BOUGHT WITH THB INCOME . FROM THE SAGE ENDOWMENT FUND THE GIFT OF 1S91 Lft.'it^^&Xo.D ......__..„ ItVi [53.. ^ ■"^ssons on number : olin,anx 3 1924 031 286 226 Cornell University Library The original of tiiis bool< is in tile Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031286226 LESSONS ON NUMBER, IN A PESTALOZZIAN SCHOOL CHEAM, SURREY. K\ft plaster's JHanuaU By C. EEINEE, TEACHER OF SIATHEMATICS AT CHEAM SCHOOL. FOURTH EDITION, CORRECTED. LONDON: WALTON AND MABEELY, UPPEU GOWEE STREET, AND TVY LANE, PATEKNOSTEB BOW. 1857. A. H-4-^C.o H^UHa) ^z&\\s. T. Name any objects of which you may say Three. P. Three chairs, &c. Adding successively a ball to the former num- ber, the teacher each time requires the pupils to point out some objects to which the appellation, four, five, six, &c., is applicable. It must be left to his discretion to determine where to stop. One child will be embarrassed by having ten or twenty objects before him, whereas another will, at one glance, ascertain their number. In the one the power of perception must be developed by a slow and gradual process; in the other, it will rapidly strengthen as larger collections of objects are suc- cessively presented. In order, however, to lead to the accurate con- ception and correct expression of number, it is desirable at this step to put before the pupils promiscuous objects, requiring them to ascertain their total number, as well as the number of those among them, which are of the same kind. Thus, for instance; putting 3 balls, 4 books, 5 slates, &c., before them, the teacher says — LESSONS ON NUMBER. 3 T. What must you do to ascertain how many objects there are here? P. We must count them. T. Count them. P. There are twelve objects. T. Twelve, then, is the number of objects before you; are they aU of the same kind? P. No; there are balls, books, and slates. T. Ascertain the number of each class of objects. P. The number of balls is three; the number of books four; and the number of slates five. Exercises of this kind may be much and inter- estingly diversified, especially for very young children, or for those whose perception is slow, by directing their attention to the number of the various parts of which one object is composed, or the number of plants, trees, birds, fishes, shells, minerals, &c.,. which they know. The teacher is referred to "Exercises on Les- sons on Number." b2 CHAPTER L— ADDITION. § 1. ADDITION or UNITS. Lesson I. To add One. The first and simplest lesson in Number is evi- dently that in which the pupil is taught to add one to each number in succession; and then to numbers taken promiscuously. To connect this operation of the mind with the idea the pupil has formed of number from the previous exercises, re- course must be had to the senses, by making use of some visible objects, either the numeral frame or the slate. This being ultimately abandoned, the exercise will become purely intellectual. In the following exercises the slate is supposed to be used. T. [Drawing one small line upon the slate, asks] How many lines are there here? | P. One. T. [Draws beneath the former two lines, and asks] Hovr many lines are there here? | i P. Two. ADDITION. 5 T. How many are one line more one line? P. Two lines. T. How many are one book more one book? — One tree more one tree? — One slate more one slate? How much are one more one? How many ones make two? T. [Drawing three lines beneath the former twOj asks] How many lines are here? ( | | P. Three. T. How many more lines are there in this row than in the one above? P. One more, T. How many lines are in the second row? P. Two, T. How many lines are two more one? P. Three lines. T. How much is two more one? P. Three. T. [Draws four lines beneath the former three, asking] How many more lines are there in this row than in the one above? 1 1 I I P. One more. T. How many are in the third row. P. Three. 2. How many are three lines more one? P. Four lines. T. How much is three more one? P. Four. T. [Draws yit)e lines beneath the former four; six, seven lines, and so on in succession, asking LESSONS ON NUMBEK. each time questions analogous to the above. These lines will stand arranged thus : — I I After due repetition, the pupil must be able to proceed as follows: [the lines on the slate being effaced.] One more one are two; Two more one are three; Three more one are four; Four more one are five; Five more one are six; Six more one are seven, &c., &c.j &c. The teacher will do well to let his pupils pro- ceed to the very limit of their distinct conception of number. After this exercise, which the whole class should repeat viv^ voce, each pupil may be required to ask the class one or more questions ADDITION. 7 promiscuously. Thus, How mucli is nine more one? — Seventeen more one? &c. It frequently happens that the teacher's class consists of several divisions of pupils, more or less advanced, and the time he can devote to each is, in consequence, limited. Exercises have, there- fore, been drawn up corresponding to this and the following: lessons, and may be given to the pupils for solution. The answers are, for the present, to be written on their slates in words. Another advantage of these exercises, and that not perhaps the least, is, that they lead each pupil in silence to think for himself, and commit the result of his thoughts to writing; — a process much calculated to sober the mind after the exciting effects of simultaneous learning. Answers to the Exercises. Lesson I. To add One. N.B. The answers to the first six questions must be inspected by the teacher. i.7. 15 Ans. 13. 74 Ans.ld. 101 8. 27 14. 79 20. 109 9. 38 15. 81 21. 127 10. 49 16. 87 22. 149 11. 56 17. 94 23. 155 12. 63 18. 97 24. 187 LESSONS ON NUMBER. Lesson II. To add Two. T. [t)rawing one line on the slate, and next to it two lines more, thus] asks How many lines are there in all? P. Three lines. T. [Drawing two lines, and next to them two lines more, three lines, four lines, &c., adding each time two lines, as below,] asks I i I How many lines are one more two? two more two? three more two lines? &c., &c. P. Two lines more two lines are four lines; Three lines more two lines are five lines; Four lines more two lines are six lines; &c., &c. After due repetition of the above, the drawing of lines on the slate should be abandoned, if possible. It is far more improving to render the operation purely intellectual, than to continue deriving assist- ance from the external senses. ADDITION. 9 To add twOj is, evident^, to add onej and tlien another one in succession. If, then. Lesson I. he perfectly known, there will he little difficulty in leading the pupil to increase any numher hy two at one step. The mode made use of to this end, is shown in the following questions : — Q. How many ones make two ? A. One more one ; that is two ones. Q. How many ones, then, make one more two? A. One more one, more one; that is three. .Q. How much is two more two? A. Two more one, more one, or four. Q How much is three more two ? A. Three more one, more one, or five. Q. How much is four more two? five more two? six more two? &c. The pupils must, after due repetition of such exercises, be able to proceed readily as follows: — One more two are three; Two more two are four; Three more two are five; Four more two are six; Five more two are seven; &c., &c. For exercise, it is recommended to require the pupils to count by twos, as they have learned to count by ones; that is, the teacher begins at any number he pleases, and calls upon the pupils to add two, and to the answer two again, and so on in succession, as far as they are able, or as may be sufficient for practice. b3 10 LESSONS ON NUMBEB. Answers to the Exercises. Lesson II. Ans.l. 19 Am. 5. 