb2 and what is their greatest common measure8--,answer1 #aaffd4 #ajj4 ,if two numbers be divisible by a third1 and if the quotients be again divisible by a fourth1 that third is not the greatest common measure4 ,for example1 #cfj and #ejd are both divisible by #d4 ,the quotients are #ij and #abf4 ,now #ij and #abf are both divisible by #i1 the quotients of which division are #aj and #ad4 ,by 7#hg71 dividing a number by #d1 and then dividing the quotient by #i1 is the same thing as dividing the number itself by #d #i1 or by #cf4 ,then1 since #cf is a common measure of #cfj and #ejd1 and is greater than #d1 #d is not the greatest common measure4 ,again1 since #aj and #ad are both divisible by #b1 #cf is not the greatest common measure4 ,it therefore follows1 that when two numbers are divided by their greatest common measure1 the quotients have no common measure except #a 7#ii74 ,otherwise1 the number which was called the greatest common measure in the last sentence is not so in reality4 #aja4 ,to find the greatest common measure of three numbers1 find the g4 c4 m4 of the first and second1 and of this and the third4 ,for since all common divisors of the first and second are contained in their g4 c4 m41 and no others1 whatever is common to the first1 second1 and third1 is common also to the third and the g4 c4 m4 of the first and second1 and no others4 ,similarly1 to find the g4 c4 m4 of four numbers1 find the g4 c4 m4 of the first1 second1 and third1 and of that and the fourth4 #ajb4 ,when a first number contains a second1 or is divisible by it without remainder1 the first is called a multiple of the second4 ,the words multiple and measure are thus connected3 ,since #d is a measure of #bd1 #bd is a multiple of #d4 ,the number #if is a multiple of #h1 #ab1 #bd1 #dh1 and several others4 ,it is therefore called a common multiple of #h1 #ab1 #bd4 #dh1 c4 ,the product of any two numbers is evidently a common multiple of both4 ,thus1 #cf #h1 or #bhh1 is a common multiple of #cf and #h4 ,but there are common multiples of #cf and #h less than #bhh2 and because it is convenient1 when a common multiple of two quantities is wanted1 to use the least of them1 ,i now shew how to find the least common multiple of two numbers4 #ajc4 ,take1 for example1 #cf and #h4 ,find their greatest common measure1 which is #d1 and observe that #cf is #i #d1 and #h is #b #d4 ,the quotients of #cf and #h1 when divided by their greatest common measure1 are therefore #i and #b4 ,multiply these quotients together1 and multiply the product by the greatest common measure1 #d1 which gives #i #b #d1 or #gb4 ,this is a multiple of #h1 or of #d #b by 7#ee72 and also of #cf or of #d #i4 ,it is also the least common multiple2 but this cannot be proved to you1 because the demonstration cannot be thoroughly understood without more practice in the use of letters to stand for numbers4 ,but you may satisfy yourself that it is the least in this case1 and that the same process will give the least common multiple in any other case which you may take4 ,it is not even necessary that you should know it is the least4 ,whenever a common multiple is to be used1 any one will do as well as the least4 ,it is only to avoid large numbers that the least is used in preference to any other4 ,when the greatest common measure is #a1 the least common multiple of the two numbers is their product4 ,the rule then is3 ,to find the least common multiple of two numbers1 find their greatest common measure1 and multiply one of the numbers by the quotient which the other gives when divided by the greatest common measure4 ,to find the least common multiple of three numbers1 find the least common multiple of the first two1 and find the least common multiple of that multiple and the third1 and so on4 ,,exercises4 ,numbers proposed4 ,least common multiple4 #ad1 #ba #db #af1 #e1 #bd #bdj #a1 #b1 #c1 #d1 #e1 #f1 #g1 #h1 #i1 #aj #bebj #f1 #h1 #aa1 #af1 #bj #bfdj #hgf1 #hfd #fcjgb #hfh1 #hed #ebidh ,a convenient mode of finding the least common multiple of several numbers is as follows1 when the common measures are easily visible3 ,pick out a number of common measures of two or more1 which have themselves no divisors greater than unity4 ,write them as divisors1 and divide every number which will divide by one or more of them4 ,bring down the quotients1 and also the numbers which will not divide by any of them4 ,repeat the process with the results1 and so on until the numbers brought down have no two of them any common measure except unity4 ,then1 for the least common multiple1 multiply all the divisors by all the numbers last brought down4 ,for instance1 let it be required to find the least common multiple of all the numbers from #aa to #ba4 #b1 #b1 #c1 #e1 #g7aa #ab #ac #ad #ae #af #ag #ah #ai #bj #ba --------------------------------- #aa #a #ac #a #a #d #ag #c #ai #a #a ,there are now no common measures left in the row1 and the least common multiple required is the product of #b1 #b1 #c1 #e1 #g1 #aa1 #ac1 #d1 #ag1 #c1 and #ai2 or #bcbgibefj4 ,,section ,v4 ,,fractions4 #ajd4 ,suppose it required to divide #di yards into five equal parts1 or1 as it is called1 to find the fifth part of #di yards4 ,if we divide #de by #e1 the quotient is #i1 and the remainder is #d2 that is 7#gb71 #di is made up of #e times #i and #d4 ,let the line ,a ,b represent #di yards3 ,a-------------------------------------- --,b ,c-------------- ,i -- ,d-------------- ,k -- ,e-------------- ,l -- ,f-------------- ,m -- ,g-------------- ,n -- ,i ,k ,l ,m ,n ,h ----- ,take #e lines1 ,c1 ,d1 ,e1 ,f1 and ,g1 each #i yards in length1 and the line ,h1 #d yards in length4 ,then1 since #di is #e nines and #d1 ,c1 ,d1 ,e1 ,f1 ,g1 and ,h1 are together equal to ,a ,b4 ,divide ,h1 which is #d yards1 into five equal parts1 ,i1 ,k1 ,l1 ,m1 and ,n1 and place one of these parts opposite to each of the lines1 ,c1 ,d1 ,e1 ,f1 and ,g4 ,it follows that the ten lines1 ,c1 ,d1 ,e1 ,f1 ,g1 ,i1 ,k1 ,l1 ,m1 ,n1 are together equal to ,a ,b1 or #di yards4 ,now ,d and ,k together are of the same length as ,c and ,i together1 and so are ,e and ,l1 ,f and ,m1 and ,g and ,n4 ,therefore1 ,c and ,i together1 repeated #e times1 will be #di yards2 that is1 ,c and ,i together make up the fifth part of #di yards4 #aje4 ,c is a certain number of yards1 viz4 #i2 but ,i is a new sort of quantity1 to which hitherto we have never come4 ,it is not an exact number of yards1 for it arises from dividing #d yards into #e parts1 and taking one of those parts4 ,it is the fifth part of #d yards1 and is called a ,,fraction of a yard4 ,it is written thus1 _/7#bc71 and is what we must add to #i yards in order to make up the fifth part of #di yards4 ,the same reasoning would apply to dividing #di bushels of corn1 or #di acres of land1 into #e equal parts4 ,we should find for the fifth part of the first1 #i bushels and the fifth part of #d bushels2 and for the second1 #i acres and the fifth part of #d acres4 ,we say1 then1 once for all1 that the fifth part of #di is #i and _/1 or #i _/2 which is usually written 7#i_/71 or if we use signs1 #di_/e 7#i_/74 ,,exercises4 ,what is the seventeenth part of #abcg8--;,answer1 7#gb-_/74 #ajjcb #ffchai #bbggccii ,what are -----1 ------1 and -------- 8 #aigd #bcgaj #bdbd #afb #bcfdi #bcdc ,answer1 7#e ----71 7#bg -----71 7#icid ----74 #aigd #bcgaj #bdbd #ajf4 ,by the term fraction is understood a part of any number1 or the sum of any of the equal parts into which a number is divided4 ,thus1 _/1 _/1 _/1 are fractions4 ,the term fraction even includes whole numbers3#ad for example1 #ag is _/1 _/1 _/1 c4 #ad ,numbers which contain an exact number of units1 such as #e1 #g1 #ajj1 c41 are called whole numbers or integers1 when we wish to distinguish them from fractions4 ,the upper number is called the numerator1 the lower number is called the denominator1 and both of these are called terms of the fraction4 ,as long as the numerator is less than the denominator1 the fraction is less than a unit3 thus1 _/ is less than a unit2 for #f divided into #f parts gives #a for each part1 and must give less when divided into #ag parts4 ,similarly1 the fraction is equal to a unit when the numerator and denominator are equal1 and greater than a unit when the numerator is greater than the denominator4 #ajg4 ,by is meant the third part of #b4 ,this is the same as twice the third part of #a4 ,to prove this1 let ,a ,b be two yards1 and divide each of the yards ,a ,c and ,c ,b into three equal parts4 ------------ ,a ,d ,e ,c ,f ,g ,b ,then1 because ,a ,e1 ,e ,f1 and ,f ,b1 are all equal to one another1 ,a ,e is the third part of #b4 ,it is therefore 4 ,but ,a ,e is twice ,a ,d1 and ,a ,d is the third part of one yard1 or 2 therefore is twice 2 that is1 in order to get the length 1 it makes no difference whether we divide two yards at once into three parts1 and take one of them1 or whether we divide one yard into three parts1 and take two of them4 ,by the same reasoning1 may be found either by dividing #e into #h parts1 and taking one of them1 or by dividing #a into #h parts1 and taking five of them4 ,in future1 of these two meanings ,i shall use that which is most convenient at the time1 as it is proved that they are the same thing4 ,this principle is the same as the following3 ,the third part of any number may be obtained by adding together the thirds of all the units of which it consists4 ,thus1 the third part of #b1 or of two units1 is made by taking one-third out of each of the units1 that is1 #b4 ,this meaning appears ambiguous when the numerator is greater than the denominator3 thus1 _/ would mean that #a is to be divided into #g parts1 and #ae of them are to be taken4 ,we should here let as many units be each divided into #g parts as will give more than #ae of those parts1 and take #ae of them4 #ajh4 ,the value of a fraction is not altered by multiplying the numerator and denominator by the same quantity4 ,take the fraction 1 multiply its numerator and denominator by #e1 and it becomes _/1 which is the same thing as 2 that is1 one-twentieth part of #ae yards is the same thing as one-fourth of #c yards3 or1 if our second meaning of the word fraction be used1 you get the same length by dividing a yard into #bj parts and taking #ae of them1 as you get by dividing it into #d parts and taking #c of them4 ,to prove this1 ,illustration let ,a ,b represent a yard2 divide it into #d equal parts1 ,a ,c1 ,c ,d1 ,d ,e1 and ,e ,b1 and divide each of these parts into #e equal parts4 ,then ,a ,e is 4 ,but the second division cuts the line into #bj equal parts1 of which ,a ,e contains #ae4 ,it is therefore _/4 ,therefore1 _/ and are the same thing4 ,again1 since is made from _/ by dividing both the numerator and denominator by #e1 the value of a fraction is not altered by dividing both its numerator and denominator by the same quantity4 ,this principle1 which is of so much importance in every part of arithmetic1 is often used in common language1 as when we say that #ad out of #ba is #b out of #c1 c4 #aji4 ,though the two fractions and _/ are the same in value1 and either of them may be used for the other without error1 yet the first is more convenient than the second1 not only because you have a clearer idea of the fourth of three yards than of the twentieth part of fifteen yards1 but because the numbers in the first being smaller1 are more convenient for multiplication and division4 ,it is therefore useful1 when a fraction is given1 to find out whether its numerator and denominator have any common divisors or common measures4 ,in 7#ih7 was given a rule for finding the greatest common measure of any two numbers2 and it was shewn that when the two numbers are divided by their greatest common measure1 the quotients have no common measure except #a4 ,find the greatest common measure of the terms of the fraction1 and divide them by that number4 ,the fraction is then said to be reduced to its lowest terms1 and is in the state in which the best notion can be formed of its magnitude4 ,,exercises4 ,with each fraction is written the same reduced to its lowest terms4 #bgid #bb #abg #bb ---- ---------- ---- #biba #bc #abg #bc #bghh #ag #afd #ag ---- ---------- ---- #dibj #cj #afd #cj #icbjh #gfd #abb #gfd ----- ---------- --- #acghf #aac #abb #aac #hhhhjj #bb #djdjj #bb -------- ------------ ----- #djceifjj #iii #djdjj #iii #iedfi #aba #ghi #aba ------ ----------- --- #ceighd #def #ghi #def #aaj4 ,when the terms of the fraction given are already in factors1#ae any one factor in the numerator may be divided by a number1 provided some one factor in the denominator is divided by the same4 ,this follows from 7#hh7 and 7#ajh74 ,in the following examples the figures altered by division are accented4 #ae ,a factor of a number is a number which divides it without remainder3 thus1 #d1 #f1 #h1 are factors of #bd1 and #f #d1 #h #c1 #b #b #b #c1 are several ways of decomposing #bd into factors4 #ab #aa #aj #c #aa #aj #a #aa #e ------------ ------------- ------------- #ee4 #b #c #d #b #c #a #a #a #a #ah #ae #ac #b #c #a #a #a #a ------------ ------------- ------------- _/4 #bj #ed #eb #d #f #d #b #b #d #bg #bh #c #d #c #b ------- --------- -------- _/4 #i #gj #a #aj #a #e #aaa4 ,as we can1 by 7#ajh71 multiply the numerator and denominator of a fraction by any number1 without altering its value1 we can now readily reduce two fractions to two others1 which shall have the same value as the first two1 and which shall have the same denominator4 ,take1 for example1 and _/2 multiply both terms of by #g1 and both terms of _/ by #c4 ,it then appears that #b #g is ------- or _/ #c #g #d #c _/ is ------- or _/4 #g #c ,here are then two fractions _/ and _/1 equal to and _/1 and having the same denominator1 #ba2 in this case1 and _/ are said to be reduced to a common denominator4 ,it is required to reduce 1 1 and _/ to a common denominator4 ,multiply both terms of the first by the product of #f and #i2 of the second by the product of #aj and #i2 and of the third by the product of #aj and #f4 ,then it appears 7#ajh7 that #a #f #i is ----------- or _/ #aj #f #i #e #aj #i is ------------ or _/ #f #aj #i #g #aj #f _/ is ------------ or _/4 #i #aj #f ,on looking at these last fractions1 we see that all the numerators and the common denominator are divisible by #f1 and 7#ajh7 this division will not alter their values4 ,on dividing the numerators and denominators of _/1 _/1 and _/ by #f1 the resulting fractions are1 _/1 _/1 and _/4 ,these are fractions with a common denominator1 and which are the same as 1 1 and _/2 and therefore these are a more simple answer to the question than the first fractions4 ,observe also that #edj is one common multiple of #aj1 #f1 and #i1 namely1 #aj #f #i1 but that #ij is the least common multiple of #aj1 #f1 and #i 7#ajc74 ,the following process1 therefore1 is better4 ,to reduce the fractions 1 1 and _/1 to others having the same value and a common denominator1 begin by finding the least common multiple of #aj1 #f1 and #i1 by the rule in 7#ajc71 which is #ij4 ,observe that #aj1 #f1 and #i are contained in #ij #i1 #ae1 and #aj times4 ,multiply both terms of the first by #i1 of the second by #ae1 and of the third by #aj1 and the fractions thus produced are _/1 _/1 and _/1 the same as before4 ,if one of the numbers be a whole number1 it may be reduced to a fraction having the common denominator of the rest1 by 7#ajf74 ,,exercises4 ,fractions proposed reduced to a common denominator4 #b #a #a #bj #f #e --- --- --- ---- --- --- #c #e #f #cj #cj #cj #a #b #c #ab #c #bh #bd #ah #dh #fc --- --- --- ---- --- --- --- --- --- --- #c #g #ad #ba #d #hd #hd #hd #hd #hd #d #e #f #cjjj #djj #ej #f #c --- ---- ---- ---- ----- ----- ---- #aj #ajj #ajjj #ajjj #ajjj #ajjj #ajjj #cc #bha #bbcda #ajfdii ---- ---- -------- ------ #cgi #fgg #befehc #befehc #aab4 ,by reducing two fractions to a common denominator1 we are able to compare them2 that is1 to tell which is the greater and which the less of the two4 ,for example1 take and _/4 ,these fractions reduced1 without alteration of their value1 to a common denominator1 are _/ and _/4 ,of these the first must be the greater1 because 7#ajg7 it may be obtained by dividing #a into #cj equal parts and taking #ae of them1 but the second is made by taking #ad of those parts4 ,it is evident that of two fractions which have the same denominator1 the greater has the greater numerator2 and also that of two fractions which have the same numerator1 the greater has the less denominator4 ,thus1 _/ is greater than _/1 since the first is a #gth1 and the last only a #ith part of #h4 ,also1 any numerator may be made to belong to as small a fraction as we please1 by sufficiently increasing the denominator4 ,thus1 _/ is _/1 _/ is _/1 and _/ is _/ 7#ajh74 ,we can now also increase and diminish the first fraction by the second4 ,for the first fraction is made up of #ae of the #cj equal parts into which #a is divided4 ,the second fraction is #ad of those parts4 ,the sum of the two1 therefore1 must be #ae #ad1 or #bi of those parts2 that is1 _/ is _/4 ,the difference of the two must be #ae-ad1 or #a of those parts2 that is1 -_/ _/4 #aac4 ,from the last two articles the following rules are obtained3 ,i4 ,to compare1 to add1 or to subtract fractions1 first reduce them to a common denominator4 ,when this has been done1 that is the greatest of the fractions which has the greatest numerator4 ,their sum has the sum of the numerators for its numerator1 and the common denominator for its denominator4 ,their difference has the difference of the numerators for its numerator1 and the common denominator for its denominator4 ,,exercises4 #a #a #a #a #ec #dd #aec #ahcbi --- --- --- - --- ---- ---- - ----- ------- #b #c #d #e #fj #c #dbg #abhb #h #c #d #ahcd #a #ab #bec #a ---- ---- ---- ---- #b - --- ---- ---- #aj #ajj #ajjj #ajjj #g #ac #ia #a #h #id #c #afc #ig #icjff --- --- ---- --- --- - ---- ------- #b #af #ahh #b #eba #hha #deijja #aad4 ,suppose it required to add a whole number to a fraction1 for example1 #f to _/4 ,by 7#ajf7 #f is _/1 and _/ _/ is _/2 that is1 #f _/1 or as it is usually written1 7#f_/71 is _/4 ,the rule in this case is3 ,multiply the whole number by the denominator of the fraction1 and to the product add the numerator of the fraction2 the sum will be the numerator of the result1 and the denominator of the fraction will be its denominator4 ,thus1 7#c7 _/1 7#bb_/7 _/1 7#gd_/7 _/4 ,this rule is the opposite of that in 7#aje74 #aae4 ,from the last rule it appears that #ijg #agbcjijg #bbe #ffgbbe #agbc ------ is --------1 #ffg ----- is ------1 #ajjjj #ajjjj #ajjj #ajjj #ii #bcjjjii and #bc ------ is -------4 #ajjjj #ajjjj ,hence1 when a whole number is to be added to a fraction whose denominator is #a followed by ciphers1 the number of which is not less than the number of figures in the numerator1 the rule is3 ,write the whole number first1 and then the numerator of the fraction1 with as many ciphers between them as the number of ciphers in the denominator exceeds the number of figures in the numerator4 ,this is the numerator of the result1 and the denominator of the fraction is its denominator4 ,if the number of ciphers in the denominator be equal to the number of figures in the numerator1 write no ciphers between the whole number and the numerator4 ,,exercises4 ,reduce the following mixed quantities to fractions3 #bcgjg #f #bii #bbaj #a ------1 #bdeg ---1 #abjg --------1 and #bcc -----4 #ajjjjj #aj #ajjjjjjj #ajjjj #aaf4 ,suppose it required to multiply by #d4 ,this by 7#dh7 is taking four times2 that is1 finding 4 ,this by 7#aab7 is _/2 so that to multiply a fraction by a whole number the rule is3 ,multiply the numerator by the whole number1 and let the denominator remain4 #aag4 ,if the denominator of the fraction be divisible by the whole number1 the rule may be stated thus3 ,divide the denominator of the fraction by the whole number1 and let the numerator remain4 ,for example1 multiply _/ by #f4 ,this 7#aaf7 is _/1 which1 since the numerator and denominator are now divisible by #f1 is 7#ajh7 the same as _/4 ,it is plain that _/ is made from _/ in the manner stated in the rule4 #aah4 ,multiplication has been defined to be the taking as many of one number as there are units in another4 ,thus1 to multiply #ab by #g is to take as many twelves as there are units in #g1 or to take #ab as many times as you must take #a in order to make #g4 ,thus1 what is done with #a in order to make #g1 is done with #ab to make #g times #ab4 ,for example1 #g is #a #a #a #a #a #a #a #g times #ab is #ab #ab #ab #ab #ab #ab #ab4 ,when the same thing is done with two fractions1 the result is still called their product1 and the process is still called multiplication4 ,there is this difference1 that whereas a whole number is made by adding #a to itself a number of times1 a fraction is made by dividing #a into a number of equal parts1 and adding one of these parts to itself a number of times4 ,this being the meaning of the word multiplication1 as applied to fractions1 what is multiplied by 8 ,whatever is done with #a in order to make must now be done with 2 but to make 1 #a is divided into #h parts1 and #g of them are taken4 ,therefore1 to make 1 must be divided into #h parts1 and #g of them must be taken4 ,now is1 by 7#ajh71 the same thing as _/4 ,since _/ is made by dividing #a into #cb parts1 and taking #bd of them1 or1 which is the same thing1 taking #c of them #h times1 if _/ be divided into #h equal parts1 each of them is _/2 and if #g of these parts be taken1 the result is _/ 7#aaf73 therefore multiplied by is _/2 and the same reasoning may be applied to any other fractions4 ,but _/ is made from and by multiplying the two numerators together for the numerator1 and the two denominators for the denominator2 which furnishes a rule for the multiplication of fractions4 #aai4 ,if this product _/ is to be multiplied by a third fraction1 for example1 by _/1 the result is1 by the same rule1 _/2 and so on4 ,the general rule for multiplying any number of fractions together is therefore3 ,multiply all the numerators together for the numerator of the product1 and all the denominators together for its denominator4 #abj4 ,suppose it required to multiply together _/ and _/4 ,the product may be written thus3 #ae #h #abj -------1 and is ----1 #af #aj #afj which reduced to its lowest terms 7#aji7 is 4 ,this result might have been obtained directly1 by observing that #ae and #aj are both measured by #e1 and #h and #af are both measured by #h1 and that the fraction may be written thus3 #c #e #h -------------4 #b #h #b #e ,divide both its numerator and denominator by #e #h 7#ajh7 and 7#hg71 and the result is at once 2 therefore1 before proceeding to multiply any number of fractions together1 if there be any numerator and any denominator1 whether belonging to the same fraction or not1 which have a common measure1 divide them both by that common measure1 and use the quotients instead of the dividends4 ,a whole number may be considered as a fraction whose denominator is #a2 thus1 #af is _/ 7#ajf72 and the same rule will apply when one or more of the quantities are whole numbers4 ,,exercises #acf #bfh #cfddh #ahbbd ---- --- ------- ------- #gdgj #iai #fhfdicj #cdcbdfe #a #b #c #d #a #b #ag #b --- --- --- --- --- ---- ---- ---- #b #c #d #e #e #ag #de #de #b #ac #bda #fbff #ac #fja #ghac --- ---- ---- ------ ---- ---- ----- #ei #g #ai #ghdg #dfa #aa #ejga ,fraction proposed4 ,square4 ,cube4 #gja #diadja #cdddgbaja ----- ------- --------- #aeh #bdifd #ciddcab #adj #aifjj #bgddjjj ----- ------ -------- #ada #aihha #bhjcbba #cee #abfjbe #ddgchhge ----- ------- --------- #aac #abgfi #addbhig ,from #ajj acres of ground1 two-thirds of them are taken away2 #ej acres are then added to the result1 and _/ of the whole is taken2 what number of acres does this produce8--,answer1 7#ei_/74 #aba4 ,in dividing one whole number by another1 for example1 #ajh by #i1 this question is asked1--,can we1 by the addition of any number of nines1 produce #ajh8 and if so1 how many nines will be sufficient for that purpose8 ,suppose we take two fractions1 for example1 and 1 and ask1 ,can we1 by dividing into some number of equal parts1 and adding a number of these parts together1 produce 8 if so1 into how many parts must we divide 1 and how many of them must we add together8 ,the solution of this question is still called the division of by 2 and the fraction whose denominator is the number of parts into which is divided1 and whose numerator is the number of them which is taken1 is called the quotient4 ,the solution of this question is as follows3 ,reduce both these fractions to a common denominator 7#aaa71 which does not alter their value 7#ajh72 they then become _/ and _/4 ,the question now is1 to divide _/ into a number of parts1 and to produce _/ by taking a number of these parts4 ,since _/ is made by dividing #a into #ae parts and taking #ab of them1 if we divide _/ into #ab equal parts1 each of these parts is _/2 if we take #aj of these parts1 the result is _/4 ,therefore1 in order to produce _/ or 7#ajh71 we must divide _/ or into #ab parts1 and take #aj of them2 that is1 the quotient is _/4 ,if we call the dividend1 and the divisor1 as before1 the quotient in this case is derived from the following rule1 which the same reasoning will shew to apply to other cases3 ,the numerator of the quotient is the numerator of the dividend multiplied by the denominator of the divisor4 ,the denominator of the quotient is the denominator of the dividend multiplied by the numerator of the divisor4 ,this rule is the reverse of multiplication1 as will be seen by comparing what is required in both cases4 ,in multiplying by _/1 ,i ask1 if out of be taken #aj parts out of #ab1 how much of a unit is taken1 and the answer is _/1 or 4 ,again1 in dividing by 1 ,i ask what part of is 1 the answer to which is _/4 #abb4 ,by taking the following instance1 we shall see that this rule can be sometimes simplified4 ,divide _/ by _/4 ,observe that #af is #d #d1 and #bh is #d #g2 #cc is #c #aa1 and #ae is #c #e2 therefore the two fractions are #d #d #d #g ------ and -----1 #c #aa #c #e and their quotient1 according to the rule1 is #d #d #c #e --------------1 #c #aa #d #g in which #d #c is found both in the numerator and denominator4 ,the fraction is therefore 7#ajh7 the same as #d #e #bj ------1 or ----4 #aa #g #gg ,the rule of the last article1 therefore1 admits of this modification3 ,if the two numerators or the two denominators have a common measure1 divide by that common measure1 and use the quotients instead of the dividends4 #abc4 ,in dividing a fraction by a whole number1 for example1 by #ae1 consider #ae as the fraction _/4 ,the rule gives _/ as the quotient4 ,therefore1 to divide a fraction by a whole number1 multiply the denominator by that whole number4 ,,exercises4 ,dividend4 ,divisor4 ,quotient4 #da #fc #da ---- ---- ----- #cc #aa #ahi #dfg #ijg #dgafg ---- ---- ------- #aea #aja #acfieg #ghac #fja #ac ----- ---- ---- #ejga #aa #dfa _/ _/ _/ - _/ _/ _/ ,what are -----------------------------------1 _/ - _/ and _/ _/ - _/ _/ ----------------------- 8 _/ - _/ #eei ,answer1 ----1 and #a4 #gbbe ,a can reap a field in #ab days1 ,b in #f1 and ,c in #d days2 in what time can they all do it together8#af--;,answer1 #b days4 #af ,the method of solving this and the following question may be shewn thus3 ,if the number of days in which each could reap the field is given1 the part which each could do in a day by himself can be found1 and thence the part which all could do together2 this being known1 the number of days which it would take all to do the whole can be found4 ,in what time would a cistern be filled by cocks which would separately fill it in #ab1 #aa1 #aj1 and #i hours8--,answer1 7#b_/7 hours4 #abd4 ,the principal results of this section may be exhibited algebraically as follows2 let a1 b1 c1 c4 stand for any whole numbers4 ,then a #a a ma 7#ajg7 ---- ---- a 7#ajh7 ---- ---- b b b mb a c ad bc 7#aaa7 --- and --- are the same as ---- and ---- b d bd bd a b a b 7#aab7 --- --- --------- c c c a b a - b --- - --- --------- c c c a c ad bc 7#aac7 --- --- ----------- b d bd a c ad - bc --- - --- ----------- b d bd a c ac 7#aah7 --- --- ---- b d bd a c a_/b ad 7#aba7 --- div4 by --- or --------- ---- b d c_/d bc #abe4 ,these results are true even when the letters themselves represent fractions4 ,for example1 take the fraction a_/b -------1 c_/d whose numerator and denominator are fractional1 and multiply its numerator and denominator by the fraction e ae_/bf ---1 which gives ----------1 f ce_/df aedf which 7#aba7 is -------1 bfce which1 dividing the numerator and denominator by ef 7#ajh71 is ad ----4 bc ,but the original fraction itself is ad a_/b 7a_/b7 7e_/f7 ----2 hence ------- --------------------- bc c_/d 7c_/d7 7e_/f7 which corresponds to the second formula#ag in 7#abd74 ,in a similar manner it may be shewn1 that the other formul of the same article are true when the letters there used either represent fractions1 or are removed and fractions introduced in their place4 ,all formul established throughout this work are equally true when fractions are substituted for whole numbers4 ,for example 7#ed71 7m n7a ma na4 ,let m1 n1 and a be respectively the fractions p r b ---1 ---1 and ---4 q s c ,then m n is p r ps qr --- ---1 or ----------- q s qs and 7m n7a is ps qr b 7ps qr7b ----------- ---1 or ---------------- qs c qsc psb qrb or -------------4 qsc psb qrb pb rb ,but this 7#aab7 is ----- -----1 which is ---- ----1 qsc qsc qc sc psb pb qrb rb since ----- ----1 and ----- ---- 7#ajc74 qsc qc qsc sc pb p b rb r b ,but ---- --- ---1 and ---- --- ---4 qc q c sc s c ,therefore 7m n7a1 or 7p r 7b p b r b 7--- --- 7--- --- --- --- ---4 7q s 7c q c s c ,in a similar manner the same may be proved of any other formula4 #ag ,a formula is a name given to any algebraical expression which is commonly used4 ,the following examples may be useful3 a c e g --- --- --- --- b d f h acfh bdeg --------------------- --------------- a e c g aedh bcfg --- --- --- --- b f d h #a b ---------- --------- #a ab #a a --- b #a #a bc #a --------------- -------------- ----------------- #a c abc a c a --------- a -------- #a bc #a b --- c #a #a #eg ,thus1 ------------ --------- --- #a #h #cej #f ------- #f --- #a #eg #g --- #h ,the rules that have been proved to hold good for all numbers may be applied when the numbers are represented by letters4 ,,section ,,vi4 ,,decimal ,,fractions4 #abf4 ,we have seen 7#aab7 7#aba7 the necessity of reducing fractions to a common denominator1 in order to compare their magnitudes4 ,we have seen also how much more readily operations are performed upon fractions which have the same1 than upon those which have different1 denominators4 ,on this account it has long been customary1 in all those parts of mathematics where fractions are often required1 to use none but such as either have1 or can be easily reduced to others having1 the same denominators4 ,now1 of all numbers1 those which can be most easily managed are such as #aj1 #ajj1 #ajjj1 c41 where #a is followed by ciphers4 ,these are called ,,decimal ,,numbers2 and a fraction whose denominator is any one of them1 is called a ,,decimal ,,fraction1 or more commonly1 a ,,decimal4 #abg4 ,a whole number may be reduced to a decimal fraction1 or one decimal fraction to another1 with the greatest ease4 ,for example1 #idj #idjj #idjjj #id is ---1 or ----1 or ----- 7#ajf72 #aj #ajj #ajjj #c #cj #cjj #cjjj ---- is ----1 or ----1 or ----- 7#ajh74 #cj #ajj #ajjj #ajjjj ,the placing of a cipher on the right hand of any number is the same thing as multiplying that number by #aj 7#eg71 and this may be done as often as we please in the numerator of a fraction1 provided it be done as often in the denominator 7#ajh74 #abh4 ,the next question is1 ,how can we reduce a fraction which is not decimal to another which is1 without altering its value8 ,take1 for example1 the fraction _/1 multiply both the numerator and denominator successively by #aj1 #ajj1 #ajjj1 c41 which will give a series of fractions1 each of which is equal to _/ 7#ajh71 viz4 _/1 _/1 _/1 _/1 c4 ,the denominator of each of these fractions can be divided without remainder by #af1 the quotients of which divisions form the series of decimal numbers #aj1 #ajj1 #ajjj1 #ajjjj1 c4 ,if1 therefore1 one of the numerators be divisible by #af1 the fraction to which that numerator belongs has a numerator and denominator both divisible by #af4 ,when that division has been made1 which 7#ajh7 does not alter the value of the fraction1 we shall have a fraction whose denominator is one of the series #aj1 #ajj1 #ajjj1 c41 and which is equal in value to _/4 ,the question is then reduced to finding the first of the numbers #gj1 #gjj1 #gjjj1 #gjjjj1 c41 which can be divided by #af without remainder4 ,divide these numbers1 one after the other1 by #af1 as follows3 #af7gj7d #af7gjj7dc #af7gjjj7dcg #af7gjjjj7dcge #fd #fd #fd #fd -- --- --- --- #f #fj #fj #fj #dh #dh #dh -- --- --- #ab #abj #abj #aab #aab --- --- #h #hj #hj -- #j ,it appears1 then1 that #gjjjj is the first of the numerators which is divisible by #af4 ,but it is not necessary to write down each of these divisions1 since it is plain that the last contains all which came before4 ,it will do1 then1 to proceed at once as if the number of ciphers were without end1 to stop when the remainder is nothing1 and then count the number of ciphers which have been used4 ,in this case1 since #gjjjj is #af #dcge1 #gjjjj #af #dcge #dcge ------1 which is ----------1 or -----1 #afjjjj #af #ajjjj #ajjjj gives the fraction required4 ,therefore1 to reduce a fraction to a decimal fraction1 annex ciphers to the numerator1 and divide by the denominator until there is no remainder4 ,the quotient will be the numerator of the required fraction1 and the denominator will be unity1 followed by as many ciphers as were used in obtaining the quotient4 ,,exercises4 ,reduce to decimal fractions 1 1 _/1 _/1 _/1 and _/4 ,answer1 _/1 _/1 _/1 _/1 _/1 and _/4 #abi4 ,it will happen in most cases that the annexing of ciphers to the numerator will never make it divisible by the denominator without remainder4 ,for example1 try to reduce _/ to a decimal fraction4 #g7ajjjjjjjjjjjjjjjjjj1 c4 ------------------- #adbhegadbhegadbheg1 c4 ,the quotient here is a continual repetition of the figures #a1 #d1 #b1 #h1 #e1 #g1 in the same order2 therefore _/ cannot be reduced to a decimal fraction4 ,but1 nevertheless1 if we take as a numerator any number of figures from the quotient #adbhegadbheg1 c41 and as a denominator #a followed by as many ciphers as were used in making that part of the quotient1 we shall get a fraction which differs very little from _/1 and which will differ still less from it if we put more figures in the numerator and more ciphers in the denominator4 ,thus1 #a is less #a #c which is not #a --- --- by --- --- #aj than #g #gj so much as #aj #ad #a #b #a --- --- --- --- #ajj #g #gjj #ajj #adb #a #f #a ---- --- ---- ---- #ajjj #g #gjjj #ajjj #adbh #a #d #a ----- --- ----- ----- #ajjjj #g #gjjjj #ajjjj #adbhe #a #e #a ------ --- ------ ------ #ajjjjj #g #gjjjjj #ajjjjj #adbheg #a #a #a ------- --- ------- ------- #ajjjjjj #g #gjjjjjj #ajjjjjj c4 c4 c4 c4 ,in the first column is a series of decimal fractions1 which come nearer and nearer to _/1 as the third column shews4 ,therefore1 though we cannot find a decimal fraction which is exactly _/1 we can find one which differs from it as little as we please4 ,this may also be illustrated thus3 ,it is required to reduce _/ to a decimal fraction without the error of say a millionth of a unit2 multiply the numerator and denominator of _/ by a million1 and then divide both by #g2 we have then #a #ajjjjjj #adbhega_/ --- ------- ----------- #g #gjjjjjj #ajjjjjj ,if we reject the fraction _/ in the numerator1 what we reject is really the #gth part of the millionth part of a unit2 or less than the millionth part of a unit4 ,therefore _/ is the fraction required4 ,,exercises4 ,make similar tables #c #ag #a with ---1 ---1 and ---4 these fractions #ia #adc #bdg #c ,the recurring --- is #cbifgj1cbifgj1 c4 quotient of #ia #ag --- #aahhha1aahhha1 c4 #adc #a --- #djdhehbiieieadagjj1djdhehb c4 #bdg #acj4 ,the reason for the recurrence of the figures of the quotient in the same order is as follows3 ,if #ajjj1 c4 be divided by the number #bdg1 the remainder at each step of the division is less than #bdg1 being either #j1 or one of the first #bdf numbers4 ,if1 then1 the remainder never become nothing1 by carrying the division far enough1 one remainder will occur a second time4 ,if possible1 let the first #bdf remainders be all different1 that is1 let them be #a1 #b1 #c1 c41 up to #bdf1 variously distributed4 ,as the #bdgth remainder cannot be so great as #bdg1 it must be one of these which have preceded4 ,from the step where the remainder becomes the same as a former remainder1 it is evident that former figures of the quotient must be repeated in the same order4 #aca4 ,you will here naturally ask1 ,what is the use of decimal fractions1 if the greater number of fractions cannot be reduced at all to decimals8 ,the answer is this3 ,the addition1 subtraction1 multiplication1 and division of decimal fractions are much easier than those of common fractions2 and though we cannot reduce all common fractions to decimals1 yet we can find decimal fractions so near to each of them1 that the error arising from using the decimal instead of the common fraction will not be perceptible4 ,for example1 if we suppose an inch to be divided into ten million of equal parts1 one of those parts by itself will not be visible to the eye4 ,therefore1 in finding a length1 an error of a ten-millionth part of an inch is of no consequence1 even where the finest measurement is necessary4 ,now1 by carrying on the table in 7#abi71 we shall see that #adbhega #a #a -------- does not differ from --- by --------2 #ajjjjjjj #g #ajjjjjjj and if these fractions represented parts of an inch1 the first might be used for the second1 since the difference is not perceptible4 ,in applying arithmetic to practice1 nothing can be measured so accurately as to be represented in numbers without any error whatever1 whether it be length1 weight1 or any other species of magnitude4 ,it is therefore unnecessary to use any other than decimal fractions1 since1 by means of them1 any quantity may be represented with as much correctness as by any other method4 ,,exercises4 ,find decimal fractions which do not differ from the following fractions by _/4 ,answer1 _/4 _/ _/4 _/ _/4 _/ _/4 #acb4 ,every decimal may be immediately reduced to a quantity consisting either of a whole number and more simple decimals1 or of more simple decimals alone1 having one figure only in each of the numerators4 ,take1 for example1 #adgcbf #adgcbf #cbf ------4 ,by 7#aae7 ------- is #adg----2 #ajjj #ajjj #ajjj and since #cbf is made up of #cjj1 and #bj1 and #f2 by 7#aab7 _/ _/ _/ _/4 ,but 7#ajh7 _/ is _/1 and _/ is _/4 ,therefore1 #f_/ is made up of #adg _/ _/ #f_/4 ,now1 take any number1 for example1 #adgcbf1 and form a number of fractions having for their numerators this number1 and for their denominators #a1 #aj1 #ajj1 #ajjj1 #ajjjj1 c41 and reduce these fractions into numbers and more simple decimals1 in the foregoing manner1 which will give the table below4 ,,decomposition ,,of ,a ,,decimal ,,fraction4 #adgcbf ------ #adgcbf #a #adgcbf #f ------ #adgcb --- #aj #aj #adgcbf #b #f ------ #adgc --- --- #ajj #aj #ajj #adgcbf #c #b #f ------ #adg --- --- ---- #ajjj #aj #ajj #ajjj #adgcbf #g #c #b #f ------ #ad --- --- ---- ----- #ajjjj #aj #ajj #ajjj #ajjjj #adgcbf #d #g #c #b #f ------ #a --- --- ---- ----- ------ #ajjjjj #aj #ajj #ajjj #ajjjj #ajjjjj #adgcbf #a #d #g #c #b #f ------- --- --- ---- ----- ------ ------- #ajjjjjj #aj #ajj #ajjj #ajjjj #ajjjjj #ajjjjjj #adgcbf #a #d #g #c #b #f -------- --- ---- ----- ------ ------- -------- #ajjjjjjj #ajj #ajjj #ajjjj #ajjjjj #ajjjjjj #ajjjjjjj ,n4,b4 ,the student should write this table himself1 and then proceed to make similar tables from the following exercises4 ,,exercises4 ,reduce the following fractions into a series of numbers and more simple fractions3 #cadaeibf #cadaeibf --------1 --------1 c4 #aj #ajj #bgjjjca #bgjjjca -------1 --------1 c4 #aj #ajj #bjgcjjj #bjgcjjj -------1 --------1 c4 #aj #ajj #cccacjc #cccacjc -------1 -------1 c4 #ajjj #ajjjj #acc4 ,if1 in this table1 and others made in the same manner1 you look at those fractions which contain a whole number1 you will see that they may be made thus3 ,mark off1 from the right hand of the numerator1 as many figures as there are ciphers in the denominator by a point1 or any other convenient mark4 ,this will give #adgcbf when the fraction is #adgcbf ------ #aj #adgcbf #adgcbf ------ #ajj #adgcbf #adgcbf ------ #ajjj c4 c4 ,the figures on the left of the point by themselves make the whole number which the fraction contains4 ,of those on its right1 the first is the numerator of the fraction whose denominator is #aj1 the second of that whose denominator is #ajj1 and so on4 ,we now come to those fractions which do not contain a whole number4 #acd4 ,the first of these is _/ which the number of ciphers in the denominator is the same as the number of figures in the numerator4 ,if we still follow the same rule1 and mark off all the figures1 by placing the point before them all1 thus1 #adgcbf1 the observation in 7#acc7 still holds good2 for1 on looking at _/ in the table1 we find it is #a #d #g #c #b #f --- --- ---- ----- ------ ------- #aj #ajj #ajjj #ajjjj #ajjjjj #ajjjjjj ,the next fraction is _/1 which we find by the table to be #a #d #g #c #b #f --- ---- ----- ------ ------- -------- #ajj #ajjj #ajjjj #ajjjjj #ajjjjjj #ajjjjjjj ,in this1 #a is not divided by #aj1 but by #ajj2 if1 therefore1 we put a point before the whole1 the rule is not true1 for the first figure on the left of the point has the denominator which1 according to the rule1 the second ought to have1 the second that which the third ought to have1 and so on4 ,in order to keep the same rule for this case1 we must contrive to make #a the second figure on the right of the point instead of the first4 ,this may be done by placing a cipher between it and the point1 thus1 #jadgcbf4 ,here the rule holds good1 for by that rule this fraction is #j #a #d #g #c #b #f --- --- ---- ----- ------ ------- -------- #aj #ajj #ajjj #ajjjj #ajjjjj #ajjjjjj #ajjjjjjj which is the same as the preceding line1 since _/ is #j1 and need not be reckoned4 ,similarly1 when there are two ciphers more in the denominator than there are figures in the numerator1 the rule will be true if we place two ciphers between the point and the numerator4 ,the rule1 therefore1 stated fully1 is this3 ,to reduce a decimal fraction to a whole number and more simple decimals1 or to more simple decimals alone if it do not contain a whole number1 mark off by a point as many figures from the numerator as there are ciphers in the denominator4 ,if the numerator have not places enough for this1 write as many ciphers before it as it wants places1 and put the point before these ciphers4 ,then1 if there be any figures before the point1 they make the whole number which the fraction contains4 ,the first figure after the point with the denominator #aj1 the second with the denominator #ajj1 and so on1 are the fractions of which the first fraction is composed4 #ace4 ,decimal fractions are not usually written at full length4 ,it is more convenient to write the numerator only1 and to cut off from the numerator as many figures as there are ciphers in the denominator1 when that is possible1 by a point4 ,when there are more ciphers in the denominator than figures in the numerator1 as many ciphers are placed before the numerator as will supply the deficiency1 and the point is placed before the ciphers4 ,thus1 #g will be used in future to denote _/1 #jg for _/1 and so on4 ,the following tables will give the whole of this notation at one view1 and will shew its connexion with the decimal notation explained in the first section4 ,you will observe that the numbers on the right of the units' place stand for units divided by #aj1 #ajj1 #ajjj1 c4 while those on the left are units multiplied by #aj1 #ajj1 #ajjj1 c4 ,the student is recommended always to write the decimal point in a line with the top of the figures or in the middle1 as is done here1 and never at the bottom4 ,the reason is1 that it is usual in the higher branches of mathematics to use a point placed between two numbers or letters which are multiplied together2 thus1 #ae4 #af1 a4 b1 7a b74 7c d7 stand for the products of those numbers or letters4 #abcd #d #d ,i4 #abcd stands for ---- or #abc--- or #abc --- #aj #aj #aj #abcd #cd #c #d #abcd ---- or #ab--- or #ab --- --- #ajj #ajj #aj #ajj #abcd #bcd #b #c #d #abcd ---- or #a---- or #a --- --- ---- #ajjj #ajjj #aj #ajj #ajjj #abcd #a #b #c #d #abcd ----- or --- --- ---- ----- #ajjjj #aj #ajj #ajjj #ajjjj #abcd #a #b #c #d #jabcd ------ or --- ---- ----- ------ #ajjjjj #ajj #ajjj #ajjjj #ajjjjj #abcd #a #b #c #d #jjabcd ------- or ---- ----- ------ ------- #ajjjjjj #ajjj #ajjjj #ajjjjj #ajjjjjj ,,ii4 #ajjc #a #c #jajjc is ------ or --- ------ #ajjjjj #ajj #ajjjjj #ajjc #a #c #ajjc is ----- or --- ----- #ajjjj #aj #ajjjj #ajjc #c #ajjc is ---- or #aj --- #ajj #ajj #ajjc #c #ajjc is ---- or #ajj --- #aj #aj ,,iii4 #a #b #h #c #abhc --- --- ---- ----- #aj #ajj #ajjj #ajjjj #a #jb #jjh #jjjc #a #jbhc #ab #jjhc #abh #jjjc #ajh #jbjc #ajjc #jbh #abjc #jjh #a is #ajjj inches #b is #bjj #c is #cj #d is #d ,,iv4 ,in #abcdefghi #e is _/ of an inch inches the #f is _/ #g is _/ #h is _/ #i is _/ #acf4 ,the ciphers on the right hand of the decimal point serve the same purpose as the ciphers in 7#aj74 ,they are not counted as any thing themselves1 but serve to shew the place in which the accompanying numbers stand4 ,they might be dispensed with by writing the numbers in ruled columns1 as in the first section4 ,they are distinguished from the numbers which accompany them by calling the latter significant figures4 ,thus1 #jjjcgdg is a decimal of seven places with four significant figures1 #cdf is a decimal of three places with three significant figures1 c4 #acg4 ,the value of a decimal is not altered by putting any number of ciphers on its right4 ,take1 for example1 #c and #cjj4 ,the first 7#ace7 is _/1 and the second _/1 which is made from the first by multiplying both its numerator and denominator by #ajj1 and 7#ajh7 is the same quantity4 #ach4 ,to reduce two decimals to a common denominator1 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 same4 ,take1 for example1 #ed and #dcbig4 ,the first is _/1 and the second _/4 ,multiply the numerator and denominator of the first by #ajj 7#ajh71 which reduces it to _/1 which has the same denominator as _/4 ,but _/ is #edjj 7#ace74 ,in whole numbers1 the decimal point should be placed at the end3 thus1 #abi should be written #abi4 ,it is1 however1 usual to omit the point2 but you must recollect that #abi and #abijjj are of the same value1 since the first is #abi and the second _/4 #aci4 ,the rules which were given in the last chapter for addition1 subtraction1 multiplication1 and division1 apply to all fractions1 and therefore to decimal fractions among the rest4 ,but the way of writing decimal fractions1 which is explained in this chapter1 makes the application of these rules more simple4 ,we proceed to the different cases4 ,suppose it required to add #dbfcd1 #debhjf1 #bjja1 and #ed4 ,by 7#aab7 these must be reduced to a common denominator1 which is done 7#ach7 by writing them as follows3 #dbfcdj1 #debhjf1 #bjjaj1 and #edjjjj4 ,these are decimal fractions1 whose numerators are #dbfcdj1 #debhjf1 #bjjaj1 and #edjjjj1 and whose common denominator is #ajjjj4 ,by 7#aab7 their sum is #dbfcdj #debhjf #bjjaj #edjjjj #adciaef --------------------------------1 which is ------- #ajjjj #ajjjj or #adciaef4 ,the simplest way of doing this is as follows3 write the decimals down under one another1 so that the decimal points may fall under one another1 thus3 #dbfcd #debhjf #bjja #ed -------- #adciaef ,add the different columns together as in common addition1 and place the decimal point under the other decimal points4 ,,exercises4 ,what are #aebg #fdgcbjid #bjjac #jjjjaigd2 #bbgfc #ajg #i #bfcagb #efgcbjja2 and #aaa #gg #jjci #jjadb #hhch8 ,answer1 #aeicgccdacgd1 #eijcefbeb1 #ifiiab4 #adj4 ,suppose it required to subtract #iajgcbd from #acgcba4 ,these fractions when reduced to a common denominator are #iajgcbd and #acgcbajj 7#ach74 ,their difference is therefore #acgcbajj - #iajgcbd #dfbdggf ------------------1 which is ------- #ajjjjj #ajjjjj or #dfbdggf4 ,this may be most simply done as follows3 write the less number under the greater1 so that its decimal point may fall under that of the greater1 thus3 #acgcba #iajgcbd --------- #dfbdggf ,subtract the lower from the upper line1 and wherever there is a figure in one line and not in the other1 proceed as if there were a cipher in the vacant place4 ,,exercises4 ,what is #abcfb - #bgdbbajg #e2 #iigfbjgcidb - #jjadcigfgbh2 and #ab #jc #jjd - #jjje8 ,answer1 #abjhhbghic1 #iigfbjeieddcbgb2 and #abcce4 #ada4 ,the multiplication of a decimal by #aj1 #ajj1 #ajjj1 c41 is performed by merely moving the decimal point to the right4 ,suppose1 for example1 #acbjgi is to be multiplied by #ajj4 ,the decimal is _/1 which multiplied by #ajj is 7#aag7 _/1 or #acbjgi4 ,again1 #acji #ajjjjj is _/ #ajjjjj1 or 7#aaf7 _/ or #acjijj4 ,from these and other instances we get the following rule3 ,to multiply a decimal fraction by a decimal number 7#abf71 move the decimal point as many places to the right as there are ciphers in the decimal number4 ,when this cannot be done1 annex ciphers to the right of the decimal 7#acg7 until it can4 #adb4 ,suppose it required to multiply #agjcf by #dbg4 ,the first of these decimals is _/1 and the second _/4 ,by 7#aah7 the product of these fractions has for its numerator the product of #agjcf and #dbg1 and for its denominator the product of #ajjj and #ajj2 therefore this product is _/1 or #gbgdcgb4 ,this may be done more shortly by multiplying the two numbers #agjcf and #dbg1 and cutting off by the decimal point as many places as there are decimal places both in #agjcf and #dbg1 because the product of two decimal numbers will contain as many ciphers as there are ciphers in both4 #adc4 ,this question now arises3 ,what if there should not be as many figures in the product as there are decimal places in the multiplier and multiplicand together8 ,to see what must be done in this case1 multiply #agb by #aja1 or _/ by _/4 ,the product of these two is _/1 or #jagcgb 7#ace74 ,therefore1 when the number of places in the product is not sufficient to allow the rule of the last article to be followed1 as many ciphers must be placed at the beginning as will make up the deficiency4 ,,additional ,,examples4 #jja #ja is #jjjja #ef #jjja is #jjef4 ,,exercises4 ,shew that #cjjb #cjjb #c #c #b #c #jjb #jjb #jjb #aaefji #ecaia #hdd #hdd - #cabji #cabji #hbag #ajjja #h #aj #h #jja #aj #bag #jja #bag4 ,fraction4 ,square4 ,cube4 #hbib #fhgegbfd #egjacebccjhh #jagc #jjjbiibi #jjjjjeagggag #adc #bjddi #bibdbjg #jji #jjjjha #jjjjjjgbi #aefbe #fd #ajjj #aefbe #fd #a #jaefbe #jjfd #jjja #aefbe #fd #a #aefbe #jfd #ajj #aefbejjj #jfd #ajjjjjj #add4 ,the division of a decimal by a decimal number1 such as #aj1 #ajj1 #ajjj1 c41 is performed by moving the decimal point as many places to the left as there are ciphers in the decimal number4 ,if there are not places enough in the dividend to allow of this1 annex ciphers to the beginning of it until there are4 ,for example1 divide #agcdbbi by #ajjj3 the decimal fraction is _/1 which divided by #ajjj 7#abc7 is _/1 or #agcdbbi4 ,if1 in the same way1 #abajf be divided by #ajjjj1 the result is #jjjabajf4 #ade4 ,before proceeding to shorten the rule for the division of one decimal fraction by another1 it will be necessary to resume what was said in 7#abh7 upon the reduction of any fraction to a decimal fraction4 ,it was there shewn that _/ is the same fraction as _/ or #dcge4 ,as another example1 convert _/ into a decimal fraction4 ,follow the same process as in 7#abh71 thus3 #abh7cjjjjjjjjjjj7bcdcge #bef ---- #ddj #chd ---- #efj #eab ---- #dhj #chd ---- #ifj #hif ---- #fdj #fdj --- #j ,since #g ciphers are used1 it appears that #cjjjjjjj is the first of the series #cj1 #cjj1 c41 which is divisible by #abh2 and therefore _/ or1 which is the same thing 7#ajh71 _/ is equal to _/ or #jbcdcge 7#ace74 ,from these examples the rule for reducing a fraction to a decimal is3 ,annex ciphers to the numerator2 divide by the denominator1 and annex a cipher to each remainder after the figures of the numerator are all used1 proceeding exactly as if the numerator had an unlimited number of ciphers annexed to it1 and was to be divided by the denominator4 ,continue this process until there is no remainder1 and observe how many ciphers have been used4 ,place the decimal point in the quotient so as to cut off as many figures as you have used ciphers2 and if there be not figures enough for this1 annex ciphers to the beginning until there are places enough4 #adf4 ,from what was shewn in 7#abi71 it appears that it is not every fraction which can be reduced to a decimal fraction4 ,it was there shewn1 however1 that there is no fraction to which we may not find a decimal fraction as near as we please4 ,thus1 _/1 _/1 _/1 _/1 _/1 c41 or #a1 #ad1 #adb1 #adbh1 #adbhe1 were shewn to be fractions which approach nearer and nearer to _/4 ,to find either of these fractions1 the rule is the same as that in the last article1 with this exception1 that1 ,i4 instead of stopping when there is no remainder1 which never happens1 stop at any part of the process1 and make as many decimal places in the quotient as are equal in number to the number of ciphers which have been used1 annexing ciphers to the beginning when this cannot be done1 as before4 ,,ii4 ,instead of obtaining a fraction which is exactly equal to the fraction from which we set out1 we get a fraction which is very near to it1 and may get one still nearer1 by using more of the quotient4 ,thus1 #adbh is very near to _/1 but not so near as #adbheg2 nor is this last1 in its turn1 so near as #adbhegadbheg1 c4 #adg4 ,if there should be ciphers in the numerator of a fraction1 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 fraction4 ,take1 for example1 _/2 annex ciphers to the numerator1 and divide by the denominator4 ,it appears that #ajjj is divisible by #abe1 and that the quotient is #h4 ,one cipher only has been annexed to the numerator1 and therefore #ajj divided by #abe is #h4 ,had the fraction been _/1 since #ajjj divided by #abe gives #h1 and three ciphers would have been annexed to the numerator1 the fraction would have been #jjh4 #adh4 ,suppose that the given fraction has ciphers at the right of its denominator2 for example1 _/4 ,then annexing a cipher to the numerator is the same thing as taking one away from the denominator2 for1 7#ajh7 _/ is the same thing as _/1 and _/ as _/4 ,the rule1 therefore1 is in this case3 ,take away the ciphers from the denominator4 ,,exercises4 ,reduce the following fractions to decimal fractions3 #a #cf #big #a ---1 ----1 ----1 and ---4 #hjj #abej #fd #abh ,answer1 #jjabe1 #jbhh1 #dfdjfbe1 and #jjghabe4 ,find decimals of #f places very near to the following fractions3 #bg #aef #bb #aid #bfcg #a #a #c --1 ---1 -----1 ---1 ----1 ----1 ---1 and ---4 #di #cc #cgjjj #ac #iijg #bijh #dff #bgg ,answer1 #eeajbj1 #dgbgbgb1 #jjjeid1 #adibcjgf1 #bffage1 #jjjcdc1 #jjbade1 and #jajhcj4 #adi4 ,from 7#aba7 it appears1 that if two fractions have the same denominator1 the first may be divided by the second by dividing the numerator of the first by the numerator of the second4 ,suppose it required to divide #aggfb by #fbe4 ,these fractions 7#ach71 when reduced to a common denominator1 are #aggfb and #fbej1 or _/ and _/4 ,their quotient is therefore _/1 which must now be reduced to a decimal fraction by the last rule4 ,the process at full length is as follows3 ,leave out the cipher in the denominator1 and annex ciphers to the numerator1 or1 which will do as well1 to the remainders1 when it becomes necessary1 and divide as in 7#ade74 #fbe7aggfb7bhdaib #abej ----- #ebfb #ejjj ----- #bfbj #bejj ----- #abjj #fbe ----- #egej #efbe ----- #abej #abej ---- #j ,here four ciphers have been annexed to the numerator1 and one has been taken from the denominator4 ,make five decimal places in the quotient1 which then becomes #bhdaib1 and this is the quotient of #aggfb divided by #fbe4 #aej4 ,the rule for division of one decimal by another is as follows3 ,equalise the number of decimal places in the dividend and divisor1 by annexing ciphers to that which has fewest places4 ,then1 further1 annex as many ciphers to the dividend#ah as it is required to have decimal places1 throw away the decimal point1 and operate as in common division4 ,make the required number of decimal places in the quotient4 #ah ,or remove ciphers from the divisor2 or make up the number of ciphers partly by removing from the divisor and annexing to the dividend1 if there be not a sufficient number in the divisor4 ,thus1 to divide #fgagc by #jad to three decimal places1 ,i first write #fgagc and #jadj1 with four places in each4 ,having to provide for three decimal places1 ,i should annex three ciphers to #fgagc2 but1 observing that the divisor #jadj has one cipher1 ,i strike that one out and annex two ciphers to #fgagc4 ,throwing away the decimal points1 then divide #fgagcjj by #jad or #ad in the usual way1 which gives the quotient #dgihjg and the remainder #b4 ,hence #dgihjg is the answer4 ,the common rule is3 ,let the quotient contain as many decimal places as there are decimal places in the dividend more than in the divisor4 ,but this rule becomes inoperative except when there are more decimals in the dividend than in the divisor1 and a number of ciphers must be annexed to the former4 ,the rule in the text amounts to the same thing1 and provides for an assigned number of decimal places4 ,but the student is recommended to make himself familiar with the rule of the characteristic given in the ,appendix1 and also to accustom himself to reason out the place of the decimal point4 ,thus1 it should be visible1 that #bfaai #gbdcf has one figure before the decimal point1 and that #bfaai #gbdcf has one cipher after it1 preceding all significant figures4 ,or the following rule may be used3 ,expunge the decimal point of the divisor1 and move that of the dividend as many places to the right as there were places in the divisor1 using ciphers if necessary4 ,then proceed as in common division1 making one decimal place in the quotient for every decimal place of the final dividend which is used4 ,thus #agcad divided by #fab is #agcad divided by #fab1 and the decimal point must precede the first figure of the quotient4 ,but #agcad divided by #ffage is #agcad by #ffage2 and since three decimal places of #agcadjjj 444 must be used before a quotient figure can be found1 that quotient figure is the third decimal place1 or the quotient is #jjb44444 ,,examples4 #ca #jjjfb ----- #abdj1 ------ #jjjifhge #jjbe #fd ,,exercises4 #aejjf #aejjf - #jjd #jjd ,shew that ----------------------------- #aejjb1 #aeja and that #ja #ja #ja #bi #bi #bi --------------------------------- #bi #bi - #bi #ja #ja #ja #bia #a #a #cfe ,what are -------1 ---------1 and ------1 as far as #f places #cadaei #bgahbhah #ahcdi of decimals8--,answer1 #cahcaj1 #cfghgi1 and #aihibjibba4 ,calculate #aj terms of each of the following series1 as far as #e places of decimals4 #a #a #a #a #a --- ----- --------- ------------- c4 #agahbd4 #b #b #c #b #c #d #b #c #d #e #a #a #a #a #a --- --- --- --- c4 #bibhie4 #b #c #d #e #hj #ha #hb #hc #hd ---- ---- ---- ---- ---- c4 #ihhbhf4 #ha #hb #hc #hd #he #aea4 ,we now enter upon methods by which unnecessary trouble is saved in the computation of decimal quantities4 ,and first1 suppose a number of miles has been measured1 and found to be #aghdfbag miles4 ,if you were asked how many miles there are in this distance1 and a rough answer were required which should give miles only1 and not parts of miles1 you would probably say #ag4 ,but this1 though the number of whole miles contained in the distance1 is not the nearest number of miles2 for1 since the distance is more than #ag miles and #h tenths1 and therefore more than #ag miles and a half1 it is nearer the truth to say1 it is #ah miles4 ,this1 though too great1 is not so much too great as the other was too little1 and the error is not so great as half a mile4 ,again1 if the same were required within a tenth of a mile1 the correct answer is #agh2 for though this is too little by #jdfbag1 yet it is not so much too little as #agi is too great2 and the error is less than half a tenth1 or _/4 ,again1 the same distance1 within a hundredth of a mile1 is more correctly #aghe than #aghd1 since the last is too little by #jjfbag1 which is greater than the half of #ja2 and therefore #aghd #ja is nearer the truth than #aghd4 ,hence this general rule3 ,when a certain number of the decimals given is sufficiently accurate for the purpose1 strike off the rest from the right hand1 observing1 if the first figure struck off be equal to or greater than #e1 to increase the last remaining figure by #a4 ,the following are examples of a decimal abbreviated by one place at a time4 #cadaei1 #cadaf1 #cadb1 #cad1 #ca1 #cj #bgahbhah1 #bgahbhb1 #bgahbh1 #bgahc1 #bgah1 #bgb1 #bg1 #cj #aiiai1 #aiib1 #aii1 #bjj1 #bj #aeb4 ,in multiplication and division it is useless to retain more places of decimals in the result than were certainly correct in the multiplier1 c41 which gave that result4 ,suppose1 for example1 that #iih and #hif are distances in inches which have been measured correctly to two places of decimals1 that is1 within half a hundredth of an inch each way4 ,the real value of that which we call #iih may be any where between #iige and #iihe1 and that of #hif may be any where between #hiee and #hife4 ,the product1 therefore1 of the numbers which represent the correct distances will lie between #iige #hiee and #iihe #hife1 that is1 taking three decimal places in the products1 between #hicbf and #hieaf4 ,the product of the actual numbers given is #hidbjh4 ,it appears1 then1 that in this case no more than the whole number #hi can be depended upon in the product1 or1 at most1 the first place of decimals4 ,the reason is1 that the error made in measuring #hif1 though only in the third place of decimals1 is in the multiplication increased at least #iige1 or nearly #aj times2 and therefore affects the second place4 ,the following simple rule will enable us to judge how far a product is to be depended upon4 ,let a be the multiplier1 and b the multiplicand2 if these be true only to the first decimal place1 the product is within 7a b7_/#bjai of the truth2 if to two decimal places1 within 7a b7_/#bjj2 if to three1 within 7a b7_/#bjjj2 and so on4 ,thus1 in the above example1 we have #iih and #hif1 which are true to two decimal places3 their sum divided by #bjj is #jidg1 and their product is #hidbjh1 which is therefore within #jidg of the truth4 ,if1 in fact1 we increase and diminish #hidbjh by #jidg1 we get #hieaee and #hicbfa1 which are very nearly the limits found within which the product must lie4 ,we see1 then1 that we cannot in this case depend upon the first place of decimals1 as 7#aea7 an error of #je cannot exist if this place be correct2 and here is a possible error of #ji and upwards4 ,it is hardly necessary to say1 that if the numbers given be exact1 their product is exact also1 and that this article applies where the numbers given are correct only to a certain number of decimal places4 ,the rule is3 ,take half the sum of the multiplier and multiplicand1 remove the decimal point as many places to the left as there are correct places of decimals in either the multiplier or multiplicand2 the result is the quantity within which the product can be depended upon4 ,in division1 the rule is3 ,proceed as in the last rule1 putting the dividend and divisor in place of the multiplier and multiplicand1 and divide by the square of the divisor2 the quotient will be the quantity within which the division of the first dividend and divisor may be depended upon4 ,thus1 if #agcbd be divided by #echji1 both being correct to the third place1 their half sum will be #ceeff1 which1 by the last rule1 is made #jceeff1 and is to be divided by the square of #echji1 or1 which will do as well for our purpose1 the square of #ej1 or #bejj4 ,the result is something less than #jjjjb1 so that the quotient of #agcbd and #echji can be depended on to four places of decimals4 #ai ,these are not quite correct1 but sufficiently so for every practical purpose4 #aec4 ,it is required to multiply two decimal fractions together1 so as to retain in the product only a given number of decimal places1 and dispense with the trouble of finding the rest4 ,first1 it is evident that we may write the figures of any multiplier in a contrary order 7for example1 #dcba instead of #abcd71 provided that in the operation we move each line one place to the right instead of to the left1 as in the following example3 #bbba #bbba #abcd #dcba ---- ---- #hhhd #bbba #fffc #dddb #dddb #fffc #bbba #hhhd ------- ------- #bgdjgad #bgdjgad ,suppose now we wish to multiply #cdhhdad by #eacjgdb1 reserving only four decimal places in the product4 ,if we reverse the multiplier1 and proceed in the manner just pointed out1 we have the following3 #cdhhdad #bdgjcae --------- #agddbjgj #cdhhdad #ajdfebdb #bddahhih #aciecfef #figfhbh ---------------- #aghihaebbbcahh ,cut off1 by a vertical line1 the first four places of decimals1 and the columns which produced them4 ,it is plain that in forming our abbreviated rule1 we have to consider only1 ,i4 all that is on the left of the vertical line2 ,,ii4 all that is carried from the first column on the right of the line4 ,on looking at the first column to the left of the line1 we see #d1 #d1 #h1 #e1 #i1 of which the first #d comes from #d #a1bj the second #d from #a #c1 the #h from #h #g1 the #e from #h #d1 and the #i from #d #b4 ,if1 then1 we arrange the multiplicand and the reversed multiplier thus1 #cdhhdad #bdgjcae 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 decimals4 ,and here observe1 that the units' figure in the multiplier #eacjgdb1 viz4 #a1 comes under #d1 the fourth decimal place in the multiplicand4 ,if there had been no carrying from the right of the vertical line1 the rule would have been3 ,reverse the multiplier1 and place it under the multiplicand1 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 preserved2 place ciphers over those figures of the multiplier which have none of the multiplicand above them1 if there be any3 proceed to multiply in the usual way1 but begin each figure of the multiplier with the figure of the multiplicand which comes above it1 taking no account of those on the right3 place the first figures of all the lines under one another4 ,to correct this rule1 so as to allow for what is carried from the right of the vertical line1 observe that this consists of two parts1 #ast1 what is carried directly in the formation of the different lines1 and #b;dly1 what is carried from the addition of the first column on the right4 ,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 multiplicand1 and carrying the tens to the next figure as usual1 but without writing down the units4 ,but both may be allowed for at once1 with sufficient correctness1 on the principle of 7#aea71 by carrying #a from #e up to #ae1 #b from #ae up to #be1 c42 that is1 by carrying the nearest ten4 ,thus1 for #cg1 #d would be carried1 #cg being nearer to #dj than to #cj4 ,this will not always give the last place quite correctly1 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 correct4 ,the rule1 then1 is as follows3 #bj ,the #a here means that the #a is in the multiplier4 #aed4 ,to multiply two decimals together1 retaining only n decimal places4 ,i4 ,reverse the multiplier1 strike out the decimal points1 and place the multiplier under the multiplicand1 so that what was its units' figure shall fall under the n decimal place of the multiplicand1 placing ciphers1 if necessary1 so that every place of the multiplier shall have a figure or cipher above it4 ,,ii4 ,proceed to multiply as usual1 beginning each figure of the multiplier with the one which is in the place to its right in the multiplicand3 do not set down this first figure1 but carry its nearest ten to the next1 and proceed4 ,,iii4 ,place the first figures of all the lines under one another2 add as usual2 and mark off n places from the right for decimals4 ,it is required to multiply #acfdjgb by #acjfji1 retaining #g decimal places4 #acfdjgbjjj #ijfjca ---------- #acfdjgbjjj #djibbafjj #hahddcb #abbgff ----------- #aghafjjgih ,in the following examples the first two lines are the multiplicand and multiplier2 and the number of decimals to be retained will be seen from the results4 #ddgafah #ccaffbdh #cdfdajaf #cggaibad #adadbacf #agcbejh #cggaibad #jccaffbdh #cdfdajafj #hafagdd #fcabdada #hjebcga -------- ---------- ------------ #aejhgfhf #ccaffbe #cdfdajafj #aejhgfh #acbffej #bdbdhgaab #bfdjcd #ccaff #ajcibcje #cggb #acbff #fibhbj #bbfc #ffc #agcbje #ch #cc #bgga #cj #aj ------------ --------- #b #fjjaehcgc #afhffeia -------- #dfijdae ,exercises may be got from article 7#adc74 #aee4 ,with regard to division1 take any two numbers1 for example1 #afhjdcgiba and #cadb1 and divide the first by the second1 as far as any required number of decimal places1 for example1 five4 ,this gives the following3 #cadb7afhjdcgiba7ecdhcj #aegaj ------- #ajidc #idbf ----- #aeagg 7,a7 #abefh ---- ------ #bfji #bfjii #bead #beacf ---- ------ #ie #ifcb #id #idbf -- ----- #a #bjfa ,now cut off by a vertical line1 as in 7#aec71 all the figures which come on the right of the first figure #b1 in the last remainder #bjfa4 ,as in multiplication1 we may obtain all that is on the left of the vertical line by an abbreviated method1 as represented at 7,a74 ,after what has been said on multiplication1 it is useless to go further into the detail2 the following rule will be sufficient3 ,to divide one decimal by another1 retaining only n places3 ,proceed one step in the ordinary division1 and determine1 by 7#aej71 in what place is the quotient so obtained2 proceed in the ordinary way1 until the number of figures remaining to be found in the quotient is less than the number of figures in the divisor3 if this should be already the case1 proceed no further in the ordinary way4 ,instead of annexing a figure or cipher to the remainder1 cut off a figure from the divisor1 and proceed one step with this curtailed divisor as usual1 remembering1 however1 in multiplying this divisor1 to carry the nearest ten1 as in 7#aed71 from the figure which was struck off2 repeat this1 striking off another figure of the divisor1 and so on1 until no figures are left4 ,since we know from the beginning in what place the first figure of the quotient is1 and also how many decimals are required1 we can tell from the beginning how many figures there will be in the whole quotient4 ,if the divisor contain more figures than the quotient1 it will be unnecessary to use them3 and they may be rejected1 the rest being corrected as in 7#aea73 if there be ciphers at the beginning of the divisor1 if it be1 for example1 #cagh #jjcagh1 since this is -----1 #ajj divide by #cagh in the usual way1 and afterwards multiply the quotient by #ajj1 or remove the decimal point two places to the right4 ,if1 therefore1 six decimals be required1 eight places must be taken in dividing by #cagh1 for an obvious reason4 ,in finding the last figure of the quotient1 the nearest should be taken1 as in the second of the subjoined examples4 ,places required1 #b #h ,divisor1 #dadcb #cadaeibg ,dividend1 #fgcadhi #bgahbhahj #dadcb #beacbgdaf -------- ---------- #behhbh #bjejjgfd #bdheib #ahhdieef ------- -------- #ajbcgba #afeabjh #hbhf #aegjgif ----- ------- #aiea #hjdab #afeg #fbhcb ----- ----- #bid #agehj #bij #aegjh --- ----- #d #ahgb #d #aega - ---- #j #cja #bhc --- #ah #ai -- ,quotient1 #afbdga #hfebeeif #ba ,this is written #g instead of #f1 because the figure which is abandoned in the dividend is #i 7#aea74 ,examples may be obtained from 7#adc7 and 7#aej74 ,,section ,,vii4 ,,on ,,the ,,extraction ,,of ,,the ,,square ,,root4 #aef4 ,we have already remarked 7#ff71 that a number multiplied by itself produces what is called the square of that number4 ,thus1 #afi1 or #ac #ac1 is the square of #ac4 ,conversely1 #ac is called the square root of #afi1 and #e is the square root of #be2 and any number is the square root of another1 which when multiplied by itself will produce that other4 ,the square root is signified by the sign or 2 thus1 #be means the square root of #be1 or #e2 7#af #i7 means the square root of #af #i1 and is #e1 and must not be confounded with #af #i1 which is #d #c1 or #g4 #aeg4 ,the following equations are evident from the definition3 a a a aa a ab ab ab 7a b7 7a b7 a a b b ab whence a b ab #aeh4 ,it does not follow that a number has a square root because it has a square2 thus1 though #e can be multiplied by itself1 there is no number which multiplied by itself will produce #e4 ,it is proved in algebra1 that no fraction#bb multiplied by itself can produce a whole number1 which may be found true in any number of instances2 therefore #e has neither a whole nor a fractional square root2 that is1 it has no square root at all4 ,nevertheless1 there are methods of finding fractions whose squares shall be as near to #e as we please1 though not exactly equal to it4 ,one of these methods gives _/1 whose square1 viz4 #aeabg #aeabg #bbhhbfabi ----- ----- or ---------1 #fgfe #fgfe #degfebbe differs from #e by only _/1 which is less than #jjjjjja3 hence we are enabled to use #e in arithmetical and algebraical reasoning3 but when we come to the practice of any problem1 we must substitute for #e one of the fractions whose square is nearly #e1 and on the degree of accuracy we want1 depends what fraction is to be used4 ,for some purposes1 _/ may be sufficient1 as its square only differs from #e by _/2 for others1 the fraction first given might be necessary1 or one whose square is even nearer to #e4 ,we proceed to shew how to find the square root of a number1 when it has one1 and from thence how to find fractions whose squares shall be as near as we please to the number1 when it has not4 ,we premise1 what is sufficiently evident1 that of two numbers1 the greater has the greater square2 and that if one number lie between two others1 its square lies between the squares of those others4 #bb ,meaning1 of course1 a really fractional number1 such as or _/1 not one which1 though fractional in form1 is whole in reality1 such as _/ or _/4 #aei4 ,let x be a number consisting of any number of parts1 for example1 four1 viz4 a1 b1 c1 and d2 that is1 let x a b c d ,the square of this number1 found as in 7#fh71 will be aa #b;a7b c d7 bb #b;b7c d7 cc #b;cd dd ,the rule there found for squaring a number consisting of parts was3 ,square each part1 and multiply all that come after by twice that part1 the sum of all the results so obtained will be the square of the whole number4 ,in the expression above obtained1 instead of multiplying #b;a by each of the succeeding parts1 b1 c1 and d1 and adding the results1 we multiplied #b;a by the sum of all the succeeding parts1 which 7#eb7 is the same thing2 and as the parts1 however disposed1 make up the number1 we may reverse their order1 putting the last first1 c42 and the rule for squaring will be3 ,square each part1 and multiply all that come before by twice that part4 ,hence a reverse rule for extracting the square root presents itself with more than usual simplicity4 ,it is3 ,to extract the square root of a number ,n1 choose a number ,a1 and see if ,n will bear the subtraction of the square of ,a2 if so1 take the remainder1 choose a second number ,b1 and see if the remainder will bear the subtraction of the square of ,b1 and twice ,b multiplied by the preceding part ,a3 if it will1 there is a second remainder4 ,choose a third number ,c1 and see if the second remainder will bear the subtraction of the square of ,c1 and twice ,c multiplied by ,a ,b3 go on in this way either until there is no remainder1 or else until the remainder will not bear the subtraction arising from any new part1 even though that part were the least number1 which is #a4 ,in the first case1 the square root is the sum of ,a1 ,b1 ,c1 c42 in the second1 there is no square root4 #afj4 ,for example1 ,i wish to know if #bjbe has a square root4 ,i choose #bj as the first part1 and find that #djj1 the square of #bj1 subtracted from #bjbe1 gives #afbe1 the first remainder4 ,i again choose #bj1 whose square1 together with twice itself1 multiplied by the preceding part1 is #bj #bj #b #bj #bj1 or #abjj2 which subtracted from #afbe1 the first remainder1 gives #dbe1 the second remainder4 ,i choose #g for the third part1 which appears to be too great1 since #g #g1 increased by #b #g multiplied by the sum of the preceding parts #bj #bj1 gives #fji1 which is more than #dbe4 ,i therefore choose #e1 which closes the process1 since #e #e1 together with #b #e multiplied by #bj #bj1 gives exactly #dbe4 ,the square root of #bjbe is therefore #bj #bj #e1 or #de1 which will be found1 by trial1 to be correct2 since #de #de #bjbe4 ,again1 ,i ask if #accdj has1 or has not1 a square root4 ,let #ajj be the first part1 whose square is #ajjjj1 and the first remainder is #ccdj4 ,let #aj be the second part4 ,here #aj #aj #b #aj #ajj is #bajj1 and the second remainder1 or #ccdj-bajj1 is #abdj4 ,let #e be the third part2 then #e #e #b #e 7#ajj #aj7 is #aabe1 which1 subtracted from #abdj1 leaves #aae4 ,there is1 then1 no square root2 for a single additional unit will give a subtraction of #a #a #b #a 7#ajj #aj #e71 or #bca1 which is greater than #aae4 ,but if the number proposed had been less by #aae1 each of the remainders would have been #aae less1 and the last remainder would have been nothing4 ,therefore #accdj-aae1 or #acbbe1 has the square root #ajj #aj #e1 or #aae2 and the answer is1 that #accdj has no square root1 and that #acbbe is the next number below it which has one1 namely1 #aae4 #afa4 ,it only remains to put the rule in such a shape as will guide us to those parts which it is most convenient to choose4 ,it is evident 7#eg7 that any number which terminates with ciphers1 as #djjj1 has double the number of ciphers in its square4 ,thus1 #djjj #djjj #afjjjjjj2 therefore1 any square number1#bc as #di1 with an even number of ciphers annexed1 as #dijjjj1 is a square number4 ,the root#bd of #dijjjj is #gjj4 ,this being premised1 take any number1 for example1 #gfagf2 setting out from the right hand towards the left1 cut off two figures2 then two more1 and so on1 until one or two figures only are left3 thus1 #g1fa1gf4 ,this number is greater than #g1jj1jj1 of which the first figure is not a square number1 the nearest square below it being #d4 ,hence1 #d1jj1jj is the nearest square number below #g1jj1jj1 which has four ciphers1 and its square root is #bjj4 ,let this be the first part chosen3 its square subtracted from #gfagf leaves #cfagf1 the first remainder2 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 #gfagf2 for #cjj is too great1 its square1 #i1jj1jj1 being greater than #gfagf3 and any denomination higher than hundreds has a square still greater4 ,it remains1 then1 to choose a second part1 as in the examples of 7#afj71 with the remainder #cfagf4 ,this part cannot be as great as #ajj1 by what has just been said2 its highest denomination is therefore a number of tens4 ,let ,n stand for a number of tens1 which is one of the simple numbers #a1 #b1 #c1 c42 that is1 let the new part be #aj,n1 whose square is #aj,n #aj,n1 or #ajj,,nn1 and whose double multiplied by the former part is #bj,n #bjj1 or #djjj,n2 the two together are #djjj,n #ajj,,nn4 ,now1 ,n must be so taken that this may not be greater than #cfagf3 still more #djjj,n must not be greater than #cfagf4 ,we may therefore try1 for ,n1 the number of times which #cfagf contains #djjj1 or that which #cf contains #d4 ,the remark in 7#hj7 applies here4 ,let us try #i tens or #ij4 ,then1 #b #ij #bjj #ij #ij1 or #ddajj1 is to be subtracted1 which is too great1 since the whole remainder is #cfagf4 ,we then try #h tens or #hj1 which gives #b #hj #bjj #hj #hj1 or #chdjj1 which is likewise too great4 ,on trying #g tens1 or #gj1 we find #b #gj #bjj #gj #gj1 or #cbijj1 which subtracted from #cfagf gives #cbgf1 the second remainder4 ,the rest of the square root can only be units4 ,as before1 let ,n be this number of units4 ,then1 the sum of the preceding parts being #bjj #gj1 or #bgj1 the number to be subtracted is #bgj #b,n ,,nn1 or #edj,n ,,nn4 ,hence1 as before1 #edj,n must be less than #cbgf1 or ,n must not be greater than the number of times which #cbgf contains #edj1 or 7#hj7 which #cbg contains #ed4 ,we therefore try if #f will do1 which gives #b #f #bgj #f #f1 or #cbgf1 to be subtracted4 ,this being exactly the second remainder1 the third remainder is nothing1 and the process is finished4 ,the square root required is therefore #bjj #gj #f1 or #bgf4 #bc ,by square number ,i mean1 a number which has a square root4 ,thus1 #be is a square number1 but #bf is not4 #bd ,the term #8root' is frequently used as an abbreviation of square root4 ,the process of forming the numbers to be subtracted may be shortened thus4 ,let ,a be the sum of the parts already found1 and ,n a new part3 there must then be subtracted #b;,,an ,,nn1 or 7#ed7 #b;,a ,n multiplied by ,n4 ,the rule1 therefore1 for forming it is3 ,double the sum of all the preceding parts1 add the new part1 and multiply the result by the new part4 #afb4 ,the process of the last article is as follows3 #g1fa1gf7bjj #g1fa1gf7bgf #d #jj #jj #gj #d ------- #f --- #djj7c1fa1gf #dg7cfa #gj7c #bi #jj #cbi ------- ----- #djj7 #cb #gf #edf7cbgf #adj7 #cb #gf #cbgf #f7 ----- ---- #j #j ,in the first of these1 the numbers are written at length1 as we found them2 in the second1 as in 7#gi71 unnecessary ciphers are struck off1 and the periods #fa1 #gf1 are not brought down1 until1 by the continuance of the process1 they cease to have ciphers under them4 ,the following is another example1 to which the reasoning of the last article may be applied4 #cd1hf1gh1dd1ja7ejjjj #cd1hf1gh1dd1ja7eijdi #be #jj #jj #jj #jj #ijjj #be -------------- #dj ---- #ajjjjj7 #i #hf #gh #dd #ja #i #aji7 #ihf #ijjj7 #i #ha #jj #jj #jj #iha ------------- ------- #ajjjjj7 #e #gh #dd #ja #aahjd7 #eghdd #ahjjj7 #d #gb #af #jj #dgbaf #dj7 ----------- -------- #ajjjjj7 #a #jf #bh #ja #aahjhi7ajfbhja #ahjjj7 #a #jf #bh #ja #ajfbhja #hj7 ---------- ------- #i7 #j #j #afc4 ,the rule is as follows3 ,to extract the square root of a number2-- ,i4 ,beginning from the right hand1 cut off periods of two figures each1 until not more than two are left4 ,,ii4 ,find the root of the nearest square number next below the number in the first period4 ,this root is the first figure of the required root2 subtract its square from the first period1 which gives the first remainder4 ,,iii4 ,annex the second period to the right of the remainder1 which gives the first dividend4 ,,iv4 ,double the first figure of the root2 see how