63 Ans.9. 103 2. 27 6. 71 10. 121 3. 49 7. 83 11. 151 4. 58 8. 91 13 301 Am. 13. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 33, &c. Ans. 14. 3, 4, 6, 8, 10, 12, 14, 16, 18, 20, 32, 34, &c. Lesson III. To add Three. Before entering upon any new lesson, the teacher must be quite certain that the preceding is firmly- fixed in the minds of the pupils. On this sup- position, the present lesson will require but little effort on their part. They are able to add two, also to add one, and are now required to add three; that is, two more one. Teacher. You have learnt to add two; if, now, instead of two, you were required to add three, what would you do? Pupils. Mrst add two, and then one more. T. How much is one more three ? P. One more two, more one, or four. T. How much is two more three ? P Two more two, more one, or five. T. How much is three more three ? ADDITION. 1 1 P. Three more two, more one, or six. T. How much is four more three? five more three? six more three ? &c., &c. After due repetition^ the pupil must be able to proceed readily thus : — One more three are four ; Two more three are five ; Three more three are six ; Pour more three are seven; &c., &c. The teacher may now call upon the pupils to give questions promiscuously to the class. Pupil. How much is seventeen more three ? Class. Twenty. P. How much is fifty-nine more three ? C. Sixty-two. &c., &c., The whole class then begin to count by adding three in succession ; thus : — C. Three, six, nine, twelve, fifteen, &c. Again, C. One, four, seven, ten, thirteen, &c. And again, C. Two, five, eight, eleven, fourteen, &c. As a repetition and preparation for the next lesson, the teacher may combine the previous two lessons with the last. Thus : — T. How much is one more three, more one? two more three, more one ? four more three, more one? five more three, more one? &c., &c. 12 LESSONS ON NUMBER. Again, T. How much is one more two^ more two? two more twOj more two ? three more two, more two ? &c., &c. And again, T. How much is one more two, more three? two more two, more three? three more three, more three? &c. Then, promiscuously : — T. How much is five more three, more two? seventeen more two, more two ? thirty-nine more two, more three? forty-five more one, more two, more three? It will much tend to enliven the pupils, if the teacher permit them to give questions, similar to the above, to the class. For instance, one pupil How much is forty-three more three, more two? Those who have found the answer, hold up one hand. The pupil who gave the question chooses among them whom he pleased to answer it; and it is he, likewise, who is to say whether the answer be right or wrong. ADDITION. 13 Answers to the Exercises. Lesson III. Ans. 1. 17 Ans. 4. 91 Ans. 7. 60 2. 32 5. 78 8. 102 3. 40 6. 101 9. 121 10. 152 Ans.U. 4, 7, 10, 13, 16, 19, &c. 12. 5, 8, 11, 14, 17, 20, &c. 13. 6, 9, 12, 15, 18, 21, &c. Ans. 14. 10 Ans. 16. 41 Ans. 18. 92 15. 23 17. 62 19. 104 20. 96 Lesson IV. To add Four. This and the following lessons are quite analo- gous to those preceding; and it is not without reason that the teacher is recommended to ob- serve a progressive order in these lessons, always beginning with the simplest form possible. Accord- ingly, the number four is added to one, then to two, next to three, and so on in succession, as far as the teacher may think proper, witMn the limit of the pupil's clear conception of number. The answers, being the numbers 6, 6, 7, &c., in their natural order, are readily found by the pupils, which serves as encouragement ; but to arrive at the aim in view, namely, to add four to any num- ber at one step, the teacher requires the pupils to 14 LESSONS ON NUMBER. count by fours; that is, he begins with adding four to one; to the answer four again is added^ and so on, which gives rise to the following pro- gression : — 1, 5, 9, 13, 17, 21, 35, 29, 33, &c. Two is then taken as the beginning of the next series; thence the following progression: 2, 6, 10, 14, 18, 22, 26, 30, 34, &c. Three now begins the series, whence the an- swers, 3, 7, 11, 15, 19, 23, 27, 31, 35, &c. And, finally, four is made the beginning from which the answers, 4, 8, 12, 16, 20, 24, 28, 32, 36, &c. It is obvious that by this mode of proceeding, four is added to each of the numbers, 1, 2, 3, 4, &c., and the subject quite exhausted. The pupils must be able to repeat these four series, with- out hesitating, before they proceed to the next lesson. Promiscuous questions for practice are left to be given by each pupil to the class. Answers to the Exercises. Lesson IV. Ans.l. 43,82,91,99,123. 2. 22 Ans.4:. 66 Ans.6. 107 3. 43 5. 95 7. 43, 47, 51, 55, 59, 63, &c. 8. 61,64,66,70,73,75,79,82,84, 88, 91, 93, 97, 100, 102, 106, &c. ADDITION. 15 Lesson V. To add Five. The difficulty of adding a number at one step increases witli tlie number itself; but it may be removed in a great measure, by inducing tbe pupils to use tbe knowledge already gained. Thus to add five is evidently to add four more onej or to add three more two; or two more two more one; ope- rations already performed by them. The teacher ought therefore to object to the pupils counting by their fingers, but constantly require them to refer to that actually known, arid to use it accordingly. The teacher may proceed thus : — Teacher. You have learnt to add four; if now it were required to add five, what would you do? Pupil. First add four and then one more. T. But if you began with adding three, what then must be done? P Add two more. T. How much then is seventeen more five ? P. Seventeen more four more one, that is twenty-two. T. Or by adding three first. P. Seventeen more three more two, which is twenty-two. T. The answer twenty-two is called the sum of seventeen and five; and whenever two or more numbers are added together, the answer is 16 LESSONS ON NUMBEK. called their sum. What is the sum of thirty-seven and five. P. Forty-two. T. What is the sum of one, two, three, four, and five? P. Fifteen. We will now learn to add five at one step, without first adding four, and then one, or three, and then two. Then follow exercises similar to those in the preceding lessons. T. How much is one more five? two more five? three more five? four more five? &c., &c. After which one is made the beginning of a series, five is added, to the sum five again, and so on. Then two is made the beginning, three next, four next, and finally five, which completes the lesson. The answers, which the pupils must be able to repeat vivd voce, are respectively as follow — Class, count 1, 6, 11, 16, 21, 26, 31, 36, 41, &c. 2,7, 12, 17, 22, 27, 32, 37,. 42, &c. 3, 8, 13, 18, 23, 28, 33, 38, 43, &c. 4, 9, 14, 19, 24, 29, 34, 39, 44, &c. 5, 10, 15, 20, 25, 30, 35, 40, 45, &c. This mode of proceeding will soon be remarked by the pupil; it will awaken his mind to reflec- ADDITION. 17 tion, and he will anticipate tlie questions of his master. Before entering upon the next lesson, each pupil gives questions to the class, now making use of the word. sum. Thus: — P. "What is the sum of fifty-eight and five? &c. Answers to the Exercises. Lesson V. Ans. 1. 48 Ans.4<. 112 Ans.7. 114 2. 69 5. 151 8. 203 3. 99 6. 162 9. 224 10. 22, 26, 29, 31, 32, 37, 41,44,46,47, 52, 56, 59, 61, 62, 67, 7 1,74, 76,77, &e. Lesson VI. To add Six. The mode of proceeding being quite in accord- ance with that detailed in the foregoing lessons, it will be sufficient to give a general outline of the exercises which follow in this paragraph. 1st, Teacher. How much is one more six? two more six? three more six? &c. &c. 2d, Class, count 1, 7, 13, 19, 25, 31, &c. 2, 8, 14, 20, 26, 32, &c. 3, 9, 15, 21, 27, 33, &c. 4, 10, 16, 22, 28, 34, &c. 5, 11, 17, 23, 29, 35, &c. FinaUy, ... 6, 12, 18, 24, 30, 36, &c. 18 LESSONS ON NUMBER. 3d. Promiscuous questions given by each pupil to the Class. Ex. 1. Pupil. How much is fifty-nine more six? Ex. 2. P. Add seventy-three and six. Ex. 3. P. What is the sum of thirty-seven and six. Ex. 4. P. What is the sum of eight, six, five, and four. N.B. — The pupils should be required to use the words, "more," "add," "sum" in order to form correct notions of their signification. Answers to the Exercises. Lesson VI. (. 1. 53 Ans. 6. 76 Ans .11. 89 2. 21 7. 126 12. 93 3. 110 8. 56 13. 110 4. 118 9. 64 14. 116 5. 147 10. 78 15. 255 Lesson VII. To add Seven. Outline of the various Exercises of this Lesson. 1st. Teacher. How much is one more seven? two more seven ? three more seven? &c., &c. ADDITION. 19 2d. Class, count 1, 8, 15, 22, 29, 36, 43, 50, &c. 2, 9, 16, 33, 30, 37, 44, 51, &c. 3, 10, 17, 24, 31, 38, 45, 52, &c. 4. 11, 18, 25, 33, 39, 46, 53, &c- 5. 12, 19, 26, 33, 40, 47, 54, &c. 6, 13, 20, 37, 34, 41, 48, 55, &c. Finally, ... 7, 14, 21, 28, 35, 42, 49, 56, &c. 3d. Promiscuous questions given by the pupils: Ex. 1. Pupil. How much is forty-three more seven? P. Increase seventy-seven by seven. P. Add eighty-nine and seven. P. Find the sum of ninety-eight and Ex. 2. Ex.3. Ex.4, seven. Ex.5. P. Find the sum of one, two, three. four, five, six, and seven. Answers to the Exercises. Lesson VII. Ans.l. 67 Ans.Q. 83 Ans .11. 66 2. 77 7. 90 12. 132 3. 111 8. 114 13. 127 4. 124 9. 122 14. 201 5. 129 10. 201 15. 326 Lesson VIII. To add Eight. Outline of this 'Lesson. 1st. How much is one more eight? two more eight? three more eight ? 20 LESSONS ON NUMBER. 2d. Class, count 1, 9, 17, 25, 33, 41, &c. 2, 10, 18, 26, 34, 43, &c. 3, 11, 19, 21, 35, 43, &c. 4, 12, 20, 28, 36, 44, &c. 5, 13, 21, 29, 37, 45, &c. 6, 14, 22, 30, 38, 46, &c. 7, 15, 23, 31, 39, 47, &c. Finally, ... 8, 16, 24, 32, 40, 48, &c. 3d. Promiscuous questions given by the pupils : Ex. 1 . How much is forty-five more eight? Ex. 2. Increase seventy-eight by eight. Ex. 3. Add eighty-seven and eight. Ex. 4. Find the sum of ninety-six and eight. Ex. 5. Find the sum of two, seven, five, and eight. Answers to the Exercises. Lesson VIII. Ans.\. 83 An. .6. 131 Ans .11. 65 2. 109 7. 138 13. 83 3. 124 8. 131 13. 89 4. 143 9. 304 14. 132 5. 251 10. 335 15. 134 Lesson IX. To add Nine. Outline of this Lesson, 1st. Teacher. How much is one more nine? two more nine ? three more nine? &c., &c. ADDITION. 21 2d. Class, count 1, 10, 19, 28, 37, &c. 2, 11, 20, 29, 38, &c. 3, 12, 21, 30, 39, &c. 4, 13, 22, 31, 40, &c. 5, 14, 23, 32, 41, &c. 6, 15, 24, 33, 42, &c. 7, 16, 25, 34, 43, &c.- 8, 17, 26, 35, 44, &c. Finally, ... 9, 18, 27, 36, 45, &c. 3d. Promiscuous questions given by the pupils: Ex. 1. Fupil. How much is sixty-three more nine? Ex. 2. P. Ex. 3. P. Ex. 4. P. and nine? Ex. 5. P. Find the sum of thirteen, nine, seven, and six. Increase forty-seven by nine. Add seventy-eight and nine. What is the sum of ninety-two Answers to the Exercises. Lesson IX. Ans.l. 50 2. 100 3. 99 4. 116 5. 174 Ans.Q. 72 7. 84 8. 129 9. 101 10. 138 .4ns. 11. 193 12. 224 13. 243 14. 262 15. 431 Lesson X. Decomposition of Numbers into Tens and Units. The decimal system of numeration being that generally adopted, it is of importance to lead the 22 LESSONS ON NUMBER. pupils, as far as their tender years allow, to the perception of its advantages: and this may be done at the present stage. Teacher. Count as far as ten. Pupils. [Counting,] One, two, ten. T. The number next above ten is eleven — that is, ten more ? P. Ten more one. T. The number following is twelve; that is ten — P. Ten more two. T. Continue counting, thirteen is ten — P. Thirteen is ten more three; Fourteen is ten more four; Fifteen is ten more five; Sixteen is ten more six; Seventeen is ten more seven; Eighteen is ten more eight; Nineteen is ten more nine; Twenty is ten more ten; Twenty-one is ten more eleven; T. You said before, that eleven was ten more one; what, then, will you say that twenty-one is? P. Ten more ten, more one. T. That is how many tens? P. Two tens more one. T. Continue, twenty-two is two tens — P. Twenty-two is two tens more two. T. Two what? P. Two ones. ADDITION. 23 T. Proceed: twenty-three is two tens — P. Twenty-three is two tens more three ones; T\7enty-four is two tens more four ones; &c., &c. Twenty-nine is two tens more nine ones. T. Instead of saying ones, it is usual to say units. Proceed. P. Thirty is two tens more ten units. T. That is how many tens? P. Three tens. Thirty-one is three tens more one unit^ &c., &c. Forty is three. tens more ten units; that is four tens. The pupils are required to continue in a similar manner as far as the teacher may judge neces- sary. T. Hence it is easy to say of how many tens and units a number consists. Thus, eighty-eight consists of how many units? P. Of eighty-eight-units. T. And of how many tens? P. Of eight tens and eight units. T. By which part of your body are you apt to count? P. By our fingers. T. And how many fingers have we? P. Ten. T. Whence, then, do you think originates our counting thus by tens? 24 LESSONS ON NUMBER. P. Probably from the circumstance of our baving ten fingers. T. How many tens are in eleven more ten? P. Two tens more one nnit. T. How many units is that? P. Twenty-one. T. Continue adding ten to twelve, thirteen, &c. P. Twelve more ten is two tens more two units, or twenty-two; Thirteen more ten is two tens more three^ or twenty-three; Fourteen more ten is two tens more four, or twenty-four; Fifteen, &c. T. Hence, if ten is to be added to a number, what must be done? P. Only increase the number of tens by one ten. T. Give an example. P. Thirty-eight more ten, is three tens more eight units, more one ten; that is, four tens more eight units, or forty-eight. It will be convenient, before proceeding further, to teach the pupil how to represent numbers by signs, not that they are essential to the succeeding lessons, which are intended as practice in mental arithmetic; but because the written exercises would, without the aid of signs, become cumbrous. In a school, then, where the teacher's time is divided among several classes, more or less advanced, the ADDITION. 