often this is contained in the number made by cutting one figure from the right of the first dividend1 attending to ,,ix41 if necessary2 use the quotient as the second figure of the root2 annex it to the right of the double of the first figure1 and call this the first divisor4 ,v4 ,multiply the first divisor by the second figure of the root2 if the product be greater than the first dividend1 use a lower number for the second figure of the root1 and for the last figure of the divisor1 until the multiplication just mentioned gives the product less than the first dividend2 subtract this from the first dividend1 which gives the second remainder4 ,,vi4 ,annex the third period to the second remainder1 which gives the second dividend4 ,,vii4 ,double the first two figures of the root2#be see how often the result is contained in the number made by cutting one figure from the right of the second dividend2 use the quotient as the third figure of the root2 annex it to the right of the double of the first two figures1 and call this the second divisor4 #be ,or1 more simply1 add the second figure of the root to the first divisor4 ,,viii4 ,get a new remainder1 as in ,v41 and repeat the process until all the periods are exhausted2 if there be then no remainder1 the square root is found2 if there be a remainder1 the proposed number has no square root1 and the number found as its square root is the square root of the proposed number diminished by the remainder4 ,,ix4 ,when it happens that the double of the figures of the root is not contained at all in all the dividend except the last figure1 or when1 being contained once1 #a is found to give more than the dividend1 put a cipher in the square root and in the divisor1 and bring down the next period2 should the same thing still happen1 put another cipher in the root and divisor1 and bring down another period2 and so on4 ,,exercises4 ,numbers proposed4 ,square roots4 #gcdda #bga #biibijj #agcj #fdadbdgiba #hjjhi #ijcfhghijfbe #iejfbe #dbdbjgdgdhbggfegf #bjeifbigf #acdbbfeicajaebdja #aaehefbja #afd4 ,since the square of a fraction is obtained by squaring the numerator and the denominator1 the square root of a fraction is found by taking the square root of both4 ,thus1 the square root of _/ is 1 since #e #e is #be1 and #h #h is #fd4 ,if the numerator or denominator1 or both1 be not square numbers1 it does not therefore follow that the fraction has no square root2 for it may happen that multiplication or division by the same number may convert both the numerator and denominator into square numbers 7#ajh74 ,thus1 _/1 which appears at first to have no square root1 has one in reality1 since it is the same as _/1 whose square root is 4 #afe4 ,we now proceed from 7#aeh71 where it was stated that any number or fraction being given1 a second may be found1 whose square is as near to the first as we please4 ,thus1 though we cannot solve the problem1 8,find a fraction whose square is #b10 we can solve the following1 8,find a fraction whose square shall not differ from #b by so much as #jjjjjjja40 ,instead of this last1 a still smaller fraction may be substituted2 in fact1 any one however small3 and in this process we are said to approximate to the square root of #b4 ,this can be done to any extent1 as follows3 ,suppose we wish to find the square root of #b within _/ of the truth2 by which ,i mean1 to find a fraction a_/b whose square is less than #b1 but such that the square of a_/b _/ is greater than #b4 ,multiply the numerator and denominator of _/ by the square of #eg1 or #cbdi1 which gives _/4 ,on attempting to extract the square root of the numerator1 ,i find 7#afc7 that there is a remainder #ih1 and that the square number next below #fdih is #fdjj1 whose root is #hj4 ,hence1 the square of #hj is less than #fdih1 while that of #ha is greater4 ,the square root of the denominator is of course #eg4 ,hence1 the square of _/ is less than _/1 or #b1 while that of _/ is greater1 and these two fractions only differ by _/2 which was required to be done4 #aff4 ,in practice1 it is usual to find the square root true to a certain number of places of decimals4 ,thus1 #adadb is the square root of #b true to four places of decimals1 since the square of #adadb1 or #aiiiifafd1 is less than #b1 while an increase of only #a in the fourth decimal place1 giving #adadc1 gives the square #bjjjbdddi1 which is greater than #b4 ,to take a more general case3 ,suppose it required to find the square root of #afcg true to four places of decimals4 ,the fraction is _/1 whose square root is to be found within #jjja1 or _/4 ,annex ciphers to the numerator and denominator1 until the denominator becomes the square of _/1 which gives _/1 extract the square root of the numerator1 as in 7#afc71 which shews that the square number nearest to it is #afcgjjjjj-acefd1 whose root is #abgid4 ,hence1 _/1 or #abgid1 gives a square less than #afcg1 while #abgie gives a square greater4 ,in fact1 these two squares are #afcfhfdcf and #afcgabjbe4 #afg4 ,the rule1 then1 for extracting the square root of a number or decimal to any number of places is3 ,annex ciphers until there are twice as many places following the units' place as there are to be decimal places in the root2 extract the nearest square root of this number1 and mark off the given number of decimals4 ,or1 more simply3 ,divide the number into periods1 so that the units' figure shall be the last of a period2 proceed in the usual way2 and if1 when decimals follow the units' place1 there is one figure on the right1 in a period by itself1 annex a cipher in bringing down that period1 and afterwards let each new period consist of two ciphers4 ,place the decimal point after that figure in forming which the period containing the units was used4 #afh4 ,for example1 what is the square root of 7#a7 to five places of decimals8 ,this is 7#ade7 #acge1 and the process is the first example over leaf4 ,the second example is the extraction of the root of #jha to seven places1 the first period being #jh1 from which the cipher is omitted as useless4 #a1cg1e7aagbfj #a --- #ba7 #cg #ba ---- #bbg7afej #aehi ------ #bcdb7 #fajj #dfhd ------ #bcddf7adafjj #adjfgf -------- #bcdeb7 #ibdjj #h1a7bhdfjdi #d --- #dh7daj #chd ------ #efd7 #bfjj #bbef ------ #efhf7 #cddjj #cdaaf --------- #efibjd7 #bhdjjjj #bbgfhaf --------- #efibjh7 #efcahdjj #jjjjjbdacfgbbba7jjaeeceii #a --- #be7 #ada #abe ----- #cje7 #afcf #aebe ------ #cajc7 #aaagb #icji ------ #cajfe7 #ahfcbb #aeecbe -------- #cajgji7 #cjiigaj #bgifcha --------- #cjccbijj #afi4 ,when more than half the decimals required have been found1 the others may be simply found by dividing the dividend by the divisor1 as in 7#aee74 ,the extraction of the square root of #ab to ten places1 which will be found in the next page1 is an example4 ,it must1 however1 be observed in this process1 as in all others where decimals are obtained by approximation1 that the last place cannot always be depended upon3 on which account it is advisable to carry the process so far1 that one or even two more decimals shall be obtained than are absolutely required to be correct4 ,a #ab7cdfdajafaeac #i --- #fd7 #cjj #bef ----- #fhf7 #ddjj #daaf ------ #fibd7 #bhdjj #bgfif ------ #fibha7 #gjdjj #fibha ---------- #fibhbja7 #aaaijjjj #fibhbja --------- #fibhbjbf7 #dbfagiijj #daefibaef ----------- #fibhbjcba7 #ajdhggddjj #fibhbjcba ----------- #fibhbjcbbe7 #ceeiedjgijj #cdfdajafabe ------------- #fibhbjcbcja7 #iedciaggejj #fibhbjcbcja ------------- #fibhbjcbcjbc7 #bfaegadeaiijj #bjghdfjifijfi ------------- #ecgbeceejhca ,b #fibhbjcbcjbf7 #ecgbeceejhca7ggedehgjedi #dhdigdbbfaah ------------ #ebbgicbdgac #dhdigdbbfaa ----------- #cghaijbajb #cdfdajafae ---------- #caghjjdhg #bggabhabi --------- #djfgbceh #cdfdajaf -------- #fjcacdb #eedbefb ------- #dhhghj #dhdigd ------ #chjf #cdfd ---- #cdb #bgg --- #fe #fb -- #c ,if from any remainder we cut off the ciphers1 and all figures which would come under or on the right of these ciphers1 by a vertical line1 we find on the left of that line a contracted division1 such as those in 7#aee74 ,thus1 after having found the root as far as #cdfdaja1 we have the remainder #dbfagii1 and the divisor #fibhbjb4 ,the figures on the left of the line are nothing more than the contracted division of this remainder by the divisor1 with this difference1 however1 that we have to begin by striking a figure off the divisor1 instead of using the whole divisor once1 and then striking off the first figure4 ,by this alone we might have doubled our number of decimal places1 and got the additional figures #faeacg1 the last #g being obtained by carrying the contracted division one step further with the remainder #ec4 ,we have1 then1 this rule3 ,when half the number of decimal places have been obtained1 instead of annexing two ciphers to the remainder1 strike off a figure from what would be the divisor if the process were continued at length1 and divide the remainder by this contracted divisor1 as in 7#aee74 ,as an example1 let us double the number of decimal places already obtained1 which are contained in #cdfdajafaeac4 ,the remainder is #ecgbeceejhca1 the divisor #fibhbjcbcjbf1 and the process is as in 7,b74 ,hence the square root of #ab is1 #cdfdajafaeacggedehgjedi2 which is true to the last figure1 and a little too great2 but the substitution of #h instead of #i on the right hand would make it too small4 ,,exercises4 ,numbers4 ,square roots4 #jjagbh #jdaefibaid #fdcd #hjbabbahe #hjgd #hiheedcid #aj #cafbbggff #aeg #abebiifdjhfadaffgghhdie ,,section ,,viii4 ,,on ,,the ,,proportion ,,of ,,numbers4 #agj4 ,when two numbers are named in any problem1 it is usually necessary1 in some way or other1 to compare the two2 that is1 by considering the two together1 to establish some connexion between them1 which may be useful in future operations4 ,the first method which suggests itself1 and the most simple1 is to observe which is the greater1 and by how much it differs from the other4 ,the connexion thus established between two numbers may also hold good of two other numbers2 for example1 #h differs from #ai by #aa1 and #ajj differs from #aaa by the same number4 ,in this point of view1 #h stands to #ai in the same situation in which #ajj stands to #aaa1 the first of both couples differing in the same degree from the second4 ,the four numbers thus noticed1 viz43 #h1 #ai1 #ajj1 #aaa1 are said to be in arithmetical#bf proportion4 ,when four numbers are thus placed1 the first and last are called the extremes1 and the second and third the means4 ,it is obvious that #aaa #h #ajj #ai1 that is1 the sum of the extremes is equal to the sum of the means4 ,and this is not accidental1 arising from the particular numbers we have taken1 but must be the case in every arithmetical proportion2 for in #aaa #h1 by 7#ce71 any diminution of #aaa will not affect the sum1 provided a corresponding increase be given to #h2 and1 by the definition just given1 one mean is as much less than #aaa as the other is greater than #h4 #bf ,this is a very incorrect name1 since the term #8arithmetical' applies equally to every notion in this book4 ,it is necessary1 however1 that the pupil should use words in the sense in which they will be used in his succeeding studies4 #aga4 ,a set or series of numbers is said to be in continued arithmetical proportion1 or in arithmetical progression1 when the difference between every two succeeding terms of the series is the same4 ,this is the case in the following series3 #a1 #b1 #c1 #d1 #e1 c4 #c1 #f1 #i1 #ab1 #ae1 c4 7#a71 #b1 7#b71 #c1 7#c71 c4 ,the difference between two succeeding terms is called the common difference4 ,in the three series just given1 the common differences are1 #a1 #c1 and 4 #agb4 ,if a certain number of terms of any arithmetical series be taken1 the sum of the first and last terms is the same as that of any other two terms1 provided one is as distant from the beginning of the series as the other is from the end4 ,for example1 let there be #g terms1 and let them be1 a b c d e f g4 ,then1 since1 by the nature of the series1 b is as much above a as f is below g 7#agj71 a g b f4 ,again1 since c is as much above b as e is below f 7#agj71 b f c e4 ,but a g b f2 therefore a g c e1 and so on4 ,again1 twice the middle term1 or the term equally distant from the beginning and the end 7which exists only when the number of terms is odd71 is equal to the sum of the first and last terms2 for since c is as much below d as e is above it1 we have c e d d #b;d4 ,but c e a g2 therefore1 a g #b;d4 ,this will give a short rule for finding the sum of any number of terms of an arithmetical series4 ,let there be #g1 viz4 those just given4 ,since a g1 b f1 and c e1 are the same1 their sum is three times 7a g71 which with d1 the middle term1 or half a g1 is three times and a half 7a g71 or the sum of the first and last terms multiplied by 7#c71 or _/1 or half the number of terms4 ,if there had been an even number of terms1 for example1 six1 viz4 a1 b1 c1 d1 e1 and f1 we know now that a f1 b e1 and c d1 are the same1 whence the sum is three times 7a f71 or the sum of the first and last terms multiplied by half the number of terms1 as before4 ,the rule1 then1 is3 ,to sum any number of terms of an arithmetical progression1 multiply the sum of the first and last terms by half the number of terms4 ,for example1 what are #ii terms of the series #a1 #b1 #c1 c48 ,the #iith term is #ii1 and the sum is #ii #ajj #ii 7#ii #a7---1 or --------1 or #diej4 #b #b ,the sum of #ej terms of the series #a #b #d #e 7 #a #ej 7 #ej ---1 ---1 #a1 ---1 ---1 #b1 c4 is 7--- ---7---1 #c #c #c #c 7 #c #c 7 #b or #ag #be1 or #dbe4 #agc4 ,the first term being given1 and also the common difference and number of terms1 the last term may be found by adding to the first term the common difference multiplied by one less than the number of terms4 ,for it is evident that the second term differs from the first by the common difference1 the third term by twice1 the fourth term by three times the common difference2 and so on4 ,or1 the passage from the first to the nth term is made by n-#a steps1 at each of which the common difference is added4 ,,exercises4 ,given4 ,to find4 ,series4 ,no4 of terms4 ,last term4 ,sum4 #d1 7#f71 #i1 c4 #cc #hd #adeb #a1 #c1 #e1 c4 #bh #ee #ghd #b1 #bj1 #ch1 c4 #ajj1jjj #agiiihd #hiiiicjjjjj #agd4 ,the sum being given1 the number of terms1 and the first term1 we can thence find the common difference4 ,suppose1 for example1 the first term of a series to be one1 the number of terms #ajj1 and the sum #aj1jjj4 ,since #aj1jjj was made by multiplying the sum of the first and last terms by _/1 if we divide by this1 we shall recover the sum of the first and last terms4 ,now1 1_/ divided by _/ is 7#abb7 #bjj1 and the first term being #a1 the last term is #aii4 ,we have then to pass from #a to #aii1 or through #aih1 by #ii equal steps4 ,each step is1 therefore1 _/1 or #b1 which is the common difference2 or the series is #a1 #c1 #e1 c41 up to #aii4 ,given4 ,to find4 ,sum4 ,no4 of terms4,first term4,last term4,common diff4 #ahjijbe #acde #a #bfhi #b #dd #aj #c _/ _/ #gjgefjj #accj #d #ajfcf #h #age4 ,we now return to 7#agj71 in which we compared two numbers together by their difference4 ,this1 however1 is not the method of comparison which we employ in common life1 as any single familiar instance will shew4 ,for example1 we say of ,a1 who has #aj thousand pounds1 that he is much richer than ,b1 who has only #c thousand2 but we do not say that ,c1 who has #ajg thousand pounds1 is much richer than ,d1 who has #ajj thousand1 though the difference of fortune is the same in both cases1 viz4 #g thousand pounds4 ,in comparing numbers we take into our reckoning not only the differences1 but the numbers themselves4 ,thus1 if ,b and ,d both received #g thousand pounds1 ,b would receive #bcc pounds and a third for every #ajj pounds which he had before1 while ,d for every #ajj pounds would receive only #g pounds4 ,and though1 in the view taken in 7#agj71 #c is as near to #aj as #ajj is to #ajg1 yet1 in the light in which we now regard them1 #c is not so near to #aj as #ajj is to #ajg1 for #c differs from #aj by more than twice itself1 while #ajj does not differ from #ajg by so much as one-fifth of itself4 ,this is expressed in mathematical language by saying1 that the ratio or proportion of #aj to #c is greater than the ratio or proportion of #ajg to #ajj4 ,we proceed to define these terms more accurately4 #agf4 ,when we use the term part of a number or fraction in the remainder of this section1 we mean1 one of the various sets of equal parts into which it may be divided1 either the half1 the third1 the fourth1 c43 the term multiple has been already explained 7#ajb74 ,by the term multiple-part of a number we mean1 the abbreviation of the words multiple of a part4 ,thus1 #a1 #b1 #c1 #d1 and #f1 are parts of #ab2 is also a part of #ab1 being contained in it #bd times2 #ab1 #bd1 #cf1 c41 are multiples of #ab2 and #h1 #i1 _/1 c4 are multiple parts of #ab1 being multiples of some of its parts4 ,and when multiple parts generally are spoken of1 the parts themselves are supposed to be included1 on the same principle that #ab is counted among the multiples of #ab1 the multiplier being #a4 ,the multiples themselves are also included in this term2 for #bd is also #dh halves1 and is therefore among the multiple parts of #ab4 ,each part is also in various ways a multiple-part2 for one-fourth is two-eighths1 and three-twelfths1 c4 #agg4 ,every number or fraction is a multiple-part of every other number or fraction4 ,if1 for example1 we ask what part #ab is of #g1 we see that on dividing #g into #g parts1 and repeating one of these parts #ab times1 we obtain #ab2 or1 on dividing #g into #ad parts1 each of which is one-half1 and repeating one of these parts #bd times1 we obtain #bd halves1 or #ab4 ,hence1 #ab is _/1 or _/1 or _/ of #g2 and so on4 ,generally1 when a and b are two whole numbers1 a_/b expresses the multiple-part which a is of b1 and b_/a that which b is of a4 ,again1 suppose it required to determine what multiple-part 7#b7 is of 7#c71 or _/ of _/4 ,these fractions1 reduced to a common denominator1 are _/ and _/1 of which the second1 divided into #aab parts1 gives _/1 which repeated #ge times gives _/1 the first4 ,hence1 the multiple-part which the first is of the second is _/1 which being obtained by the rule given in 7#aba71 shews that a_/b1 or a divided by b1 according to the notion of division there given1 expresses the multiple-part which a is of b in every case4 #agh4 ,when the first of four numbers is the same multiple-part of the second which the third is of the fourth1 the four are said to be geometrically#bg proportional1 or simply proportional4 ,this is a word in common use2 and it remains to shew that our mathematical definition of it1 just given1 is1 in fact1 the common notion attached to it4 ,for example1 suppose a picture is copied on a smaller scale1 so that a line of two inches long in the original is represented by a line of one inch and a half in the copy2 we say that the copy is not correct unless all the parts of the original are reduced in the same proportion1 namely1 that of #b to 7#a74 ,since1 on dividing two inches into #d parts1 and taking #c of them1 we get 7#a71 the same must be done with all the lines in the original1 that is1 the length of any line in the copy must be three parts out of four of its length in the original4 ,again1 interest being at #e per cent1 that is1 #e being given for the use of #ajj1 a similar proportion of every other sum would be given2 the interest of #gj1 for example1 would be just such a part of #gj as #e is of #ajj4 #bg ,the same remark may be made here as was made in the note on the term #8arithmetical proportion1' page #aja4 ,the word #8geometrical' is1 generally speaking1 dropped1 except when we wish to distinguish between this kind of proportion and that which has been called arithmetical4 ,since1 then1 the part which a is of b is expressed by the fraction a_/b1 or any other fraction which is equivalent to it1 and that which c is of d by c_/d1 it follows1 that when a1 b1 c1 and d1 are proportional1 a_/b c_/d4 ,this equation will be the foundation of all our reasoning on proportional quantities2 and in considering proportionals1 it is necessary to observe not only the quantities themselves1 but also the order in which they come4 ,thus1 a1 b1 c1 and d1 being proportionals1 that is1 a being the same multiple-part of b which c is of d1 it does not follow that a1 d1 b1 and c are proportionals1 that is1 that a is the same multiple-part of d which b is of c4 ,it is plain that a is greater than1 equal to1 or less than b1 according as c is greater than1 equal to1 or less than d4 #agi4 ,four numbers1 a1 b1 c1 and d1 being proportional in the order written1 a and d are called the extremes1 and b and c the means1 of the proportion4 ,for convenience1 we will call the two extremes1 or the two means1 similar terms1 and an extreme and a mean1 dissimilar terms4 ,thus1 a and d are similar1 and so are b and c2 while a and b1 a and c1 d and b1 d and c1 are dissimilar4 ,it is customary to express the proportion by placing dots between the numbers1 thus3 a 3 b c 3 d #ahj4 ,equal numbers will still remain equal when they have been increased1 diminished1 multiplied1 or divided1 by equal quantities4 ,this amounts to saying that if a b and p q1 a p b q1 a - p b - q1 ap bq1 a b and --- ---4 p q ,it is also evident1 that a p-p1 a -p p1 ap_/p1 and a_/p p1 are all equal to a4 #aha4 ,the product of the extremes is equal to the product of the means4 ,let a_/b c_/d1 and multiply these equal numbers by the product bd4 ,then1 a abd --- bd ----- 7#aaf7 ad1 b b c cbd and --- bd ----- cb3 hence 7#ahj71 ad bc4 d d ,thus1 #f1 #h1 #ba1 and #bh1 are proportional1 since #f #c #c #g #ba --- --- ------ --- 7#ahj72 #h #d #d #g #bh and it appears that #f #bh #h #ba1 since both products are #afh4 #ahb4 ,if the product of two numbers be equal to the product of two others1 these numbers are proportional in any order whatever1 provided the numbers in the same product are so placed as to be similar terms2 that is1 if ab pq1 we have the following proportions3-- a 3 p q 3 b a 3 q p 3 b b 3 p q 3 a b 3 q p 3 a p 3 a b 3 q p 3 b a 3 q q 3 a b 3 p q 3 b a 3 p ,to prove any one of these1 divide both ab and pq by the product of its second and fourth terms2 for example1 to shew the truth of a3 q p3 b1 divide both ab and pq by bq4 ,then1 ab a pq p ---- ---1 and ---- ---2 hence 7#ahj71 bq q bq b a p --- ---1 or a 3 q p 3 b4 q b ,the pupil should not fail to prove every one of the eight cases1 and to verify them by some simple examples1 such as #a #f #b #c1 which gives #a3 #b #c3 #f1 #c3 #a #f3 #b1 c4 #ahc4 ,hence1 if four numbers be proportional1 they are also proportional in any other order1 provided it be such that similar terms still remain similar4 ,for since1 when a c --- ---1 b d it follows 7#aha7 that ad bc1 all the proportions which follow from ad bc1 by the last article1 follow also from a c --- ---1 b d #ahd4 ,from 7#aad7 it follows that a b a #a --- ---------1 b b a and if --- be less than #a1 b a b - a #a - --- ---------1 b b a while if --- be greater than #a1 b a a - b --- - #a ---------4 b b a b a - b ,also 7#abb71 if --------- be divided by --------- b b a b the result is ---------4 a - b ,hence1 a1 b1 c1 and d1 being proportionals1 we may obtain other proportions1 thus3 a c ,let --- --- b d a c ,then 7#aad7 #a --- #a --- b d a b c d or --------- --------- b d or a b3 b c d3 d ,that is1 the sum of the first and second is to the second as the sum of the third and fourth is to the fourth4 ,for brevity1 we shall not state in words any more of these proportions1 since the pupil will easily supply what is wanting4 ,resuming the proportion a3 b c3 d a c or --- --- b d a c a #a - --- #a - ---1 if --- be less than #a1 b d b b - a d - c or --------- --------- b d that is1 b-a3 b d-c3 d or1 a-b3 b c-d3 d1 a if --- be greater than #a4 b a b c d ,again1 since --------- --------- b d a - b c - d a and --------- --------- 7--- being greater than #a7 b d b a b c d dividing the first by the second we have --------- ----------1 a - b c - d or a b 3 a - b c d 3 c - d and also a b 3 b - a c d 3 d - c1 a if --- be less than #a4 b #ahe4 ,many other proportions might be obtained in the same manner4 ,we will1 however1 content ourselves with writing down a few which can be obtained by combining the preceding articles4 a b 3 a c d 3 c a 3 a - b c 3 c - d a c 3 a - c b d 3 b - d4 ,in these and all others it must be observed1 that when such expressions as a-b and c-d occur1 it is supposed that a is greater than b1 and c greater than d4 #ahf4 ,if four numbers be proportional1 and any two dissimilar terms be both multiplied1 or both divided by the same quantity1 the results are proportional4 ,thus1 if a3 b c3 d1 and m and n be any two numbers1 we have also the following3 ma 3 b mc 3 d a 3 mb c 3 md a c --- 3 mb --- 3 md n n ma 3 nb mc 3 nd a b c d --- 3 --- --- 3 --- m m m m a b c d --- 3 --- --- 3 --- m m n n and various others4 ,to prove any one of these1 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 means4 ,take the third of those just given2 the product of its extremes is a mad --- md1 or -----1 n n c mbc while that of the means is mb ---1 or -----4 n n ,but since a 3 b c 3 d1 by 7#aha7 ad bc1 mad mbc whence1 by 7#ahj71 mad mbc1 and ----- -----4 n n a c ,hence1 ---1 mb1 ---1 and md1 are proportionals4 n n #ahg4 ,if the terms of one proportion be multiplied by the terms of a second1 the products are proportional2 that is1 if a3 b c3 d1 and p3 q r3 s1 it follows that ap3 bq cr3 ds4 ,for1 since ad bc1 and ps qr1 by 7#ahj7 adps bcqr1 or ap ds bq cr1 whence 7#ahb7 ap3 bq cr3 ds4 #ahh4 ,if four numbers be proportional1 any similar powers of these numbers are also proportional2 that is1 if a 3 b c 3 d ,then aa 3 bb cc 3 dd aaa 3 bbb ccc 3 ddd c4 c4 ,for1 if we write the proportion twice1 thus1 a 3 b c 3 d a 3 b c 3 d by 7#ahg7 aa 3 bb cc 3 dd ,but a 3 b c 3 d ,whence 7#ahg7 aaa 3 bbb ccc 3 ddd2 and so on4 #ahi4 ,an expression is said to be homogeneous with respect to any two or more letters1 for instance1 a1 b1 and c1 when every term of it contains the same number of letters1 counting a1 b1 and c only4 ,thus1 maab nabc rccc is homogeneous with respect to a1 b1 and c2 and of the third degree1 since in each term there is either a1 b1 and c1 or one of these repeated alone1 or with another1 so as to make three in all4 ,thus1 #h;aaabc1 #ab;abccc1 maaaaa1 naabbc1 are all homogeneous1 and of the fifth degree1 with respect to a1 b1 and c only2 and any expression made by adding or subtracting these from one another1 will be homogeneous and of the fifth degree4 ,again ma mnb is homogeneous with respect to a and b1 and of the first degree2 but it is not homogeneous with respect to m and n1 though it is so with respect to a and n4 ,this being premised1 we proceed to a theorem1#bh which will contain all the results of 7#ahd71 7#ahe71 and 7#ahh74 #bh ,a theorem is a general mathematical fact3 thus1 that every number is divisible by four when its last two figures are divisible by four1 is a theorem2 that in every proportion the product of the extremes is equal to the product of the means1 is another4 #aij4 ,if any four numbers be proportional1 and if from the first two1 a and b1 any two homogeneous expressions of the same degree be formed2 and if from the last two1 two other expressions be formed1 in precisely the same manner1 the four results will be proportional4 ,for example1 if a3 b c3 d1 and if #b;aaa #c;aab and bbb abb be chosen1 which are both homogeneous with respect to a and b1 and both of the third degree2 and if the corresponding expressions #b;ccc #c;ccd and ddd cdd be formed1 which are made from c and d precisely in the same manner as the two former ones from a and b1 then will #b;aaa #c;aab 3 bbb abb #b;ccc #c;ccd 3 ddd cdd a ,to prove this1 let --- be called x4 b a a c ,then1 since --- x1 and --- ---1 b b d c it follows that --- x4 d ,but since a divided by b gives x1 x multiplied by b will give a1 or a bx4 ,for a similar reason1 c dx4 ,put bx and dx instead of a and c in the four expressions just given1 recollecting that when quantities are multiplied together1 the result is the same in whatever order the multiplications are made2 that1 for example1 bxbxbx is the same as bbbxxx4 ,hence1 #b;aaa #c;aab #b;bxbxbx #c;bxbxb #b;bbbxxx #c;bbbxx which is bbb multiplied by #bxxx #cxx or bbb 7#bxxx #cxx7#bi ,similarly1 #b;ccc #c;ccd ddd 7#bxxx #cxx7 ,also1 bbb abb bbb bxbb bbb multiplied by #a x or bbb7#a x7 ,similarly1 ddd cdd ddd 7#a x7 ,now1 bbb 3 bbb ddd 3 ddd #bi ,if bx be substituted for a in any expression which is homogeneous with respect to a and b1 the pupil may easily see that b must occur in every term as often as there are units in the degree of the expression3 thus1 aa ab becomes bxbx bxb or bb7xx x72 aaa bbb becomes bxbxbx bbb or bbb7xxx #a72 and so on4 ,whence 7#ahf71 bbb7#bxxx #cxx73 bbb7#a x7 ddd7#bxxx #cxx73 ddd7#a x71 which1 when instead of these expressions their equals just found are substituted1 becomes #b;aaa #c;aab3 bbb abb #b;ccc #c;ccd3 ddd cdd4 ,the same reasoning may be applied to any other case1 and the pupil may in this way prove the following theorems3 ,if a 3 b c 3 d #b;a #c;b 3 b #b;c #c;d 3 d aa bb 3 aa - bb cc dd 3 cc - dd mab 3 #b;aa bb mcd 3 #b;cc dd #aia4 ,if the two means of a proportion be the same1 that is1 if a 3 b b3 c1 the three numbers1 a1 b1 and c1 are said to be in continued proportion1 or in geometrical progression4 ,the same terms are applied to a series of numbers1 of which any three that follow one another are in continued proportion1 such as #a #b #d #h #af #cb #fd c4 #b #b #b #b #b #b #b --- --- --- --- ---- ---- c4 #c #i #bg #ha #bdc #gbi ,which are in continued proportion1 since #b #b #b #a 3 #b #b 3 #d #b 3 --- --- 3 --- #c #c #i #b #b #b #b #b 3 #d #d 3 #h --- 3 --- --- 3 --- #c #i #i #bg c4 c4 #aib4 ,let a1 b1 c1 d1 e be in continued proportion2 we have then a b a 3 b b 3 c or --- --- or ac bb b c b c b 3 c c 3 d --- --- bd cc c d c d c 3 d d 3 e --- --- ce dd d e ,each term is formed from the preceding1 by multiplying it by the same number4 ,thus1 b c b --- a 7#ahj72 c --- b2 a b a b b c b and since --- ---1 --- --- or c --- b4 b c a b a d d c b ,again1 d --- c1 but --- ---1 which is ---2 c c b a b therefore1 d --- c1 and so on4 c b ,if1 then1 --- a 7which is called the common ratio of the series7 be denoted by r1 we have b ar c br arr d cr arrr and so on2 whence the series a b c d c4 is a ar arr arrr c4 ,hence a 3 c a 3 arr 7#ahf7 aa 3 aarr aa 3 bb because1 b being ar1 bb is arar or aarr4 ,again1 a 3 d a 3 arrr 7#ahf7 aaa 3 aaarrr aaa 3 bbb ,also a 3 e aaaa 3 bbbb1 and so on2 that is1 the first bears to the n term from the first the same proportion as the n power of the first to the n power of the second4 #aic4 ,a short rule may be found for adding together any number of terms of a continued proportion4 ,let it be first required to add together the terms #a1 r1 rr1 c4 where r is greater than unity4 ,it is evident that we do not alter any expression by adding or subtracting any numbers1 provided we afterwards subtract or add the same4 ,for example1 p p-q q-r r- s s ,let us take four terms of the series1 #a1 r1 rr1 c4 or1 #a r rr rrr ,it is plain that rrrr-#a rrrr-rrr rrr-rr rr-r r-#a ,now 7#ed71 rr-r r7r-#a71 rrr -rr rr7r-#a71 rrrr-rrr rrr7r-#a71 and the above equation becomes rrrr -#a rrr7r-#a7 rr 7r-#a7 r 7r-#a7 r-#a2 which is 7#ed7 rrr rr r #a taken r-#a times4 ,hence1 rrrr-#a divided by r-#a will give #a r rr rrr1 the sum of the terms required4 ,in this way may be proved the following series of equations3 rr - #a #a r -------- r - #a rrr - #a #a r rr --------- r - #a rrrr - #a #a r rr rrr ---------- r - #a rrrrr - #a #a r rr rrr rrrr ----------- r - #a ,if r be less than unity1 in order to find #a r rr rrr1 observe that #a - rrrr #a - r r - rr rr - rrr rrr - rrrr #a - r r7#a - r7 rr7#a - r7 rrr7#a - r72 whence1 by similar reasoning1 #a r rr rrr is found by dividing #a-rrrr by #a-r2 and equations similar to these just given may be found1 which are1 #a - rr #a r -------- #a - r #a - rrr #a r rr --------- #a - r #a - rrrr #a r rr rrr ---------- #a - r #a - rrrrr #a r rr rrr rrrr ----------- #a - r ,the rule is3 ,to find the sum of n terms of the series1 #a r rr c41 divide the difference between #a and the 7n #a7 term by the difference between #a and r4 #aid4 ,this may be applied to finding the sum of any number of terms of a continued proportion4 ,let a1 b1 c1 c4 be the terms of which it is required to sum four1 that is1 to find a b c d1 or 7#aib7 a ar arr arrr1 or 7#ed7 a7#a r rr rrr71 which 7#aic7 is rrrr - #a #a - rrrr ---------- a1 or ---------- a1 r - #a #a - r according as r is greater or less than unity4 ,the first fraction is arrrr - a e - a -------------1 or 7#aib7 ---------4 r - #a r - #a a - e ,similarly1 the second is ---------4 #a - r ,the rule1 therefore1 is3 ,to sum n terms of a continued proportion1 divide the difference of the 7n #a7 and first terms by the difference between unity and the common measure4 ,for example1 the sum of #aj terms of the series #a #c #i #bg c4 is required4 ,the eleventh term is #eijdi1 and _/ is #biebd4 ,again1 the sum of #ah terms of the series #b #a c4 of which the nineteenth term is _/1 is #a #b - ------ #acajgb #acajgj ----------- #c ------4 #a - #acajgb ,,examples4 #i terms of #a #d #af c4 are #hgcha #f #ab #hdgdbbfge #aj 444444 #c --- ---- c4 444 --------- #g #di #bjagfhjce #a #a #a #ajdhege #bj 444444 --- --- --- c4 444 ------- #b #d #h #ajdhegf #aie4 ,the powers of a number or fraction greater than unity increase2 for since #b is greater than #a1 #b #b is #b taken more than once1 that is1 is greater than #b1 and so on4 ,this increase goes on without limit2 that is1 there is no quantity so great but that some power of #b is greater4 ,to prove this1 observe that every power of #b is made by multiplying the preceding power by #b1 or by #a #a1 that is1 by adding to the former power that power itself and its half4 ,there will1 therefore1 be more added to the #ajth power to form the #aath1 than was added to the #ith power to form the #ajth4 ,but it is evident that if any given quantity1 however small1 be continually added to #b1 the result will come in time to exceed any other quantity that was also given1 however great2 much more1 then1 will it do so if the quantity added to #b be increased at each step1 which is the case when the successive powers of #b are formed4 ,it is evident1 also1 that the powers of #a never increase1 being always #a2 thus1 #a #a #a1 c4 ,also1 if a be greater than m times b1 the square of a is greater than mm times the square of b4 ,thus1 if a #b;b c1 where a is greater than #b;b1 the square of a1 or aa1 which is 7#fh7 #d;bb #d;bc cc is greater than #d;bb1 and so on4 #aif4 ,the powers of a fraction less than unity continually decrease2 thus1 the square of 1 or 1 is less than 1 being only two-fifths of it4 ,this decrease continues without limit2 that is1 there is no quantity so small but that some power of is less4 ,for if #e #b #a #a #a --- x1 --- ---1 and the powers of are ----1 -----1 #b #e x xx xxx and so on4 ,since x is greater than #a 7#aie71 some power of x may be found which shall be greater than a given quantity4 ,let this be called m2 then #a_/m is the corresponding power of 2 and a fraction whose denominator can be made as great as we please1 can itself be made as small as we please 7#aab74 #aig4 ,we have1 then1 in the series #a r rr rrr rrrr c4 ,i4 ,a series of increasing terms1 if r be greater than #a4 ,,ii4 ,of terms having the same value1 if r be equal to #a4 ,,iii4 ,a series of decreasing terms1 if r be less than #a4 ,in the first two cases1 the sum #a r rr rrr c4 may evidently be made as great as we please1 by sufficiently increasing the number of terms4 ,but in the third this may or may not be the case2 for though something is added at each step1 yet1 as that augmentation diminishes at every step1 we may not certainly say that we can1 by any number of such augmentations1 make the result as great as we please4 ,to shew the contrary in a simple instance1 consider the series1 #a _/ c4 ,carry this series to what extent we may1 it will always be necessary to add the last term in order to make as much as #b4 ,thus1 7#a 7 #a #a #a #b 7#a 7 #b4 7#a _/7 _/ #b1 c4 ,but in the series1 every term is only the half of the preceding2 consequently no number of terms1 however great1 can be made as great as #b by adding one more4 ,the sum1 therefore1 of #a1 1 1 c4 continually approaches to #b1 diminishing its distance from #b at every step1 but never reaching it4 ,hence1 #b is celled the limit of #a c4 ,we are not1 therefore1 to conclude that every series of decreasing terms has a limit4 ,the contrary may be shewn in the very simple series1 #a c4 which may be written thus3 #a 7 7 7 444 up to 7 7 444 up to _/7 7_/ 444 up to _/7 c4 ,we have thus divided all the series1 except the first two terms1 into lots1 each containing half as many terms as there are units in the denominator of its last term4 ,thus1 the fourth lot contains #af or _/#b terms4 ,each of these lots may be shewn to be greater than 4 ,take the third1 for example1 consisting of 1 _/1 _/1 _/1 _/1 _/1 _/1 and _/4 ,all except _/1 the last1 are greater than _/2 consequently1 by substituting _/ for each of them1 the amount of the whole lot would be lessened2 and as it would then become _/1 or 1 the lot itself is greater than 4 ,now1 if to #a 1 be continually added1 the result will in time exceed any given number4 ,still more will this be the case if1 instead of 1 the several lots written above be added one after the other4 ,but it is thus that the series #a 1 c4 is composed1 which proves what was said1 that this series has no limit4 #aih4 ,the series #a r rr rrr c4 always has a limit when r is less than #a4 ,to prove this1 let the term succeeding that at which we stop be a1 whence 7#aid7 the sum is #a - a #a a -------1 or 7#aab7 ------- - ------4 #a - r #a - r #a - r ,the terms decrease without limit 7#aif71 whence we may take a term so far distant from the beginning1 that a1 and therefore a -------1 #a - r shall be as small as we please4 ,but it is evident that in this case #a a ------- - ------- though always less than #a - r #a - r #a #a -------- may be brought as near to ------- #a - r #a - r as we please2 that is1 the series #a r rr c4 continually approaches to the limit #a --------4 #a - r ,thus #a c4 where r 1 continually approaches to #a ----- or #b1 as was shewn in the last article4 #a - ,,exercises4 #b #b ,the limit of #b --- --- c4 #c #i #a #a or #b7a --- --- c47 is #c #c #i #i #ha 444 #a --- ---- c4 444 #aj #aj #ajj #ae #de 444 #e ---- ---- c4 444 #h #g #di #aii4 ,when the fraction a_/b is not equal to c_/d1 but greater1 a is said to have to b a greater ratio than c has to d2 and when a_/b is less than c_/d1 a is said to have to b a less ratio than c has to d4 ,we propose the following questions as exercises1 since they follow very simply from this definition4 ,i4 ,if a be greater than b1 and c less than or equal to d1 a will have a greater ratio to b than c has to d4 ,,ii4 ,if a be less than b1 and c greater than or equal to d1 a has a less ratio to b than c has to d4 ,,iii4 ,if a be to b as c is to d1 and if a have a greater ratio to b than c has to x1 d is less than x2 and if a have a less ratio to b than c to x1 d is greater than x4 ,,iv4 a has to b a greater ratio than ax to bx y1 and a less ratio than ax to bx- y4 #bjj4 ,if a have to b a greater ratio than c has to d1 a c has to b d a less ratio than a has to b1 but a greater ratio than c has to d2 or1 in other words1 if a_/b be the greater of the two fractions a_/b and c_/d1 a c --------- b d will be greater than c_/d1 but less than a_/b4 ,to shew this1 observe that 7mx ny7_/7m n7 must lie between x and y1 if x and y be unequal3 for if x be the less of the two1 it is certainly greater than mx nx ----------- or than x2 m n and if y be the greater of the two1 it is certainly less than my ny -----------1 or than y4 m n ,it therefore lies between x and y4 ,now let a_/b be x1 and let c_/d be y3 then a bx1 c dy4 ,now bx dy ----------- b d is something between x and y1 as was just proved2 therefore a c --------- b d is something between a_/b and c_/d4 ,again1 since a_/b and c_/d are respectively equal to ap_/bp and cq_/dq1 and since1 as has just been proved1 ap cq ----------- bp dq lies between the two last1 it also lies between the two first2 that is1 if p and q be any numbers or fractions whatsoever1 ap cq ----------- bp dq lies between a_/b and c_/d4 #bja4 ,by the last article we may often form some notion of the value of an expression too complicated to be easily calculated4 ,thus1 #a x #a x #a -------- lies between --- and ----1 or #a and ---2 #a xx #a xx x ax by ax by -------------- lies between ----- and ------1 axx bbyy axx bbyy that is1 between #a_/x and #a_/;by4 ,and it has been shewn that 7a b7_/#b lies between a and b1 the denominator being considered as #a #a4 #bjb4 ,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 ---1 ---1 ---1 and ---1 p q r s that is1 is less than the greatest of them1 and greater than the least4 ,let these fractions be arranged in order of magnitude2 that is1 let a_/p be greater than b_/q1 b_/q be greater than c_/r1 and c_/r greater than d_/s4 ,then by 7#bjj7 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 evident4 #bjc4 ,it is usual to signify 8a is greater than b0 by a b and 8a is less than b0 by a b2 the opening of ,v being turned towards the greater quantity4 ,the pupil is recommended to make himself familiar with these signs4 ,,section ,,ix4 ,,on ,,permutations ,,and ,,combinations4 #bjd4 ,if a number of counters1 distinguished by different letters1 be placed on the table1 and any number of them1 say four1 be taken away1 the question is1 to determine in how many different ways this can be done4 ,each way of doing it gives what is called a combination of four1 but which might with more propriety be called a selection of four4 ,two combinations or selections are called different1 which differ in any way whatever2 thus1 abcd and abce are different1 d being in one and e in the other1 the remaining parts being the same4 ,let there be six counters1 a1 b1 c1 d1 e1 and f2 the combinations of three which can be made out of them are twenty in number1 as follow3 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 number1 namely1 abcd abde acde adef bcef abce abdf acdf bcde bdcf abcf abef acef bcdf cdef and so on4 #bje4 ,each of these combinations may be written in several different orders2 thus1 abcd may be disposed in any of the following ways3 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 order4 ,each of these is said to be a distinct permutation of abcd4 ,considered as a combination1 they are all the same1 as each contains a1 b1 c1 and d4 #bjf4 ,we now proceed to find how many permutations1 each containing one given number1 can be made from the counters in another given number1 six1 for example4 ,if we knew how to find all the permutations containing four counters1 we might make those which contain five thus3 ,take any one which contains four1 for example1 abcf in which d and e are omitted2 write d and e successively at the end1 which gives abcfd1 abcfe1 and repeat the same process with every other permutation of four2 thus1 dabc gives dabce and dabcf4 ,no permutation of five can escape us if we proceed in this manner1 provided only we know those of four2 for any given permutation of five1 as dbfea1 will arise in the course of the process from dbfe1 which1 according to our rule1 furnishes dbfea4 ,neither will any permutation be repeated twice1 for dbfea1 if the rule be followed1 can only arise from the permutation dbfe4 ,if we begin in this way to find the permutations of two out of the six1 a b c d e f each of these gives five2 thus1 a gives ab ac ad ae af b 444 ba bc bd be bf and the whole number is #f #e1 or #cj4 ,again1 ab gives abc abd abe abf ac 444 acb acd ace acf and here are #cj1 or #f #e permutations of #b1 each of which gives #d permutations of #c2 the whole number of the last is therefore #f #e #d1 or #abj4 ,again1 abc gives abcd abce abcf abd 444 abdc abde abdf and here are #abj1 or #f #e #d1 permutations of three1 each of which gives #c permutations of four2 the whole number of the last is therefore #f #e #d #c1 or #cfj4 ,in the same way1 the number of permutations of #e is #f #e #d #c #b1 and the number of permutations of six1 or the number of different ways in which the whole six can be arranged1 is #f #e #d #c #b #a4 ,the last two results are the same1 which must be2 for since a permutation of five only omits one1 it can only furnish one permutation of six4 ,if instead of six we choose any other number1 x1 the number of permutations of two will be x7x-#a71 that of three will be x7x-#a77x-#b71 that of four x7x -#a77x-#b77x-#c71 the rule being3 ,multiply the whole number of counters by the next less number1 and the result by the next less1 and so on1 until as many numbers have been multiplied together as there are to be counters in each permutation3 the product will be the whole number of permutations of the sort required4 ,thus1 out of #ab counters1 permutations of four may be made to the number of #ab #aa #aj #i1 or #aahhj4 ,,exercises4 #bjg4 ,in how many different ways can eight persons be arranged on eight seats8 ,answer1 #djcbj4 ,in how many ways can eight persons be seated at a round table1 so that all shall not have the same neighbours in any two arrangements8#cj ,answer1 #ejdj4 #cj ,the difference between this problem and the last is left to the ingenuity of the pupil4 ,if the hundredth part of a farthing be given for every different arrangement which can be made of fifteen persons1 to how much will the whole amount8 ,answer1 #acfbafjh4 ,out of seventeen consonants and five vowels1 how many words can be made1 having two consonants and one vowel in each8 ,answer1 #djhj4 #bjh4 ,if two or more of the counters have the same letter upon them1 the number of distinct permutations is less than that given by the last rule4 ,let there be a1 a1 a1 b1 c1 d1 and1 for a moment1 let us distinguish between the three as thus1 a1 a1 a4 ,then1 abcaad1 and abcaad are reckoned as distinct permutations in the rule1 whereas they would not have been so1 had it not been for the accents4 ,to compute the number of distinct permutations1 let us make one with b1 c1 and d1 leaving places for the as1 thus1 7 7 bc 7 7 7 7 d4 ,if the as had been distinguished as a1 a1 a1 we might have made #c #b #a distinct permutations1 by filling up the vacant places in the above1 all which six are the same when the as are not distinguished4 ,hence1 to deduce the number of permutations of a1 a1 a1 b1 c1 d1 from that of aaabcd1 we must divide the latter by #c #b #a1 or #f1 which gives #f #e #d #c #b #a --------------------- or #abj4 #c #b #a ,similarly1 the number of permutations of aaaabbbcc is #i #h #g #f #e #d #c #b #a --------------------------------- #d #c #b #a #c #b #a #b #a4 ,,exercise4 ,how many variations can be made of the order of the letters in the word antitrinitarian8 ,answer1 #abfabfjjj4 #bji4 ,from the number of permutations we can easily deduce the number of combinations4 ,but1 in order to form these combinations independently1 we will shew a method similar to that in 7#bjf74 ,if we know the combinations of two which can be made out of a1 b1 c1 d1 e1 we can find the combinations of three1 by writing successively at the end of each combination of two1 the letters which come after the last contained in it4 ,thus1 ab gives abc1 abd1 abe2 ad gives ade only4 ,no combination of three can escape us if we proceed in this manner1 provided only we know the combinations of two2 for any given combination of three1 as acd1 will arise in the course of the process from ac1 which1 according to our rule1 furnishes acd4 ,neither will any combination be repeated twice1 for acd1 if the rule be followed1 can only arise from ac1 since neither ad nor cd furnishes it4 ,if we begin in this way to find the combinations of the five1 a b c d e a gives ab ac ad ae b bc bd be c cd ce d de ,of these1 ab gives abc abd abe ac acd ace ad ade bc bcd bce bd bde cd cde ae be ce and de give none4 ,of these1 abc gives abcd abce abd abde acd acde bcd bcde ,those which contain e give none1 as before4 ,of the last1 abcd gives abcde1 and the others none1 which is evidently true1 since only one selection of five can be made out of five things4 #baj4 ,the rule for calculating the number of combinations is derived directly from that for the number of permutations4 ,take #g counters2 then1 since the number of permutations of two is #g #f1 and since two permutations1 ba and ab1 are in any combination ab1 the number of combinations is half that of the permutations1 or 7#g #f7_/b4 ,since the number of permutations of three is #g #f #e1 and as each combination abc has #c #b #a permutations1 the number of combinations of three is #g #f #e ----------4 #a #b #c ,also1 since any combination of four1 abcd1 contains #d #c #b #a permutations1 the number of combinations of four is #g #f #e #d -------------1 #a #b #c #d and so on4 ,the rule is3 ,to find the number of combinations1 each containing n counters1 divide the corresponding number of permutations by the product of #a1 #b1 #c1 c4 up to n4 ,if x be the whole number1 the number of combinations of two is x7x - #a7 -------------2 #a #b that of three is x7x - #a77x - #b7 ---------------------2 #a #b #c that of four is x7x - #a77x - #b77x - #c7 ------------------------------2 and so on4 #a #b #c #d #baa4 ,the rule may in half the cases be simplified1 as follows4 ,out of ten counters1 for every distinct selection of seven which is taken1 a distinct combination of #c is left4 ,hence1 the number of combinations of seven is as many as that of three4 ,we may1 therefore1 find the combinations of three instead of those of seven2 and we must moreover expect1 and may even assert1 that the two formul for finding these two numbers of combinations are the same in result1 though different in form4 ,and so it proves2 for the number of combinations of seven out of ten is #aj #i #h #g #f #e #d --------------------------1 #a #b #c #d #e #f #g in which the product #g #f #e #d occurs in both terms1 and therefore may be removed from both 7#ajh71 leaving #aj #i #h ----------1 #a #b #c which is the number of combinations of three out of ten4 ,the same may be shewn in other cases4 ,,exercises4 ,how many combinations of four can be made out of twelve things8 ,answer1 #die4 ,what number #f #h #bh of combinations #d out of #aa ,answer1 #ccj can be made of #bf #bh #cgh #f #ae #ejje ,how many combinations can be made of #ac out of #eb2 or how many different hands may a person hold at the game of whist8 ,answer1 #fcejaceeifjj4 ,,book ,,ii4 ,,commercial ,,arithmetic4 ,,section ,i4 ,,weights1 ,,measures1 ,c4 #bab4 ,in making the calculations which are necessary in commercial affairs1 no more processes are required than those which have been explained in the preceding book4 ,but there is still one thing wanted--not to insure the accuracy of our calculations1 but to enable us to compare and judge of their results4 ,we have hitherto made use of a single unit 7#ae71 and have treated of other quantities which are made up of a number of units1 in ,sections ,,ii41 ,,iii41 and ,,iv41 and of those which contain parts of that unit in ,sections ,v4 and ,,vi4 ,thus1 if we are talking of distances1 and take a mile as the unit1 any other length may be represented1#ca either by a certain number of miles1 or a certain number of parts of a mile1 and 7#a meaning one mile7 may be expressed either by a whole number or a fraction4 ,but we can easily see that in many cases inconveniences would arise4 ,suppose1 for example1 ,i say1 that the length of one room is _/ of a mile1 and of another _/ of a mile1 what idea can we form as to how much the second is longer than the first8 ,it is necessary to have some smaller measure2 and if we divide a mile into #agfj equal parts1 and call each of these parts a yard1 we shall find that the length of the first room is #i yards and _/ of a yard1 and that of the second #aj yards and _/ of a yard4 ,from this we form a much better notion of these different lengths1 but still not a very perfect one1 on account of the fractions _/ and _/4 ,to get a clearer idea of these1 suppose the yard to be divided into three equal parts1 and each of these parts to be called a foot2 then _/ of a yard contains #b feet1 and _/ of a yard contains _/ of a foot1 or a little more than of a foot4 ,therefore the length of the first room is now #i yards1 #b feet1 and of a foot2 that of the second is #aj yards and a little more than of a foot4 ,we see1 then1 the convenience of having large measures for large quantities1 and smaller measures for small ones2 but this is done for convenience only1 for it is possible to perform calculations upon any sort of quantity1 with one measure alone1 as certainly as with more than one2 and not only possible1 but more convenient1 as far as the mere calculation is concerned4 #ca ,it is not true1 that if we choose any quantity as a unit1 any other quantity of the same kind can be exactly represented either by a certain number of units1 or of parts of a unit4 ,to understand how this is proved1 the pupil would require more knowledge than he can be supposed to have2 but we can shew him that1 for any thing he knows to the contrary1 there may be quantities which are neither units nor parts of the unit4 ,take a mathematical line of one foot in length1 divide it into ten parts1 each of those parts into ten parts1 and so on continually4 ,if a point ,a be taken at hazard in the line1 it does not appear self-evident that if the decimal division be continued ever so far1 one of the points of division must at last fall exactly on ,a3 neither would the same appear necessarily true if the division were made into sevenths1 or elevenths1 or in any other way4 ,there may then possibly be a part of a foot which is no exact numerical fraction whatever of the foot2 and this1 in a higher branch of mathematics1 is found to be the case times without number4 ,what is meant in the words on which this note is written1 is1 that any part of a foot can be represented as nearly as we please by a numerical fraction of it2 and this is sufficient for practical purposes4 ,the measures which are used in this country are not those which would have been chosen had they been made all at one time1 and by a people well acquainted with arithmetic and natural philosophy4 ,we proceed to shew how the results of the latter science are made useful in our system of measures4 ,whether the circumstances introduced are sufficiently well known to render the following methods exact enough for the recovery of astronomical standards1 may be matter of opinion2 but no doubt can be entertained of their being amply correct for commercial purposes4 ,it is evidently desirable that weights and measures should always continue the same1 and that posterity should be able to replace any one of them when the original measure is lost4 ,it is true that a yard1 which is now exact1 is kept by the public authorities2 but if this were burnt by accident1#cb how are those who shall live #ejj years hence to know what was the length which their ancestors called a yard8 ,to ensure them this knowledge1 the measure must be derived from something which cannot be altered by man1 either from design or accident4 ,we find such a quantity in the time of the daily revolution of the earth1 and also in the length of the year1 both of which1 as is shewn in astronomy1 will remain the same1 at least for an enormous number of centuries1 unless some great and totally unknown change take place in the solar system4 ,so long as astronomy is cultivated1 it is impossible to suppose that either of these will be lost1 and it is known that the latter is #cfebdbbd mean solar days1 or about #cfe of the average interval which elapses between noon and noon1 that is1 between the times when the sun is highest in the heavens4 ,our year is made to consist of #cfe days1 and the odd quarter is allowed for by adding one day to every fourth year1 which gives what we call leap-year4 ,this is the same as adding of a day to each year1 and is rather too much1 since the excess of the year above #cfe days is not #be but #bdbbd of a day4 ,the difference is #jjggf of a day1 which is the quantity by which our average year is too long4 ,this amounts to a day in about #abh years1 or to about #c days in #d centuries4 ,the error is corrected by allowing only one out of four of the years which close the centuries to be leap-years4 ,thus1 ,a4,d4 #ahjj and #aijj are not leap-years1 but #bjjj is so4 #cb ,since this was first written1 the accident has happened4 ,the standard yard was so injured as to be rendered useless by the fire at the ,houses of ,parliament4 #bac4 ,the day is therefore the first measure obtained1 and is divided into #bd parts or hours1 each of which is divided into #fj parts or minutes1 and each of these again into #fj parts or seconds4 ,one second1 marked thus1 #a1cc is therefore the #hfdjj part of a day1 and the following is the ,,measure ,,of ,,time4#cd #fj seconds are #a minute #a m4 #fj minutes 0 #a hour #a h4 #bd hours 0 #a day #a d4 #g days 0 #a week #a wk4 #cfe days 0 #a year #a yr4 #bad4 ,the second having been obtained1 a pendulum can be constructed which shall1 when put in motion1 perform one vibration in exactly one second1 in the latitude of ,greenwich4#ce ,if we were inventing measures1 it would be convenient to call the length of this pendulum a yard1 and make it the standard of all our measures of length4 ,but as there is a yard already established1 it will do equally well to tell the length of the pendulum in yards4 ,it was found by commissioners appointed for the purpose1 that this pendulum in ,london was #ciacic inches1 or about one yard1 three inches1 and _/ of an inch4 ,the following is the division of the yard4 ,,measures ,,of ,,length4 ,the lowest measure is a barleycorn4#cf #c barleycorns are #a inch #a in4 #ab inches #a foot #a ft4 #c feet #a yard #a yd4 #e yards #a pole #a po4 #dj poles or #bbj yards #a furlong #a fur4 #h furlongs or #agfj yards #a mile #a mi4 ,also #f feet #a fathom #a fth4 #fi miles #a degree #a deg4 or #a4 #cc ,the minute and second are often marked thus1 #a1 #a3 but this notation is now almost entirely appropriated to the minute and second of angular measure4 #cd ,the measures in italics are those which it is most necessary that the student should learn by heart4 #ce ,the lengths of the pendulums which will vibrate in one second are slightly different in different latitudes4 ,greenwich is chosen as the station of the ,royal ,observatory4 ,we may add1 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 it4 #cf ,the inch is said to have been originally obtained by putting together three grains of barley4 ,a geographical mile is _/th of a degree1 and three such miles are one nautical league4 ,in the measurement of cloth or linen the following are also used3 #b inches are #a nail #a nl4 #d nails #a quarter 7of a yard7 #a qr4 #c quarters #a ,flemish ell #a ,fl4 e4 #e quarters #a ,english ell #a ,e4 e4 #f quarters #a ,french ell #a ,fr4 e4 #bae4 ,,measures ,,of ,,surface1 ,,or ,,superficies4 ,all surfaces are measured by square inches1 square feet1 c42 the square inch being a square whose side is an inch in length1 and so on4 ,the following measures may be deduced from the last1 as will afterwards appear4 #add square inches are #a square foot #a sq4 ft4 #i square feet #a square yard #a sq4 yd4 #cj square yards #a square pole #a sq4 p4 #dj square poles #a rood #a rd4 #d roods #a acre #a ac4 ,thus1 the acre contains #dhdj square yards1 which is ten times a square of #bb yards in length and breadth4 ,this #bb yards is the length which land-surveyors' chains are made to have1 and the chain is divided into #ajj links1 each #bb of a yard or #gib inches4 ,an acre is then #aj square chains4 ,it may also be noticed that a square whose side is #fi_/ yards is nearly an acre1 not exceeding it by of a square foot4 #baf4 ,,measures ,,of ,,solidity ,,or ,,capacity4#cg ,cubes are solids having the figure of dice4 ,a cubic inch is a cube each of whose sides is an inch1 and so on4 #agbh cubic inches are #a cubic foot #a c4 ft4 #bg cubic feet #a cubic yard #a c4 yd4 #cg #8,capacity' is a term which cannot be better explained than by its use4 ,when one measure holds more than another1 it is said to be more capacious1 or to have a greater capacity4 ,this measure is not much used1 except in purely mathematical questions4 ,in the measurements of different commodities various measures were used1 which are now reduced1 by act of parliament1 to one4 ,this is commonly called the imperial measure1 and is as follows3 ,,measure ,,of ,,liquids ,,and ,,of ,,all ,,dry ,,goods4 #d gills are #a pint #a pt4 #b pints #a quart #a qt4 #d quarts #a gallon #a gall4 #b gallons #a peck#ch #a pk4 #d pecks #a bushel #a bu4 #h bushels #a quarter #a qr4 #e quarters #a load #a ld4 ,the gallon in this measure is about #bggbgd cubic inches2 that is1 very nearly #bgg cubic inches4#ci #bag4 ,the smallest weight in use is the grain1 which is thus determined4 ,a vessel whose interior is a cubic inch1 when filled with water1#dj has its weight increased by #bebdeh grains4 ,of the grains so determined1 #gjjj are a pound averdupois1 and #egfj a pound troy4 ,the first pound is always used1 except in weighing precious metals and stones1 and also medicines4 ,it is divided as follows3 #ch ,this measure1 and those which follow1 are used for dry goods only4 #ci ,since the publication of the third edition1 the heaped measure1 which was part of the new system1 has been abolished4 ,the following paragraph from the third edition will serve for reference to it3 8,the other imperial measure is applied to goods which it is customary to sell by heaped measure1 and is as follows3 #b gallons #a peck #d pecks #a bushel #c bushels #a sack #ab sacks #a chaldron4 ,the gallon and bushel in this measure hold the same when only just filled1 as in the last4 ,the bushel1 however1 heaped up as directed by the act of parliament1 is a little more than one-fourth greater than before40 #dj ,pure water1 cleared from foreign substances by distillation1 at a temperature of #fb ,fahr4 ,,averdupois ,,weight4 #bg_/ grains are #a dram #a dr4 #f drams1 or drachms #a ounce#da #a oz4 #af ounces #a pound #a lb4 #bh pounds #a quarter #a qr4 #d quarters #a hundred-weight #a cwt4 #bj hundred-weight #a ton #a ton4 #da ,it is more common to divide the ounce into four quarters than into sixteen drams4 ,the pound averdupois contains #gjjj grains4 ,a cubic foot of water weighs #fbcbajfjf pounds averdupois1 or #iigacfifia ounces4 ,for the precious metals and for medicines1 the pound troy1 containing #egfj grains1 is used1 but is differently divided in the two cases4 ,the measures are as follow3 ,,troy ,,weight4 #bd grains are #a pennyweight #a dwt4 #bj pennyweights #a ounce #a oz4 #ab ounces #a pound #a lb4 ,the pound troy contains #egfj grains4 ,a cubic foot of water weighs #gegcgd pounds troy1 or #ijhhdhh ounces4 ,,apothecaries' ,,weight4 #bj grains are #a scruple #c scruples #a dram #h drams #a ounce #ab ounces #a pound lb #bah4 ,the standard coins of copper1 silver1 and gold1 are1--the penny1 which is #aj drams of copper2 the shilling1 which weighs #c pennyweights #ae grains1 of which #c parts out of #dj are alloy1 and the rest pure silver2 and the sovereign1 weighing #e pennyweights and #c grains1 of which #a part out of #ab is copper1 and the rest pure gold4 ,,measures ,,of ,,money4 ,the lowest coin is a farthing1 which is marked thus1 1 being one fourth of a penny4 #b farthings are #a halfpenny d4 #b halfpence #a penny #a;d4 #ab pence #a shilling #as4 #bj shillings #a pound#db or sovereign #a #ba shillings #a guinea4#dc #bai4 ,when any quantity is made up of several others1 expressed in different units1 such as #a4 #ad4 #f1 or #b;cwt4 #aqr4 #clbs41 it is called a compound quantity4 ,from these tables it is evident that any compound quantity of any substance can be measured in several different ways4 ,for example1 the sum of money which we call five pounds four shillings is also #ajd shillings1 or #abdh pence1 or #diib farthings4 ,it is easy to reduce any quantity from one of these measurements to another2 and the following examples will be sufficient to shew how to apply the same process1 usually called ,,reduction1 to all sorts of quantities4 ,i4 ,how many farthings are there in #ah4 #ab4 #f8dd #db ,the ,english pound is generally called a pound sterling1 which distinguishes it from the weight called a pound1 and also from foreign coins4 #dc ,the coin called a guinea is now no longer in use1 but the name is still given1 from custom1 to #ba shillings4 ,the pound1 which was not a coin1 but a note promising to pay #bj shillings to the bearer1 is also disused for the present1 and the sovereign supplies its place2 but the name pound is still given to #bj shillings4 #dd ,farthings are never written but as parts of a penny4 ,thus1 three farthings being #c_/d of a penny1 is written #c_/d1 or 4 ,one halfpenny may be written either as #b_/d or 2 the latter is most common4 ,since there are #bj shillings in a pound1 there are1 in #ah1 #ah #bj1 or #cfj shillings2 therefore1 #ah4 #ab is #cfj #ab1 or #cgb shillings4 ,since there are #ab pence in a shilling1 in #cgb shillings there are #cgb #ab1 or #ddfd pence2 and1 therefore1 in #ah4 #ab4 #f there are #ddfd #f1 or #ddgj pence4 ,since there are #d farthings in a penny1 in #ddgj pence there are #ddgj #d1 or #aghhj farthings2 and1 therefore1 in #ah4 #ab4 #f there are #aghhj #c1 or #aghhc farthings4 ,the whole of this process may be written as follows3 #ah 4 #ab 4 #f #bj -------- #cfj #ab #cgb #ab ----- #ddfd #f #ddgj #d ----- #aghhj #c #aghhc ,,ii4 ,in #aghhc farthings1 how many pounds1 shillings1 pence1 and farthings are there8 ,since #aghhc1 divided by #d1 gives the quotient #ddgj1 and the remainder #c1 #aghhc farthings are #ddgj pence and #c farthings 7#bah74 ,since #ddgj1 divided by #ab1 gives the quotient #cgb1 and the remainder #f1 #ddgj pence is #cgb shillings and #f pence4 ,since #cgb1 divided by #bj1 gives the quotient #ah1 and the remainder #ab1 #cgb shillings is #ah pounds and #ab shillings4 ,therefore1 #aghhc farthings is #ddgj;d41 which is #cgbs4 #f;d41 which is #ah4 #ab4 #f4 ,the process may be written as follows3 #d7aghhc ----- #ab7ddgj 444 #c ---- #bj7cgb 444 #f #ah 4 #ab 4 #f ,,exercises4 ,a has #ajj4 #d4 #aa1 and ,b has #fdcib farthings4 ,if ,a receive #adib farthings1 and ,b #a4 #b4 #c1 which will then have the most1 and by how much8--,answer1 ,a will have #cc4 #ab4 #c more than ,b4 ,in the following table the quantities written opposite to each other are the same3 each line furnishes two exercises4 #ae 4 #ah 4 #i #aecjb farthings4 #aae #a #h #ffcjgb grains4 #c #ad #i #ajja drams4 #c #adi yds #b #i in #aiedgg inches4 #ai #b #a gall #b q #abfj pints4 #af #bc #dg #eijbg seconds4 #bbj4 ,the same may be done where the number first expressed is fractional4 ,for example1 how many shillings and pence are there in _/ of a pound8 ,now1 _/ of a pound is _/ of #bj shillings2 _/ of #bj is #d #bj #d #d #af ------1 or ----- 7#aaj71 or ---1 #ae #c #c or 7#aje7 #e of a shilling4 ,again1 of a shilling is of #ab pence1 or #d pence4 ,therefore1 _/ #es4 #d;d4 ,also1 #bc of a day is #bc #bd in hours1 or #eeb2 and #eb of an hour is #eb #fj in minutes1 or #cb2 and #b of a minute is #b #fj in seconds1 or #ab2 whence #bc of a day is #e #ca #ab4 ,again1 suppose it required to find what part of a pound #fs4 #h;d4 is4 ,since #fs4 #h;d4 is #hj pence1 and since the whole pound contains #bj #ab or #bdj pence1 #fs4 #h;d4 is made by dividing the pound into #bdj parts1 and taking #hj of them4 ,it is therefore _/ 7#ajg71 but _/ 7#ajh72 therefore1 #fs4 #h;d4 4 ,,exercises4 of a day is #i #cf #abhda of a day #c #d #edfbdde #beg of a cwt4 #bh #ab #hgjd #adicf #b #aa #cchef #de ,when a decimal follows a whole number1 the decimal is always of the same unit as the whole number4 ,thus1 #ee is five seconds and five-tenths of a second4 ,thus1 #je means five-tenths of a second2 #jc1 three-tenths of an hour4 #bba1 #bbb4 ,i have thought it best to refer the mode of converting shillings1 pence1 and farthings into decimals of a pound to the ,appendix 7,see ,appendix ,on ,decimal ,money74 ,i should strongly recommend the reader to make himself perfectly familiar with the modes given in that ,appendix4 ,to prevent the subsequent sections from being altered in their numbering1 ,i have numbered this paragraph as above4 #bbc4 ,the rule of addition#df of two compound quantities of the same sort will be evident from the following example4 ,suppose it required to add #aib4 #ad4 #b to #fd4 #ac4 #aa4 ,the sum of these two is the whole of that which arises from adding their several parts4 ,now d4 d4 _/d4 #j 4 #j 4 #a 7#bai7 #aa;d4 #b;d4 #ac;d4 #j 4 #a 4 #a #acs4 #ads4 #bgs4 #a 4 #g 4 #j #fd #aib #bef 4 #j 4 #j ----------- ,the sum of all of which is #beg4 #h 4 #b ,this may be done at once1 and written as follows3 #aib 4 #ad 4 #b #fd 4 #ac 4 #aa ---------------- #beg 4 #h 4 #b #df ,before reading this article and the next1 articles 7#bi7 and 7#db7 should be read again carefully4 ,begin by adding together the farthings1 and reduce the result to pence and farthings4 ,set down the last only1 carry the first to the line of pence1 and add the pence in both lines to it4 ,reduce the sum to shillings and pence2 set down the last only1 and carry the first to the line of shillings1 and so on4 ,the same method must be followed when the quantities are of any other sort2 and if the tables be kept in memory1 the process will be easy4 #bbd4 ,,subtraction is performed on the same principle as in 7#dj71 namely1 that the difference of two quantities is not altered by adding the same quantity to both4 ,suppose it required to subtract #ai 4 #ac4 #aj from #bd4 #e4 #g4 ,write these quantities under one another thus3 #bd4 #e4 #g #ai4 #ac4 #aj ,since cannot be taken from or _/1 add #a;d4 to both quantities1 which will not alter their difference2 or1 which is the same thing1 add #d farthings to the first1 and #a;d4 to the second4 ,the pence and farthings in the two lines then stand thus3 #g_/;d4 and #aa;d4 ,now subtract from _/1 and the difference is which must be written under the farthings4 ,again1 since #aa;d4 cannot be subtracted from #g;d41 add #as4 to both quantities by adding #ab;d4 to the first1 and #as4 to the second4 ,the pence in the first line are then #ai1 and in the second #aa1 and the difference is #h1 which write under the pence4 ,since the shillings in the lower line were increased by #a1 there are now #ads4 in the lower1 and #es4 in the upper one4 ,add #bjs4 to the upper and #a to the lower line1 and the subtraction of the shillings in the second from those in the first leaves #aas4 ,again1 there are now #bj in the lower1 and #bd in the upper line1 the difference of which is #d2 therefore the whole difference of the two sums is #d4 #aa4 #h4 ,if we write down the two sums with all the additions which have been made1 the process will stand thus3 #bd 4 #be 4 #ai_/ #bj 4 #ad 4 #aa ------------------ ,difference #d 4 #aa 4 #h #bbe4 ,the same method may be applied to any of the quantities in the tables4 ,the following is another example3 ,from #g cwt4 #b qrs4 #ba lbs4 #ad oz4 ,subtract #b cwt4 #c qrs4 #bg lbs4 #ab oz4 ,after alterations have been made similar to those in the last article1 the question becomes3 ,from #g cwt4 #f qrs4 #di lbs4 #ad oz4 ,subtract #c cwt4 #d qrs4 #bg lbs4 #ab oz4 ---------------------------- ,the difference is #d cwt4 #b qrs4 #bb lbs4 #b oz4 ,in this example1 and almost every other1 the process may be a little shortened in the following way4 ,here we do not subtract #bg lbs4 from #ba lbs41 which is impossible1 but we increase #ba lbs4 by #a qr4 or #bh lbs4 and then subtract #bg lbs4 from the sum4 ,it would be shorter1 and lead to the same result1 first to subtract #bg lbs4 from #a qr4 or #bh lbs4 and add the difference to #ba lbs4 #bbf4 ,,exercises4 ,a man has the following sums to receive3 #aic4 #ad4 #aa1 #bb4 #j4 #f1 #fdgc4 #j4 #j1 and #di4 #ad4 #d2 and the following debts to pay3 #bjj 4 #ai4 #f1 #cje4 #af4 #aa1 #bb1 and #ai4 #f4 #j4 ,how much will remain after paying the debts8 ,answer1 #faij4 #g4 #d4 ,there are four towns1 in the order ,a1 ,b1 ,c1 and ,d4 ,if a man can go from ,a to ,b in #e #bj #cc1 from ,b to ,c in #f #di #b and from ,a to ,d in #ai #j #ag1 how long will he be in going from ,b to ,d1 and from ,c to ,d8 ,answer1 #ac #ci #dd1 and #f #ej #db4 #bbg4 ,in order to perform the process of ,,multiplication1 it must be recollected that1 as in 7#eb71 if a quantity be divided into several parts1 and each of these parts be multiplied by a number1 and the products be added1 the result is the same as would arise from multiplying the whole quantity by that number4 ,it is required to multiply #g4 #ac4 #f by #ac4 ,the first quantity is made up of #g pounds1 #ac shillings1 #f pence1 and #a farthing4 ,and #a farth4 #ac is #ac farth4 or #j 4 #j 4 #c 7#bai7 #f pence #ac is #gh pence1 or #j 4 #f 4 #f #ac shill4 #ac is #afi shill4 or #h 4 #i 4 #j #g pounds #ac is #ia pounds1 or #ia 4 #j 4 #j -------------- ,the sum of all these is #ii 4 #ae 4 #i which is therefore #g4 #ac4 #f #ac4 ,this process is usually written as follows3 #g 4 #ac 4 #f #ac ------------- #ii 4 #ae 4 #i #bbh4 ,,division is performed upon the same principle as in 7#gd71 viz4 that if a quantity be divided into any number of parts1 and each part be divided by any number1 the different quotients added together will make up the quotient of the whole quantity divided by that number4 ,suppose it required to divide #ii4 #ae4 #i by #ac4 ,since #ii divided by #ac gives the quotient #g1 and the remainder #h1 the quantity is made up of #ac #g1 or #ia1 and #h4 #ae4 #i4 ,the quotient of the first1 #ac being the divisor1 is #g3 it remains to find that of the second4 ,since #h is #afjs41 #h4 #ae4 #i is #ages4 #i;d41 and #age divided by #ac gives the quotient #ac1 and the remainder #f2 that is1 #ages4 #i;d4 is made up of #afis4 and #fs4 #i;d41 the quotient of the first of which is #acs41 and it remains to find that of the second4 ,since #fs4 is #gb;d41 #fs4 #i;d4 is #ha;d41 and #ha divided by #ac gives the quotient #f and remainder #c2 that is1 #ha;d4 is #gh;d4 and #c;d41 of the first of which the quotient is #f;d4 ,again1 since #c;d4 is _/1 or #ab farthings1 #c;d4 is #ac farthings1 the quotient of which is #a farthing1 or 1 without remainder4 ,we have then divided #ii4 #ae4 #i into four parts1 each of which is divisible by #ac1 viz4 #ia1 #afis41 #gh;d41 and #ac farthings2 so that the thirteenth part of this quantity is #g4 #ac4 #f4 ,the whole process may be written down as follows2 and the same sort of process may be applied to the exercises which follow3 s4 d4 s4 d4 #ac7ii #ae #i7 #g #ac #f #ia -- #h #bj --- #afj #ae #age #ac --- #de #ci -- #f #ab -- #gb #i #ha #gh -- #c #d -- #ab #a #ac #ac -- #j ,here1 each of the numbers #ii1 #age1 #ha1 and #ac1 is divided by #ac in the usual way1 though the divisor is only written before the first of them4 ,,exercises4 #b cwt4 #a qr4 #ba lbs4 #g oz4 #ec #abi cwt4 #a qr4 #af lbs4 #c oz4 #b #d #c #bg #aji #bcf #aj #af #c #bg 4 #aj 4 #h #efi #aefff 4 #i 4 #d #g 4 #d 4 #h #abc #hhi 4 #ad #aff _/ #dj 4 #d 4 #aj_/ #ahg 4 #f 4 #g _/ #e 4 #ab 4 #d _/ #ds4 #f;d4 #aaba #bed 4 #aa 4 #b #ds4 #d;d4 #dbfj #fs4 #f;d4 #bhdj #bbi4 ,suppose it required to find how many times #as4 #d;d4 is contained in #c4 #ai4 #aj4 ,the way to do this is to find the number of farthings in each4 ,by #bai1 in the first there are #fe1 and in the second #chce farthings4 ,now1 #chce contains #fe #ei times2 and therefore the second quantity is #ei times as great as the first4 ,in the case1 however1 of pounds1 shillings1 and pence1 it would be best to use decimals of a pound1 which will give a sufficiently exact answer4 ,thus #as4 #d;d4 is #jfg1 and #c4 #ai4 #aj is #ciid1 and #ciid divided by #jfg is #ciid by #fg1 or #ei_/4 ,this is an extreme case1 for the smaller the divisor1 the greater the effect of an error in a given place of decimals4 ,,exercises4 ,how many times does #f cwt4 #b qrs4 contain #a qr4 #ad lbs4 #a oz48 and #a #b #j #dg contain #c #df8 ,answer1 #agcjgeh and #dadcfgbeg4 ,if #b cwt4 #c qrs4 #a lb4 cost #aej4 #ac4 #aj1 how much does #a lb4 cost8 ,answer1 #is4 #i;d4 _/4 ,a grocer mixes #b cwt4 #ae lbs4 of sugar at #aa;d4 per pound with #ad cwt4 #c lbs4 at #e;d4 per pound4 ,at how much per pound must he sell the mixture so as not to lose by mixing them8 ,answer1 #e;d4 _/4 #bcj4 ,there is a convenient method of multiplication called ,,practice4 ,suppose ,i ask1 ,how much do #aec tons cost if each ton cost #b4 #ae4 #g8 ,it is plain that if this sum be multiplied by #aec1 the product is the price of the whole4 ,but this is also evident1 that1 if ,i buy #aec tons at #b4 #ae4 #g each ton1 payment may be made by first putting down #b for each ton1 then #ajs4 for each1 then #es41 then #f;d41 and then #a;d4 ,these sums together make up #b4 #ae4 #g1 and the reason for this separation of #b4 #ae 4 #g into different parts will be soon apparent4 ,the process may be carried on as follows3 #a4 #aec tons1 at #b each ton1 will cost #cjf #j #j #b4 ,since #ajs4 is 1 #aec tons1 at #ajs4 each1 will cost #ae_/1 which is #gf #aj #j #c4 ,since #es4 is of #ajs41 #aec tons1 at #es41 will cost half as much as the same number at #ajs4 each1 that is1 of #gf 4 #aj1 which is #ch #e #j #d4 ,since #f;d4 is of #es41 #aec tons1 at #f;d4 each1 will cost of what the same number costs at #es4each1 that is1 of #ch 4 #e1 which is #c #af #f #e4 ,since #a or #c halfpence is of #f;d4 or #ab halfpence1 #aec tons1 at #a;d4 each1 will cost of what the same number costs at #f;d4 each1 that is1 of#c 4 #af 4 #f1 which is #j #ai #a ----------- ,the sum of all these quantities is #dbe #aj #g which is1 therefore1 #b 4 #ae 4 #g #aec4 ,the whole process may be written down as follows3 or what #aec tons would #aec #j #j cost at #a per ton4 ----------- ----------- #b is #b #a #cjf #j #j #b #j #j #ajs4 is of #a #gf #aj #j #j #aj #j #es4 is of #ajs4 #ch #e #j #j #e #j #f;d4 is of #es4 #c #af #f #j #j #f #a;d4 is of #f;d4 #j #ai #a #j #j #a ------------ ---------- ,sum #dbe #aj #g #b #ae #g ,,another ,,example4 ,what do #agce lbs4 cost at #is4 #aj;d4 per lb48 ,the price #is4 #aj;d4 is made up of #es41 #ds41 #aj;d41 d41 and d42 of which #es4 is of #a1 #ds4 is of #a1 #aj;d4 is of #es41 d4 is _/ of #aj;d41 and d4 is of d4 ,follow the same method as in the last example1 which gives the following3 or what #agce tons would #agce #j #j cost at #a per lb4 ------------ ----------- #es4 is of #a #dcc #ae #j #j #e #j #ds4 is of #a #cdg #j #j #j #d #j #aj;d4 is of #es4 #gb #e #aj #j #j #aj d4 is _/ of #aj;d4 #c #ab #c #j #j #j d4 is of d4 #a #af #a #j #j #j ------------- ------------ by addition 444 #heh #i #c #j #i #aj ,in all cases1 the price must first be divided into a number of parts1 each of which is a simple fraction#dg of some one which goes before4 ,no rule can be given for doing this1 but practice will enable the student immediately to find out the best method for each case4 ,when that is done1 he must find how much the whole quantity would cost if each of these parts were the price1 and then add the results together4 #dg ,any fraction of a unit1 whose numerator is unity1 is generally called an aliquot part of that unit4 ,thus1 #bs4 and #ajs4 are both aliquot parts of a pound1 being and 4 ,,exercises4 ,what is the cost of #bdc cwt4 at #ad 4 #ah 4 #h per cwt48--,answer1 #cfbi 4 #a 4 #j4 #afi bushels at #b 4 #a 4 #c per bushel8--,answer1 #cdh 4 #ad 4 #i4 #bgc qrs4 at #ais4 #b;d4 per quarter8--,answer1 #bfa 4 #ab4 #f4 #bfbg sacks at #gs4 #h;d4 per sack8--,answer1 #ajab 4 #i 4 #i4 #bca4 ,throughout this section it must be observed1 that the rules can be applied to cases where the quantities given are expressed in common or decimal fractions1 instead of the measures in the tables4 ,the following are examples3 ,what is the price of #bgbcdgi cwt4 at #b4 #a4 #c per cwt48 ,answer1 #efbbhdi1 or #efb4 #e4 #h4 #fflbs4 at #as4 #d;d4 per lb4 cost #d4 #aa4 #e4 ,how many pounds1 shillings1 and pence1 will #bgicja acres let for if each acre lets for #cajgf8--;,answer1 #hfgieeh1 or #hfg4 #ai4 #a4 ,what does of _/ of #ag bush4 cost at of of #ag4 #ad per bushel8 ,answer1 #bcadf1 or #b4 #f4 #c4 ,what is the cost of #ailbs4 #hoz4 #ab;dwt4 #h;gr4 at #d4 #d4 #f per ounce8--,answer1 #iii4 #ad4 #a 4 #bcb4 ,it is often required to find to how much a certain sum per day will amount in a year4 ,this may be shortly done1 since it happens that the number of days in a year is #bdj #abj #e2 so that a penny per day is a pound1 half a pound1 and #e pence per year4 ,hence the following rule3 ,to find how much any sum per day amounts to in a year1 turn it into pence and fractions of a penny2 to this add the half of itself1 and let the pence be pounds1 and each farthing five shillings2 then add five times the daily sum1 and the total is the yearly amount4 ,for example1 what does #abs4 #c;d4 amount to in a year8 ,this is #adg;d41 and its half is #gc;d41 which added to #adg;d4 gives #bba;d41 which turned into pounds is #bba4 #ab4 #f4 ,also1 #abs4 #c;d4 #e is #c4 #a4 #f1 which added to the former sum gives #bbd4 #ad4 #j for the yearly amount4 ,in the same way the yearly amount of #bs4 #c;d4 is #da4 #af4 #e2 that of #f;d4 is #aj4 #e4 #c2 and that of #aa;d4 is #af4 #ad4 #g4 #bcc4 ,an inverse rule may be formed1 sufficiently correct for every purpose1 in the following way3 ,if the year consisted of #cfj days1 or _/ of #bdj1 the subtraction of one-third from any sum per year would give the proportion which belongs to #bdj days2 and every pound so obtained would be one penny per day4 ,but as the year is not #cfj1 but #cfe days1 if we divide each day's share into #cfe parts1 and take #e away1 the whole of the subtracted sum1 or #cfj #e such parts1 will give #cfj parts for each of the #e days which we neglected at first4 ,but #cfj such parts are left behind for each of the #cfj first days2 therefore1 this additional process divides the whole annual amount equally among the #cfe days4 ,now1 #e parts out of #cfe is one out of #gc1 or the #gc;d part of the first result must be subtracted from it to produce the true result4 ,unless the daily sum be very large1 the #gb;d part will do equally well1 which1 as #gb farthings are #ah pence1 is equivalent to subtracting at the rate of one farthing for #ah;d41 or d4 for #cs41 or #aj;d4 for #c4 ,the rule1 then1 is as follows3 ,to find how much per day will produce a given sum per year1 turn the shillings1 c4 in the given sum into decimals of a pound 7#bba72 subtract one-third2 consider the result as pence2 and diminish it by one farthing for every eighteen pence1 or ten pence for every #c4 ,for example1 how much per day will give #bbd4 #ad4 #j per year8 ,this is #bbdgjc1 and its third is #gdija1 which subtracted from #bbdgjc1 gives #adihjb1 which1 if they be pence1 amounts to #abs4 #ehjb;d41 in which #as4 #f;d4 is contained #h times4 ,subtract #h farthings1 or #b;d41 and we have #abs4 #chjb;d41 which differs from the truth