25 necessity of the exercises will be felt; and for sxich it is recommended to have recourse to written arithmetic : whereas in a family of a few children, no such necessity existing, it will be better to de- fer it, and continue the exercises purely mental, until the pupils' minds^ from the following lessons are more developed. How to represent Units hy Signs. I. — Units. Teacher. We will now learn to represent num- bers by signs [writing the following on the large school-slate:] 1. 2. 3. 4. 5. 6. 7. 8. 9. [Poiating to the lines, and their respective repre- sentatives, says, and pupils repeat,] One, two, three, &c. These signs are cdiAeA. figures ; imitate them on your own slates. 26 LESSONS ON NUMBER. T. What numbers to these figures represent? Pupils. Ones, or units. T. Hence the figure 2 is equal in value to which two figures added together? P. To 1 more 1. T. I will teach you now another sign, generally- used to denote the word "more;" it is this, +. If, then, we wish to write down what we have just said, we would, instead of writing thus, 2 is equal in value to 1 more 1, write, 2 is equal in value to 1 + 1. For the words, "is equal in value," or "are equal in value," it has been agreed to substitute the sign =; so that 2 is equal in value to 1 more 1, is written, 2 = 1 + 1. What advantage has this sort of short-hand writing? P. It saves time and room. T. Let one of you come here to the slate, and, by means of these signs> write which three num- bers the figure 3 is equal to. P. [Writing.] 3=1 + 1 + 1. T. And which two figures is 3 equal to? P. \_Writing.] 3=2 + 1. T. Proceed in a similar manner with the figure 4. P. [Writing.] 4=1 + 1 + 1 + 1. 4=2 + 1 + 1. 4=2 + 2. 4=3 + 1. ADDITION. 27 T. Letj now, every one of you proceed in the same way with the remaining figures, b, 6, 7, 8, and 9, on your own slates. II.— Tens. T. What is the number next above nine? P. Ten. T. Represent ten by the addition of two of the figures you have learnt. P. 9 + 1; 8 + 3j 7 + 3, &c. T. How many ones, or units, are there in ten? P. Ten units. T. And how many tens? P. One ten. T. Which of the figures represents one? P. 1. T. If, now, we wish to represent one ten, will it be sufficient to write 1 ? P. No ; because we could not tell whether one was meant or ten. T. In order to raise this difficulty, I wiU place this sign, 0, called zero, next to the 1, showing that the 1 is meant for 1 ten; thus, 10. Tell me, now, which place occupies with respect to thel. P. It stands behind the 1; it stands before the 1. T. Some of you said, "behind," others, "be- fore;" which is right? c3 28 ' LESSONS ON NUMBER. P. "Behind;" because we read and write from the left to the right. T. So we do with the letters of the alphabet; but are you sure it is the same when signs are used instead of letters? You mentioned "left" and "right;" which of those places does occupy in respect to one? P. It stands to the right of it. T. Well; tell me^ now, how many units there are in ten. P. Ten units. T. And how many tens ? P. One ten. T. So that if we suppose these ten units col- lected, there would be one ten, and no more units. "Which, now, of the figures 10, indicates one ten, and which no units ? P. The one in 10 indicates one ten, and the zero no units. T. And you have before remarked that the figure zero stands where? P. To the right of one. T. Hence we shall for the future agree, that, of two figures, that which stands to the right of the other indicates what? P. Units. T. And what to the left? P. Ten. T. How am I to represent two tens, or twenty? ADDITION. 29 P. Write the figure two, and to the right of it zero. T. Let one of you come here and Trnte three tens, or thirty; four tens, or forty; fifty, sixty, &e., ninety. T. How many tens are there in one hun- dred? F. Ten tens. T. You have learnt to write ten; we have agreed upon the place which the tens should occupy; — ^try and express one hundred. It must he left to the pupils to represent that number correctly; when done, the teacher pro- ceeds : — T. How many figures have you used to repre- sent one hundred? P. Three figures. T. How many units are there in one hun- dred? P. One hundred units. T. How many tens? P. Ten tens. T. And how many hundreds? P. One hundred. T. So that if we suppose these hundred units collected into tens, there would be how many of them? P. Ten, and no units. 30 LESSONS ON NDMBER. T. And again; if we suppose these tens col- lected into hundredsj there would be p. One of them and no tens. r. Which, now, of the figures 100, indicates no units, which no tens, and which one hun- dred? P. The last zero to the right indicates no units ; the next, to the left of the former, indicates no tens; and the one to the left of this indicates one hundred. T. Hence we will for the future agree, that, of three figures, that which stands to the right of the two others, shall indicate P. Units. T. And that which stands next to the left of the former, shall indicate P. Tens. T. Finally, that which stands to the left of this shall iadicate — P. Hundreds. T. Now represent on your slates two hundred, three hundred, four hundred, &c. From this it will be readily perceived how the pupils may be taught to represent thousands, tens of thousands, &c. ; the teacher remembering that the miad of the pupil will be strengthened by pur- suing these exercises so long as he conceives with clearness, but no longer. ADDITION. 31 As before with units^ so now the pupils are re- quired to proceed with tens. Thus : — 20=10+10. 30=10+10 + 10=30+10. 40=10+10+10+10=20+10 + 10. =30+20=30+10. &c., &c. III. — Tens and Units. Teacher. Mention the numbers which are be- tween ten and twenty. Pupils. Eleven, twelve, &c nineteen. T. Bearing in mind what places it was agreed that units and tens should occupy, try to repre- sent the number eleven. The pupils must be left to represent that number correctly. The right or the wrong the class must decide J for that purpose it is best to call each in turn to the school slate. Should eleven be written thus 101, the teacher asks — T. We agreed that the last figure to the right shall represent units; the next to the left of it tens; the third hundreds, &c. Does this number represent eleven, &c., &c., until the correct answer I ] is found. T. Here then are two ones, the one to the right of the other indicates? P. One unit. 32 LESSONS ON NUMBER. T. . That figure is then said to stand in the units' •place; the other 1 to the left of the former indi- cates P. One ten. T. It is said to stand in the tens' place; now re- present twelve, thirteen, &c. In a manner quite analogous to the above, the pupils are led to represent the numbers, twenty- one, twenty-two, &c., &c., which will not be at- tended with any diflSculty, if the preceding has been understood. The pupils must discover them- selves that all numbers between 10 and 100 are pro- perly represented by 2 figures, those between 100 and 1000 by 3 figures, and so on. For practice, it is recommended to require the pupils to write in figures the questions and their answers according to the following model: thus, beginning at — Lesson I, Quest. 7. The pupils -write on their slates — 14-1-1=15. 26+1=27. 37+1=88. &c. Lesson II. Quest. 1. 17+2=19. 25 + 2=27. 47 + 2=49. &c. ADDITION. 33 Lesson III. Quest. 1. 14+3=17. 29+3=32. 37+3=40. &c., &c. And so on witli the lessons which follow: — Thus, Lesson IX. Quest. 