only about _/ of a farthing4 ,in the same way1 #ajj per year is #es4 #e;d4 per day4 #bcd4 ,the following connexion between the measures of length and the measures of surface is the foundation of the application of arithmetic to geometry4 ,illustration ,suppose an oblong figure1 ,a1 ,b1 ,c1 ,d1 as here drawn 7which is called a rectangle in geometry71 with the side ,a ,b #f inches1 and the side ,a ,c #d inches4 ,divide ,a ,b and ,c ,d 7which are equal7 each into #f inches by the points a1 b1 c1 l1 m1 c42 and ,a ,c and ,b ,d 7which are also equal7 into #d inches by the points f1 g1 h1 x1 y1 and z4 ,join a and l1 b and m1 c41 and f and x1 c4 ,then1 the figure ,a ,b ,c ,d is divided into a number of squares2 for a square is a rectangle whose sides are equal1 and therefore ,a a f ,e is square1 since ,a a is of the same length as ,a f1 both being #a inch4 ,there are also four rows of these squares1 with six squares in each row2 that is1 there are #f #d1 or #bd squares altogether4 ,each of these squares has its sides #a inch in length1 and is what was called in 7#bae7 a square inch4 ,by the same reasoning1 if one side had contained #f yards1 and the other #d yards1 the surface would have contained #f #d square yards2 and so on4 ,illustration #bce4 ,let us now suppose that the sides of ,a ,b ,c ,d1 instead of being a whole number of inches1 contain some inches and a fraction4 ,for example1 let ,a ,b be #c inches1 or 7#aad7 _/ of an inch1 and let ,a ,c contain #b inches1 or _/ of an inch4 ,draw ,a ,e twice as long as ,a ,b1 and ,a ,f four times as long as ,a ,c1 and complete the rectangle ,a ,e ,f ,g4 ,the rest of the figure needs no description4 ,then1 since ,a ,e is twice ,a ,b1 or twice _/ inches1 it is #g inches4 ,and since ,a ,f is four times ,a ,c1 or four times _/ inches1 it is #i inches4 ,therefore1 the whole rectangle ,a ,e ,f ,g contains1 by 7#bcd71 #g #i or #fc square inches4 ,but the rectangle ,a ,e ,f ,g contains #h rectangles1 all of the same figure as ,a ,b ,c ,d2 and therefore ,a ,b ,c ,d is one-eighth part of ,a ,e ,f ,g1 and contains _/ square inches4 ,but _/ is made by multiplying _/ and _/ together 7#aah74 ,from this and the last article it appears1 that1 whether the sides of a rectangle be a whole or a fractional number of inches1 the number of square inches in its surface is the product of the numbers of inches in its sides4 ,the square itself is a rectangle whose sides are all equal1 and therefore the number of square inches which a square contains is found by multiplying the number of inches in its side by itself4 ,for example1 a square whose side is #ac inches in length contains #ac #ac or #afi square inches4 #bcf4 ,,exercises4 ,what is the content1 in square feet and inches1 of a room whose sides are #db ft4 #e inch4 and #ca ft4 #i inch48 and supposing the piece from which its carpet is taken to be three quarters of a yard in breadth1 what length of it must be cut off8--,answer1 ,the content is #acdf square feet #aje square inches1 and the length of carpet required is #eih feet #f_/ inches4 ,the sides of a rectangular field are #bec yards and a quarter of a mile2 how many acres does it contain8--,answer1 #bc4 ,what is the difference between #ah square miles1 and a square of #ah miles long1 or #ah miles square8--,answer1 #cjf square miles4 #bcg4 ,it is by this rule that the measure in 7#bae7 is deduced from that in 7#bad72 for it is evident that twelve inches being a foot1 the square foot is #ab #ab or #add square inches1 and so on4 ,in a similar way it may be shewn that the content in cubic inches of a cube1 or parallelepiped1#dh may be found by multiplying together the number of inches in those three sides which meet in a point4 ,thus1 a cube of #f inches contains #f #f #f1 or #baf cubic inches2 a chest whose sides are #f1 #h1 and #e feet1 contains #f #h #e1 or #bdj cubic feet4 ,by this rule the measure in 7#baf7 was deduced from that in 7#bad74 #dh ,a parallelepiped1 or more properly1 a rectangular parallelepiped1 is a figure of the form of a brick2 its sides1 however1 may be of any length2 thus1 the figure of a plank has the same name4 ,a cube is a parallelepiped with equal sides1 such as is a die4 ,,section ,,ii4 ,,rule ,,of ,,three4 #bch4 ,suppose it required to find what #aef yards will cost1 if #bb yards cost #ags4 #d;d4 ,this quantity1 reduced to pence1 is #bjh;d42 and if #bb yards cost #bjh;d41 each yard costs _/d4 ,but #aef yards cost #aef times the price of one yard1 and therefore cost #bjh #bjh #aef ---- #aef pence1 or --------- pence 7#aag74 #bb #bb ,again1 if #be ,french francs be #bj shillings sterling1 how many francs are in #bj4 #ae8 ,since #be francs are #bj shillings1 twice the number of francs must be twice the number of shillings2 that is1 #ea francs are #dj shillings1 and one shilling is the fortieth part of #ea francs1 or _/ francs4 ,but #bj #aes4 contain #dae shillings 7#bai72 and since #a shilling is _/ francs1 #dae shillings is #ea #dae _/ #dae francs1 or 7#aag7 -------- francs4 #dj #bci4 ,such questions as the last two belong to the most extensive rule in ,commercial ,arithmetic1 which is called the ,,rule ,,of ,,three1 because in it three quantities are given1 and a fourth is required to be found4 ,from both the preceding examples the following rule may be deduced1 which the same reasoning will shew to apply to all similar cases4 ,it must be observed1 that in these questions there are two quantities which are of the same sort1 and a third of another sort1 of which last the answer must be4 ,thus1 in the first question there are #bb and #aef yards and #bjh pence1 and the thing required to be found is a number of pence4 ,in the second question there are #bj and #dae shillings and #be francs1 and what is to be found is a number of francs4 ,write the three quantities in a line1 putting that one last which is the only one of its kind1 and that one first which is connected with the last in the question4#di ,put the third quantity in the middle4 ,in the first question the quantities will be placed thus3 #bb yds4 #aef yds4 #ags4 #d;d4 ,in the second question they will be placed thus3 #bjs4 #bj #aes4 #be francs4 #di ,this generally comes in the same member of the sentence4 ,in some cases the ingenuity of the student must be employed in detecting it4 ,the reasoning of 7#bch7 is the best guide4 ,the following may be very often applied4 ,if it be evident that the answer must be less than the given quantity of its kind1 multiply that given quantity by the less of the other two2 if greater1 by the greater4 ,thus1 in the first question1 #aef yards must cost more than #bb2 multiply1 therefore1 by #aef4 ,reduce the first and second quantities1 if necessary1 to quantities of the same denomination4 ,thus1 in the second question1 #bj #aes4 must be reduced to shillings 7#bai74 ,the third quantity may also be reduced to any other denomination1 if convenient2 or the first and third may be multiplied by any quantity we please1 as was done in the second question2 and1 on looking at the answer in 7#bch71 and at 7#ajh71 it will be seen that no change is made by that multiplication4 ,multiply the second and third quantities together1 and divide by the first4 ,the result is a quantity of the same sort as the third in the line1 and is the answer required4 ,thus1 to the first question the answer is 7#bch7 #bjh #aef #ags4 #d;d4 #aef ----------pence1 or1 which is the same thing1 -------------------4 #bb #bb #bdj4 ,the whole process in the first question is as follows3#ej yds4 yds4 s4 d4 #bb 3 #aef #ag 4 #d #ab --- #bjh pence4 #aef ---- #abdh #ajdj #bjh ----- #bb7cbddh7adgd;d4 and _/1 or _/ of a farthing1 #bb or 7#bai7 #f 4 #b 4 #aj-_/4 --- #ajd #hh ---- #afd #aed ---- #ajh #hh -- #bj 7#bbh7 #d -- #hj #ff -- #ad #ej ,it is usual to place points1 in the manner here shewn1 between the quantities4 ,those who have read ,section ,,viii4 will see that the ,rule of ,three is no more than the process for finding the fourth term of a proportion from the other three4 ,the question might have been solved without reducing #ags4 #d;d4 to pence1 thus3 yds4 yds4 s4 d4 #bb 3 #aef #ag 4 #d #aef 7#bbg7 ---------- #bb7ace 4 #d 4 #j7f 4 #b 4 #aj-_/ 7#bbh7 #acb --- #c #bj #d #fd #dd -- #bj #ab #bdj #bbj --- #bj #d #hj #ff -- #ad ,the student must learn by practice which is the most convenient method for any particular case1 as no rule can be given4 #bda4 ,it may happen that the three given quantities are all of one denomination2 nevertheless it will be found that two of them are of one1 and the third of another sort4 ,for example3 ,what must an income of #djj pay towards an income-tax of #ds4 #f;d4 in the pound8 ,here the three given quantities are1 #djj1 #ds4 #f;d41 and #a1 which are all of the same species1 viz4 money4 ,nevertheless1 the first and third are income2 the second is a tax1 and the answer is also a tax2 and therefore1 by 7#aeb71 the quantities must be placed thus3 #a 3 #djj #ds4 #f;d4 #bdb4 ,the following exercises either depend directly upon this rule1 or can be shewn to do so by a little consideration4 ,there are many questions of the sort1 which will require some exercise of ingenuity before the method of applying the rule can be found4 ,,exercises4 ,if #ae cwt4 #b qrs4 cost #aih4 #ae4 #d1 what does #a qr4 #bb lbs4 cost8 ,answer1 #e 4 #ad 4 #e _/4 ,if a horse go #ad m4 #c fur4 #bg yds4 in #c #bf #ab1 how long will he be in going #bc miles8 ,answer1 #e #bi #cd7_/74 ,two persons1 ,a and ,b1 are bankrupts1 and owe exactly the same sum2 ,a can pay #aes4 #d;d4 in the pound1 and ,b only #gs4 7#f7;d4 ,at the same time ,a has in his possession #acjd4 #ag more than ,b2 what do the debts of each amount to8 ,answer1 #ccdj 4 #h 4 #c _/4 ,for every 7#ab7 acres which one country contains1 a second contains 7#ef74 ,the second country contains #ag1cjj square miles4 ,how much does the first contain8 ,again1 for every #c people in the first1 there are #e in the second2 and there are in the first #bg people on every #bj acres4 ,how many are there in each country8--,answer1 ,the number of square miles in the first is #chdd_/1 and its population #c1cba1fjj2 and the population of the second is #e1ecf1jjj4 ,if 7#db7 yds4 of cloth1 #ah in4 wide1 cost #ei4 #ad4 #b1 how much will 7#aah7 yds4 cost1 if the width be #a yd48 ,answer1 #ccb4 #e4 7#b_/74 ,if #i4 #c4 #f last six weeks1 how long will #ajj last8 ,answer1 7#fe_/7 weeks4 ,how much sugar1 worth 7#i;d74 a pound1 must be given for #b cwt4 of tea1 worth #aj;d4 an ounce8 ,answer1 #cb cwt4 #c qrs4 #g lbs4 _/4 #bdc4 ,suppose the following question asked3 ,how long will it take #ae men to do that which #de men can finish in #aj days8 ,it is evident that one man would take #de #aj1 or #dej days1 to do the same thing1 and that #ae men would do it in one-fifteenth part of the time which it employs one man1 that is1 in 7#dej #ae7 or #cj days4 ,by this and similar reasoning the following questions can be solved4 ,,exercises4 ,if #ae oxen eat an acre of grass in #ab days1 how long will it take #bf oxen to eat #ad acres8 ,answer1 7#if_/7 days4 ,if #bb masons build a wall #e feet high in #f days1 how long will it take #dc masons to build #aj feet8 ,answer1 7#f_/7 days4 #bdd4 ,the questions in the preceding article form part of a more general class of questions1 whose solution is called the ,,double ,,rule ,,of ,,three1 but which might1 with more correctness1 be called the ,rule of ,five1 since five quantities are given1 and a sixth is to be found4 ,the following is an example3 ,if #e men can make #cj yards of cloth in #c days1 how long will it take #d men to make #fh yards8 ,the first thing to be done is to find out1 from the first part of the question1 the time it will take one man to make one yard4 ,now1 since one man1 in #c days1 will do the fifth part of what #e men can do1 he will in #c days make _/1 or #f yards4 ,he will1 therefore1 make one yard in _/#f or 7#c #e7_/cj of a day4 ,from this we are to find how long it will take #d men to make #fh yards4 ,since one man makes a yard in #c #e #c #e ----- of a day1 he will make #fh yards in ----- #fh days1 #cj #cj #c #e #fh or 7#aaf7 in ---------- days2 and #d men will do this in one-fourth #cj #c #e #fh of the time1 that is 7#abc71 in ---------- days1 or in #h days4 #cj #d ,again1 suppose the question to be3 ,if #e men can make #cj yards in #c days1 how much can #f men do in #ab days8 ,here we must first find the quantity one man can do in one day1 which appears1 on reasoning similar to that in the last example1 to be #cj_/7c #e7 yards4 ,hence1 #f men1 in one day1 will make #f #cj #ab #f #cj ------ yards1 and in #ab days will make ----------- or #add yards4 #e #c #e #c ,from these examples the following rule may be drawn4 ,write the given quantities in two lines1 keeping quantities of the same sort under one another1 and those which are connected with each other1 in the same line4 ,in the two examples above given1 the quantities must be written thus3 ,illustration ,,second ,,example4 ,illustration ,draw a curve through the middle of each line1 and the extremities of the other4 ,there will be three quantities on one curve and two on the other4 ,divide the product of the three by the product of the two1 and the quotient is the answer to the question4 ,if necessary1 the quantities in each line must be reduced to more simple denominations 7#bai71 as was done in the common ,rule of ,three 7#bch74 ,,exercises4 ,if #f horses can1 in #b days1 plough #ag acres1 how many acres will #ic horses plough in #d days8 ,answer1 #eib4 ,if #bj men1 in #c days1 can dig #g rectangular fields1 the sides of each of which are #dj and #ej yards1 how long will #cg men be in digging #ec fields1 the sides of each of which are #ij and #abe yards8 #bdea ,answer1 #ge----- days4 #bjgbj ,if the carriage of #fj cwt4 through #bj miles cost #ad #ajs41 what weight ought to be carried #cj miles for #e4 #h4 #i8 ,answer1 #ae cwt4 ,if #ajj gain #e in a year1 how much will #hej gain in #c years and #h months8 ,answer1 #aee4 #af4 #h4 ,,section ,,iii4 ,,interest1 ,,etc4 #bde4 ,in the questions contained in this ,section1 almost the only process which will be employed is the taking a fractional part of a sum of money1 which has been done before in several cases4 ,suppose it required to take #g parts out of #dj from #af1 that is1 to divide #af into #dj equal parts1 and take #g of them4 ,each of these parts is #af #af #af #g ----1 and #g of them make ---- #g1 or ------ pounds 7#aaf74 #dj #dj #dj ,the process may be written as below3 #af #g ----- #dj7aab7b 4 #afs4 #hj -- #cb #bj --- #fdj #dj --- #bdj #bdj --- #j ,suppose it required to take #ac parts out of a hundred from #ef4 #ac4 #g4 #ef 4 #ac 4 #g #ac ---------------- #ajj7 #gcf 4 #ag 4 #a 7 #g 4 #g 4 #d _/ #gjj --- #cf #bj #ag #gcg #gjj --- #cg #ab #a #dde #djj --- #de #d #b #ahb #ajj --- #hb ,let it be required to take #b parts out of a hundred from #c #abs4 ,the result1 by the same rule is #c #abs4 #b #e -------------------1 or #abc #c #abs4 ---2 #ajj #bjj so that taking #b out of a hundred is the same as taking #e parts out of #bjj4 ,,exercises4 ,take #g parts out of #ec from #a #ajs4 #abi ,answer1 #ds4 #a---;d4 #aei ,take #e parts out of #ajj from #ajg #acs4 #d;d4 ,answer1 #e4 #g4 #h and _/ of a farthing4 #ef #cs4 #b;d4 is equally divided among #cb persons4 ,how much does the share of #bc of them exceed that of the rest8 ,answer1 #bd4 #aa4 #d 4 #bdf4 ,it is usual1 in mercantile business1 to mention the fraction which one sum is of another1 by saying how many parts out of a hundred must be taken from the second in order to make the first4 ,thus1 instead of saying that #af #abs4 is the half of #cc #ds41 it is said that the first is #ej per cent of the second4 ,thus1 #e is #b per cent of #bjj2 because1 if #bjj be divided into #ajj parts1 #b of those parts are #e4 ,also1 #ac is #aej per cent of #h4 #ac4 #d1 since the first is the second and half the second4 ,suppose it asked1 ,how much per cent is #bc parts out of #ef of any sum8 ,the question amounts to this3 ,if he who has #ef gets #ajj for them1 how much will he who has #bc receive8 ,this1 by #bch1 is #bc _/ or _/ or #da_/4 ,hence1 #bc out of #ef is #da_/ per cent4 ,similarly #af parts out of #ah is #af _/1 or #hh_/ per cent1 and #b parts out of #e is #b _/1 or #dj per cent4 ,from which the method of reducing other fractions to the rate per cent is evident4 ,suppose it asked1 ,how much per cent is #f4 #ab4 #b of #ab4 #c8 ,since the first contains #aehf;d41 and the second #biaf;d41 the first is #aehf out of #biaf parts of the second2 that is1 by the last rule1 it is _/1 or #ed_/1 or #ed4 #g4 #i per cent1 very nearly4 ,the more expeditious way of doing this is to reduce the shillings1 c4 to decimals of a pound4 ,three decimal places will give the rate per cent to the nearest shilling1 which is near enough for all practical purposes4 ,for instance1 in the last example1 which is to find how much #ffjh is of #abae1 #ffjh #ajj is #ffjh1 which divided by #abae gives #edch1 or #ed4 #g4 ,greater correctness may be had1 if necessary1 as in the ,appendix4 ,,exercises4 ,how much per cent is #aih out of #bcc parts8--,ans4 #he4 #a4 #h4 ,goods which are bought for #aic4 #ab1 are sold for #baf4 #ac4 #d2 how much per cent has been gained by them8 ,answer1 ,a little less than #aa4 #ah4 #f4 ,a sells goods for ,b to the amount of #bcj4 #ab1 and is allowed a commission#ea of #c per cent2 what does that amount to8 ,answer1 #f 4 #ah4 #d _/4 #ea ,commission is what is allowed by one merchant to another for buying or selling goods for him1 and is usually a per-centage on the whole sum employed4 ,brokerage is an allowance similar to commission1 under a different name1 principally used in the buying and selling of stock in the funds4 ,insurance is a per-centage paid to those who engage to make good to the payers any loss they may sustain by accidents from fire1 or storms1 according to the agreement1 up to a certain amount which is named1 and is a per-centage upon this amount4 ,tare1 tret1 and cloff1 are allowances made in selling goods by wholesale1 for the weight of the boxes or barrels which contain them1 waste1 c42 and are usually either the price of a certain number of pounds of the goods for each box or barrel1 or a certain allowance on each cwt4 ,a stockbroker buys #agjj stock1 brokerage being at per cent2 what does he receive8--,answer1 #b4 #b4 #f4 ,a ship whose value is #ae1dbc is insured at #ai per cent2 what does the insurance amount to8--,answer1 #cjcc4 #c4 #i _/4 #bdg4 ,in reckoning how much a bankrupt is able to pay his creditors1 as also to how much a tax or rate amounts1 it is usual to find how many shillings in the pound is paid4 ,thus1 if a person who owes #ajj can only pay #ej1 he is said to pay #ajs4 in the pound4 ,the rule is easily derived from the same reasoning as in #bdf4 ,for example1 #ej out of #hb is #ej #ejbj ---- out of #a1 or ----- shillings1 #hb #hb or #abs4 #b _/ in the pound4 #bdh4 ,,interest is money paid for the use of other money1 and is always a per-centage upon the sum lent4 ,it may be paid either yearly1 half-yearly1 or quarterly2 but when it is said that #ajj is lent at #d per cent1 it must be understood to mean #d per cent per annum2 that is1 that #d pounds are paid every year for the use of #ajj4 ,the sum lent is called the principal1 and the interest upon it is of two kinds4 ,if the borrower pay the interest as soon as1 from the agreement1 it becomes due1 it is evident that he has to pay the same sum every year2 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 years4 ,but if he do not pay the interest at once1 but keeps it in his hands until he returns the principal1 he will then have more of his creditor's money in his hands every year1 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 due4 ,in the first case1 the interest is called simple1 and in the second compound4 ,the interest and principal together are called the amount4 #bdi4 ,what is the simple interest of #ajdi4 #af4 #f for #f years and one-third1 at #d per cent8 ,this interest must be #f times the interest of the same sum for one year1 which 7#bde7 is found by multiplying the sum by #d1 and dividing by #ajj4 ,the process is as follows3 7#bcj7 7a7 #ajdi 4 #af 4 #f -------------- a #d #daii 4 #f 4 #j a #ebd 4 #ah 4 #c -------------- 7#hb7 #ajj7 #dg1bd 4 #d 4 #c7dg 4 #d 4 #aj_/ #bj ---- 7#bbh7 #d1hdeb #ab ------ #aj1aaec 7b7 #dg 4 #d 4 #aj_/ ,int4 for one yr4 ------------------ b #f #bhc 4 #i 4 #j_/ b #ae 4 #ad 4 #aa_/ --------------------- #bii 4 #d 4 #j_/ ,int4 for #f yrs4 #eb ,here the #ds4 from the dividend is taken in4 #ec ,here the #c;d4 from the dividend is taken in4 ,,exercises4 ,what is the interest of #aje4 #f4 #b for #ai years and #g weeks at #c per cent8 ,answer1 #fj4 #i1 very nearly4 ,what is the difference between the interest of #ej4 #ai for #g years at #c per cent1 and for #h years at #b per cent8 ,answer1 #ajs4 7#b7;d4 ,what is the interest of #aeg4 #ag4 #f for one year at #e per cent8 ,answer1 #g4 #ag4 #aj4 ,shew that the interest of any sum for #i years at #d per cent is the same as that of the same sum for #d years at #i per cent4 #bej4 ,in order to find the interest of any sum at compound interest1 it is necessary to find the amount of the principal and interest at the end of every year2 because in this case 7#bdh7 it is the amount of both principal and interest at the end of the first year1 upon which interest accumulates during the second year4 ,suppose1 for example1 it is required to find the interest1 for #c years1 on #ajj1 at #e per cent1 compound interest4 ,the following is the process3 #ajj ,first principal4 #e ,first year's interest4 --- #aje ,amount at the end of the first year4 7#bdi7 #e 4 #e ,interest for the second year on #aje4 -------- #aaj 4 #e ,amount at the end of two years4 #e 4 #aj 4 #c ,interest due for the third year4 ------------ #aae 4 #ae 4 #c ,amount at the end of three years4 #ajj 4 #j 4 #j ,first principal4 ------------ #ae 4 #ae 4 #c ,interest gained in the three years4 ,when the number of years is great1 and the sum considerable1 this process is very troublesome2 on which account tables#ed are constructed to shew the amount of one pound1 for different numbers of years1 at different rates of interest4 ,to make use of these tables in the present example1 look into the column headed 8#e per cent20 and opposite to the number #c1 in the column headed 8,number of years10 is found #aaegfbe2 meaning that #a will become #aaegfbe in #c years4 ,now1 #ajj must become #ajj times as great2 and #aaegfbe #ajj is #aaegfbe 7#ada72 but 7#bba7 #gfbe is #aes4 #c;d42 therefore the whole amount of #ajj is #aae4 #ae4 #c1 as before4 #ed ,sufficient tables for all common purposes are contained in the article on ,interest in the ,penny ,cyclopdia2 and ample ones in the ,treatise on ,annuities and ,reversions1 in the ,library of ,useful ,knowledge4 #bea4 ,suppose that a sum of money has lain at simple interest #d years1 at #e per cent1 and has1 with its interest1 amounted to #cej2 it is required to find what the sum was at first4 ,whatever the sum was1 if we suppose it divided into #ajj parts1 #e of those parts were added every year for #d years1 as interest2 that is1 #bj of those parts have been added to the first sum to make #cej4 ,if1 therefore1 #cej be divided into #abj parts1 #ajj of those parts are the principal which we want to find1 and #bj parts are interest upon it2 that is1 the principal is 7#cej #ajj7_/aej1 or #bia4 #ac4 #d4 #beb4 ,suppose that ,a was engaged to pay ,b #cej at the end of four years from this time1 and that it is agreed between them that the debt shall be paid immediately2 suppose1 also1 that money can be employed at #e per cent1 simple interest2 it is plain that ,a ought not to pay the whole sum1 #cej1 because1 if he did1 he would lose #d years' interest of the money1 and ,b would gain it4 ,it is fair1 therefore1 that he should only pay to ,b as much as will1 with interest1 amount in four years to #cej1 that is 7#bea71 #bia4 #ac4 #d4 ,therefore1 #eh4 #f4 #h must be struck off the debt in consideration of its being paid before the time4 ,this is called ,,discount2#ee and #bia4 #ac4 #d is called the present value of #cej due four years hence1 discount being at #e per cent4 ,the rule for finding the present value of a sum of money 7#bea7 is3 ,multiply the sum by #ajj1 and divide the product by #ajj increased by the product of the rate per cent and number of years4 ,if the time that the debt has yet to run be expressed in years and months1 or months only1 the months must be reduced to the equivalent fraction of a year4 #ee ,this rule is obsolete in business4 ,when a bill1 for instance1 of #ajj having a year to run1 is discounted 7as people now say7 at #e per cent1 this means that #e per cent of #ajj1 or #e1 is struck off4 ,,exercises4 ,what is the discount on a bill of #ach4 #ad4 #d1 due #b years hence1 discount being at #d per cent8 ,answer1 #aa4 #i4 #a4 ,what is the present value of #ajca4 #ag1 due #f months hence1 interest being at #c per cent8 ,answer1 #ajaf4 #ab4 #bec4 ,if we multiply by a b1 or by a-b1 when we should multiply by a1 the result is wrong by the fraction b b --- b1 or ---------1 a a - b of itself3 being too great in the first case1 and too small in the second4 ,again1 if we divide by a b1 where we should have divided by a1 the result is too small by the fraction b_/a of itself2 while1 if we divide by a-b instead of a1 the result is too great by the same fraction of itself4 ,thus1 if we divide by #bj instead of #ag1 the result is _/ of itself too small2 and if we divide by #cfj instead of #cfe1 the result is too great by _/1 or _/ of itself4 ,if1 then1 we wish to find the interest of a sum of money for a portion of a year1 and have not the assistance of tables1 it will be found convenient to suppose the year to contain only #cfj days1 in which case its #gc;d part 7the #gb;d part will generally do7 must be subtracted from the result1 to make the alteration of #cfj into #cfe4 ,the number #cfj has so large a number of divisors1 that the rule of ,practice 7#bcj7 may always be readily applied4 ,thus1 it is required to find the portion which belongs to #bgd days1 the yearly interest being #ah4 #i4 #aj1 or #ahdia4 #bgd #ahdia ------ #ahj is of #cfj #ibdf --- #id #ij is of #ahj #dfbc -- #d is _/ of #cfj #bje ------ #i7adjgd ------ #h7aefd ----- #aif #achgh #ac 4 #ag 4 #g ,answer4 ,but if the nearest farthing be wanted1 the best way is to take #b-tenths of the number of days as a multiplier1 and #gc as a divisor2 since m #cfe is #bm #gcj1 or 7_/7m #gc4 ,thus1 in the preceding instance1 we multiply by #edh and divide by #gc2 and #edh #ahdia #ajaccjfh1 which divided by #gc gives #achha1 very nearly agreeing with the former1 and giving #ac4 #ag4 #g1 which is certainly within a farthing of the truth4 #bed4 ,suppose it required to divide #ajj among three persons in such a way that their shares may be as #f1 #e1 and #i2 that is1 so that for every #f which the first has1 the second may have #e1 and the third #i4 ,it is plain that if we divide the #ajj into #f #e #i1 or #bj parts1 the first must have #f of those parts1 the second #e1 and the third #i4 ,therefore 7#bde7 their shares are respectively1 #ajj #f #ajj #e #ajj #i -------1 ------- and -------1 or #cj1 #be1 and #de4 #bj #bj #bj ,,exercises4 ,divide #cid4 #ab among four persons1 so that their shares may be as #a1 #f1 #g1 and #ah4--;,answer1 #ab4 #f4 #g2 #gc4 #ai4 #i2 #hf4 #f4 #d2 #bba4 #ai4 #c4 ,divide #bj among #f persons1 so that the share of each may be as much as those of all who come before put together4--,answer1 ,the first two have #abs4 #f;d42 the third #a4 #e2 the fourth #b4 #aj2 the fifth #e2 and the sixth #aj4 #bee4 ,when two or more persons employ their money together1 and gain or lose a certain sum1 it is evidently not fair that the gain or loss should be equally divided among them all1 unless each contributed the same sum4 ,suppose1 for example1 ,a contributes twice as much as ,b1 and they gain #ae1 ,a ought to gain twice as much as ,b2 that is1 if the whole gain be divided into #c parts1 ,a ought to have two of them and ,b one1 or ,a should gain #aj and ,b #e4 ,suppose that ,a1 ,b1 and ,c engage in an adventure1 in which ,a embarks #bej1 ,b #acj1 and ,c #de4 ,they gain #ajjj4 ,how much of it ought each to have8 ,each one ought to gain as much for #a as the others4 ,now1 since there are #bej #acj #de1 or #dbe pounds embarked1 which gain #ajjj1 for each pound there is a gain of _/4 ,therefore ,a should gain #ajjj _/ pounds1 ,b should gain #ajjj _/ pounds1 and ,c #ajjj _/ pounds4 ,on these principles1 by the process in 7#bde71 the following questions may be answered4 ,a ship is to be insured1 in which ,a has ventured #aibh1 and ,b #difc4 ,the expense of insurance is #dgd4 #aj4 #b4 ,how much ought each to pay of it8 ,answer1 ,a must pay #acb4 #ae4 7#b74 ,a loss of #adi is to be made good by three persons1 ,a1 ,b1 and ,c4 ,had there been a gain1 ,a would have gained #d times as much as ,b1 and ,c as much as ,a and ,b together4 ,how much of the loss must each bear8 ,answer1 ,a pays #ei4 #ab1 ,b #ad4 #ah1 and ,c #gd4 #aj4 #bef4 ,it may happen that several individuals employ several sums of money together for different times4 ,in such a case1 unless there be a special agreement to the contrary1 it is right that the more time a sum is employed1 the more profit should be made upon it4 ,if1 for example1 ,a and ,b employ the same sum for the same purpose1 but ,a's money is employed twice as long as ,b's1 ,a ought to gain twice as much as ,b4 ,the principle is1 that one pound employed for one month1 or one year1 ought to give the same return to each4 ,suppose1 for example1 that ,a employs #c for #f months1 ,b #d for #g months1 and ,c #ab for #b months1 and the gain is #ajj2 how much ought each to have of it8 ,now1 since ,a employs #c for six months1 he must gain #f times as much as if he employed it one month only2 that is1 as much as if he employed #f #c1 or #ah1 for one month2 also1 ,b gains as much as if he had employed #d #g for one month2 and ,c as if he had employed #ab #b for one month4 ,if1 then1 we divide #ajj into #f #c #d #g #ab #b1 or #gj parts1 ,a must have #f #c1 or #ah1 ,b must have #d #g1 or #bh1 and ,c #ab #b1 or #bd of those parts4 ,the shares of the three are1 therefore1 #f #c #ajj #d #g #ajj ----------------------1 ----------------------1 #f #c #d #g #ab #b #f #c #d #g #ab #b #ab #b #ajj and ----------------------4 #f #c #d #g #ab #b ,,exercises4 ,a1 ,b1 and ,c embark in an undertaking2 ,a placing #c4 #f for #b years1 ,b #ajj for #a year1 and ,c #ab for #a years4 ,they gain #dbgf4 #g ,how much must each receive of the gain8 ,answer1 ,a #bbf4 #aj4 #d2 ,b #cdcb4 #a4 #c2 ,c #fag4 #ae4 #e4 ,a1 ,b1 and ,c rent a house together for #b years1 at #aej per annum4 ,a remains in it the whole time1 ,b #af months1 and ,c #d months1 during the occupancy of ,b4 ,how much must each pay of the rent8#ef ,answer1 ,a should pay #aij4 #ab4 #f2 ,b #ij4 #ab4 #f2 ,c #ah4 #ae4 #ef ,this question does not at first appear to fall under the rule4 ,a little thought will serve to shew that what probably will be the first idea of the proper method of solution is erroneous4 #beg4 ,these are the principal rules employed in the application of arithmetic to commerce4 ,there are others1 which1 as no one who understands the principles here laid down can fail to see1 are virtually contained in those which have been given4 ,such is what is commonly called the ,rule of ,exchange1 for such questions as the following3 ,if #bj shillings be worth #be francs1 in ,france1 what is #afj worth8 ,this may evidently be done by the ,rule of ,three4 ,the rules here given are those which are most useful in common life2 and the student who understands them need not fear that any ordinary question will be above his reach4 ,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 passed4 ,there is no such thing as a set of rules which are at once most convenient for a butcher and a banker's clerk1 a grocer and an actuary1 a farmer and a bill-broker2 but a person with a good knowledge of the principles laid down in this work1 will be able to examine and meet his own future wants1 or1 at worst1 to catch with readiness the manner in which those who have gone before him have done so for themselves4 ,,appendix ,,to ,,the ,,fifth ,,edition ,,of ,,de ,,morgan',s ,,elements ,,of ,,arithmetic4 ,i4 ,,on ,,the ,,mode ,,of ,,computing4 ,the rules in the preceding work are given in the usual form1 and the examples are worked in the usual manner4 ,but if the student really wish to become a ready computer1 he should strictly follow the methods laid down in this ,appendix2 and he may depend upon it that he will thereby save himself trouble in the end1 as well as acquire habits of quick and accurate calculation4 ,i4 ,in numeration learn to connect each primary decimal number1 #aj1 #ajj1 #ajjj1 c4 not with the place in which the unit falls1 but with the number of ciphers following4 ,call ten a one-cipher number1 a hundred a two-cipher number1 a million a six-cipher number1 and so on4 ,if five figures be cut off from a number1 those that are left are hundred-thousands2 for #ajj1jjj is a five-cipher number4 ,learn to connect tens1 hundreds1 thousands1 tens of thousands1 hundreds of thousands1 millions1 c4 with #a1 #b1 #c1 #d1 #e1 #f1 c4 in the mind4 ,what is a seventeen-cipher number8 ,for every #f in seventeen say million1 for the remaining #e say hundred-thousand3 the answer is a hundred thousand millions of millions4 ,if twelve places be cut off from the right of a number1 what does the remaining number stand for8--,answer1 ,as many millions of millions as there are units in it when standing by itself4 ,,ii4 ,after learning to count forwards and backwards with rapidity1 as in #a1 #b1 #c1 #d1 c4 or #cj1 #bi1 #bh1 #bg1 c41 learn to count forwards or backwards by twos1 threes1 c4 up to nines at least1 beginning from any number4 ,thus1 beginning from four and proceeding by sevens1 we have #d1 #aa1 #ah1 #be1 #cb1 c41 along which series you must learn to go as easily as along the series #a1 #b1 #c1 #d1 c42 that is1 as quick as you can pronounce the words4 ,the act of addition must be made in the mind without assistance3 you must not permit yourself to say1 #d and #g are #aa1 #aa and #g are #ah1 c42 but only #d1 #aa1 #ah1 c4 ,and it would be desirable1 though not so necessary1 that you should go back as readily as forward2 by sevens for instance1 from sixty1 as in #fj1 #ec1 #df1 #ci1 c4 ,,iii4 ,seeing a number and another both of one figure1 learn to catch instantly the number you must add to the smaller to get the greater4 ,seeing #c and #h1 learn by practice to think of #e without the necessity of saying #c from #h and there remains #e4 ,and if the second number be the less1 as #h and #c1 learn also by practice how to pass up from #h to the next number which ends with #c 7or #ac71 and to catch the necessary augmentation1 five1 without the necessity of formally undertaking in words to subtract #h from #ac4 ,take rows of numbers1 such as #d #b #f #j #e #j #a #h #f #d and practise this rule upon every figure and the next1 not permitting yourself in this simple case ever to name the higher one4 ,thus1 say #d and #h 7#d first1 #b second1 #d from the next number that ends with #b1 or #ab1 leaves #h71 #b and #d1 #f and #d1 #j and #e1 #e and #e1 #j and #a1 #a and #g1 #h and #h1 #f and #h4 ,,iv4 ,study the same exercise as the last one with two figures and one4 ,thus1 seeing #bg and #f1 pass from #bg up to the next number that ends with #f 7or #cf71 catch the #i through which you have to pass1 and allow yourself to repeat as much as 8#bg and #i are #cf40 ,thus1 the row of figures #aggbifchaji will give the following practice3 #ag and #j are #ag2 #gg and #e are #hb2 #gb and #g are #gi2 #bi and #g are #cf2 #if and #g are #ajc2 #fc and #e are #fh2 #ch and #c are #da2 #ha and #i are #ij2 #aj and #i are #ai4 ,v4 ,in a number of two figures1 practise writing down the units at the moment that you are keeping the attention fixed upon the tens4 ,in the preceding exercise1 for instance1 write down the results1 repeating the tens with emphasis at the instant of writing down the units4 ,,vi4 ,learn the multiplication table so well as to name the product the instant the factors are seen2 that is1 until #h and #g1 or #g and #h1 suggest #ef at once1 without the necessity of saying 8#g times #h are #ef40 ,thus looking along a row of numbers1 as #cigjfedh1 learn to name the products of every successive pair of digits as fast as you can repeat them1 namely1 #bg1 #fc1 #j1 #j1 #cj1 #bj1 #cb4 ,,vii4 ,having thoroughly mastered the last exercise1 learn further1 on seeing three numbers1 to augment the product of the first and second by the third without any repetition of words4 ,practise until #c1 #h1 #d1 for instance1 suggest #c times #h and #d1 or #bh1 without the necessity of saying 8#c times #h are #bd1 and #d is #bh40 ,thus1 #agibcfdjh will suggest the following practice1 #af1 #fe1 #ba1 #ab1 #bb1 #bd1 #h4 ,,viii4 ,now1 carry the last still further1 as follows3 ,seeing four figures1 as #b1 #g1 #f1 #i1 catch up the product of the first and second1 increased by the third1 as in the last1 without a helping word2 name the result1 and add the next figure1 name the whole result1 laying emphasis upon the tens4 ,thus1 #b1 #g1 #f1 #i1 must immediately suggest 8#bj and #i are #bi40 ,the row of figures #ggcfihigd will give the instances #eb and #f are #eh2 #bg and #i are #cf2 #bg and #h are #ce2 #fb and #i are #ga2 #ha and #g are #hh2 #gi and #d are #hc4 ,,ix4 ,having four numbers1 as #b1 #d1 #g1 #i1 vary the last exercise as follows3 ,catch the product of the first and second1 increased by the third2 but instead of adding the fourth1 go up to the next number that ends with the fourth1 as in exercise ,,iv4 ,thus1 #b1 #d1 #g1 #i1 are to suggest 8#ae and #d are #ai40 ,and the row of figures #agbcifhibi will afford the instances #i and #d are #ac2 #ag and #b are #ai2 #ae and #a are #af2 #cc and #e are #ch2 #fb and #g are #fi2 #eg and #e are #fb2 #gd and #e are #gi4 ,x4 ,learn to find rapidly the number of times a digit is contained in given units and tens1 with the remainder4 ,thus1 seeing #h and #ec1 arrive at and repeat 8#f and #e over40 ,common short division is the best practice4 ,thus1 in dividing #bcfdajgib by #g1 #g7bcfdajgib --------- #ccggbigj1 remainder #b4 ,all that is repeated should be #c and #b2 #c and #e2 #g and #e2 #g and #b2 #b and #f2 #i and #d2 #g and #j2 #j and #b4 ,in performing the several rules1 proceed as follows3 ,,addition4 ,not one word more than repeating the numbers written in the following process3 the accented figure is the one to be written down2 the doubly accented figure is carried 7and don't say 8carry #c10 but do it74 #dgifc #aeih #bfcaf #edgib #hai #ffhf ------ #achagd #f1 #ae1 #ag1 #bc1 #ca1 #c #d2 #aa1 #ab1 #ba1 #bb1 #ca1 #cg2 #i1 #ag1 #bd1 #bg1 #cb1 #da2 #aj1 #ad1 #bj1 #ba1 #bh2 #g1 #i1 #ac4 ,in verifying additions1 instead of the usual way of omitting one line1 adding without it1 and then adding the line omitted1 verify each column by adding it both upwards and downwards4 ,,subtraction4 ,the following process is enough4 ,the carriages1 being always of one1 need not be mentioned4 ,from #gidcfbehaij ,take #ehfdeifbgch ----------- #bjgijbiedeb #h and #b1 #d and #e1 #g and #d1 #c and #e1 #f and #i1 #aj and #b1 #f and #j1 #d and #i1 #g and #g1 #i and #j1 #e and #b4 ,it is useless to stop and say1 #h and #b make #aj2 for as soon as the #b is obtained1 there is no occasion to remember what it came from4 ,,multiplication4 ,the following1 put into words1 is all that need be repeated in the multiplying part2 the addition is then done as usual4 ,the unaccented figures are carried4 #fgjchc #ihgf ------- #djbbbih #ah1 #di1 #bb1 #b1 #db1 #dj1 #dfibfha #ba1 #eh1 #bf1 #b1 #di1 #df1 #ecfcjfd #bd1 #ff1 #cj1 #c1 #ef1 #ec1 #fjccddg #bg1 #gd1 #cd1 #c1 #fc1 #fj4 ---------- #ffbjgjbejh ,verify each line of the multiplication and the final result by casting out the nines4 7,appendix ,,ii4 p4 #aff47 ,it would be almost as easy1 for a person who has well practised the #hth exercise1 to add each line to the one before in the process1 thus3 #fgjchc #ihgf ------- #djbbbih #ejidiajh #ehgbeeejh #ffbjgjbejh #h2 #ba and #i are #cj2 #ei and #b are #fa2 #bg and #b are #bi2 #b and #b are #d2 #di and #j are #di2 #df and #d are #ej4 ,on the right is all the process of forming the second line1 which completes the multiplication by #gf1 as the third line completes that by #hgf1 and the fourth line that by #ihgf4 ,,division4 ,make each multiplication and the following subtraction in one step1 by help of the process in the #ith exercise1 as follows3 #bgfic7ddaigbhjiffb7aeieigcj #afejdb #bfeggh #afedaj #bfidei #bjbbbf #hcgef #fggb ,the number of words by which #bfegg is obtained from #afedjb 7the multiplier being #e7 is as follows3 #ae and #g are #bb2 #dg and #g are #ed2 #ce and #e are #dj2 #ci and #f are #de2 #ad and #b are #af4 ,the processes for extracting the square root1 and for the solution of equations 7,appendix ,,xi471 should be abbreviated in the same manner as the division4#eg #eg ,the teacher will find further remarks on this subject in the ,companion to the ,almanac for #ahdd1 and in the ,supplement to the ,penny ,cyclopdia1 article ,computation4 ,,appendix ,,ii4 ,,on ,,verification ,,by ,,casting ,,out ,,nines ,,and ,,elevens4 ,the process of casting out the nines1 as it is called1 is one which the young computer should learn and practise1 as a check upon his computations4 ,it is not a complete check1 since if one figure were made too small1 and another as much too great1 it would not detect this double error2 but as it is very unlikely that such a double error should take place1 the check furnishes a strong presumption of accuracy4 ,the proposition upon which this method depends is the following3 ,if a1 b1 c1 d be four numbers1 such that a bc d1 and if m be any other number whatsoever1 and if a1 b1 c1 d1 severally divided by m1 give the remainders p1 q1 r1 s1 then p and qr s give the same remainder when divided by m 7and perhaps are themselves equal74 ,for instance1 #ccd #ag #ai #aa2 divide these four numbers by #g1 the remainders are #e1 #c1 #e1 and #d4 ,and #e and #e #c #d1 or #e and #ai1 both leave the remainder #e when divided by #g4 ,any number1 therefore1 being used as a divisor1 may be made a check upon the correctness of an operation4 ,to provide a check which may be most fit for use1 we must take a divisor the remainder to which is most easily found4 ,the most convenient divisors are #c1 #i1 and #aa1 of which #i is far the most useful4 ,as to the numbers #c and #i1 the remainder is always the same as that of the sum of the digits4 ,for instance1 required the remainder of #bdfabjcgg divided by #i4 ,the sum of the digits is #b #d #f #a #b #j #c #g #g1 or #cb1 which gives the remainder #e4 ,but the easiest way of proceeding is by throwing out nines as fast as they arise in the sum4 ,thus1 repeat #b1 #f 7#b #d71 #ab 7#f #f71 say #c 7throwing out #i71 #d1 #f1 #i 7throw this away71 #g1 #ad1 7or throwing out the #i7 #e4 ,this is the remainder required1 as would appear by dividing #bdfabjcgg by #i4 ,a proof may be given thus3 ,it is obvious that each of the numbers1 #a1 #aj1 #ajj1 #ajjj1 c4 divided by #i1 leaves a remainder #a1 since they are #a1 #i #a1 #ii #a1 c4 ,consequently1 #b1 #bj1 #bjj1 c4 leave the remainder #b2 #c1 #cj1 #cjj1 the remainder #c2 and so on4 ,if1 then1 we divide1 say #agfd by #i in parcels1 #ajjj will be one more than an exact number of nines1 #gjj will be seven more1 and #fj will be six more4 ,so1 then1 from #a1 #g1 #f1 #d1 put together1 and the nines taken out1 comes the only remainder which can come from #agfd4 ,to apply this process to a multiplication3 ,it is asserted1 in page #cb1 that #ajjjdefi #cafc #cafdddeagdg4 ,in casting out the nines from the first1 all that is necessary to repeat is1 one1 five1 ten1 one1 seven2 in the second1 three1 four1 ten1 one1 four2 in the third1 three1 four1 ten1 one1 five1 nine1 four1 nine1 eight1 twelve1 three1 ten1 one4 ,the remainders then are1 #g1 #d1 #a4 ,now1 #g #d is #bh1 which1 casting out the nines1 gives #a1 the same as the product4 ,again1 in page #dc1 it is asserted that #bcgifdhd #acjjjj #ahc #fdhd4 ,cast out the nines from #acjjj1 #ahc1 #fdhd1 and we have #d1 #c1 and #d4 ,now1 #d #c #d1 with the nines cast out1 gives #g2 and so does #bcgifdhd4 ,to avoid having to remember the result of one side of the equation1 or to write it down1 in order to confront it with the result of the other side1 proceed as follows3 ,having got the remainder of the more complicated side1 into which two or more numbers enter1 subtract it from #i1 and carry the remainder into the simple side1 in which there is only one number4 ,then the remainder of that side ought to be #j4 ,thus1 having got #g from the left-hand of the preceding1 take #b1 the rest of #i1 forget #g1 and carry in #b as a beginning to the left-hand side1 giving #b1 #d1 #g1 #ad1 #e1 #aa1 #b1 #f1 #ad1 #e1 #i1 #j4 ,practice will enable the student to cast out nines with great rapidity4 ,this process of casting out the nines does not detect any errors in which the remainder to #i happens to be correct4 ,if a process be tedious1 and some additional check be desirable1 the method of casting out elevens may be followed after that of casting out the nines4 ,observe that #aj #a1 #ajj-a1 #ajjj #a1 #ajjjj-a1 c4 are all divisible by eleven4 ,from this the following rule for the remainder of division by #aa may be deduced1 and readily used by those who know the algebraical process of subtraction4 ,for those who have not got so far1 it may be doubted whether the rule can be made easier than the actual division by #aa4 ,subtract the first figure from the second1 the result from the third1 the result from the fourth1 and so on4 ,the final result1 or the rest of #aa if the figure be negative1 is the remainder required4 ,thus1 to divide #afdbiae by #aa1 and find the remainder1 we have #a from #f1 #e2 #e from #d1-a2-a from #b1 #c2 #c from #i1 #f2 #f from #a1-e2-e from #e1 #aj2 and #aj is the remainder4 ,but #afd gives-#a1 and #aj is the remainder2 #afdbia gives-#e1 and #f is the remainder4 ,with very little practice these remainders may be read as rapidly as the number itself4 ,thus1 for #abgfaihccdbd need only be repeated1 #a1 #f1 #j1 #a1 #h1 #j1 #c1 #j1 #d1-b1 #f1 and #f is the remainder4 ,when a question has been tried both by nines and elevens1 there can be no error unless it be one which makes the result wrong by a number of times #ii exactly4 ,,appendix ,,iii4 ,,on ,,scales ,,of ,,notation4 ,we are so well accustomed to #aj1 #ajj1 c41 as standing for ten1 ten tens1 c41 that we are not apt to remember that there is no reason why #aj might not stand for five1 #ajj for five fives1 c41 or for twelve1 twelve twelves1 c4 ,because we invent different columns of numbers1 and let units in the different columns stand for collections of the units in the preceding columns1 we are not therefore bound to allow of no collections except in tens4 ,if #aj stood for #b1 that is1 if every column had its unit double of the unit in the column on the right1 what we now represent by #a1 #b1 #c1 #d1 #e1 #f1 c41 would be represented by #a1 #aj1 #aa1 #ajj1 #aja1 #aaj1 #aaa1 #ajjj1 #ajja1 #ajaj1 #ajaa1 #aajj1 c4 ,this is the binary scale4 ,if we take the ternary scale1 in which #aj stands for #c1 we have #a1 #b1 #aj1 #aa1 #ab1 #bj1 #ba1 #bb1 #ajj1 #aja1 #ajb1 #aaj1 c4 ,in the quinary scale1 in which #aj is five1 #bcd stands for #b twenty-fives1 #c fives1 and #d1 or sixty-nine4 ,if we take the duodenary scale1 in which #aj is twelve1 we must invent new symbols for ten and eleven1 because #aj and #aa now stand for twelve and thirteen2 use the letters t and e4 ,then #agf means #a twelve-twelves1 #g twelves1 and #f1 or two hundred and thirty-four2 and #ate means two hundred and seventy-five4 ,the number which #aj stands for is called the radix of the scale of notation4 ,to change a number from one scale into another1 divide the number1 written as in the first scale1 by the number which is to be the radix of the new scale2 repeat this division again and again1 and the remainders are the digits required4 ,for example1 what1 in the quinary scale1 is that number which1 in the decimal scale1 is #agjcf8 #e7agjcf ----- #e7cdjg ,rem4 #a ---- #e7fha #b --- #e7acf #a --- #e7bg #a -- #e7e #b - #e7a #j - #j #a ,answer #ajbaaba ,quinary4 ,decimal4 ,verification1 #ajjjjjj means #aefbe #bjjjj #abej #ajjj #abe #ajj #be #bj #aj #a #a ------ ----- #ajbaaba #agjcf ,the reason of this rule is easy4 ,our process of division is nothing but telling off #agjcf into #cdjg fives and #a over2 we then find #cdjg fives to be #fha fives of fives and #b fives over4 ,next we form #fha fives of fives into #acf fives of fives of fives and #a five of fives over2 and so on4 ,it is a useful exercise to multiply and divide numbers represented in other scales of notation than the common or decimal one4 ,the rules are in all respects the same for all systems1 the number carried being always the radix of the system4 ,thus1 in the quinary system we carry fives instead of tens4 ,i now give an example of multiplication and division3 ,quinary4 ,decimal4 #dbadc means #bgih #abcd #aid ------ ----- #cbdbcb #aaaib #bcbjcd #beahb #acdcda #bgih #dbadc --------- ------ #aadccbbbb #edbhab ,duodecimal4 ,decimal4 #dt#i7gft#d;e#jh7affhg #gje7bbfajgdd7cbjga #dt#i #adfj ----- #ejgd #bhad #acid #bedf #fhi ----- #bhte #bedf ------ #cfej #ccbj ----- #ccjh #bt#cc ------ #die ,another way of turning a number from one scale into another is as follows3 ,multiply the first digit by the old radix in the new scale1 and add the next digit2 multiply the result again by the old radix in the new scale1 and take in the next digit1 and so on to the end1 always using the radix of the scale you want to leave1 and the notation of the scale you want to end in4 ,thus1 suppose it required to turn #affhg 7duodecimal7 into the decimal scale1 and #afdcb 7septenary7 into the quaternary scale3 #affhg #afdcb ,duodecimals into ,decimals4 ,septenaries into ,quaternaries4 #a #ab #f #ah #a #g #f #ca #ab #f #g #d --- ---- #bbb #aacc #ab #h #g #c ---- ----- #bfgb #bbacj #ab #g #g #b ------ ------- ,answer #cbjga #ajbajab ,owing to our division of a foot into #ab equal parts1 the duodecimal scale often becomes very convenient4 ,let the square foot be also divided into #ab parts1 each part is #ab square inches1 and the #abth of the #abth is one square inch4 ,suppose1 now1 that the two sides of an oblong piece of ground are #agf feet #i inches #g-abths of an inch1 and #fe feet #aa inches #e-abths of an inch4 ,using the duodecimal scale1 and duodecimal fractions1 these numbers are #abhig and #ee;e#e4 ,their product1 the number of square feet required1 is thus found3 #abhig #ee;e#e --------- #fag;ee #aafjie #fag;ee #fag;ee ------------ #fh;e#hadd;e ,answer1 #fh;e#hadd;e 7duod47 square feet1 or #aaffj square feet #af square inches _/ and _/ of a square inch4 ,it would1 however1 be exact enough to allow #b-;hundredths of a foot for every quarter of an inch1 an additional hundredth for every #c inches1#eh and #a-;hundredth more if there be a #abth or #b-abths above the quarter of an inch4 ,thus1 #i_/ inches should be #gf #jc #ja1 or #hj1 and #aa_/ would be #ie2 and the preceding might then be found decimally as #agfh #feie as #aafeiif square feet1 near enough for every practical purpose4 #eh ,and at discretion one hundredth more for a large fraction of three inches4 ,,appendix ,,iv4 ,,on ,,the ,,definition ,,of ,,fractions4 ,the definition of a fraction given in the text shews that _/1 for instance1 is the ninth part of seven1 which is shewn to be the same thing as seven-ninths of a unit4 ,but there are various modes of speech under which a fraction may be signified1 all of which are more or less in use4 #a4 ,in _/ we have the #ith part of #g4 #b4 #g-iths of a unit4 #c4 ,the fraction which #g is of #i4 #d4 ,the times and parts of a time 7in this case part of a time only7 which #g contains #i4 #e4 ,the multiplier which turns nines into sevens4 #f4 ,the ratio of #g to #i1 or the proportion of #g to #i4 #g4 ,the multiplier which alters a number in the ratio of #i to #g4 #h4 ,the #dth proportional to #i1 #a1 and #g4 ,the first two views are in the text4 ,the third is deduced thus3 ,if we divide #i into #i equal parts1 each is #a1 and #g of the parts are #g2 consequently the fraction which #g is of #i is _/4 ,the fourth view follows immediately3 ,for a time is only a word used to express one of the repetitions which take place in multiplication1 and we allow ourselves1 by an easy extension of language1 to speak of a portion of a number as being that number taken a part of a time4 ,the fifth view is nothing more than a change of words3 ,a number reduced to _/ of its amount has every #i converted into a #g1 and any fraction of a #i which may remain over into the corresponding fraction of #g4 ,this is completely proved when we prove the equation _/ of a #g times a_/#i4 ,the sixth1 seventh1 and eighth views are illustrated in the chapter on proportion4 ,when the student comes to algebra1 he will find that1 in all the applications of that science1 fractions such as a_/b most frequently require that a and b should be themselves supposed to be fractions4 ,it is1 therefore1 of importance that he should learn to accommodate his views of a fraction to this more complicated case4 #b ,suppose we take -----4 #d_/ ,we shall find that we have1 in this case1 a better idea of the views from and after the third inclusive1 than of the first and second1 which are certainly the most simple ways of conceiving _/4 ,we have no notion of the 7#d_/7th part of #b1 #a 7 #c 7 nor of #b ---7#d---7ths #b 7 #e 7 of a unit2 indeed1 we coin a new species of adjective when we talk of the 7#d_/7th part of anything4 ,but we can readily imagine that #b is some fraction of #d_/2 that the first is some part of a time the second2 that there must be some multiplier which turns every #d_/ in a number into #b2 and so on4 ,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 above4 ,we can easily imagine a fourth part of a length1 and a fifth part1 meaning the lines of which #d and #e make up the length in question2 and there is also in existence a length of which four lengths and two-fifths of a length make up the original length in question4 ,for instance1 we might say that #f1 #f1 #b is a division of #ad into #b equal parts--#b equal parts1 #f1 #f1 and a third of a part1 #b4 ,so we might agree to say1 that the 7#b7th1 or 7#b7rd1 or 7#b7st 7the reader may coin the adjective as he pleases7 part of #ad is #f4 ,if we divide the line ,a ,b into eleven equal parts in ,c1 ,d1 ,e1 c41 we must then say that ,a ,c is the #aath part1 ,illustration ,a ,d the 7#e7th1 ,a ,e the 7#c7th1 ,a ,f the 7#b7th1 ,a ,g the 7#b7th1 ,a ,h the 7#a7th1 ,a ,i the 7#a_/7th1 ,a ,k the 7#a7th1 ,a ,l the 7#a_/7th1 ,a ,m the 7#a7th1 and ,a ,b itself the #ast part of ,a ,b4 ,the reader may refuse the language if he likes 7though it is not so much in defiance of etymology as talking of multiplying by 72 but when ,a ,b is called #a1 he must either call ,a ,f #a_/7b71 or make one definition of one class of fractions and another of another4 ,whatever abbreviations they may choose1 all persons will agree that a_/b is a direction to find such a fraction as1 repeated b times1 will give #a1 and then to take that fraction a times4 ,so1 to get #b_/d1 the simplest way is to divide the whole unit into #df parts2 #aj of these parts1 repeated #d times1 give the whole4 ,the ,illustration #dth is then _/1 and #b such parts is _/1 or ,a ,c4 ,the student should try several examples of this mode of interpreting complex fractions4 ,but what are we to say when the denominator itself is less than unity1 as in #c_/8 ,are we to have a 77th part of a unit8 and what is it8 ,had there been a #e in the denominator1 we should have taken the part of which #e will make a unit4 ,as there is in the denominator1 we must take the part of which will be a unit4 ,that part is larger than a unit2 it is #b units2 #b is that of which is #a4 ,the above fraction then directs us to repeat #b units #c times4 ,by extending our word #8multiplication' to the taking of a part of a time1 all multiplications are also divisions1 and all divisions multiplications1 and all the terms connected with either are subject to be applied to the results of the other4 ,if #b yards cost #c shillings1 how much does one yard cost8 ,in such a case as this1 the student looks at a more simple question4 ,if #e yards cost #aj shillings1 he sees that each yard costs _/1 or #b shillings1 and1 concluding that the same process will give the true result when the data are fractional1 he forms #c_/b1 reduces it by rules to _/ or #a1 and concludes that #a yard costs #ah pence4 ,the answer happens to be correct2 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 demonstration4 ,in the above question we want money which1 repeated #b times1 shall give #c shillings4 ,if we divide the shilling into #ad equal parts1 #f of these parts repeated #b times give the shilling4 ,to get #c times as much by the same repetition1 we must take #c of these #f parts at each step1 or #ba parts4 ,hence1 _/1 or #a1 is the number of shillings in the price4 ,,appendix ,v4 ,,on ,,characteristics4 ,when the student comes to use logarithms1 he will find what follows very useful4 ,in the mean while1 ,i give it merely as furnishing a rapid rule for finding the place of a decimal point in the quotient before the division is commenced4 ,when a bar is written over a number1 thus1 #g let the number be called negative1 and let it be thus used3 ,let it be augmented by additions of its own species1 and diminished by subtractions2 thus1 #g and #b give #i1 and let #g with #b subtracted give #e4 ,but let the addition of a number without the bar diminish the negative number1 and the subtraction increase it4 ,thus1 #g and #d are #c1 #g and #ab make #e1 #g with #h subtracted is #ae4 ,in fact1 consider #a1 #b1 #c1 c41 as if they were gains1 and #a1 #b1 #c1 as if they were losses3 let the addition of a gain or the removal of a loss be equivalent things1 and also the removal of a gain and the addition of a loss4 ,thus1 when we say that #d diminished by #aa gives #g1 we say that a loss of #d incurred at the moment when a loss of #aa is removed1 is1 on the whole1 equivalent to a gain of #g2 and saying that #d diminished by #b is #f1 we say that a loss of #d1 accompanied by the removal of a gain of #b1 is altogether a loss of #f4 ,by the characteristic of a number understand as follows3 ,when there are places before the decimal point1 it is one less than the number of such places4 ,thus1 #cbad1 #ajjhc1 #h 7which is #hjj 4447 #iiii1 all have #j for their characteristics4 ,but #agcb1 #dh1 #icaaf1 all have #a2 #abfjc and #abf have #b2 #aaicgbfdfff has #g4 ,but when there are no places before the decimal point1 look at the first decimal place which is significant1 and make the characteristic negative accordingly4 ,thus1 #fab1 #aba1 #ijjd1 in all of which significance begins in the first decimal place1 have the characteristic #a2 but #jah and #jii have #b2 #jjjag has #d2 #jjjjjjjja has #i4 ,to find the characteristic of a quotient1 subtract the characteristic of the divisor from that of the dividend1 carrying one before subtraction if the first significant figures of the divisor are greater than those of the dividend4 ,for instance1 in dividing #adfjh by #jjbgi4 ,the characteristics are #b and #c2 and #b with #c removed would be #e4 ,but on looking1 we see that the first significant figures of the divisor1 #bg1 taken by themselves1 and without reference to their local value1 mean a larger number than #ad1 the first two figures of the dividend4 ,consequently1 to #c we carry #a before subtracting1 and it then becomes #b1 which1 taken from #b1 gives #d4 ,and this #d is the characteristic of the quotient1 so that the quotient has #e places before the decimal point4 ,or1 if abcdef be the first figures of the quotient1 the decimal point must be thus placed1 abcdef4 ,but if it had been to divide #jjbgi by #adfjh1 no carriage would have been required2 and #c diminished by #b is #e2 that is1 the first significant figure of the quotient is in the #eth place4 ,the quotient1 then1 has #jjjj before any significant figure4 ,a few applications of this rule will make it easy to do it in the head1 and thus to assign the meaning of the first figure of the quotient even before it is found4 ,,appendix ,,vi4 ,,on ,,decimal ,,money4 ,of all the simplifications of commercial arithmetic1 none is comparable to that of expressing shillings1 pence1 and farthings as decimals of a pound4 ,the rules are thereby put almost upon as good a footing as if the country possessed the advantage of a real decimal coinage4 ,any fraction of a pound sterling may be decimalised by rules which can be made to give the result at once4 ,two shillings is #ajj ,one shilling is #jej ,sixpence is #jbe ,one farthing is #jja #jd ,thus1 every pair of shillings is a unit in the first decimal place2 an odd shilling is a #ej in the second and third places2 a farthing is so nearly the thousandth part of a pound1 that to say one farthing is #jja1 two farthings is #jjb1 c41 is so near the truth that it makes no error in the first three decimals till we arrive at sixpence1 and then #bd farthings is exactly #jbe or #be thousandths4 ,but #be farthings is #jbf1 #bf farthings is #jbg1 c4 ,hence the rule for the first three places is ,one in the first for every pair of shillings2 #ej in the second and third for the odd shilling1 if any2 and #a for every farthing additional1 with #a extra for sixpence4 ,thus1 #js4 #c;d4 #jad #js4 #g;d4 #jcb #as4 #b;d4 #jfj #as4 #aa;d4 #jif #bs4 #f;d4 #abe #bs4 #i;d4 #aci #cs4 #b;d4 #afa #acs4 #aj;d4 #fid ,in the fourth and fifth places1 and those which follow1 it is obvious that we have no produce from any farthings except those above sixpence4 ,for at every sixpence1 #jjjjd is converted into #jja1 and this has been already accounted for4 ,consequently1 to fill up the fourth and fifth places1 ,take #d for every farthing#ei above the last sixpence1 and an additional #a for every six farthings1 or three halfpence4 #ei ,the student should remember all the multiples of #d up to #d #be1 or #ajj4 ,the remaining places arise altogether from #jjjjj for every farthing above the last three halfpence2 for at every three halfpence complete1 #jjjjj is converted into #jjjja1 and has been already accounted for4 ,consequently1 to fill up all the places after the fifth1 ,let the number of farthings above the last three halfpence be a numerator1 #f a denominator1 and annex the figures of the corresponding decimal fraction4 ,it may be easily remembered that ,the figures of _/ are #afffff444 0 _/ 444 #cccccc444 0 _/ 444 #e 0 _/ 444 #ffffff444 0 _/ 444 #hccccc444 #js4 #c;d4 #jadehcccc444 #js4 #g;d4 #jcbbiafff444 #as4 #b;d4 #jfjdaffff444 #as4 #aa;d4 #jifhge #bs4 #f;d4 #abejjjjjj444 #bs4 #i;d4 #aciehcccc444 #cs4 #b;d4 #afadehcccc444 #acs4 #aj;d4 #fidgiafff444 ,the following examples will shew the use of this rule1 if the student will also work them in the common way4 ,to turn pounds1 c41 into farthings3 ,multiply the pounds by #ifj1 or by #ajjj-dj1 or by #ajjj7a-_/72 that is1 from #ajjj times the pounds subtract #d per cent of itself4 ,thus1 required the number of farthings in #affc4 #aa4 #i4 #affceijfbe #ajjj #affceijfbe #d per cent of this1 #ffedcfbe ----------- ,no4 of farthings required1 #aeigjdg ,what is #dg per cent of #aff4 #ac4 #aj and #fadh of #biga4 #af4 #i8 #afffia -------- #dj p4 c4 #fffgfd #e p4 c4 #hccdf #b p4 c4 #dafgc -------- #giaghc #gi4c4f #bigahcg --------- #f #aghcajbb #ja #bigahd #jjd #aahhgc #jjjh #bcgge --------- #ahbgjhed #ahbg4a4h ,the inverse rule for turning the decimal of a pound into shillings1 pence1 and farthings1 is obviously as follows3 ,a pair of shillings for every unit in the first place2 an odd shilling for #ej 7if there be #ej7 in the second and third places2 and a farthing for every thousandth left1 after abating #a if the number of thousandths so left exceed #bd4 ,the direct rule 7with three places7 gives too little1 the inverse rule too much1 except at the end of a sixpence1 when both are accurate4 ,thus1 #ahc is rather less than #cs4 #h;d41 and #fs4 #d;d4 is rather greater than #cai2 or when the two do not exactly agree1 the common money is the greatest4 ,but #abe and #ce are exactly #bs4 #f;d4 and #gs4 ,required the price of #ag cwt4 #ha lb4 #ac oz4 at #c4aa4i per cwt4 true to the hundredth of a farthing4 #ceijfbe #ag --------- #fajdjfbe lb4 #ef #agiecac #af #eabidf #g #bbddad #b #jfdaah oz4 #h #jafjbi #d #jjhjae #a #jjbjjd #jjajjb --------- #fcffddff #fc4ac4c ,three men1 ,a1 ,b1 ,c1 severally invest #aia4ab4g1 #fa4ad4h1 and #abb4a4i in an adventure which yields #eaa4ab4f4 ,how ought the proceeds to be divided among them8 ,a1 #aiafcbbi ,b1 #fagcccc ,c1 #abbjhieh ,produce of #a4 --------- #cgedeebj7eaafbgjh7acfbfhf #acfagahh #bcececb #ajjhja #begaj #cahc #ahj #acfbfhf #acfbfhf #acfbfhf #ibbcfaia #ccccgaf #heihjbba --------- --------- --------- #acfbfhf #hagfab #acfbfhf #abbfdag #acfbg #bgbecg #acfbg #iech #bgbed #hagf #dji #ajij #dji #da #abb #bg #d #g #c -------- #a #a #hdabca --------- --------- #affcfig #bfaacdf #bfaacdf 444 ,a's share #bfa4b4h #hdabca 444 ,b's 444 #hd4b4e #affcfig 444 ,c's 444 #aff4g4d -------- ------------- #eaafbgd #eaa4b4f ,if ever the fraction of a farthing be wanted1 remember that the coinage-result is larger than the decimal of a pound1 when we use only three places4 ,from #ajjj times the decimal take #d per cent1 and we get the exact number of farthings1 and we need only look at the decimal then left to set the preceding right4 ,thus1 in #acdf #abca #cfig #ech #dib #adgi ----- ------ ------ #bb #ah #ia we see that 7if we use four decimals only7 the pence of the above results are nearly #h;d4 #bb of a farthing1 #e;d4 #ah1 and #d;d4 #ia4 ,a man can pay #bcgf4 #d4 #d1 his debts being #cbic4 #aa4 #j4 ,how much per cent can he pay1 and how much in the pound8 #cbiceec7bcgfbahj7gbadgef #gjgcji #dheih #aeffb #bdhh #ahc #ah ,answer1 #gb4 #b4aa per cent4 #j4ad4e per pound4 ,,appendix ,,vii4 ,,on ,,the ,,main ,,principle ,,of ,,book-,,keeping4 ,a brief notice of the principle on which accounts are kept 7when they are properly kept7 may perhaps be useful to students who are learning book-keeping1 as the treatises on that subject frequently give too little in the way of explanation4 ,any person who is engaged in business must desire to know accurately1 whenever an investigation of the state of his affairs is made4 #a1 ,what he had at the commencement of the account1 or immediately after the last investigation was made2 #b1 ,what he has gained and lost in the interval in all the several branches of his business2 #c1 ,what he is now worth4 ,from the first two of these things he obviously knows the third4 ,in the interval between two investigations1 he may at any one time desire to know how any one account stands4 ,an account is a recital of all that has happened1 in reference to any class of dealings1 since the last investigation4 ,it can only consist of receipts and expenditures1 and so it is said to have two sides1 a debtor and a creditor side4 ,all accounts are kept in money4 ,if goods be bought1 they are estimated by the money paid for them4 ,if a debtor give a bill of exchange1 being a promise to pay a certain sum at a certain time1 it is put down as worth that sum of money4 ,all the tools1 furniture1 horses1 c4 used in the business are rated at their value in money4 ,all the actual coin1 bank-notes1 c41 which are in or come in1 being the only money in the books which really is money1 is called cash4 ,the accounts are kept as if every different sort of account belonged to a separate person1 and had an interest of its own1 which every transaction either promotes or injures4 ,if the student find that it helps him1 he may imagine a clerk to every account3 one to take charge of1 and regulate1 the actual cash2 another for the bills which the house is to receive when due2 another for those which it is to pay when due2 another for the cloth 7if the concern deal in cloth72 another for the sugar 7if it deal in sugar72 one for every person who has an account with the house2 one for the profits and losses2 and so on4 ,all these clerks 7or accounts7 belonging to one merchant1 must account to him in the end--must either produce all they have taken in charge1 or relieve themselves by shewing to whom it went4 ,for all that they have received1 for every responsibility they have undertaken to the concern itself1 they are bound1 or are debtors2 for everything which has passed out of charge1 or about which they are relieved from answering to the concern1 they are unbound1 or are creditors4 ,these words must be taken in a very wide sense by any one to whom book-keeping is not to be a mystery4 ,thus1 whenever any account assumes responsibility to any parties out of the concern1 it must be creditor in the books1 and debtor whenever it discharges any other parties of their responsibility4 ,but whenever an account removes responsibility from any other account in the same books it is debtor1 and creditor whenever it imposes the same4 ,to whom are all these parties1 or accounts1 bound1 and from whom are they released8 ,undoubtedly the merchant himself1 or1 more properly1 the balance-clerk1 presently mentioned4 ,but it is customary to say that the accounts are debtors to each other1 and creditors by each other4 ,thus1 cash debtor to bills receivable1 means that the cash account 7or the clerk who keeps it7 is bound to answer for a sum which was paid on a bill of exchange due to the house4 ,at full length it would be3 8,mr4 ,c 7who keeps the cash-box7 has received1 and is answerable for1 this sum which has been paid in by ,mr4 ,a1 when he paid his bill of exchange40 ,on the other hand1 the corresponding entry in the account of bills receivable runs--bills receivable1 creditor by cash4 ,at full length3 8,mr4 ,b 7who keeps the bills receivable7 is freed from all responsibility for ,mr4 ,a's bill1 which he once held1 by handing over to ,mr4 ,c1 the cash-clerk1 the money with which ,mr4 ,a took it up40 ,bills receivable creditor by cash is intelligible1 but cash debtor to bills receivable is a misnomer4 ,the cash account is debtor to the merchant by the sum received for the bill1 and it should be cash debtor by bill receivable4 ,the fiction of debts1 not one of which is ever paid to the party to whom it is said to be owing1 though of no consequence in practice1 is a stumbling-block to the learner2 but he must keep the phrase1 and remember its true meaning4 ,the account which is made debtor1 or bound1 is said to be debited2 that which is made creditor1 or released1 is said to be credited4 ,all who receive must be debited2 all who give must be credited4 ,no cancel is ever made4 ,if cash received be afterwards repaid1 the sum paid is not struck off the receipts 7or debtor-side of the cash account71 but a discharge1 or credit1 is written on the expenditure 7or credit7 side4 ,the book in which the accounts are kept is called a ledger4 ,it has double columns1 or else the debtor-side is on one page1 and the creditor side on the opposite1 of each account4 ,the debtor-side is always the left4 ,other books are used1 but they are only to help in keeping the ledger correct4 ,thus there may be a waste-book1 in which all transactions are entered as they occur1 in common language2 a journal1 in which the transactions described in the waste-book are entered at stated periods1 in the language of the ledger4 ,the items entered in the journal have references to the pages of the ledger to which they are carried1 and the items in the ledger have also references to the pages of the journal from which they come2 and by this mode of reference it is easy to make a great deal of abbreviation in the ledger4 ,thus1 when it happens1 in making up the journal to a certain date1 that several different sums were paid or received at or near the same time1 the totals may be entered in the ledger1 and the cash account may be made debtor to1 or creditor by1 sundry accounts1 or sundries2 the sundry accounts being severally credited or debited for their shares of the whole4 ,the only book that need be explained is the ledger4 ,all the other books1 and the manner in which they are kept1 important as they may be1 have nothing to do with the main principle of the method4 ,let us1 then1 suppose that all the items are entered at once in the ledger as they arise4 ,it has appeared that every item is entered twice4 ,if ,a pay on account of ,b1 there is an entry1 8,a1 creditor by ,b20 and another1 8,b1 debtor to ,a40 ,this is what is called double-entry2 and the consequence of it is1 that the sum of all the debtor items in the whole book is equal to the sum of all the creditor items4 ,for what is the first set but the second with the items in a different order8 ,if it were convenient1 one entry of each sum might be made a double-entry4 ,the multiplication table is called a table of double-entry1 because #db1 for instance1 though it occurs only once1 appears in two different aspects1 namely1 as #f times #g and as #g times #f4 ,suppose1 for example1 that there are five accounts1 ,a1 ,b1 ,c1 ,d1 ,e1 and that each account has one transaction of its own with every other account2 and let the debits be in the columns1 the credits in the rows1 as follows3 ---------------------------------------- --- ,debtor ,a ,b ,c ,d ,e ---------------------------------------- --- ,a1 ,creditor #bc #ai #cb #d ------------------------------ ,b1 ,creditor #ag #f #aa #be ------------------------------ ,c1 ,creditor #i #da #aj #b ------------------------------ ,d1 ,creditor #ad #bh #af #c ------------------------------ ,e1 ,creditor #ae #d #fj #a ------------------------------ ,here the #af is supposed to appear in ,d's account as ,d creditor by ,c1 and in ,c's account as ,c debtor to ,d4 ,and to say that the sum of debtor items is the same as that of creditor items1 is merely to say that the preceding numbers give the same sum1 whether the rows or the columns be first added up4 ,if it be desired to close the ledger when it stands as above1 the following is the way the accounts will stand3 the lines in italics will presently be explained4 ,a1 ,debtor4 ,a1 ,creditor4 ,b1 ,debtor4 ,b1 ,creditor4 ,to ,b #ag ,by ,b #bc ,to ,a #bc ,by ,a #ag ,to ,c #i ,by ,c #ai ,to ,c #da ,by ,c #f ,to ,d #ad ,by ,d #cb ,to ,d #bh ,by ,d #aa ,to ,e #ae ,by ,e #d ,to ,e #d ,by ,e #be ,to ,balance #bc ,by ,balance #cg -- -- -- -- #gh #gh #if #if ---------------------------------------- ---------------------- ,c1 ,debtor4 ,c1 ,creditor4 ,d1 ,debtor4 ,d1 ,creditor4 ,to ,a #ai ,by ,a #i ,to ,a #cb ,by ,a #ad ,to ,b #f ,by ,b #da ,to ,b #aa ,by ,b #bh ,to ,d #af ,by ,d #aj ,to ,c #aj ,by ,c #af ,to ,e #fj ,by ,e #b ,to ,e #a ,by ,e #c ,by ,balance #ci ,to ,balance #g --- --- -- -- #aja #aja #fa #fa ---------------------------------------- ------------------------ ,e1 ,debtor4 ,e1 ,creditor4 ,balance1 ,debtor4 ,balance1 ,cred4 ,to ,a #d ,by ,a #ae ,to ,b #cg ,by ,a #bc ,to ,b #be ,by ,b #d ,to ,c #ci ,by ,d #g ,to ,c #b ,by ,c #fj ,by ,e #df -- -- ,to ,d #c ,by ,d #a #gf #gf ,to ,balance #df -- -- #hj #hj ,in all the part of the above which is printed in ,roman letters we see nothing but the preceding table repeated4 ,but when all the accounts have been completed1 and no more entries are left to be made1 there remains the last process1 which is termed balancing the ledger4 ,to get an idea of this1 suppose a new clerk1 who goes round all the accounts1 collecting debts and credits1 and taking them all upon himself1 that he alone may be entitled to claim the debts and to be responsible for the assets of the concern4 ,to this new clerk1 whom ,i will call the balance-clerk1 every account gives up what it has1 whether the same be debt or credit4 ,the cash-clerk gives up all the cash2 the clerks of the two kinds of bills give up all their documents1 whether bills receivable or entries of bills payable 7remember that any entry against which there is money set down in the books counts as money when given up1 that is1 as money due or money owing72 the clerks of the several accounts of goods give up all their unsold remainders at cost prices2 the clerks of the several personal accounts give up vouchers for the sums owing to or from the several parties2 and so on4 ,but where more has been paid out than received1 the balance-clerk adjusts these accounts by giving instead of receiving2 in fact1 he so acts as to make the debtor and creditor sides of the accounts he visits equal in amount4 ,for instance1 the ,a account is indebted to the concern #ee1 while payments or discharges to the amount of #gh have been made by it4 ,the balance-clerk accordingly hands over #bc to that account1 for which it becomes debtor1 while the balance enters itself as creditor to the same amount4 ,but in the ,b account there is #if of receipt1 and only #ei of payment or discharge4 ,the balance-clerk then receives #cg from this account1 which is therefore credited by balance1 while the balance acknowledges as much of debt4 ,the balance account must1 of course1 exactly balance itself1 if the accounts be all right2 for of all the equal and opposite entries of which the ledger consists1 so far as they do not balance one another1 one goes into one side of the balance account1 and the other into the other4 ,thus the balance account becomes a test of the accuracy of one part of the work3 if its two sides do not give the same sums1 either there have been entries which have not had their corresponding balancing entries correctly made1 or else there has been error in the additions4 ,but since the balance account must thus always give the same sum on both sides1 and since balance debtor implies what is favourable to the concern1 and balance creditor what is unfavourable1 does it not appear as if this system could only be applied to cases in which there is neither loss nor gain8 ,this brings us to the two accounts in which are entered all that the concern began with1 and all that it gains or loses--the stock account1 and the profit-and-loss account4 ,in order to make all that there was to begin with a matter of double entry1 the opening of the ledger supposes the merchant himself to put his several clerks in charge of their several departments4 ,in the stock account1 stock1 which here stands for the owner of the books1 is made creditor by all the property1 and debtor by all the liabilities2 while the several accounts are made debtors for all they take from the stock1 and creditors by all the responsibilities they undertake4 ,suppose1 for instance1 there are #ejj in cash at the commencement of the ledger4 ,there will then appear that the merchant has handed over to the cash-box #ejj1 and in the stock account will appear1 8,stock creditor by cash1 #ejj20 while in the cash account will appear1 8,cash debtor to stock1 #ejj40 ,suppose that at the beginning there is a debt outstanding of #ej to ,smith and ,co41 then there will appear in the stock account1 8,stock debtor to ,smith and ,co4 #ej10 and in ,smith and ,co4's account will appear 8,smith and ,co4 creditors by stock1 #ej40 ,thus there is double entry for all that the concern begins with by this contrivance of the stock account4 ,the account to which everything is placed for which an actual equivalent is not seen in the books is the profit-and-loss account4 ,this profit-and-loss account1 or the clerk who keeps it1 is made answerable for every loss1 and the supposed cause of every gain4 ,this account1 then1 becomes debtor for every loss1 and creditor by every gain4 ,if goods be damaged to the amount of #bj by accident1 and a loss to that amount occur in their sale1 say they cost #hj and sell for #fj cash1 it is clear that there is an entry 8,cash debtor to goods #fj10 and 8,goods creditor by cash #fj40 ,now1 there is an entry of #hj somewhere to the debit of the goods for cash laid out1 or bills given1 for the whole of the goods4 ,it would affect the accuracy of the accounts to take no notice of this2 for when the balance-clerk comes to adjust this account1 he would find he receives #bj less than he might have reckoned upon1 without any explanation of the reason2 and there would be a failure of the principle of double-entry4 ,since it is convenient that the balance account of the goods should merely represent the stock in hand at the close1 the account of goods therefore lays the responsibility of #bj upon the profit-and-loss account1 or there is the entry 8,goods creditor by profit-and-loss1 #bj10 and also 8,profit-and-loss debtor to goods1 #bj40 ,again1 in all payments which are not to bring in a specific return1 such as house and trade expenses1 wages1 c4 these several accounts are supposed to adjust matters with the profit-and-loss account before the balance begins4 ,thus1 suppose the outgoings from the mere premises occupied exceed anything those premises yield by #bjj1 or the debits of the house account exceed its credits by #bjj1 the account should be balanced by transferring the responsibility to the profit-and-loss account1 under the entries 8,house expenses creditor by profit-and-loss1 #bjj01 8,profit-and-loss debtor to house expenses1 #bjj40 ,in this way the profit-and-loss account steps in from time to time before the balance account commences its operations1 in order that that same balance account may consist of nothing but the necessary matters of account for the next year's ledger4 ,this transference of accounts1 or transfusion of one account into another1 requires attentive consideration4 ,the receiving account becomes creditor for the credits1 and debtor for the debits1 of the transmitting account4 ,the rule1 therefore1 is3 ,make the transmitting account balance itself1 and1 on whichever side it is necessary to enter a balancing sum1 make the account debtor or creditor1 as the case may be1 to the receiving account1 and the latter creditor or debtor to the former4 ,thus1 suppose account ,a is to be transferred to account ,b1 and the latter is to arrange with the balance account4 ,if the two stand as in ,roman letters1 the processes in ,italic letters will occur before the final close4 ,a1 ,debtor4 ,a1 ,creditor4 ,b1 ,debtor4 ,b1 ,creditor4 ,to sundries #ajj;,by sundries #ejj,to sundries #fjj;,by sundries #djj ,to ,b 4 4 4 #djj ,to ,balance #bjj;,by ,a 4 4 4 #djj ---- ---- ---- ---- #ejj #ejj #hjj #hjj ,and the entry in the balance account will be1 8,creditor by ,b1 #bjj10 shewing that1 on these two accounts1 the credits exceed the debits by #bjj4 ,still1 before the balance account is made up1 it is desirable that the profit-and-loss account should be transferred to the stock account2 for the profit and loss of this year is of no moment as a part of next year's ledger1 except in so far as it affects the stock at the commencement of the latter4 ,let this be done1 and the balance account may then be made in the form required4 ,the stock account and the profit-and-loss account1 the latter being the only direct channel of alteration for the former1 differ in a peculiar manner#fj from the other preliminary accounts1 and the balance account is a species of umpire4 ,they represent the merchant3 their interests are his