1. 5+9+4+9+3+9+2+9=50. The mental exercises are now resumed; but those which the pupils have to perform, as practice on their slates, are now written in figures instead of words. § 2. ADDITION OF TENS. Lesson I. To add Ten. Teacher. How much is one more one? Pupils. Two. T. How much is one ten more one ten ? P. Two tens. T. That is— P. Twenty units. T. How much is two more one? P. Three. T. How much is two tens more one ten? P. Three tens; that is, thirty. 3 34 LESSONS ON NUMBER. T. How much is three more one? P. Four. T. How much is three tens more one ten? P. Four tens, or forty. The teacher thus continues, first, by asking how much is four more one, then four tens more one ten; five more one, then five tens more one ten ; and so on, until the pupils are able to proceed readily. Thus: — Pupils count: ten more ten are twenty'; twenty more ten are thirty; thirty more ten are forty ; &c., &c. On comparing this lesson with Lesson I, Ad- dition of Units, it will be found to correspond with it in every step ; with this difference, that there is an addition of tens instead of units. The lessons which foUow in this paragraph, namely, those which refer to the addition of two tens or twenty, three tens or thirty, &c., precisely follow the same course as those which had for their object to add two, three, &c.; and treated in this manner, present but little or no difficulty. From this consideration, it is thought sufficient to give an outline of one or two lessons on this subject, and leave it to the teacher to supply those which remain. Answers to the Exercises. Lesson I. To add Ten. The teacher, on examining the slates, must find ADDITION. 35 the questions aad their answers drawn up a,s fol- lows : — Ans. 1. 40 + 10=50. Ans. 6. 150 + 10=160. 2. 70 + 10=80. 7. 160+10=170. 3. 80+10=90. 8. 190+10=200. 4. 90+10=100. 9. 280+10=290. 5. 120+10=130. 10. 360 4-10=370. Lesson II. To add. Twenty. Teacher. How much is one more two? Pupils. Three. T. How much is one ten more two tens? P. Three tens, that is thirty. T. How much is two more two? P. Four. T. How much is two tens more two tens ? P. Four tensj that is forty. The teacher continues questioning, in a similar manner, until the pupils are able to proceed readily. Thus:— Class count: ten more twenty are thirty; twenty more twenty are forty; thirty more twenty are fifty; &c., &c., &c. Then foUow promiscuous questions, given as usual by the pupils to the class. Thus : — Pupil. 1. How much is ninety more twenty? 2. Find the sum of 270 more 20. 3. Add 380 and 20. 36 LESSONS ON NUMBER. Answers to the Exercises. Lesson II. Model how the exercises are to be written by the pupUs: — Ans. . 1. 50 + 20=70 3. 90 + 20 + 10= =120. Ans. 3. 130 Ans. 7. 180 4. 150 8. 210 5. 210 9. 280 6. 120 10. 390 Lesson III. To add Thirty. Teacher. How much is one more three ? How much is one ten more three tens? How much is ten more thirty ? The questions being answered satisfactorily, the teacher proceeds : — T. How much is two more three? How much is two tens more three tpns ? How much is twenty more thirty? And so on with thirty more thirty, forty more thirty, &c.; and it will be necessary to begin always with the addition of units, and then trans- fer the same to the corresponding tens. The pupils must, at the end of this lesson, be able to proceed. Thus : — Class count: ten more thirty are forty; twenty more thirty are fifty; thirty more thirty are sixty; &c., &c., &c. ADDITION. 37 The following promiscuous questions given by the pupils : — Pupils 1. Add 90 to 30. 2. Find the sum of 50^ 30, and 30. 3. Increase 180 by 30. Answers to the Exercises. Lesson III. Model. Ans. 1. 70+20 + 30+10=130 Ans.2. 200 Ans.Q. 180 Ans.lO. 500 3. 270 7. 340 11. 530 4. 300 8. 130 12. 650 5. 140 9. 370 Lesson IV. To add Forty. Answers to the Exercises. Lesson IV. 1.1. 100 Ai IS. 6. 310 2. 210 7. 390 3. 280 8. 590 4. 470 9. 670 5. 570 10. 760 Lesson V. To add Fifty. Answer to the Exercises. Lesson V. Ans. 1. 150 Ans.Q. 250 2. 230 7. 450 3. 300 8. 600 4. 380 9. 710 5. 420 10. 1090 38 LESSONS ON NUMBER. Lesson VI. To add Sixty. Answers to the Exercises. Lesson VI. ns.\. 310 Ans. 6. 540 2. 350 7. 550 3. 350 8. 730 4. 330 9. 810 5. 410 10. 960 Lesson VIL To add Seventy. Answers to the Exercises. Lesson VII. Ans.l. 380 Ans.4!. 390 3. 400 5. 640 3. 390 6. 1090 Lesson VIII. To add Eighty. Answers to the Exercises. Lesson VIII. ^ns.l. 360 J.WS.4. 730 2. 490 5. 770 3. 580 6. 1190 Lesson IX. To add Ninety. Answers to the Exercises. Lesson IX. Ans.\. 440 Ans. A. 600 3. 500 5. 860 3. 460 6. 1310 ADDITION. 39 § 3. ADDITION OF TENS AND UNITS. Lesson I. To add 10, 11, 13 19 Thus prepared, the pupils must now begin to add numbers consisting of tens and units, commencing again with the simplest form ; thus : — Teacher. How much is 1 more 10 ? 2 more 10 ? 3 more 10 ? &c., &c. 10 more 10? 11 more 10? Pupils. 10 more 1, more 10 j that is, 21. T. How much is 12 more 10? P. 10 more 2, more 10 ; that is, 22. T. How much is 13 more 10 ? P. 10 more 3, more 10 j that is, 23. T. How much is 14 more 10? 15 more 10? &c. 20 more 10? 2] more 10? P. 20 more 1, more 10, or 31. T. How much is 23 more 10? P. 20 more 2, more 10, or 82. T. How much is 23 more 10 ? 24 more 10? 25 more 10? &c. 31 more 10? 40 LESSONS ON NUMBER. P. 30 more 1, more 10, or 41. T. How much is 33 more 10 ? 33 more 10? 34 more 10? &C.J &c. And so on with the numbers which follow. The pupils will soon be able to proceed readily thus : — Class count : 44 more 10 are 54 ; 45 more 10 are 55 ; &c. 73 more 10 are 83 ; Sec., &c. T. You have learnt to add 10 to every number ; if now, it were required to add 11, what would you do ? P. First add 10, and then 1 more. T. How much is 15 more 11 ? P. 15 more 10, more 1, or 26. T. How much is 39 more 11? P. 39 more 10, more 1, or 50. T. And if it were required to add 13, what then would you do ? P. First add 10, and then 3 more. T. How much is 57 more 13 ? P. 57 more 10, more 3, or 69. T. How much is 85 more 13 ? P. 85 more 10, more 3, or 97. ADDITION. 41 T. If it were required to add 13^ what would you do? P. First add 10, and then 3 more. T. Add 88 and 13. P. 88 more 13, are 88 more 10, more 3, or 101. T. If it were required to add 14, what would you do ? P. First add 10, and then 4 more. T. How much is 137 more 14? P. 137 more 10, more 4, or 151. To add the numbers, 15, 16, 17, 18, and 19, similar questions are asked. The above must be considered as a mere outline of the mode of pro- ceeding, not as being sufficient practice in the addition of these numbers. Each number may be taken for the subject of one lesson; and before the pupUs have recourse to the written exercises, verbal solutions must be given of questions re- ferring to that number. The pupils should give questions to the class. Answers to the Exercises. Lesson I. Models. 57 + 10= 50+7-f-10= 67. 83-Hl= 83 + 10 + 1= 94. 149 + 13=149 + 10 + 2=161. 217+19=217+10 + 9=336. 42 LESSONS ON NUMBER. To add 10. ToaddW. To add 12. Ans. 1. 59 Ans. 1. 94 Am.\. 61 2. 68 2. 68 2. 65 3. 86 3. 106 3. 90 4 105 4. 114 4. 98 5. 111 5. 128 5. 105 6. 137 6. 165 6. 117 7. 163 7. 160 7. 170 8. 210 8. 187 8. 247 To add 13. To add 14. To add 15. ^ws. 1. 61 Ans. 1. 51 Ans. 1. 79 2. 70 2. 72 2. 