interests2 he is solvent upon the excess of their credits over their debits1 insolvent upon the excess of their debits over their credits4 ,it is exactly the reverse in all the other accounts4 ,if a malicious person were to get at the ledger1 and put on a cipher to the pounds in various items1 with a view of making the concern appear worse than it really is1 he would make his alterations on the debtor sides of the stock and profit-and-loss accounts1 and on the creditor sides of all the others4 ,accordingly1 in the balance account1 the net stock1 after the incorporation of the profit-and-loss account1 appears on the creditor side 7if not1 it should be called amount of insolvency1 not stock71 and the debts of the concern appear on the same side4 ,but on the debit side of the balance account appear all the assets of the concern 7for which the balance-clerk is debtor to the clerks from whom he has taken them74 #fj ,the treatises on book-keeping have described this difference in as peculiar a manner4 ,they call these accounts the fictitious accounts4 ,now they represent the merchant himself2 their credits are gain to the business1 their debits losses or liabilities4 ,if the terms real and fictitious are to be used at all1 they are the real accounts1 end all the others are as fictitious as the clerks whom we have supposed to keep them4 ,the young student must endeavour to get the enlarged view of the words debtor and creditor which is requisite1 and must then learn by practice 7for nothing else will give it7 facility in allotting the actual entries in the waste-book to the proper sides of the proper accounts4 ,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-keeping1 which he will probably find to contain little more than examples4 ,,appendix ,,viii4 ,,on ,,the ,,reduction ,,of ,,fractions ,,to ,,others ,,of ,,nearly ,,equal ,,value4 ,there is a useful method of finding fractions which shall be nearly equal to a given fraction1 and with which the computer ought to be acquainted4 ,proceed as in the rule for finding the greatest common measure of the numerator and denominator1 and bring all the quotients into a line4 ,then write down1 #a #bnd ,quot4 -------- ------------------------ #ast ,quot4 #ast ,quot4 #b;d ,quot4 #a ,then take the third quotient1 multiply the numerator and denominator of the second by it1 and add to the products the preceding numerator and denominator4 ,form a third fraction with the results for a numerator and denominator4 ,then take the fourth quotient1 and proceed with the third and second fractions in the same way2 and so on till the quotients are exhausted4 ,for example1 let the fraction be _/4 #iaca7acabh7a1 #b #aacg #ciig7c1 #a #eea #ehf7a1 #ae #bja #ce7a1 #b #bf #i7a1 #h #h #a ,this is the process for finding the greatest common measure of #iaca and #acabh in its most compact form1 and the quotients and fractions are3 #a #b #c #a #a #ae #a #b #a #h #a #b #g #i #af #bdi #bfe #ggi #ajdd #iaca --- --- --- --- ---- ----- ----- ----- ------ ------ #a #c #aj #ac #bc #ceh #cha #aabj #aeja #acabh ,it will be seen that we have thus a set of fractions ending with the original fraction itself1 and formed by the above rule1 as follows3 #a #a #ast ,fraction -------- --- #ast ,quot4 #a #b;d ,quot4 #b #b;d ,fraction ------------------------ --- #ast ,quot4 #b;d ,quot4 #a #c #b;d ,num4 #c;d ,quot4 #ast ,num4 #b #c #a #g #c;d ,fraction ------------------------------- --------- --- #b;d ,den4 #c;d ,quot4 #ast ,den4 #c #c #a #aj #c;d ,num4 #dth ,quot4 #b;d ,num4 #g #a #b #i #dth ,fraction ------------------------------- --------- ---2 #c;d ,den4 #dth ,quot4 #b;d ,den4 #aj #a #c #ac and so on4 ,but we have done something more than merely reascend to the original fraction by means of the quotients4 ,the set of fractions1 _/1 _/1 _/1 _/1 c4 are continually approaching in value to the original fraction1 the first being too great1 the second too small1 the third too great1 and so on alternately1 but each one being nearer to the given fraction than any of those before it4 ,thus1 _/ is too great1 and _/ is too small2 but _/ is not so much too small as _/ is too great4 ,and again1 _/1 though too great1 is not so much too great as _/ is too small4 ,moreover1 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 denominator4 ,thus1 _/ does not err by so much as _/1 nor _/ by so much as _/1 nor _/ by so much as _/1 nor _/ by so much as _/1 c4 ,lastly1 no fraction of a less numerator and denominator can come so near to the given fraction as any one of the fractions in the list4 ,thus1 no fraction with a less numerator than #bdi1 and a less denominator than #ceh1 can come so near to #iaca #bdi ----- as ---4 #acabh #ceh ,the reader may take any example for himself1 and the test of the accuracy of the process is the ultimate return to the fraction begun with4 ,another test is as follows3 ,the numerator of the difference of any two consecutive approximating fractions ought to be unity4 ,thus1 in our instance1 we have _/ and _/1 which1 with a common denominator1 #bc #ceh1 have #egbh and #egbg for their numerators4 ,as another example1 let us examine this question3 ,the length of the year is #cfebdbbd days1 which is called in common life #cfe days4 ,take the fraction _/1 and proceed as in the rule4 #bdbbd7ajjjjj7d1 #g1 #a1 #d1 #i1 #b #bdif #cajd #fd #fjh #j #cb #a #g #h #ci #cei #geg --- --- --- ---- ---- ---- #d #bi #cc #afa #adhb #cabe and _/ is #bdbbd in its lowest terms4 ,hence1 it appears that the excess of the year over #cfe days amounts to about #a day in #d years1 which is not wrong by so much as #a day in #aaf years2 more accurately1 to #g days in #bi years1 which is not wrong by so much as #a day in #ieg years2 more accurately still1 to #h days in #cc years1 which is not wrong by so much as #a day in #ecac years2 and so on4 ,this method may be applied to finding fractions nearly equal to the square roots of integers1 in the following manner3 #dc #f 444 #f #a #e #d #e #e #d #e #a #f #f #a #e #d1 c4 #a #g #f #c #i #b #i #c #f #g #a #g #f #c1 c4 ------------------------------ #f #a #a #c #a #e #a #c #a #a #a #ba #a #c1 c4 ,set down the number whose square root is wanted1 say #dc4 ,this square root is #f and a fraction4 ,set down the integer #f in the first and third row1 and #a in the second row always4 ,form the successive rows each from the one before1 in the following manner3 ,one row ,the next row has b1 a1 c1 formed in this order1 being thus1 a a excess of bc1 already formed1 over a4 b b quotient of #dc - a divided by b4 c c integer in the quotient of #f a divided by b4 ,thus the second row is formed from the first1 as under3 #fa excess of #g #a 7both just found7 over #f4 #ag #dc - #f #f divided by #a4 ---- #fa integer of #f #f divided by #g 7just found74 ,the third row is formed from the second1 thus3 #a #e excess of #a #f over #a4 #g #f #dc - #a #a divided by #g4 #a #a integer of #f #a divided by #f2 and so on4 ,in process of time the second column1 #a1 #g1 #a1 occurs again1 after which the several columns are repeated in the same order4 ,as a final process1 take the set in the lowest line 7excluding the first1 #f71 namely1 #a1 #a1 #c1 #a1 #e1 #a1 #c1 c4 and use them by the rule given at the beginning of this article1 as follows3 #a #a #c #a #e #a #c #a #a1 c4 #a #a #d #e #bi #cd #aca #afe #bif --- --- --- --- ---- ---- ---- ---- ---- #a #b #g #i #eb #fa #bce #bif #eca ,hence1 #f_/ is very near the square root of #dc1 not erring by so much as #a ---------4 #bif #eca ,if we try it1 we shall find 7_/7 to be _/1 the square of which is _/1 or #dc_/4 ,this rule is of use when it is frequently wanted to use one square root1 and therefore desirable to ascertain whether any easy approximation exists by means of a common fraction4 ,for example1 #b is often used4 #b #a 444 #aa #a #aa #a #ab #b #b #b #b #b #a #b #e #ab #bi #gj --- --- --- --- --- ---1 c4 #b #e #ab #bi #gj #afi ,here it appears that #bi #a #ii #ajj - #a #a---- does not err by --------2 consequently1 ---- or ------- is1 #gj #gj #afi #gj #gj considering the ease of the operation1 a fair approximation4 ,in fact1 _/ is #adadbheg 444 the truth being #adadbace 444 ,the following is an additional example3 #ai #d 444 #d #b #c #c #b #d #d #b #a #c #e #b #e #c #a #c #d #b #a #c #a #b #h #b #a #c #a #b1 c4 #a #a #d #e #ad --- --- --- --- ---1 c4 #b #c #aa #ad #ci ,,appendix ,,ix4 ,,on ,,some ,,general ,,properties ,,of ,,numbers4 ,,prop4 #a4 ,if a fraction be reduced to its lowest terms1 so called1#fa that is1 if neither1 numerator nor denominator be divisible by any integer greater than unity1 then no fraction of a smaller numerator and denominator can have the same value4 #fa ,this theorem shews that what is called reducing a fraction to its lowest terms 7namely1 dividing numerator and denominator by their greatest common measure71 is correctly so called4 ,let a_/b be a fraction in which a and b have no common measure greater than unity3 and1 if possible1 let c_/d be a fraction of the same value1 c being less than a1 and d less than b4 ,now1 since a c a b --- ---1 we have --- ---2 b d c d let m be the integer quotient of these last fractions 7which must exist1 since a c1 b d71 and let e and f be the remainders4 ,then a mc e c mc --- or ---------- --- ---- b md f d md ,hence1 e mc --- and ---- must be equal1 for if not1 f md mc e mc e ---------- would lie between ---- and ---1 md f md f instead of being equal to the former4 ,hence1 a e --- ---2 b f so that if a fraction whose numerator and denominator have no common measure greater than unity1 be equal to a fraction of lower numerator and denominator1 it is equal to another in which the numerator and denominator are still lower4 ,if we proceed with a e --- --- in a similar manner1 we find b f a g --- --- where g e1 h f1 b h and so on4 ,now1 if there be any process which perpetually diminishes the terms of a fraction by one or more units at every step1 it must at last bring either the numerator or denominator1 or both1 to #j4 ,let a v --- --- b w be one of the steps1 and let a kv x1 b kw y2 so that kv x v ---------- ---4 kw y w ,now1 if x #j but not y1 this is absurd1 for it gives kv kv ---------- ----4 kw y kw ,a similar absurdity follows if y be #j1 but not x2 and if both x and y be #j1 then a kv1 b kw1 or a and b have a common measure1 k4 ,now k must be greater than #a1 for v and w are less than c and d1 which by hypothesis are less than a and b4 ,consequently a and b have a common measure k greater than #a1 which by hypothesis they have not4 ,if1 then1 a and b be integers not divisible by any integer greater than #a1 the fraction a_/b is really in its lowest terms4 ,also a and b are said to be prime to one another4 ,,prop4 #b4 ,if the product ab be divisible by c1 and if c be prime to b1 it must divide a4 ,let ab b d ---- d1 then --- ---4 c c c ,now b_/c is in its lowest terms2 therefore1 by the last proposition1 d and a must have a common measure4 ,let the greatest common measure be k1 and let a kl1 d km4 ,then b km m m --- ---- ---1 and --- c kl l l is also in its lowest terms2 but so is b_/c2 therefore we must have m b1 l c1 for otherwise a fraction in its lowest terms would be equal to another of lower terms4 ,therefore a kc1 or a is divisible by c4 ,and from this it follows1 that if a number be prime to two others1 it is prime to their product4 ,let a be prime to b and c1 then no measure of a can measure either b or c1 and no such measure can measure the product bc2 for any measure of bc which is prime to one must measure the other4 ,,prop4 #c4 ,if a be prime to b1 it is prime to all the powers of b4 ,every measure#fb of a is prime to b1 and therefore does not divide b4 ,hence1 by the last1 no measure of a divides b2 hence1 a is prime to b1 and so is every measure of it2 therefore1 no measure of a divides bb1 consequently a is prime to b1 and so on4 ,hence1 if a be prime to b1 a cannot divide without remainder any power of b4 ,this is the reason why no fraction can be made into a decimal unless its denominator be measured by no prime#fc numbers except #b and #e4 ,for if a c --- ---1 b #aj which last is the general form of a decimal fraction1 let a #aj;a --- be in its lowest terms2 then ------ b b is an integer1 whence 7,prop4 #b7 b must divide #aj1 and so must all the divisors of b4 ,if1 then1 among the divisors of b there be any prime numbers except #b and #e1 we have a prime number 7which is of course a number prime to #aj7 not dividing #aj1 but dividing one of its powers1 which is absurd4 #fb ,for that which measures a measure is itself a measure2 so that if a measure of a could have a measure in common with b1 a itself would have a common measure with b4 #fc ,a prime number is one which is prime to all numbers except its own multiples1 or has no divisors except #a and itself4 ,,prop4 #d4 ,if b be prime to a1 all the multiples of b1 as b1 #b;b1 444 up to 7a-#a7;b must leave different remainders when divided by a4 ,for if1 m being greater than n1 and both less than a1 we have mb and nb giving the same remainder1 it follows that mb-nb1 or 7m-n7b1 is divisible by a2 whence 7,prop4 #b71 a divides m-n1 a number less than itself1 which is absurd4 99 99 99 99 99 ,if a number be divided into its prime factors1 or reduced to a product of prime numbers only 7as in #cfj #b #b #b #c #c #e71 and if a1 b1 c1 c4 be the prime factors1 and 1 1 1 c4 the number of times they severally enter1 so that the number is a b c c41 then this can be done in only one way3 ,for any prime number v1 not included in the above list1 is prime to a1 and therefore to a1 to b and therefore to b and therefore to a b ,proceeding in this way1 we prove that v is prime to the complete product above1 or to the given number itself4 ,the number of divisors which the preceding number abc 444 can have1 #j and itself included1 is 7 #a77 #a77 #a74444 ,for a as the divisors #a1 a1 a 444 a and no others1 #a in all4 ,similarly1 b has #a divisors1 and so on4 ,now as all the divisors are made by multiplying together one out of each set1 their number 7page #bjb7 is 7 #a77 #a77 #a74444 ,if a number1 n1 be divisible by certain prime numbers1 say #c1 #e1 #g1 #aa1 then the third part of all the numbers up to n is divisible by #c1 the fifth part by #e1 and so on4 ,but more than this3 when the multiples of #c are omitted1 exactly the fifth part of those which remain are divisible by #e2 for the fifth part of the whole are divisible by #e1 and the fifth part of those which are removed are divisible by #e1 therefore the fifth part of those which are left are divisible by #e4 ,again1 because the seventh part of the whole are divisible by #g1 and the seventh part of those which are divisible by #c1 or by #e1 or by #ae1 it follows that when all those which are multiples of #c or #e1 or both1 are removed1 the seventh part of those which remain are divisible by #g2 and so on4 ,hence1 the number of numbers not exceeding n1 which are not divisible by #c1 #e1 #g1 or #aa1 is _/ of _/ of _/ of _/ of n4 ,proceeding in this way1 we find that the number of numbers which are prime to n1 that is1 which are not divisible by any one of its prime factors1 a1 b1 c1 444 is a - #a b - #a c - #a n ------- ------- ------- 444 a b c or a-#a - #a;bc 444 7a - #a77;b - #a77;c - #a74444 ,thus1 #cfj being #bce1 its number of divisors is #d #c #b1 or #bd1 and there are #bc4a4b4d or #if numbers less than #cfj which are prime to it4 ,,prop4 #e4 ,if a be prime to b1 then the terms of the series1 a1 a1 a1 444 severally divided by b1 must all leave different remainders1 until #a occurs as a remainder1 after which the cycle of remainders will be again repeated4 ,let a b give the remainder r 7not unity72 then a b gives the same remainder as ra b1 which 7,prop4 #d7 cannot be r3 let it be s4 ,then a b gives the same remainder as sa b1 which 7,prop4 #d7 cannot be either r or s1 unless s be #a3 let it be t4 ,then a b gives the same remainder as ta b2 if t be not #a1 this cannot be either r1 s1 or t3 let it be u4 ,so we go on getting different remainders1 until #a occurs as a remainder2 after which1 at the next step1 the remainder of a b is repeated4 ,now1 #a must come at last2 for division by b cannot give any remainders but #j1 #a1 #b1 444 b- #a2 and #j never arrives 7,prop4 #c71 so that as soon as b-#b different remainders have occurred1 no one of which is unity1 the next1 which must be different from all that precede1 must be #a4 ,if not before1 then at a we must have a remainder #a2 after which the cycle will obviously be repeated4 ,thus1 #g1 #g1 #g1 #g1 c4 will1 when divided by #e1 be found to give the remainders #b1 #d1 #c1 #a1 c4 ,,prop4 #f4 ,the difference of two mth powers is always divisible without remainder by the difference of the roots2 or a -b is divisible by a-b2 for a - b a - ab ab - b a7a - b7 b7a - b7 ,from which1 if a-b is divisible by a -b1 so is a-b4 ,but a-b is divisible by a-b2 so therefore is a- b2 so therefore is a-b2 and so on4 ,therefore1 if a and b1 divided by c1 leave the same remainder1 a and b1 a and b1 c4 severally divided by c1 leave the same remainders2 for this means that a-b is divisible by c4 ,but a - b is divisible by a-b1 and therefore by every measure of a-b1 or by c2 but a-b cannot be divisible by c1 unless a and b1 severally divided by c1 give the same remainder4 ,,prop4 #g4 ,if b be a prime number1 and a be not divisible by b1 then a and 7a-#a7 #a leave the same remainder when divided by b4 ,this proposition cannot be proved here1 as it requires a little more of algebra than the reader of this work possesses4#fd #fd ,expand 7a-#a7 by the binomial theorem2 shew that when b is a prime number every coefficient which is not unity is divisible by b2 and the proposition follows4 ,,prop4 #h4 ,in the last case1 a divided by b leaves a remainder #a4 ,from the last1 a-a leaves the same remainder as 7a-#a7 #a-;a or 7a-#a7- 7a-#a72 that is1 the remainder of a-a is not altered if a be reduced by a unit4 ,by the same rule1 it may be reduced another unit1 and so on1 still without any alteration of the remainder4 ,at last it becomes #a-a1 or #j1 the remainder of which is #j4 ,accordingly1 a-a1 which is a7a- #a71 is divisible by b2 and since b is prime to a1 it must 7,prop4 #b7 divide a-#a2 that is1 a1 divided by b1 leaves a remainder #a1 if b be a prime number and a be not divisible by b4 ,from the above it appears 7,prop4 #e and #g71 that if a be prime to b1 the set #a1 a1 a1 a1 c4 successively divided by b1 give a set of remainders beginning with #a1 and in which #a occurs again at a1 if not before1 and at a certainly 7whether before or not71 if b be a prime number4 ,from the point at which #a occurs1 the cycle of remainders recommences1 and #a is always the beginning of a cycle4 ,if1 then1 a be the first power which gives #a for remainder1 m must either be b-#a1 or a measure of it1 when b is a prime number4 ,but if we divide the terms of the series m1 ma1 ma1 ma1 c4 by b1 m being less than b1 we have cycles of remainders beginning with m4 ,if #a1 r1 s1 t1 c4 be the first set of remainders1 then the second set is the set of remainders arising from m1 mr1 ms1 mt1 c4 ,if #a never occur in the first set before a 7except at the beginning71 then all the numbers under b-#a inclusive are found among the set #a1 r1 s1 t1 c42 and if m be prime to b 7,prop4 #d71 all the same numbers are found1 in a different order1 among the remainders of m1 mr1 c4 ,but should it happen that the set #a1 r1 s1 t1 c4 is not complete1 then m1 mr1 ms1 c4 may give a different set of remainders4 ,all these last theorems are constantly verified in the process for reducing a fraction to a decimal fraction4 ,if m be prime to b1 or the fraction m_/b in its lowest terms1 the process involves the successive division of m1 m #aj1 m #aj1 c4 by b4 ,this process can never come to an end unless some power of #aj1 say #aj1 is divisible by b2 which cannot be1 if b contain any prime factors except #b and #e4 ,in every other case the quotient repeats itself1 the repeating part sometimes commencing from the first figure1 sometimes from a later figure4 ,thus1 _/ yields #adbhegadbheg1 c41 but _/ gives #jg7adbheg77adbheg71 c41 and _/ gives #jc7egadbh77egadbh71 c4 ,in m_/b1 the quotient always repeats from the very beginning whenever b is a prime number and m is less than b2 and the number of figures in the repeating part is then always b-#a1 or a measure of it4 ,that it must be so1 appears from the above propositions4 ,before proceeding farther1 we write down the repeating part of a quotient1 with the remainders which are left after the several figures are formed4 ,let the fraction be _/1 we have #jehhbcebidaagfdg ,this may be read thus3 #aj by #ag1 quotient #j1 remainder #aj2 #aj by #ag1 quotient #je1 remainder #ae2 #aj by #ag1 quotient #jeh1 remainder #ad2 and so on4 ,it thus appears that #aj by #ag leaves a remainder #a1 which is according to the theorem4 ,if we multiply #jehh1 c4 by any number under #ag1 the same cycle is obtained with a different beginning4 ,thus1 if we multiply by #ac1 we have #gfdgjehhbcebidaa beginning with what comes after remainder #ac in the first number4 ,if we multiply by #g1 we have #daag1 c4 ,the reason is obvious3 _/ #ac1 or _/1 when turned into a decimal fraction1 starts with the divisor #acj1 and we proceed just as we do in forming _/1 when within four figures of the close of the cycle4 ,it will also be seen1 that in the last half of the cycle the quotient figures are complements to #i of those in the first half1 and that the remainders are complements to #ag4 ,thus1 in #jehh1 c4 and #idaa1 c4 we see #j #i #i1 #e #d #i1 #h #a #i1 c41 and #aj #g #ag1 #ae #b #ag1 #ad #c #ag1 c4 ,we may shew the necessity of this as follows3 ,if the remainder #a never occur till we come to use a1 then1 b being prime1 b-#a is even2 let it be #bk4 ,accordingly1 a-#a is divisible by b2 but this is the product of a-#a and a #a1 one of which must be divisible by b4 ,it cannot be a-#a1 for then a power of a preceding the 7b-#a7th would leave remainder #a1 which is not the case in our instance3 it must then be a #a1 so that a divided by b leaves a remainder b-#a2 and the kth step concludes the first half of the process4 ,accordingly1 in our instance1 we see1 b being #ag and a being #aj1 that remainder #af occurs at the #hth step of the process4 ,at the next step1 the remainder is that yielded by #aj7;b-#a71 or #i;b b-#aj1 which gives the remainder b-#aj4 ,but the first remainder of all was #aj1 and #aj 7b-#aj7 b4 ,if ever this complemental character occur in any step1 it must continue1 which we shew as follows3 ,let r be a remainder1 and b-r a subsequent remainder1 the sum being b4 ,at the next step after the first remainder1 we divide #ajr by b1 and1 at the next step after the second remainder1 we divide #aj;b-#ajr by b4 ,now1 since the sum of #ajr and #aj;b-#ajr is divisible by b1 the two remainders from these new steps must be such as added together will give b1 and so on2 and the quotients added together must give #i1 for the sum of the remainders #ajr and #aj;b-#ajr yields a quotient #aj1 of which the two remainders give #a4 ,if _/ and _/ be taken1 the repeating parts will be found to contain #eh and #fj figures4 ,of these we write down only the first halves1 as the reader may supply the rest by the complemental property just given4 #jafidiaebedbcgbhhaceeicbbjcch1 c4 #jafcicddbfbbiejhaifgbacaadgedj1 c4 ,here1 then1 are two numbers1 the first of which multiplied by any number under #ei1 and the second by any number under #fa1 can have the products formed by carrying certain of the figures from one end to the other4 ,but1 b being still prime1 it may happen that remainder #a may occur before b-#a figures are obtained2 in which case1 as shewn1 the number of figures must be a measure of b-#a4 ,for example1 take _/4 ,the repeating quotient1 written as above1 has only #e figures1 and #e measures #da-a4 #jbdci ,now1 this period1 it will be found1 has its figures merely transposed1 if we multiply by #aj1 #ah1 #af1 or #cg4 ,but if we multiply by any other number under #da1 we convert this period into the period of another fraction whose denominator is #da4 ,the following are #h periods which may be found4 #jbdci #abaie #jdhgh #adfcd #jgcag #bfhbi #jigef #cfehe ,to find m_/#da1 look out for m among the remainders1 and take the period in which it is1 beginning after the remainder4 ,thus1 _/ is #hbibfhbibf1 c41 and _/ is #cfehecfehe1 c4 ,these periods are complemental1 four and four1 as #jbdci and #igefj1 #jgcag and #ibfhb1 c4 ,and if the first number1 #jbdci1 be multiplied by any number under #da1 look for that number among the remainders1 and the product is found in the period of that remainder by beginning after the remainder4 ,thus1 #jbdci multiplied by #bc gives #efjig1 and by #f gives #adfcd4 ,the reader may try to decipher for himself how it is that1 with no more figures than the following1 we can extend the result of our division4 ,the fraction of which the period is to be found is _/4 #hg7ajj7jaadidbe #acj #dcj #hbj #jaadidbe #be #cgj #bhgcefbe #be #bbj #gahcijfbe #be #dfj #agieigfefbe #be #be #ddhiidadjfbe #jaadidbebhgcefbe #gahcijfbe #agieigf #efbe #ddhiid ---------------------------------- #jaadidbebhgcefcbahcijhjdeiggjaadid ,,appendix ,x4 ,,on ,,combinations4 ,there are some things connected with combinations which ,i place in an appendix1 because ,i intend to demonstrate them more briefly than the matters in the text4 ,suppose a number of boxes1 say #d1 in each of which there are counters1 say #e1 #g1 #c1 and #aa severally4 ,in how many ways can one counter be taken out of each box1 the order of going to the boxes not being regarded4 ,answer1 in #e #g #c #aa ways4 ,for out of the first box we may draw a counter in #e different ways1 and to each such drawing we may annex a drawing from the second in #g different ways--giving #e #g ways of making a drawing from the first two4 ,to each of these we may annex a drawing from the third box in #c ways--giving #e #g #c drawings from the first three2 and so on4 ,the following statements may now be easily demonstrated1 and similar ones made as to other cases4 ,if the order of going to the boxes make a difference1 and if a1 b1 c1 d be the numbers of counters in the several boxes1 there are #d #b #c #a a b c d distinct ways4 ,if we want to draw1 say #b out of the first box1 #c out of the second1 #a out of the third1 and #c out of the fourth1 and if the order of the boxes be not considered1 the number of ways is a - #a b - #a b - #b d - #a d - #b a------- b------- -------- c d------- ------- #b #b #c #b #c ,if the order of going to the boxes be considered1 we must multiply the preceding by #d #c #b #a4 ,if the order of the drawings out of the boxes makes a difference1 but not the order of the boxes1 then the number of ways is a7a-#a7;b7b-#a77;b-#b7;cd7d-#a77;d-#b7 ,the nth power of a1 or a1 represents the number of ways in which a counters differently marked can be distributed in n boxes1 order of placing them in each box not being considered4 ,suppose we want to distribute #d differently-marked counters among #g boxes4 ,the first counter may go into either box1 which gives #g ways2 the second counter may go into either2 and any of the first #g allotments may be combined with any one of the second #g1 giving #g #g distinct ways2 the third counter varies each of these in #g different ways1 giving #g #g #g in all2 and so on4 ,but if the counters be undistinguishable1 the problem is a very different thing4 ,required the number of ways in which a number can be compounded of other numbers1 different orders counting as different ways4 ,thus1 #a #c #a and #a #a #c are to be considered as distinct ways of making #e4 ,it will be obvious1 on a little examination1 that each number can be composed in exactly twice as many ways as the preceding number4 ,take #h for instance4 ,if every possible way of making #g be written down1 #h may be made either by increasing the last component by a unit1 or by annexing a unit at the end4 ,thus1 #a #c #b #a may yield #a #c #b #b1 or #a #c #b #a #a3 and all the ways of making #h will thus be obtained2 for any way of making #h1 say a b c d1 must proceed from the following mode of making #g1 a b c 7d-#a74 ,now1 7d-#a7 is either #j--that is1 d is unity and is struck out--or 7d-#a7 remains1 a number #a less than d4 ,hence it follows that the number of ways of making n is #b4 ,for there is obviously #a way of making #a1 #b of making #b2 then there must be1 by our rule1 #b ways of making #c1 #b ways of making #d2 and so on4 #a #a #a #a #a #a #a #a #a #a #a #b #a #b #a #b #a #a #a #c #b #a #b #a #a #b #b #b #c #c #a #d ,this table exhibits the ways of making #a1 #b1 #c1 and #d4 ,hence it follows 7which ,i leave the reader to investigate7 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 b2 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 c2 and so on4 ,also1 in summing numbers which make up a b1 there are ways in which a is a rest1 and ways in which it is not1 and as many of one as of the other4 ,required the number of ways in which a number can be compounded of odd numbers1 different orders counting as different ways4 ,if a be the number of ways in which n can be so made1 and b the number of ways in which n #a can be made1 then a b must be the number of ways in which n #b can be made2 for every way of making #ab out of odd numbers is either a way of making #aj with the last number increased by #b1 or a way of making #aa with a #a annexed4 ,thus1 #a #e #c #c gives #ab1 formed from #a #e #c #a giving #aj4 ,but #a #i #a #a is formed from #a #i #a giving #aa4 ,consequently1 the number of ways of forming #ab is the sum of the number of ways of forming #aj and of forming #aa4 ,now1 #a can only be formed in #a way1 and #b can only be formed in #a way2 hence #c can only be formed in #a #a or #b ways1 #d in only #a #b or #c ways4 ,if we take the series #a1 #a1 #b1 #c1 #e1 #h1 #ac1 #ba1 #cd1 #ee1 #hi1 c4 in which each number is the sum of the two preceding1 then the nth number of this set is the number of ways 7orders counting7 in which n can be formed of odd numbers4 ,thus1 #aj can be formed in #ee ways1 #aa in #hi ways1 c4 ,shew that the number of ways in which mk can be made of numbers divisible by m 7orders counting7 is #b4 ,in the two series1 #a #a #a #b #c #d #f #i #ac #ai #bh1 c4 #j #a #j #a #a #a #b #b #c #d #e1 c41 the first has each new term after the third equal to the sum of the last and last but two2 the second has each new term after the third equal to the sum of the last but one and last but two4 ,shew that the nth number in the first is the number of ways in which n can be made up of numbers which1 divided by #c1 leave a remainder #a2 and that the nth number in the second is the number of ways in which n can be made up of numbers which1 divided by #c1 leave a remainder #b4 ,it is very easy to shew in how many ways a number can be made up of a given number of numbers1 if different orders count as different ways4 ,suppose1 for instance1 we would know in how many ways #ab can be thus made of #g numbers4 ,if we write down #ab units1 there are #aa intervals between unit and unit4 ,there is no way of making #ab out of #g numbers which does not answer to distributing #f partition-marks in the intervals1 #a in each of #f1 and collecting all the units which are not separated by partition-marks4 ,thus1 #a #a #c #b #a #b #b1 which is one way of making #ab out of #g numbers1 answers to #a #a #aaa #aa #a #aa #aa in which the partition-marks come in the #ast1 #b;d1 #eth1 #gth1 #hth1 and #ajth of the #aa intervals4 ,consequently1 to ask in how many ways #ab can be made of #g numbers1 is to ask in how many ways #f partition-marks can be placed in #aa intervals2 or1 how many combinations or selections can be made of #f out of #aa4 ,the answer is1 #aa #aj #i #h #g #f -----------------------1 or #dfb4 #a #b #c #d #e #f ,let us denote by m the number of ways in which m things can be taken out of n things1 so that m is the abbreviation for n - #a n - #b n - m #a n ------ ------- 444 as far as ------------- #b #c m ,then m also represents the number of ways in which m #a numbers can be put together to make n #a4 ,what we proved above is1 that #f is the number of ways in which we can put together #g numbers to make #ab4 ,there will now be no difficulty in proving the following3 #b #a #a #b #c 444 n ,in the preceding question1 #j did not enter into the list of numbers used4 ,thus1 #c #a #j #j was not considered as one of the ways of putting together four numbers to make #e4 ,but let us now ask1 what is the number of ways of putting together #g numbers to make #ab1 allowing #j to be in the list of numbers4 ,there can be no more 7nor fewer7 ways of doing this than of putting #g numbers together1 among which #j is not included1 to make #ai4 ,take every way of making #ab 7#j included71 and put on #a to each number1 and we get a way of making #ai 7#j not included74 ,take any way of making #ai 7#j not included71 and strike off #a from each number1 and we have one of the ways of making #ab 7#j included74 ,accordingly1 #f is the number of ways of putting together #g numbers 7#j being allowed7 to make #ab4 ,and 7m- #a7 is the number of ways of putting together m numbers to make n1 #j being included4 ,this last amounts to the solution of the following3 ,in how many ways can n counters 7undistinguishable from each other7 be distributed into m boxes8 ,and the following will now be easily proved3 ,the number of ways of distributing c undistinguishable counters into b boxes is 7b - #a7;b c - #a1 if any box or boxes may be left empty4 ,but if there must be #a at least in each box1 the number of ways is 7b - #a7;c - #a2 if there must be #b at least in each box1 it is 7b - #a7;c- b-#a2 if there must be #c at least in each box1 it is 7b - #a7;c - #b;b - #a2 and so on4 ,the number of ways in which m odd numbers can be put together to make n1 is the same as the number of ways in which m even numbers 7#j included7 can be put together to make n-m2 and this is the number of ways in which m numbers 7odd or even1 #j included7 can be put together to make 7n-m74 ,accordingly1 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-#a things out of 7n-m7 m-#a1 or 7n m7-#a4 ,unless n and m be both even or both odd1 the problem is evidently impossible4 ,there are curious and useful relations existing between numbers of combinations1 some of which may readily be exhibited1 under the simple expression of m to stand for the number of ways in which m things may be taken out of n4 ,suppose we have to take #e out of #ab3 ,let the #ab things be marked ,a1 ,b1 ,c1 c4 and set apart one of them1 ,a4 ,every collection of #e out of the #ab either does or does not include ,a4 ,the number of the latter sort must be #e2 the number of the former sort must be #d1 since it is the number of ways in which the other four can be chosen out of all but ,a4 ,consequently1 #e must be #e #d1 and thus we prove in every case1 m m 7m - #a7 #j and n both are #a2 for there is but one way of taking none1 and but one way of taking all4 ,and again m and 7n-m7 are the same things4 ,and if m be greater than n1 m is #j2 for there are no ways of doing it4 ,we make one of our preceding results more symmetrical if we write it thus1 #b #j #a #b 444 n ,if we now write down the table of symbols in which the 7m #a7th #j #a #b #c1 c4 -------------------------------------- #a #j #a #b #c1 c4 #b #j #a #b #c1 c4 #c #j #a #b #c1 c4 c4 c4 c4 c4 c4 number of the nth row represents m1 the number of combinations of m out of n1 we see it proved above that the law of formation of this table is as follows3 ,each number is to be the sum of the number above it and the number preceding the number above it4 ,now1 the first row must be #a1 #a1 #j1 #j1 #j1 c4 and the first column must be #a1 #a1 #a1 #a1 c4 so that we have a table of the following kind1 which may be carried as far as we please3 #j #a #b #c #d #e #f #g #h #i #aj ---------------------------------------- ------------ #a #a #a #j #j #j #j #j #j #j #j #j #b #a #b #a #j #j #j #j #j #j #j #j #c #a #c #c #a #j #j #j #j #j #j #j #d #a #d #f #d #a #j #j #j #j #j #j #e #a #e #aj #aj #e #a #j #j #j #j #j #f #a #f #ae #bj #ae #f #a #j #j #j #j #g #a #g #ba #ce #ce #ba #g #a #j #j #j #h #a #h #bh #ef #gj #ef #bh #h #a #j #j #i #a #i #cf #hd #abf #abf #hd #cf #i #a #j #aj #a #aj #de #abj #baj #beb #baj #abj #de #aj #a ,thus1 in the row #i1 under the column headed #d1 we see #abf1 which is #i #h #g #f 7#a #b #c #d71 the number of ways in which #d can be chosen out of #i1 which we represent by #d-i4 ,if we add the several rows1 we have #a #a or #b1 #a #b #a or #b1 next #a #c #c #a or #b1 c4 which verify a theorem already announced2 and the law of formation shews us that the several columns are formed thus3 #a #a #a #b #a #a #c #c #a #a #a #a #b #a #a #c #c #a ----- ------- --------- #a #b #a #a #c #c #a #a #d #f #d #a1 c4 so that the sum in each row must be double of the sum in the preceding4 ,but we can carry the consequences of this mode of formation further4 ,if we make the powers of #a x by actual algebraical multiplication1 we see that the process makes the same oblique addition in the formation of the numerical multipliers of the powers of x4 #a x #a x ------- #a x x x --------------- #a #bx x #a #bx x #a x --------------- #a #bx x x #bx x ----------------------- #a #cx #cx x ,here are the second and third powers of #a x3 the fourth1 we can tell beforehand from the table1 must be #a #dx #fx #dx x2 and so on4 ,hence we have 7#a x7 #j #ax #bx #cx 444 nx which is usually written with the symbols #j1 #a1 c4 at length1 thus1 n - #a n - #a n - #b 7#a x7 #a nx n-------x n------- -------x c4 #b #b #c ,this is the simplest case of what in algebra is called the binomial theorem4 ,if instead of #a x we use x a1 we get 7xa7 x#a;ax#b;ax#c;ax444 na ,we can make the same table in another form4 ,if we take a row of ciphers beginning with unity1 and setting down the first1 add the next1 and then the next1 and so on1 and then repeat the process with one step less1 and then again with one step less1 we have the following3 #a #j #j #j #j #j #j #a #a #a #a #a #a #a #a #b #c #d #e #f #a #c #f #aj #ae #a #d #aj #bj #a #e #ae #a #f #a ,in the oblique columns we see #a #a1 #a #b #a1 #a #c #c #a1 c4 the same as in the original table1 and formed by the same additions4 ,if1 before making the additions1 we had always multiplied by a1 we should have got the several components of the powers of #a a1 thus1 #a #j #j #j #j #a a a a a #a #b;a #c;a #d;a #a #c;a #f;a #a #d;a #a where the oblique columns #a a1 #a #b;a a1 #a #c;a #c;a a1 c41 give the several powers of #a a4 ,if instead of beginning with #a1 #j1 #j1 c4 we had begun with p1 #j1 #j1 c4 we should have got p1 p #d;a1 p #f;a1 c4 at the bottom of the several columns2 and if we had written at the top x1 x1 x1 x1 #a1 we should have had all the materials for forming p7x a7 by multiplying the terms at the top and bottom of each column together1 and adding the results4 ,suppose we follow this mode of forming p7x a7 q7x a7 r7x a7 s4 x x x #a x x #a x #a #a p #j #j #j q #j #j r #j #c p pa pa pa q qa qa r ra p #bpa #cpa q #bqa r p #cpa q p px #cpax #cpax pa qx #bqax qa rx ra s px 7#cpa q7x 7#cpa #bqa r7x pa qa ra s ,now1 observe that all this might be done in one process1 by entering q1 r1 and s under their proper powers of x in the first process1 as follows x x x #a p q r s p pa q pa qa r pa qa ra s p #bpa q #cpa #bqa r p #cpa q p ,this process#fe is the one used in ,appendix ,,xi41 with the slight alteration of varying the sign of the last letter1 and making subtractions instead of additions in the last column4 ,as it stands1 it is the most convenient mode of writing x a instead of x in a large class of algebraical expressions4 ,for instance1 what does #bx x #cx #gx #i become when x #e is written instead of x8 ,the expression1 made complete1 is1 #bx #ax #jx #cx #gx #i #a #j #c #g #i #b #aa #ee #bgh #acig #fiid #b #ba #afj #ajgh #fghg #b #ca #cae #bfec #b #da #ebj #b #ea ,answer1 #bx #eax #ebjx #bfecx #fghgx #fiid4 #fe ,the principle of this mode of demonstration of ,horner's method was stated in ,young's ,algebra 7#ahbc71 being the earliest elementary work in which that method was given4 ,,appendix ,,xi4 ,,on ,,horner',s ,,method ,,of ,,solving ,,equations4 ,the rule given in this chapter is inserted on account of its excellence as an exercise in computation4 ,the examples chosen will require but little use of algebraical signs1 that they may be understood by those who know no more of algebra than is contained in the present work4 ,to solve an equation such as #bx x - #cx #dafgic1 or1 as it is usually written1 #bx x - #cx - #dafgic #j1 we must first ascertain by trial not only the first figure of the root1 but also the denomination of it3 if it be a #b1 for instance1 we must know whether it be #b1 or #bj1 or #bjj1 c41 or #b1 or #jb1 or #jjb1 c4 ,this must be found by trial2 and the shortest way of making the trial is as follows3 ,write the expression in its complete form4 ,in the preceding case the form is not complete1 and the complete form is #bx #jx #ax - #cx - #dafgic4 ,to find what this is when x is any number1 for instance1 #cjjj1 the best way is to take the first multiplier 7#b71 multiply it by #cjjj1 and take in the next multiplier 7#j71 multiply the result by #cjjj1 and take in the next multiplier 7#a71 and so on to the end1 as follows3 #b #cjjj #j #fjjj2 #fjjj #cjjj #a #ahjjjjja #ahjjjjja #cjjj - #c #edjjjjjbiig #edjjjjjbiig #cjjj - #dafgic #afbjjjjjhegdbjg ,now try the value of the above when x #cj4 ,we have then1 for the steps1 #fj 7#b #cj #j71 #ahja1 #edjbg1 and lastly1 #afbjhaj-dafgic1 or x #cj makes the first terms greater than #dafgic4 ,now try x #bj which