90 ' 3. 99 3. 83 3. 102 4. 108 4. 87 4. 111 5. 130 5. 100 5. 132 6. 172 6. 111 6. 138 7. 201 7. 128 7. 153 8. 206 8. 150 8. 157 9. 218 9. 171 9. 166 10. 251 10. 293 10. 193 To add 16. Ans. 1. 74 J.ns. 4. 114 Ans. 7. 165 2. 91 5. 133 8. 169 3. 105 6. 140 9. 10. 203 213 ADDITION. 43 To add 17 To add 18. Ans.l. 84 Ans.l. 77 2. 103 3. 94 3. 114 3. 113 4. 153 4. 131 5. 160 5. 155 6. 194 6. 176 7. 312 7. 147 8. 348 8. 204 9. 375 9. 315 la. 306 10. 336 To add 19. Ans..\. 77 Ans. 5. 146 Ans. 9. 336 2. 106 6. 151 10. 378 3. 113 7. 194 11. 337 4. 133 8. 303 13. 408 Prom the mode hitherto pursued, the teacher will perceive how he should treat the addition of the numbers which follow. We have only to guard him against introducing any ready mode of solving the exercises ; as, for instance, that of arranging the numbers according to their local values. These exercises are only valuable in as much as they are solved mentally. The work upon the slate must exhibit the process of the mind : thus, if it be 4<4! LESSONS ON NUMBER. required to add 79 and 48, the pupils proceed mentally, in the following manner : — 79 more 48, are 79 more 40, more 8, which are 119 more 8; that is, 127. And upon the slate the same has the following form : — 79+48=79+40+8=119+8=137. And this mode of showing the work upon the slate it is advisable to continue, till the pupils have acquired sufficient practice in numeration; after which it may be abandoned, the pupils simply writing down the answer in the usual manner. Lesson II. To add 20, 21, 22 29. Answers to the Exercises. Ans. 1. 57 Ans. 9. 207 Ans. 17. 405 2, 66 10. 226 18. 412' 3. 98 11. 274 19. 425 4. 111 12. 296 20. 440 5. 121 13. 318 21. 449 6. 138 14. 330 22. 66 7. 154 15. 363 23. 75 8. 165 16. 394 24. 84 ADDITION. 45 Lesson III. To add 30,31,32 . .39. Ans.l. 88 Ans.8. 271 Ans. 15. 2- 118 9. 304 16 3- 131 10. 386 17 4. 150 11. 480 18 5. 170 13. 500 19 6. 309 13. 522 20 7. 233 14. 540 562 93 99 103 108 114 A, 1. 129 Ans. 6. 365 Ans .11. 178 2. 138 7. 375 12. 186 3. 190 8. 404 13. 193 4. 282 9. 166 14. 130 5. 320 10. 170 Ans. 1 Le 3S0N V. To add 50, 51, 52 59 1. 133 Ans. 6. 333 Ans. 11. 169 2. 134 7. 341 12. 176 3. 149 8. 292 13. 170 4. 172 9. 376 14. 201 5. 190 10. 423 46 LESSONS ON NUMBER. Ans. 1. 127 Ans. 6. 340 ■ • • • ■ »-»t/« Ans. 11. 216 2. 146 7. 384 12. 241 3. 189 8. 414 13. 272 4, 241 9. 453 14. 387 5. 258 10. 496 Lesson VII . To add 70, 71 79. Ans. 1. 128 Ans. 5. 330 Ans. 9. 665 2. 151 6. 464 10. 733 3. 169 7. 514 11. 207 4. 216 8. 521 13. 238 Lesson VIII. To add SO, 81 89. Ans. 1. 156 Ans. 5. 297 Ans. 9. 752 2. 178 6. 480 10. 865 3. 227 7. 544 • 11. 290 4. 261 8. 624 12. 224 Lesson. IX . To add 90, 91 , 99. Ans. 1. 177 Ans. 5. 569 Ans. 9. 917 2. 234 6. 632 10. 1042 3. 350 7. 783 11. 264 4. 479 8. 871 12. 256 ADDITION. , 47 Lesson X. Promiscuous Questions. Note. The numbers 2\, 41, 61, 81, and 79, are not included in tlie first fi\re questions. Ans.l. 210 Ans.A. 755 Ans.7. 792 2. 355 5. 870 8. 940 3. 555 6. 381 8. 2082 10. 2492 48 LESSONS ON NUMBER. CHAPTER II.- SUBTRACTION. Lesson I. To subtract \, 10 and 11. The mode of proceeding in subtraction may be anticipated from the manner in whicli addition has been treated ; according to which, 1 would be subtracted, then 2, next 3, and so on. But since the pupils are, at this stage, acquainted with the distribution of the numbers into tens, it will be in no wise unsuitable to avail ourselves of their know- ledge, and to apply it to the subtraction of tens simultaneously with that of units. Thus, then, if the pupils once know how to subtract one, they will require but a slight effort to subtract one ten, one hundred, one thousand, &c., and even to sub- tract one ten more one, that is, 11. Agaia, if the subtraction of 2 be known, that of 2 tens, 2 hun- dreds, &c., of 10-1-2, and of 2 tens -1-2, may be attempted. A similar plan is to be adopted with regard to the numbers, 3, 30, 300, &c.; 13, 23, and 33; and with the numbers fol- lowing, as will be shown in the lessons. SUBTRACTION. 49 Teacher. Prom 1 take away 1 ; what remains? Pupils. Nothing. T. Instead of "take away," it is usual to say subtract; hence from 1 subtract 1 what re- mains? P. Nothing. T. What word did we use when we had to add 1 and 1 ? P. We said 1 more 1. T. HerCj now, we have to take away or to sub- tract 1 ; what word will express this shortly? P. 1 less 1. T. In your written exercises you substituted a sign for the word more; what was it? P. The sign +. T. Instead of the word "less/' we wiU use an- other sign, viz.. this, — . Express, now, on your slates the result obtained by taking 1 from 1. P. 1 — 1=0. T. What is the result called, when two or more numbers are added? P. Their sum. T. What would you call the result obtained by taking one number from another number? P. Their difference. T. From 2 take 1 ; what remains ? From 3 subtract 1; what remains? What is the difference between 2 and 1 ? From 3 take 1 ; what remains? From 3 subtract 1: what remains? 50 LESSONS ON NUMBER. What is the difference between 3 and 1 ? How much is 4 less 1 ? ... 5 less 1 ? &c., &c. The result of these exercises must be, that the pupils are able readily to proceed thus : — Class count: 1 less 1 is 0; 2 less 1 is 1 ; 3 less 1 is 3; 4 less 1 is 3 ; &c.> &c. 100 less 1 is 99. To subtract 10. Teacher. From what numbers can ten be subr tracted? Pupils. From 10, and those numbers which are more than 10. T. From 10 take 10; what remains? P. Nothing. T. Decompose 11 into tens and units. P. 11 is 10 more 1. T. If, then, from 11, 10 be taken, away, what must remain? P. 1. T. Decompose 12 into tens and units. P. 12 is 10 more 2. SUBTRACTION. 51 T. li, then, from 12, 10 be taken away, what remains? P. 2. T. [Proceeds in a similar manner with 13, 14, 15, &c.] Decompose 19 into tens and units. If, now, 10 be taken from 19, what remains? P. 9. T. Decompose 20 in tens. P. 20 is 2 tens. T. If, then, from 20, or 2 tens, 1 ten be taken, what remains? P. 10. T. Decompose 2 1 into tens and units. Take 10 from 21, what remains? Decompose 22 into tens and units. Take 10 from 22, what remains? The pupils will experience no difficultj- in sub- tracting 10 from the numbers which follow, if the subject be treated in ' this manner. The result must be, that the class will be able to proceed readily thus : Class count: 10 less 10 is 0; 11 less 10 is 1; , 12 less 10 is 2; 13 less 10 is 3; &c., &c.* 78 less 10 is 68; 100 less 10 is 90; &c., &c. D 2 52 LESSONS ON NUMBER. To subtract 11. Teacher. You have learnt to subtract \, and likewise 10; can you teU what other number yon are now able to subtract? Pupils. 11. T. Why? P. Because 11 is 10 more 1. T. From which numbers can 11 be subtracted? P. From 11, and those numbers which are more than 11. T. If, then, it were required to take 11 from any of these numbers, what would you do? P. First take away 10 and then 1. T. Subtract 11 from 47. P. 47 less 11 is 47 less 10, less 1; 47 less 10 is 37 ; 37 less 1 is 36, the answer. The pupils are now required to give questions to the class; after which the exercises, Part II., are taken up for further practice. Answers to the Exercises. Lesson I. Model. 100-1=99. 57-10=47. 304-11=304-10-1=294-1=293. Ans. 1. , 99, 98, 97, &c, 2. 203, 193, 183, 173, &c. 3. 