gives #dj1 #hja1 #afjag1 and lastly1 #cbjcdj-dafgic1 or x #bj makes the first terms less than #dafgic4 ,between #bj and #cj1 then1 must be a value of x which makes #bx x-#cx equal to #dafgic4 ,and this is the preliminary step of the process4 ,having got thus far1 write down the coefficients #b1 #j1 #a1-c1 and -#dafgic1 each with its proper algebraical sign1 except the last1 in which let the sign be changed4 ,this is the most convenient way when the last sign is-4 ,but if the last sign be 1 it may be more convenient to let it stand1 and change all which come before4 ,thus1 in solving x-#abx #a #j1 we might write -#a #j #ab #a whereas in the instance before us1 we write #b #j #a -#c #dafgic ,having done this1 take the highest figure of the root1 properly named1 which is #b tens1 or #bj4 ,begin with the first column1 multiply by #bj1 and join it to the number in the next column2 multiply that by #bj1 and join it to the number in the next column2 and so on4 ,but when you come to the last column1 subtract the product which comes out of the preceding column1 or join it to the last column after changing its sign4 ,when this has been done1 repeat the process with the numbers which now stand in the columns1 omitting the last1 that is1 the subtracting step2 then repeat it again1 going only as far as the last column but two1 and so on1 until the columns present a set of rows of the following appearance3 a b c d e f g h i k l m n o p to the formation of which the following is the key3 f #bj;a b1 g #bj;f c1 h #bj;g d1 i e - #bj;h1 k #bj;a f1 l #bjk g1 m #bjl h1 n #bj;a k1 o #bjn l1 p #bj;a n4 ,we call this ,horner's ,process1 from the name of its inventor4 ,the result is as follows3 #b #j #a -#c #dafgic 7#bj #dj #hja #afjag #ifdec #hj #bdja #fdjcg #abj #dhja #afj ,we have now before us the row #b #afj #dhja #fdjcg #ifdec which furnishes our means of guessing at the next1 or units' figure of the root4 ,call the last column the dividend1 the last but one the divisor1 and all that come before antecedents4 ,see how often the dividend contains the divisor2 this gives the guess at the next figure4 ,the guess is a true one1#ff if1 on applying ,horner's process1 the divisor result1 augmented as it is by the antecedent processes1 still go as many times in the dividend4 ,for example1 in the case before us1 #ifdec contains #fdjcg once2 let #a be put on its trial4 ,horner's process is found to succeed1 and we have for the second process1 #b #afj #dhja #fdjcg #ifdec #afb #difc #fijjj #bgdec #afd #eabg #gdabg #aff #ebic #afh ,as soon as we come to the fractional portion of the root1 the process assumes a more#fg methodical form4 ,the equation being of the fourth degree1 annex four ciphers to the dividend1 three to the divisor1 two to the antecedent1 and one to the previous antecedent1 leaving the first column as it is2 then find the new figure by the dividend and divisor1 as before1#fh and apply ,horner's process4 ,annex ciphers to the results1 as before1 and proceed in the same way4 ,the annexing of the ciphers prevents our having any thing to do with decimal points1 and enables us to use the quotient-figures without paying any attention to their local values4 ,the following exhibits the whole process from the beginning1 carried as far as it is here intended to go before beginning the contraction1 which will give more figures1 as in the rule for the square root4 ,the following1 then1 is the process as far as one decimal place3 #ff ,various exceptions may arise when an equation has two nearly equal roots4 ,but ,i do not here introduce algebraical difficulties2 and a student might give himself a hundred examples1 taken at hazard1 without much chance of lighting upon one which gives any difficulty4 #fg ,this form might be also applied to the integer portions2 but it is hardly needed in such instances as usually occur4 ,see the article ,involution and ,evolution in the ,supplement to the ,penny ,cyclopdia4 #fh ,after the second step1 the trial will rarely fail to give the true figure4 #b #j #a -#c #dafgic7bac #dj #hja #afjag #ifdec #hj #bdja #fdjcg ----- #abj #dhja ----- #bgdecjjjj #afj ---- #fijjj #dgcciggh --- #difc #gdabgjjj --------- #afb #eabg -------- #afd #ebicjj #gegcjjgd #aff ------ #ggcdhcgf #afhj #ecdceh ---- #ecidcd #afhf #eddebh #afib #afih #agjd ---- ,if we now begin the contraction1 it is good to know beforehand on what number of additional root-figures we may reckon4 ,we may be pretty certain of having nearly as many as there are figures in the divisor when we begin to contract--one less1 or at least two less4 ,thus1 there being now eight figures in the divisor1 we may conclude that the contraction will give us at least six more figures4 ,to begin the contraction1 let the dividend stand1 cut off one figure from the divisor1 two from the column before that1 three from the one before that1 and so on4 ,thus1 our contraction begins with #jjjb #agjd #eddebh #ggcdhcgf #dgcciggh ,the first column is rendered quite useless here4 ,conduct the process as before1 using only the figures which are not cut off4 ,but it will be better to go as far as the first figure cut off1 carrying from the second figure cut off4 ,we shall then have as follows3 #agjd #eddebh #ggcdhcgf #dgcciggh7f #edeee #ggfgegjf #gcdced #edfeg #ghjjcfdh #edgei ,at the next contraction the column #agjd becomes #jjagjd1 and is quite useless4 ,the next step1 separately written 7which is not1 however1 necessary in working71 is #edgei #ghjjcfdh #gcdced7j ,here the dividend #gcdced does not contain the divisor #ghjjcf1 and we1 therefore1 write #j as a root figure and make another contraction1 or begin with #edgei #ghjjcfdh #gcdced7i #ghjjhe #cbbgg #ghjacd ,at the next contraction the first column becomes #jjedgei1 and is quite useless1 so that the remainder of the process is the contracted division4 #ghjacd7cbbgg7dacg #ajgb #bib #eh #c and the root required is #bacfjidacg4 ,i now write down the complete process for another equation1 one root of which lies between #c and #d3 it is x - #ajx #a #j #a #j -#aj -#a7caaajcijebjgcjiijgif #c -#a #bjjj #f #agjj #bjijjj #i #j #agia #aigfijjj #i #a #ahhcjj #gdccfijjjjjj #i #b #ahibca #agbcaagajbgcjjj #i #cj #aijafcjj #iiabdgddgfha #i #ca #aijbefca #cidfbhgedbj #i #cb #aijcdifcjj #j #j #aciadiaeei #i #cc #j #aijcebdbii #j #i #ehiicabc #i #cc #a #aijceebbih #b #g #j #j #ahhfjdg #i #cc #b #aijcefjfih #j #eia #agbhce #i #cc #cj #j #aijcefijig #h #efc #aeae #i #cc #cj #c #aijcefiadd #e #bb #ahc #i #cc #cj #f #aijcefiaia #ahh #ab #i #cc #cj #ij #aijcefiaic #jf #a #i #cccjii #aijcefiaidic #i #cccajh #jicccaag ,the student need not repeat the rows of figures so far as they come under one another3 thus1 it is not necessary to repeat #aijcef4 ,but he must use his own discretion as to how much it would be safe for him to omit4 ,i have set down the whole process here as a guide4 ,the following examples will serve for exercise3 #a4 #bx - #ajjx - #g #j x #gajehaacc4 #b4 x x x x #fjjj x #hecadcggbf4 #c4 x #cx - #dx - #aj #j x #ahiefidiafejd4 #d4 x #ajjx - #ex - #bagc #j x #dehbbdfjgajehdfd4 #e4 #b #abeiibajdihidhgcafdgfgbajfjgbgh4fi #f4 x - #fx #ajj x #ejgaceagdhgca4 #g4 x #bx #cx #cjj x #eieebeifgabbcih4 #h4 x x #ajjj x #iifffffgi4 #i4 #bgjjjx #bgjjjx #bfiiiiii x #iiffffff44444 #aj4 x - #fx #ajj x #ejgaceagdhg4 #aa4 x - #dx #gx - #hfc #j x #deaieejg4 #ab4 x - #bjx #h #j x #dffjjcgficjjjhgbgh4 #ac4 x x x - #aj #j x #agcgcgjbcc4 #ad4 x - #dfx - #cfx #ah #j x #dfgfafcjahdg1 or x #cdgafbcaib4 #ae4 x #dfx - #cfx - #ah #j x #aajhgibejcg4 #af4 #hiiax - #afbhchx #gdfbgax - #hajjj #j x #aaabbbcccdddeee4444 #ag4 #gbix - #dhfx #iix - #f #j x #aaaa4441 or #bbbb4441 or #cccc4444 #ah4 #bx #cx - #dx #ejj x #eicdhagifbcaeaebgi4 #ai4 x #bx x - #aej #j x #dffhdjijadeedaihcbecgdbiiabjagjehii4 #bj4 x x x #ejj x #hbdjifceehaddhehebfifc4 #ba4 x #bx #cx - #ajjjj #j x #bjhebijeebfjji4 #bb4 x - #dx - #bjjj #j x #dehadjjcfb4 #bc4 #ajx - #ccx - #aax - #ajj #j x #dadfgighjhehdbghghe4 #bd4 x x x x #abgfid x #ahfddhbcgcjie4 #be4 #ajx #aax #abx #ajjjjj x #baafeeiieeedejhhje4 #bf4 x x #ac x #bbjigeccjabjhhdi4 #bg4 x x - #dx - #afjj #j x #aadhbhcgaeg4 #bh4 x - #bx #e x #bjideeadhaedbcbfeiadhbchfedjegicjbifche gcjfajefbhbci4 #bi4 x - #hjx #bdx - #fx - #hjcgifci #j x #abc4gj #cj4 x - #bdbx - #fcaex #beggjif #j x #abc4ga #ca4 #bx - #cx #fx - #h #j x #adadbacefbcgcjiejdhhjc4gb #cb4 x - #aix #acbx - #cjbx #bjj #j x #ajbhjd1 or #d1 or #fegfec1 or #gciedcgc4 #cc4 #gx - #aax #fx #ex #bae x #bgjfdhjdichegia4gd #cd4 #gx #fx #ex #dx #cx #aa x #ggjgfhhaifbbfehebbcgibifeje4ge #ce4 #dx #gx #ix #fx #ex #cx #gib x #bjebjdbagfhgifjecfebadjdcdjabhabjaigcdf jbgeeiie #deedagbdbad4gf #cf4 #bahgx - #bdcjx #idex - #aejx #h #j x #aaaa44441 or #bbbb44441 or #cccc44441 or #dddd4444 #fi ,the solution of x #jx #jx-#b #j4 #gj ,taken from a paper on the subject1 by ,mr4 ,peter ,gray1 in the ,mechanics' ,magazine4 #ga ,taken from a paper on the subject1 by ,mr4 ,peter ,gray1 in the ,mechanics' ,magazine4 #gb ,taken from a paper on the subject1 by ,mr4 ,peter ,gray1 in the ,mechanics' ,magazine4 #gc ,taken from the late ,mr4 ,peter ,nicholson's ,essay on ,involution and ,evolution4 #gd ,taken from the late ,mr4 ,peter ,nicholson's ,essay on ,involution and ,evolution4 #ge ,taken from the late ,mr4 ,peter ,nicholson's ,essay on ,involution and ,evolution4 #gf ,taken from the late ,mr4 ,peter ,nicholson's ,essay on ,involution and ,evolution4 ,,appendix ,,xii4 ,,rules ,,for ,,the ,,application ,,of ,,arithmetic ,,to ,,geometry4 ,the student should make himself familiar with the most common terms of geometry1 after which the following rules will present no difficulty4 ,in them all1 it must be understood1 that when we talk of multiplying one line by another1 we mean the repetition of one line as often as there are units of a given kind1 as feet or inches1 in another4 ,in any other sense1 it is absurd to talk of multiplying a quantity by another quantity4 ,all quantities of the same kind should be represented in numbers of the same unit2 thus1 all the lines should be either feet and decimals of a foot1 or inches and decimals of an inch1 c4 ,and in whatever unit a length is represented1 a surface is expressed in the corresponding square units1 and a solid in the corresponding cubic units4 ,this being understood1 the rules apply to all sorts of units4 ,to find the area of a rectangle4 ,multiply together the units in two sides which meet1 or multiply together two sides which meet2 the product is the number of square units in the area4 ,thus1 if #f feet and #e feet be the sides1 the area is #f #e1 or #cj square feet4 ,similarly1 the area of a square of #f feet long is #f #f1 or #cf square feet 7#bcd74 ,to find the area of a parallelogram4 ,multiply one side by the perpendicular distance between it and the opposite side2 the product is the area required in square units4 ,to find the area of a trapezium4#gg ,multiply either of the two sides which are not parallel by the perpendicular let fall upon it from the middle point of the other4 #gg ,a four-sided figure1 which has two sides parallel1 and two sides not parallel4 ,to find the area of a triangle4 ,multiply any side by the perpendicular let fall upon it from the opposite vertex1 and take half the product4 ,or1 halve the sum of the three sides1 subtract the three sides severally from this half sum1 multiply the four results together1 and find the square root of the product4 ,the result is the number of square units in the area2 and twice this1 divided by either side1 is the perpendicular distance of that side from its opposite vertex4 ,to find the radius of the internal circle which touches the three sides of a triangle4 ,divide the area1 found in the last paragraph1 by half the sum of the sides4 ,given the two sides of a right-angled triangle1 to find the hypothenuse4 ,add the squares of the sides1 and extract the square root of the sum4 ,given the hypothenuse and one of the sides1 to find the other side4 ,multiply the sum of the given lines by their difference1 and extract the square root of the product4 ,to find the circumference of a circle from its radius1 very nearly4 ,multiply twice the radius1 or the diameter1 by #cadaeibg1 taking as many decimal places as may be thought necessary4 ,for a rough computation1 multiply by #bb and divide by #g4 ,for a very exact computation1 in which decimals shall be avoided1 multiply by #cee and divide by #aac4 ,see 7#aca71 last example4 ,to find the arc of a circular sector1 very nearly1 knowing the radius and the angle4 ,turn the angle into seconds1#gh multiply by the radius1 and divide the product by #bjfbfe4 ,the result will be the number of units in the arc4 #gh ,the right angle is divided into #ij equal parts called degrees1 each degree into #fj equal parts called minutes1 and each minute into #fj equal parts called seconds4 ,thus1 #b #ae #dj means #b degrees1 #ae minutes1 and #dj seconds4 ,to find the area of a circle from its radius1 very nearly4 ,multiply the square of the radius by #cadaeibg4 ,to find the area of a sector1 very nearly1 knowing the radius and the angle4 ,turn the angle into seconds1 multiply by the square of the radius1 and divide by #bjfbfe #b1 or #dabecj4 ,to find the solid content of a rectangular parallelopiped4 ,multiply together three sides which meet3 the result is the number of cubic units required4 ,if the figure be not rectangular1 multiply the area of one of its planes by the perpendicular distance between it and its opposite plane4 ,to find the solid content of a pyramid4 ,multiply the area of the base by the perpendicular let fall from the vertex upon the base1 and divide by #c4 ,to find the solid content of a prism4 ,multiply the area of the base by the perpendicular distance between the opposite bases4 ,to find the surface of a sphere4 ,multiply #d times the square of the radius by #cadaeibg4 ,to find the solid content of a sphere4 ,multiply the cube of the radius by #cadaeibg _/1 or #dahhgi4 ,to find the surface of a right cone4 ,take half the product of the circumference of the base and slanting side4 ,to find the solid content1 take one-third of the product of the base and the altitude4 ,to find the surface of a right cylinder4 ,multiply the circumference of the base by the altitude4 ,to find the solid content1 multiply the area of the base by the altitude4 ,the weight of a body may be found1 when its solid content is known1 if the weight of one cubic inch or foot of the body be known4 ,but it is usual to form tables1 not of the weights of a cubic unit of different bodies1 but of the proportion which these weights bear to some one amongst them4 ,the one chosen is usually distilled water1 and the proportion just mentioned is called the specific gravity4 ,thus1 the specific gravity of gold is #aicfb1 or a cubic foot of gold is #aicfb times as heavy as a cubic foot of distilled water4 ,suppose now the weight of a sphere of gold is required1 whose radius is #d inches4 ,the content of this sphere is #d #d #d #dahhh1 or #bfhjhcb cubic inches2 and since1 by 7#bag71 each cubic inch of water weighs #bebdeh grains1 each cubic inch of gold weighs #bebdeh #aicfb1 or #dhhhjia grains2 so that #bfhjhcb cubic inches of gold weigh #bfhjhcb #dhhhjia grains1 or #bbg pounds troy nearly4 ,tables of specific gravities may be found in most works of chemistry and practical mechanics4 ,the cubic foot of water is #ijhhdhh troy ounces1 #gegcgd troy pounds1 #iigacfifia averdupois ounces1 and #fbcbajfjf averdupois pounds4 ,for all rough purposes it will do to consider the cubic foot of water as being #ajjj common ounces1 which reduces tables of specific gravities to common terms in an obvious way4 ,thus1 when we read of a substance which has the specific gravity #daagb1 we may take it that a cubic foot of the substance weighs #daag ounces4 ,for greater correctness1 diminish this result by #c parts out of a thousand4 ,,the ,,end4 ,,walton ,,and ,,maberly',s ,,catalogue ,,of ,,educational ,,works1 ,,and ,,works ,,in ,,science ,,and ,,general ,,literature4 ,,english4 ,dr4 ,r4 ,g4 ,latham4 ,the ,english ,language4 ,fourth ,edition4 #b vols4 #hvo4 #a #hs4 cloth4 ,latham's ,elementary ,english ,grammar1 for the ,use of ,schools4 ,eighteenth thousand4 ,small #hvo4 #ds4 #f;d4 cloth4 ,latham's ,hand-book of the ,english ,language1 for the ,use of ,students of the ,universities and higher ,classes of ,schools4 ,third ,edition4 ,small #hvo4 #gs4 #f;d4 cloth4 ,latham's ,logic in its ,application to ,language4 #abmo4 #fs4 cloth4 ,latham's ,first ,english ,grammar1 adapted for general use4 ,by ,,dr4 ,r4 ,g4 ,,latham and an ,experienced ,teacher4 ,fcap4 #hvo4 ,latham's ,history and ,etymology of the ,english ,language1 for the ,use of ,classical ,schools4 ,second ,edition1 revised4 ,fcap4 #hvo4 #as4 #f;d4 cl4 ,mason's ,english ,grammar1 including the ,principles of ,grammatical ,analysis4 #abmo4 #cs4 #f;d4 cloth4 ,mason's ,cowper's ,task ,book ,i4 7the ,sofa71 with ,notes on the ,analysis and ,parsing4 ,crown #hvo4 #as4 #f;d4 cloth4 ,abbott's ,first ,english ,reader4 ,third ,edition4 #abmo41 with ,illustrations4 #as4 cloth1 limp4 ,abbott's ,second ,english ,reader4 ,third ,edition4 #abmo4 #as4 #f;d4 cloth1 limp4 ,,greek4 ,greenwood's ,greek ,grammar1 including ,accidence1 ,irregular ,verbs1 and ,principles of ,derivation and ,composition2 adapted to the ,system of ,crude ,forms4 ,small #hvo4 #es4 #f;d4 cloth4 ,khner's ,new ,greek ,delectus2 being ,sentences for ,translation from ,greek into ,english1 and ,english into ,greek2 arranged in a systematic ,progression4 ,translated and ,edited by the late ,,dr4 ,,alexander ,,allen4 ,fourth ,edition1 revised4 #abmo4 #ds4 cloth4 ,gillespie's ,greek ,testament ,roots1 in a ,selection of ,texts1 giving the power of ,reading the whole ,greek ,testament without difficulty4 ,with ,grammatical ,notes1 and a ,parsing ,lexicon associating the ,greek ,primitives with ,english ,derivatives4 ,post #hvo4 #gs4 #f;d4 cloth4 ,robson's ,constructive ,exercises for ,teaching the ,elements of the ,greek ,language1 on a system of ,analysis and ,synthesis1 with ,greek ,reading ,lessons and copious ,vocabularies4 #abmo41 pp4 #djh4 #gs4 #f;d4 cloth4 ,robson's ,first ,greek ,book4 ,exercises and ,reading ,lessons with ,copious ,vocabularies4 ,being the ,first ,part of the 8,constructive ,greek ,exercises40 #abmo4 #cs4 #f;d4 cloth4 ,the ,london ,greek ,grammar4 ,designed to exhibit1 in small ,compass1 the ,elements of the ,greek ,language4 ,sixth ,edition4 #abmo4 #as4 #f;d4 cloth4 ,hardy and ,adams's ,anabasis of ,xenophon4 ,expressly for ,schools4 ,with ,notes1 ,index of ,names1 and a ,map4 #abmo4 #ds4 #f;d4 cloth4 ,smith's ,plato4 ,the ,apology of ,socrates1 ,the ,crito1 and part of the ,phaedo2 with ,notes in ,english from ,stallbaum1 ,schleiermacher's ,introductions1 and his ,essay on the ,worth of ,socrates as a ,philosopher4 ,edited by ,dr4 ,,wm4 ,,smith1 ,editor of the ,dictionary of ,greek and ,roman ,antiquities1 c4 ,third ,edition4 #abmo4 #es4 cloth4 ,,latin4 ,new ,latin ,reading ,book2 consisting of ,short ,sentences1 ,easy ,narrations1 and ,descriptions1 selected from ,caesar's ,gallic ,war2 in ,systematic ,progression4 ,with a ,dictionary4 ,second ,edition1 revised4 #abmo4 #bs4 #f;d4 ,allen's ,new ,latin ,delectus2 being ,sentences for ,translation from ,latin into ,english1 and ,english into ,latin2 arranged in a systematic ,progression4 ,fourth ,edition1 revised4 #abmo4 #ds4 cloth4 ,the ,london ,latin ,grammar2 including the ,eton ,syntax and ,prosody in ,english1 accompanied with ,notes4 ,sixteenth ,edition4 #abmo4 #as4 #f;d4 ,robson's ,constructive ,latin ,exercises1 for teaching the ,elements of the ,language on a ,system of ,analysis and ,synthesis2 with ,latin ,reading ,lessons and ,copious ,vocabularies4 ,third and ,cheaper ,edition1 thoroughly revised4 #abmo4 #ds4 #f;d4 cloth4 ,robson's ,first ,latin ,reading ,lessons4 ,with ,complete ,vocabularies4 ,intended as an ,introduction to ,caesar4 #abmo4 #bs4 #f;d4 cloth4 ,smith's ,tacitus2 ,,germania1 ,,agricola1 ,,and ,,first ,,book ,,of ,,the ,,annals4 ,with ,english ,notes1 original and selected1 and ,btticher's remarks on the style of ,tacitus4 ,edited by ,dr4 ,,wm4 ,,smith1 ,editor of the ,dictionary of ,greek and ,roman ,antiquities1 etc4 ,third ,edition1 greatly improved4 #abmo4 #es4 ,caesar4 ,civil ,war4 ,book ,i4 ,with ,english ,notes for the ,use of ,students preparing for the ,cambridge ,school ,examination4 #abmo4 #as4 #f;d4 ,terence4 ,andria4 ,with ,english ,notes1 ,summaries1 and ,life of ,terence4 ,by ,,newenham ,,travers1 ,b4,a41 ,assistant-,master in ,university ,college ,school4 ,fcap4 #hvo4 #cs4 #f;d4 ,,hebrew4 ,hurwitz's ,grammar of the ,hebrew ,language4 ,fourth ,edition4 #hvo4 #acs4 cloth4 ,or in ,two ,parts1 sold separately3-- ,,elements4 #ds4 #f;d4 cloth4 ,,etymology ,,and ,,syntax4 #is4 cloth4 ,,french4 ,merlet's ,french ,grammar4 ,by ,p4 ,f4 ,,merlet1 ,professor of ,french in ,university ,college1 ,london4 ,new ,edition4 #abmo4 #es4 #f;d4 bound4 ,or sold in ,two ,parts3--,,pronunciation and ,,accidence1 #cs4 #f;d42 ,,syntax1 #cs4 #f;d4 7,,key1 #cs4 #f;d47 ,merlet's ,le ,traducteur2 ,selections1 ,historical1 ,dramatic1 and ,miscellaneous1 from the best ,french ,writers1 on a plan calculated to render reading and translation peculiarly serviceable in acquiring the ,french ,language2 accompanied by ,explanatory ,notes1 a ,selection of ,idioms1 etc4 ,fourteenth ,edition4 #abmo4 #es4 #f;d4 bound4 ,merlet's ,exercises on ,french ,composition4 ,consisting of ,extracts from ,english ,authors to be turned into ,french2 with ,notes indicating the ,differences in ,style between the two ,languages4 ,a ,list of ,idioms1 with ,explanations1 ,mercantile ,terms and ,correspondence1 ,essays1 etc4 #abmo4 #cs4 #f;d4 ,merlet's ,french ,synonymes1 explained in ,alphabetical ,order4 ,copious ,examples 7from the 8,dictionary of ,difficulties074 #abmo4 #bs4 #f;d4 ,merlet's ,aperu de la ,litterature ,franaise4 #abmo4 #bs4 #f;d4 ,merlet's ,stories from ,french ,writers2 in ,french and ,english ,interlinear 7from ,merlet's 8,traducteur074 ,second ,edition4 #abmo4 #bs4 cl4 ,lombard ,de ,luc's ,classiques ,franais1 l',usage de la ,jeunesse ,protestante2 or1 ,selections from the best ,french ,classical ,works1 preceded by ,sketches of the ,lives and ,times of the ,writers4 #abmo4 #cs4 #f;d4 cloth4 ,,italian4 ,smith's ,first ,italian ,course2 being a ,practical and ,easy ,method of ,learning the ,elements of the ,italian ,language4 ,edited from the ,german of ,,filippi1 after the method of ,dr4 ,,ahn4 #abmo4 #cs4 #f;d4 cloth4 ,,interlinear ,,translations4 ,locke's ,system of ,classical ,instruction4 ,,interlinear ,,translations #as4 #f;d4 each4 ,latin4 ,phaedrus's ,fables of sop4 ,virgil's neid4 ,book ,i4 ,parsing ,lessons to ,virgil4 ,caesar's ,invasion of ,britain4 ,greek4 ,lucian's ,dialogues4 ,selections4 ,homer's ,iliad4 ,book ,i4 ,xenophon's ,memorabilia4 ,book ,i4 ,herodotus's ,histories4 ,selections4 ,french4 ,sismondi2 the ,battles of ,cressy and ,poictiers4 ,german4 ,stories from ,german ,writers4 ,also1 to accompany the ,latin and ,greek ,series4 ,the ,london ,latin ,grammar4 #abmo4 #as4 #f;d4 ,the ,london ,greek ,grammar4 #abmo4 #as4 #f;d4 ,,history1 ,,mythology1 ,,and ,,antiquities4 ,creasy's 7,professor7 ,history of ,england4 ,with ,illustrations4 ,one ,volume4 ,small #hvo4 ,uniform with ,schmitz's 8,history of ,rome10 and ,smith's 8,history of ,greece40 7,preparing74 ,schmitz's ,history of ,rome1 from the ,earliest ,times to the ,,death ,,of ,,commodus1 ,a4,d4 #aib4 ,ninth ,edition4 ,one ,hundred ,engravings4 #abmo4 #gs4 #f;d4 cloth4 ,smith's ,history of ,greece1 from the ,earliest ,times to the ,roman ,conquest4 ,with ,supplementary ,chapters on the ,history of ,literature and ,art4 ,new ,edition4 ,one ,hundred ,engravings on ,wood4 ,large #abmo4 #gs4 #f;d4 cloth4 ,smith's ,smaller ,history of ,greece4 ,with ,illustrations4 ,fcp4 #hvo4 #cs4 #f;d4 cloth4 ,smith's ,dictionary of ,greek and ,roman ,antiquities4 ,by various ,writers4 ,second ,edition4 ,illustrated by ,several ,hundred ,engravings on ,wood4 ,one thick volume1 medium #hvo4 #b #bs4 cloth4 ,smith's ,smaller ,dictionary of ,greek and ,roman ,antiquities4 ,abridged from the larger ,dictionary4 ,new ,edition4 ,crown #hvo41 #gs4 #f;d4 cloth4 ,smith's ,dictionary of ,greek and ,roman ,biography and ,mythology4 ,by various ,writers4 ,medium #hvo4 ,illustrated by numerous ,engravings on ,wood4 ,complete in ,three ,volumes4 #hvo4 #e #aes4 #f;d4 cloth4 ,smith's ,new ,classical ,dictionary of ,biography1 ,mythology1 and ,geography4 ,partly based on the 8,dictionary of ,greek and ,roman ,biography and ,mythology40 ,third ,edition4 #gej ,illustrations4 #hvo4 #ahs4 cloth4 ,smith's ,smaller ,classical ,dictionary of ,biography1 ,mythology1 and ,geography4 ,abridged from the larger ,dictionary4 ,illustrated by #bjj ,engravings on ,wood4 ,new ,edition4 ,crown #hvo4 #gs4 #f;d4 cloth4 ,smith's ,dictionary of ,greek and ,roman ,geography4 ,by various ,writers4 ,illustrated with ,woodcuts of ,coins1 ,plans of ,cities1 etc4 ,two ,volumes #hvo4 #d4 cloth4 ,niebuhr's ,history of ,rome4 ,from the ,earliest ,times to the ,first ,punic ,war4 ,fourth ,edition4 ,translated by ,,bishop ,,thirlwall1 ,,archdeacon ,,hare1 ,,dr4 ,,smith1 and ,,dr4 ,,schmitz4 ,three ,vols4 #hvo4 #a #afs4 ,niebuhr's ,lectures on the ,history of ,rome4 ,from the ,earliest ,times to the ,first ,punic ,war4 ,edited by ,,dr4 ,,schmitz4 ,third ,edition4 #hvo4 #hs4 ,newman 7,f4 ,w47 ,the ,odes of ,horace4 ,translated into ,unrhymed ,metres1 with ,introduction and ,notes4 ,crown #hvo4 #es4 cloth4 ,newman 7,f4 ,w47 ,the ,iliad of ,homer1 ,faithfully translated into ,unrhymed ,metre4 #a vol4 crown #hvo4 #fs4 #f;d4 cloth4 ,akerman's ,numismatic ,manual1 or ,guide to the ,collection and ,study of ,greek1 ,roman1 and ,english ,coins4 ,many ,engravings4 #hvo4 #a #as4 ,ramsay's 7,george7 ,principles of ,psychology4 #hvo4 #a,os4 #f;d4 ,,pure ,,mathematics4 ,de ,morgan's ,elements of ,arithmetic4 ,fifteenth ,thousand4 ,royal #abmo4 #es4 cloth4 ,de ,morgan's ,trigonometry and ,double ,algebra4 ,royal #abmo4 #gs4 #f;d4 cloth4 ,ellenberger's ,course of ,arithmetic1 as taught in the ,pestalozzian ,school1 ,worksop4 ,post #hvo4 #es4 cloth4 ,the ,answers to the ,questions in this ,volume are now ready1 price #as4 #f;d4 ,mason's ,first ,book of ,euclid4 ,explained to ,beginners4 ,fcap4 #hvo4 #as4 #i;d4 ,reiner's ,lessons on ,form2 or1 ,an ,introduction to ,geometry1 as given in a ,pestalozzian ,school1 ,cheam1 ,surrey4 #abmo4 #cs4 #f;d4 ,reiner's ,lessons on ,number1 as given in a ,pestalozzian ,school1 ,cheam1 ,surrey4 ,master's ,manual1 #es4 ,scholar's ,praxis1 #bs4 ,tables of ,logarithms ,common and ,trigonometrical to ,five ,places4 ,under the ,superintendence of the ,society for the ,diffusion of ,useful ,knowledge4 ,fcap4 #hvo4 #as4 #f;d4 ,four ,figure ,logarithms and ,anti-,logarithms4 ,on a ,card4 ,price #as4 ,barlow's ,tables of ,squares1 ,cubes1 ,square ,roots1 ,cube ,roots1 and ,reciprocals of all ,integer ,numbers1 up to #aj1jjj4 ,royal #abmo4 #hs4 ,,mixed ,,mathematics4 ,potter's ,treatise on ,mechanics1 for ,junior ,university ,students4 ,by ,,richard ,,potter1 ,m4,a41 ,professor of ,natural ,philosophy in ,university ,college1 ,london4 ,third ,edition4 #hvo4 #hs4 #f;d4 ,potter's ,treatise on ,optics4 ,part ,i4 ,all the requisite ,propositions carried to ,first ,approximations1 with the construction of ,optical ,instruments1 for ,junior ,university ,students4 ,second ,edition4 #hvo4 #is4 #f;d4 ,potter's ,treatise on ,optics4 ,part ,,ii4 ,the ,higher ,propositions1 with their application to the more perfect forms of ,instruments4 #hvo4 #abs4 #f;d4 ,potter's ,physical ,optics2 or1 the ,nature and ,properties of ,light4 ,a ,descriptive and ,experimental ,treatise4 #ajj ,illustrations4 #hvo4 #fs4 #f;d4 ,newth's ,mathematical ,examples4 ,a graduated series of ,elementary ,examples1 in ,arithmetic1 ,algebra1 ,logarithms1 ,trigonometry1 and ,mechanics4 ,crown #hvo4 ,with ,answers4 #hs4 #f;d4 cloth4 ,sold also in separate ,parts1 without ,answers3-- ,arithmetic1 #bs4 #f;d4 ,algebra1 #bs4 #f;d4 ,trigonometry and ,logarithms1 #bs4 #f;d4 ,mechanics1 #bs4 #f;d4 ,newth's ,elements of ,mechanics1 including ,hydrostatics1 with numerous ,examples4 ,by ,,samuel ,,newth1 ,m4,a41 ,fellow of ,university ,college1 ,london4 ,second ,edition4 ,large #abmo4 #gs4 #f;d4 cloth4 ,newth's ,first ,book of ,natural ,philosophy2 or an ,introduction to the ,study of ,statics1 ,dynamics1 ,hydrostatics1 and ,optics1 with numerous ,examples4 #abmo4 #cs4 #f;d4 cloth4 ,,natural ,,philosophy1 ,,astronomy1 ,etc4 ,lardner's ,museum of ,science and ,art4 ,complete in #ab ,single ,volumes1 #ahs41 ornamental boards2 or #f ,double ,ones1 #a #as41 cl4 lettered4 ,also1 handsomely half-bound morocco1 #f volumes1 #a #aas4 #f;d4 ,,contents3--,the ,planets2 are they inhabited ,worlds8 ,weather ,prognostics4 ,popular ,fallacies in ,questions of ,physical ,science4 ,latitudes and ,longitudes4 ,lunar ,influences4 ,meteoric ,stones and ,shooting ,stars4 ,railway ,accidents4 ,light4 ,common ,things4--,air4 ,locomotion in the ,united ,states4 ,cometary ,influences4 ,common ,things4--,water4 ,the ,potter's ,art4 ,common ,things4--,fire4 ,locomotion and ,transport1 their ,influence and ,progress4 ,the ,moon4 ,common ,things4--,the ,earth4 ,the ,electric ,telegraph4 ,terrestrial ,heat4 ,the ,sun4 ,earthquakes and ,volcanoes4 ,barometer1 ,safety ,lamp1 and ,whitworth's ,micrometric ,apparatus4 ,steam4 ,the ,steam ,engine4 ,the ,eye4 ,the ,atmosphere4 ,time4 ,common ,things4--,pumps4 ,common ,things4--,spectacles--,the ,kaleidoscope4 ,clocks and ,watches4 ,microscopic ,drawing and ,engraving4 ,the ,locomotive4 ,thermometer4 ,new ,planets4--,leverrier and ,adams's ,planet4 ,magnitude and ,minuteness4 ,common ,things4--,the ,almanack4 ,optical ,images4 ,how to ,observe the ,heavens4 ,common ,things4--,the ,looking ,glass4 ,stellar ,universe4 ,the ,tides4 ,colour4 ,common ,things4--,man4 ,magnifying ,glasses4 ,instinct and ,intelligence4 ,the ,solar ,microscope4 ,the ,camera ,lucida4 ,the ,magic ,lantern4 ,the ,camera ,obscura4 ,the ,microscope4 ,the ,white ,ants2 their ,manners and ,habits4 ,the ,surface of the ,earth1 or ,first ,notions of ,geography4 ,science and ,poetry4 ,the ,bee4 ,steam ,navigation4 ,electro-,motive ,power4 ,thunder1 ,lightning1 and the ,aurora ,borealis4 ,the ,printing ,press4 ,the ,crust of the ,earth4 ,comets4 ,the ,stereoscope4 ,the ,pre-,adamite ,earth4 ,eclipses4 ,sound4 ,lardner's ,animal ,physics1 or the ,body and its ,functions familiarly ,explained4 #ebj ,illustrations4 #a vol41 small #hvo4 #abs4 #f;d4 cloth4 ,lardner's ,animal ,physiology for ,schools 7chiefly taken from the 8,animal ,physics074 #aij ,illustrations4 #abmo4 #cs4 #f;d4 cloth4 ,lardner's ,hand-,book of ,mechanics4 #ceg ,illustrations4 #a vol41 small #hvo41 #es4 ,lardner's ,hand-,book of ,hydrostatics1 ,pneumatics1 and ,heat4 #bib ,illustrations4 #a vol41 small #hvo41 #es4 ,lardner's ,hand-,book of ,optics4 #bij ,illustrations4 #a vol41 small #hvo41 #es4 ,lardner's ,hand-,book of ,electricity1 ,magnetism1 and ,acoustics4 #cie ,illustrations4 #a vol41 small #hvo4 #es4 ,lardner's ,hand-,book of ,astronomy and ,meteorology1 forming a companion work to the 8,hand-,book of ,natural ,philosophy40 #cg ,plates1 and upwards of #bjj ,illustrations on ,wood4 #b vols41 each #es41 cloth lettered4 ,lardner's ,natural ,philosophy for ,schools4 #cbh ,illustrations4 #a vol41 large #abmo41 #cs4 #f;d4 cloth4 ,lardner's ,chemistry for ,schools4 #agj ,illustrations4 #a vol41 large #abmo4 #cs4 #f;d4 cloth4 ,pictorial ,illustrations of ,science and ,art4 ,large ,printed ,sheets1 each containing from #ej to #ajj ,engraved ,figures4 ,part ,i4 #as4 #f;d4 #a4 ,mechanic ,powers4 #b4 ,machinery4 #c4 ,watch and ,clock ,work4 ,part ,,ii4 #as4 #f;d4 #d4 ,elements of ,machinery4 #e4 ,motion and ,force4 #f4 ,steam ,engine4 ,part ,,iii4 #as4 #f;d4 #g4 ,hydrostatics4 #h4 ,hydraulics4 #i4 ,pneumatics4 ,lardner's ,popular ,geology4 7,from 8,the ,museum of ,science and ,art407 #bja ,illustrations4 #bs4 #f;d4 ,lardner's ,common ,things ,explained4 ,containing3 ,air--,earth--,fire--,water--,time--,the ,almanack--,clocks and ,watches--,spectacles--,colour--,kaleido scope--,pumps--,man--,the ,eye--,the ,printing ,press--,the ,potter's ,art--,locomotion and ,transport--,the ,surface of the ,earth1 or ,first ,notions of ,geography4 7,from 8,the ,museum of ,science and ,art407 ,with #bcc ,illustrations4 ,complete1 #es41 cloth lettered4 ,sold also in ,two ,series1 #bs4 #f;d4 each4 ,lardner's ,popular ,physics4 ,containing3 ,magnitude and ,minuteness--,atmosphere--,thunder and ,lightning--,terrestrial ,heat--,meteoric ,stones--,popular ,fallacies--,weather ,prognostics--,thermometer--,barometer-- ,safety ,lamp--,whitworth's ,micrometric ,apparatus--,electro-,motive ,power--,sound--,magic ,lantern--,camera ,obscura--,camera ,lucida--,looking ,glass--,stereoscope--,science and ,poetry4 7,from 8,the ,museum of ,science and ,art407 ,with #he ,illustrations4 #bs4 #f;d4 cloth lettered4 ,lardner's ,popular ,astronomy4 ,containing3 ,how to ,observe the ,heavens--,latitudes and ,longitudes --,the,earth--,the ,sun--,the ,moon--,the ,planets3 are they ,inhabited8--,the ,new ,planets--,leverrier and ,adams's ,planet--,the ,tides--,lunar ,influences--and the ,stellar ,universe--,light--,comets--,cometary ,influences--,eclipses--,terrestrial ,rotation--,lunar ,rotation--,astronomical ,instruments4 7,from 8,the ,museum of ,science and ,art407 #ahb ,illustrations4 ,complete1 #ds4 #f;d4 cloth lettered4 ,sold also in ,two ,series1 #bs4 #f;d4 and #bs4each4 ,lardner on the ,microscope4 7,from 8,the ,museum of ,science and ,art407 #a vol4 #adg ,engravings4 #bs4 ,lardner on the ,bee and ,white ,ants2 their ,manners and ,habits2 with ,illustrations of ,animal ,instinct and ,intelligence4 7,from 8,the ,museum of ,science and ,art407 #a vol4 #ace ,illustrations4 #bs41 cloth lettered4 ,lardner on ,steam and its ,uses2 including the ,steam ,engine and ,locomotive1 and ,steam ,navigation4 7,from 8,the ,museum of ,science and ,art407 #a vol41 with #hi ,illustrations4 #bs4 ,lardner on the ,electric ,telegraph1 ,popularised4 ,with #ajj ,illustrations4 7,from 8,the ,museum of ,science and ,art407 #abmo41 #bej pages4 #bs41 cloth lettered4 ,the following ,works from 8,lardner's ,museum of ,science and ,art10 may also be had arranged as described1 handsomely half bound morocco1 cloth sides4 ,common ,things4 ,two series in one vol4 #gs4 #f;d4 ,popular ,astronomy4 ,two series in one vol4 #gs4 #j;d4 ,electric ,telegraph1 with ,steam and its ,uses4 ,in one vol4 #gs4 #j;d4 ,microscope and ,popular ,physics4 ,in one vol4 #gs4 #j;d4 ,popular ,geology1 and ,bee and ,white ,ants4 ,in one vol4 #gs4 #f;d4 ,lardner on the ,steam ,engine1 ,steam ,navigation1 ,roads1 and ,railways4 ,explained and ,illustrated4 ,eighth ,edition4 ,with numerous ,illustrations4 #a vol4 large #abmo4 #hs4 #f;d4 ,a ,guide to the ,stars for every ,night in the ,year4 ,in ,eight ,planispheres4 ,with an ,introduction4 #hvo4 #es41 cloth4 ,minasi's ,mechanical ,diagrams4 ,for the ,use of ,lecturers and ,schools4 #ae ,sheets of ,diagrams1 coloured1 #aes41 illustrating the following subjects3 #a and #b4 ,composition of ,forces4--#c4 ,equilibrium4--#d and #e4 ,levers4--#f4 ,steelyard1 ,brady ,balance1 and ,danish ,balance4--#g4 ,wheel and ,axle4--#h4 ,inclined ,plane4--#i1 #aj1 #aa4 ,pulleys4--#ab4 ,hunter's ,screw4--#ac and #ad4 ,toothed ,wheels4--#ae4 ,combination of the ,mechanical ,powers4 ,,logic4 ,de ,morgan's ,formal ,logic2 or1 ,the ,calculus of ,inference1 ,necessary and ,probable4 #hvo4 #fs4 #f;d4 ,neil's ,art of ,reasoning3 a ,popular ,exposition of the ,principles of ,logic1 ,inductive and ,deductive2 with an ,introductory ,outline of the ,history of ,logic1 and an ,appendix on recent ,logical ,developments1 with ,notes4 ,crown #hvo4 #ds4 #f;d41 cloth4 ,,english ,,composition4 ,neil's ,elements of ,rhetoric2 a ,manual of the ,laws of ,taste1 including the ,theory and ,practice of ,composition4 ,crown #hvo4 #ds4 #f;d41 cl4 ,,drawing4 ,lineal ,drawing ,copies for the earliest ,instruction4 ,comprising upwards of #bjj subjects on #bd sheets1 mounted on #ab pieces of thick pasteboard1 in a ,portfolio4 ,by the ,author of 8,drawing for ,young ,children40 #es4 #f;d4 ,easy ,drawing ,copies for ,elementary ,instruction4 ,simple ,outlines without ,perspective4 #fg subjects1 in a ,portfolio4 ,by the ,author of 8,drawing for ,young ,children40 #fs4 #f;d4 ,sold also in ,two ,sets4 ,,set ,i4 ,twenty-six ,subjects mounted on thick pasteboard1 in a ,portfolio4 #cs4 #f;d4 ,,set ,,ii4 ,forty-one ,subjects mounted on thick pasteboard1 in a ,portfolio4 #cs4 #f;d4 ,the copies are sufficiently large and bold to be drawn from by forty or fifty children at the same time4 ,,singing4 ,a ,musical ,gift from an ,old ,friend1 containing ,twenty-four ,new ,songs for the ,young4 ,by ,w4 ,e4 ,,hickson1 author of the ,moral ,songs of 8,the ,singing ,master40 #hvo4 #bs4 #f;d4 ,the ,singing ,master4 ,containing ,first ,lessons in ,singing1 and the ,notation of ,music2 ,rudiments of the ,science of ,harmony2 ,the ,first ,class ,tune ,book2 ,the ,second ,class ,tune ,book2 and the ,hymn ,tune ,book4 ,sixth ,edition4 #hvo4 #fs41 cloth lettered4 ,sold also in ,five ,parts1 any of which may be had separately4 ,i4--,first ,lessons in ,singing and the ,notation of ,music4 ,containing ,nineteen ,lessons in the ,notation and ,art of ,reading ,music1 as adapted for the ,instruction of ,children1 and especially for ,class ,teaching1 with ,sixteen ,vocal ,exercises1 arranged as simple two-part harmonies4 #hvo4 #as41 sewed4 ,,ii4--,rudiments of the ,science of ,harmony or ,thorough ,bass4 ,containing a general view of the principles of ,musical ,composition1 the ,nature of ,chords and ,discords1 mode of applying them1 and an ,explanation of ,musical ,terms connected with this branch of ,science4 #hvo4 #as41 sewed4 ,,iii4--,the ,first ,class ,tune ,book4 ,a ,selection of ,thirty ,single and ,pleasing ,airs1 arranged with suitable words for young children4 #hvo4 #as41 sewed4 ,,iv4--,the ,second ,class ,tune ,book4 ,a ,selection of ,vocal ,music adapted for youth of different ages1 and arranged 7with suitable words7 as two or three-part harmonies4 #hvo1 #as4 #f;d4 ,v4--,the ,hymn ,tune ,book4 ,a ,selection of ,seventy popular ,hymn and ,psalm ,tunes1 arranged with a view of facilitating the progress of ,children learning to sing in parts4 #hvo4 #as4 #f;d4 ,the ,vocal ,exercises1 ,moral ,songs1 and ,hymns1 with the ,music1 may also be had1 printed on ,cards1 price ,twopence each ,card1 or ,twenty-five for ,three ,shillings4 ,,chemistry4 ,gregory's ,hand-,book of ,chemistry4 ,for the use of ,students4 ,by ,,william ,,gregory1 ,m4,d41 late ,professor of ,chemistry in the ,university of ,edinburgh4 ,fourth ,edition1 revised and enlarged4 ,illustrated by ,engravings on ,wood4 ,complete in ,one ,volume4 ,large #abmo4 #ahs4 cloth4 ,the ,work may also be had in two ,volumes1 as under4 ,,inorganic ,,chemistry4 ,fourth ,edition1 revised and enlarged4 #fs4 #f;d4 cloth4 ,,organic ,,chemistry4 ,fourth ,edition1 very carefully revised1 and greatly enlarged4 #abs41 cloth4 7,sold separately47 ,chemistry for ,schools4 ,by ,,dr4 ,,lardner4 #aij ,illustrations4 ,large #abmo4 #cs4 #f;d4 cloth4 ,liebig's ,familiar ,letters on ,chemistry1 in its ,relations to ,physiology1 ,dietetics1 ,agriculture1 ,commerce1 and ,political ,economy4 ,fourth ,edition1 revised and enlarged1 with additional ,letters4 ,edited by ,,dr4 ,,blyth4 ,small #hvo4 #gs4 #f;d4 cloth4 ,liebig's ,letters on ,modern ,agriculture4 ,small #hvo4 #fs4 ,liebig's ,principles of ,agricultural ,chemistry2 with ,special ,reference to the late ,researches made in ,england4 ,small #hvo4 #cs4 #f;d41 cloth4 ,liebig's ,chemistry in its ,applications to ,agriculture and ,physiology4 ,fourth ,edition1 revised4 #hvo4 #fs4 #f;d41 cloth4 ,liebig's ,animal ,chemistry2 or1 ,chemistry in its ,application to ,physiology and ,pathology4 ,third ,edition4 ,part ,i4 7the first half of the work74 #hvo4 #fs4 #f;d41 cloth4 ,liebig's ,hand-,book of ,organic ,analysis2 containing a detailed ,account of the various ,methods used in determining the ,elementary ,composition of ,organic ,substances4 ,illustrated by #he ,woodcuts4 #abmo4 #es41 cloth4 ,bunsen's ,gasometry2 comprising the ,leading ,physical and ,chemical ,properties of ,gases1 together with the ,methods of ,gas ,analysis4 ,fifty-nine ,illustrations4 #hvo4 #hs4 #f;d41 cloth4 ,whler's ,hand-,book of ,inorganic ,analysis2 ,one ,hundred and ,twenty-two ,examples1 illustrating the most important processes for determining the ,elementary composition of ,mineral substances4 ,edited by ,,dr4 ,a4 ,w4 ,,hofmann1 ,professor in the ,royal ,college of ,chemistry4 ,large #abmo4 ,parnell on ,dyeing and ,calico ,printing4 7,reprinted from ,parnell's 8,applied ,chemistry in ,manufactures1 ,arts1 and ,domestic ,economy1 #ahdd407 ,with ,illustrations4 #hvo4 #gs41 cloth4 ,,general ,,literature4 ,de ,morgans ,book of ,almanacs4 ,with an ,index of ,reference by which the ,almanac may be found for every ,year1 whether in ,old ,style or ,new1 from any ,epoch1 ,ancient or ,modern1 up to ,a4,d4 #bjjj4 ,with means of finding the ,day of ,new or ,full ,moon1 from ,b4,c4 #bjjj to ,a4,d4 #bjjj4 #es41 cloth lettered4 ,guesses at ,truth4 ,by ,two ,brothers4 ,new ,edition4 ,with an ,index4 ,complete #a vol4 ,small #hvo4 ,handsomely bound in cloth with red edges4 #ajs4 #f;d4 ,rudall's ,memoir of the ,rev4 ,james ,crabb2 late of ,southampton4 ,with ,portrait4 ,large #abmo41 #fs41 cloth4 ,herschell 7,r4 ,h47 ,the ,jews2 a brief ,sketch of their ,present ,state and ,future ,expectations4 ,fcap4 #hvo4 #as4 #f;d41 cloth4