336, 325, 314, 303, 292, &c. SUBTRACTION. 53 nsA. 140 Ans.6. 879 Ans. 8. 990 5. 769 7. 893 Lesson IL 9. 10. 1034 1293 To subtract 2, 12, 20, and 22. In this lesson the pupils have actually to learn to subtract 2 from the units, and from 10 and 11; this being known, they must be led to refer all further subtractions of 2, 12, 20, and 22 to these results. Teacher. If 2 were to be taken from any number what would you do? Pupils. First take away 1, and then 1 again. T. Put on your slates the numbers 2, 3 .... 11 ; from each take away 2, and commit the answers to memory. This done. Class counts: 2 less 2 is 0; 3 less 2 is 1 ; 4 less 2 is 2 ; &c. 11 less 2 is 9. T. Do you think it necessary to learn by heart the answers obtained by subtracting 2 from 12, 13, 14, &c., and all other numbers? F. No; 12 is 10 more 2; 13 is 10 more 3; 14 is 10 more 4; &c. and we know what remains if 2 be taken from 2 ; 54 LESSONS ON NUMBER. hence we know too what remains if 2 be taken from 10 more 2, that is from 12; and so on with the other numbers. T. Take 2 from 21; how will you proceed in this instance? p. 21 is twenty more 1, or 10 more 11; and 2 taken from 11 is 9; therefore 2 taken from 21 is 19. T. Name another number which would cause you to proceed in a similar manner. P. 31, 41, 51, &c. T. Give questions to the class. To subtract 12, 20, 21, and 22. Teacher. Since you know now how to subtract 2, can you teU what other numbers you are now able to subtract? Pupils. 10 more 2 or 12. T. Besides 12 there are other numbers which you will be able to subtract very easily; which are they? P. 2 tens or 20; 20 more 1 or 21; and 20 more 2 or 22. T. How much is 57 less 12? P. 57 less 12 is 57 less 10 less 2; 57 less 10 is 47; 47 less 2 is 45; therefore 57 less 12 is 45. T, How much is 89 less 20? SUBTRACTION. 55 P. 89 less 20 is 80 more 9 less 20; or, 8 tens less 2 tens, more 9, wliich are 6 tens more 9, or 69. T. How much is 90 less 21? P. 90 less 21 is 90 less 20 less 1; 90 less 20 is 70; 70 less 1 is 69, the answer. T. How much is 71 less 22? P. 71 less 22 is 71 less 30 less 2; 71 less 20 is 51; 51 less 2 is 49, the answer. The pupils are required to give similar solutions to the questions. The teacher is recommended to proceed slowly, and to dwell upon each subject until the pupils have acquired sufiScient facility. Answers to the Exercises. Lesson II. Ans.l. 98, 96, 94, 92, 80, &c. 3. 188, 176, 164, 152, 140, 128, &c. 3, 387, 367, 347, 837, 307, &c. 4, 359, 338, 317, 396, 275, 254, 233, 212, 191, 170, 149, &c. 5, 329, 307, 285, 263, 241, 219, 197, 175, 153, 131, 109, 87, 65, 43, 21. Ans. 6. 49 ^ns. 11. 159 AnsA%. 549 7. 75 12. 178 17. 595 8. 73 13. 325 18. 680 9. 83 14. 395 19. 751 0. 97 15. 487 20. 21. 839 879 56 LESSONS ON NUMBER. Lesson III. To subtract 3, 13, 23, 30, 33, and 33. What tlie pupils have actually to learn is, to subtract 3 from 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12, This being known, but a slight effort on their part will enable them to subtract 3 from the numbers above 12. Again, they know how to subtract 10; they like- wise know how to subtract 3 , they will experience no difficulty in subtracting 10 + 3, or 13. The same remark applies to the subtraction of 20 + 3, or 23. Also, if the pupils know how to subtract 3, they must, for the same reason, be able to subtract 3 tens, or 30; and, finally, combining this and the previous lessons with the subtraction of 30, to apply the same reasoning to subtract 30 + 2, and 30+3. The mode to which the teacher has to resort, in order to obtain from his pupils the above result, has, it is thought, been sufficiently shown in the previous lessons. There is nothing to prevent applying the same reasoning to the subtraction of 300, 303, 313, 323, 330, 333, and 333, and even thousands. The questions, however, have only SUBTBACTION. 57 been adapted to tens, from the consideration that these exercises are intended for young children. Answers to the Exercises. Lesson III. Ans. 1. 84, 81, 78, 75, 72, 69, 66, 63, 60, 57, 54, &c. 2. 92, 79, 66, 53, 40, 27 14, 1. 3. 148, 125, 102, 79, 56, 33, 10. 4. 171, 141, 111, 81, 51, 21. 5. 239, 307, 175, 143, 111, 79, 47, 15. 6. 309, 376, 343, 310, 177, 144, 111, 78, 45, 13. Ans.7. 155 Ans.W. 385 Am.l5. 578 8. 158 13. 447 16. 668 9. 139 13. 487 17. 771 10. 375 14. 539 18. 773 Lesson IV. To subtract 4, 14, 34, 34, 40, 41, 43, 43, and 44. What the pupils have actually to learn in this lesson is, to subtract 4 from 4, 5, 6, &c 13; and when this is known, the pupils must be left to discover that, 1. They are now able to subtract 4 from the sucpeeding numbers 14, 15, &c. 3. Also 10+4 from 14, 15, &c. 3. Also 30+4 from 34, 35, &c. d5 58 LESSONS ON NUMBEK. 4s. Likewise 30+4 from 34, &c. 5. Again, 40 from 40, 41, &c. 6. Similarly, 40 + 1, 40+2, 40 + 3, and, finally, 40+4 from 41, &c.; 43, &c.j 43. &c.; and 44, &c. It is well that the pupils should find this out for themselves, otherwise they will not feel interested. It is also necessary to dwell upon each subject, to give questions, and let each of the pupils give questions, which are to be solved mentally, before the exercises, Part II., are taken up. Answers to the Zeroises. Lesson IV. Ans. 1. 93, 89, 85, 81, 77, 73, 69, 65, 61, 57, 53, &c. 2. 97, 83, 69, 55, 41, 27, 13. 3. L99, 175, 151, 127, 103, 79, 55, 31, 7. 4. 169, 135, 101, 67, 32. 5. 247,207, 167, 127,87,47,7. 6. 329, 288, 247, 206, 165, 124, 83, 42, 1. 7. 359, 317, 275, 233, 191, 149, 107, 65, 23. 8. 489, 446, 403, 360, 317, 274, 231, 188, 145, 102, 59, 16. 9. 559, 515, 471, 427, 383, 339, 295, 251, 207, 163, 119, 75, 31. SUBTRACTION. 59 By subtracting the lesser number from the greater. Ans. 10. 57 Ans. 14. 89 Ans.lS. 529 11. 118 15. 159, 19. 599 12. 49 16. 189 20. ,668 13. 67 17. 279 21. 779 Lesson V. To subtract 5, 15, 25, 35, 45, 50, 51, 52, 53, 54, and 55. In this lesson, again, the pupils have only to learn to subtract 5 from 5, 6, 7, &c 14, the teacher then proceeds: Teacher. When in our last lesson you learnt to subtract 4, from what numbers did you find it sufficient to know how to subtract 4? Fupils. From 4, 5, 6, 7, 8, 9, 10, 11, 12, and 13. T. When you knew this, what followed? P. We then could take away 4 very easily from every other number. T. Can you tell from which numbers you must learn to subtract 5, so as to be able to subtract 5 readily from every number? P. From 5, 6, 7, 8, 9, 10, 11, 12, 13, and 14. 60 LESSONS ON NUMBER. P. If you know this can you then subtract 5 from every number? P. Yes; for 15 is 10 + 5; and if we know how to subtract 5 from 5, we likewise know how to sub- tract 5 from 10 + 5, or from 20 + 5, 30 + 5, &c. And if we know how to subtract 5 from 6, we like- wise know how to subtract 5 from 10+6, 30 + 6, 30+6, &c., and so on with the other numbers. T. Write on your slates the numbers you have mentioned ; subtract 5 from each, and learn it by heart. This done, questions are given by the teacher, and afterwards by the pupils, for practice. Next, T. Is this all you have learnt by committing these answers to memory, or are there yet other numbers which you are now able to subtract as readily as 5? P. We can likewise subtract 10 + 5, 20+5, 30+5, 40 + 5, 50, 51, 52, 53, 54, and 55. •T. Explain how it is, that by knowing how to subtract 5, you are able to subtract 10 + 5? P. We have learnt to subtract 10 in the first lesson, and now we have learnt to subtract 5, therefore we must be able to subtract 10+5. T Explain how it is you are able to subtract 25, 35, 45, &c. P. [Give similar reasons as above.] SUBTRACTION. 61 Questions are now given by tlie teacher, and afterward by the pupils, the results required to be explainedj and finally, the exercises. Part II, are taken up for further practice. Answers to the Exercises. Lesson V. Am. 1. 78, 73, 68, 63, 58, 53, 48, 43, 38, 33, &c. 2. 109, 94, 79, 64, 49, 34, 19, 4. 3. 206, 181, 156, 131, 106, 81, 56, 31, 6. 4. 337, 302, 267, 232, 197, 162, 127, 92, 57, 22. 5. 441, 396,351,306,261, 216, 171, 126, 81, 36. 6. 534, 484, 434, 384, 334, 284, 234, 184, 134, 84. 7. 546, 495, 444, 393, 342, 291, 240, 189, 138, 87, 36. 8. 596, 544, 492, 440, 388, 336, 284, 232, 180, 128. 7Q, 24. g. 844, 791, 738, 685, 632, 579, 526,473, 420,367,314,261,208,155,102,49. 10. 939,885,831, 717, 723, 669, 615, 561, 507, 453, 399, 345, 291, 237, 183, 129, 75, 21. 63 LESSONS ON NUMBER. Ans. 11. 946, 891, 836, 781, 726, 671, 616, 561, 506, 451, 396, 341, 286, 231. 176, 121, 66, 11. 12. By adding the mambers. Model. 13. 103 + 15=118, sum; 103—15=103—10—5 =93—5=88, difference, 14. 233, sum; 183, difference. 15. 416 16. 462 17. 633 18. 701 19. 763 20. 879 21. 989 22. 1178 346 372 533 599 659 769 879 1068 Lesson VI. To subtract 6, 16, 26, 36, 46, 56, 60, 61, 62, 63, 64, 65, and 66. The pupils have to learn in this lesson to sub- tract 6 from 6, 7, 8 15; the rest is obvious. Answers to the Exercises. Lesson VI. Ans. 1. 94, 88, 82, 76, 70, 64, 58, 52, 46, 40, &c. SUBTRACTION. 63 Am. 2. 122, 106, 90, 74, 58, 42, 26, 10. 3. 153, 127, 101, 75, 49, 23. 4. 177, 141, 105, 69, 33. 5. 239, 193, 147, 101, 55, 9. 6. 263, 207, 151, 95, 39. 7. 334, 274, 214, 154, 94, 34. 8. 426, 365, 304, 243, 182, 121, 60. 9. 511 , 449, 387, 325, 263, 201, 139, 77, 15. 10. 578, 515, 452, 389, 326, 263, 200, 137, 74, 11. 11. 718, 654, 590, 526, 462, 398, 334, 270, 206, 142, 78, 14. 12. 768, 703, 638, 573, 508, 443, 378, 313, 248, 183, 118, 53. 13. 851, 785, 719, 653, 587, 521, 455, 389, 323, 257, 191, 125, 59. Am. 14. 523 Am. 16. 787 Am. 18. 946 15. 647 17. 878 19. 1042 Lesson VII. To subtract 7, 17, 27, 37, 47, 57, 67, 70, 71, 72, 73, 74, 75, 76, and 77. Answers to the Exercises. Am. 1. 97, 90, 83, 76, 69, 62, 55, 48, 41, 34, 27, &c. 2. 142, 125, 108, 91, 74, 57, 40, 23. 64 LESSONS ON NUMBER. Ans. 3. 166, 139, 112, 85, 58, 31, 4. 4. 177, 140, 103, 66, 29. 5. 225, 178, 131, 84, 37. 6. 284,227,170,113,56. 7. 325, 258, 191, 124, 57. 8. 373, 303, 233, 163, 93, 23. 9. 422, 351, 280, 209, 138, 67. 10. 502, 430, 358, 286, 214, 142, 70. 11. 588,515,442, 369, 296, 223, 150, 77, 4. 12. 674, 600, 526, 452, 378, 304, 230, 156. 82,8. 13. 751, 676, 601, 526, 451, 376, 301, 226, 151, 76, 1. 14. 857, 781, 705, 629, 553, 477, 401, 325, 249, 173, 97, 21. 15. 923, 846, 769, 692, 615, 538, 461, 384, 307, 230, 153, 76. 16. Sum = 391 ; difference = 237. 17. Sum = 530; difference = 376. Lesson VIII. To subtract 8, 18, 28, 38, 48, 58, 68, 78, 80, 81, 82, 83, 84, 85, 86, 87, and 88. Answers to the Exercises. Ans. 1. 113, 105, 97, 89, 81, 73, 65, 57, 49, 41, 33, 25, 17, 9, 1. SUBTRACTION. 65 Am.2. 175, 157, 139, 121, 103, 85, 67, 49, 31, 13. 3. 185, 157, 139, 101, 73, 45, 17. 4. 233, 195, 157, 119, 81, 43, 5. 5. 248, 200, 152, 104, 56, 8. 6. 275, 217, 159, 101, 43. 7. 319, 251, 183, 115, 47. 8. 366, 288, 210, 132, 54. 9. 470, 390, 310, 230, 150, 70. 10. 586, 505, 424, 343, 262, 181, 100, 19. 11. 691, 609, 527, 445, 363, 281, 199, 117, 35. 12. 754, 671, 588, 505, 422, 339, 256, 173, 90,7. 13. 804, 720, 636, 552, 468, 384, 300, 216, 132, 48. 14. 858, 773, 688, 603, 518, 433, 348, 263, 178, 93, 8. 15. 906, 820, 734, 648, 562, 476, 390, 304, 218, 132, 46. 16. 913, 826, 739, 652, 565, 478, 391, 304, 217, 130, 43. 17. 1023, 935, 847, 759, 671, 583, 495, 407, 319, 231, 143, 55. Lesson IX. To subtract 9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, and 99. 66 LESSONS ON NCMBER. Answers to the Exercises. Am. 1. 92, 83, 74, 65, 56, 47, 38, 29, 20, 11, 2. 2. 163, 144, 125, 106, 87, 68, 49, 30, 11. 3. 175, 146, 117, 88, 59, 30, 1. 4. 231,192,153,114,75,36. 5. 307, 258, 209, 160, 111, 62, 13. 6. 324, 265, 206, 147, 88, 29. 7. 378, 309, 240, 171, 102, 33. 8. 474, 395, 316, 237, 158, 79, 0. 9. 576, 487, 398, 309, 220, 131, 42. 10. 641, 551, 461, 371, 281, 191, 101, 11. 11. 828, 737, 646, 555, 464, 373, 282, 191, 100, 9. 12. 857, 765, 673, 581, 489, 397, 305, 213, 121, 29. 13. 852, 759, 666, 573, 480, 387, 294, 201, 108, 15. 14. 906, 8 12, 718, 634, 530, 436, 342, 248, 154, 60. 15. 1007, 912, 817, 722, 627, 532, 437, 343, 247, 153, 57. 16. 1088, 993, 896, 800, 704, 608, 513, 416, 320, 234, 128, 32. 17. 1116, 1019, 922, 825, 738, 631, 534, 437, 340, 343, 146, 49. 18. 1143, 1045, 947, 849, 751, 653, 555, 457, 359, 361, 163, 65. 19. 1238, 1139, 1040, 941, 842, 743, 644, 545, 446, 347, 248, 149, 50. SUBTRACTION. 67 N.B. In the subtraction of 9, 19, 29, &c., the pupils must be led to observe that it is easier to subtract 10, 20, 30, &c., and then to add 1 to the result. Sum = 107; difference = Ans.20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 107 143 134 208 335 428 547 678 711 866 897 1001 1005 914 862 759 658 677 600 593 693 1655 31. 29. 44. 22. 139. 254. 349. 484. 515. 688. 739. 867. 755. 628. 464. 347. 224. 89. 28. 313. 185. 179. 68 CHAPTER III— MULTIPLICATION. § 1. MULTIPLICATION BY UNITS. Lesson I. To multiply by 1 and by 2. The subject of multiplication might be treated like that of addition; and accordingly be begun by taking or repeating a number once, 1 ten times, 10x1 times, 1 hundred times, Sec. The next step would be to repeat a number twice, 2 ten times, 20+1 times, 20 + 3 times, 2 hundred times, &c.; and so on with the numbers 3, 4, 5, &c. For, developed as the pupils' minds must.be by the fore- going, little effort is required by them to conceive a number repeated ten times; since to repeat 1 10 times is evidently making 1 unit 1 ten; again, to repeat 2 10 times is to make 2, 2 tens, and to repeat or multiply 37 10 times, is to make 37, 37 tens. This once understood, the multiplication by 11 or 10 + 1 will immediately result from it. Thus, to multiply a number by 11 is to repeat it 10 times, (which is known) , and once more, (which is also known); again, if it be known how to mul- tiply the numbers 1, 2, 3, 10, by 2, it MULTIPLICATION. 69 follows, that the multiplication of 10+1, 10+3, 10 + 3, &c., by 2, is likewise kxiown; and this ob- tained, the multiplication of any number by 10 + 2, i.e. by 12, is immediately deduced. Nothing pre- vents them from applying the same to the multi- plication by 20 or 2 tens, for it is known how to multiply by 2; hence to multiply a number by 2 tens is to take it twice; but then the results are not units but tens. To multiply by 20+1 and 10 + 2 are the immediate results from this consi- deration. If, then, the teacher find his class suf- ficiently developed, it will much more tend to their further development, if they pursue this course, than if they simply begin with multiplying a number by 2, next by 3, then by 4, &c. This healthy state of mind is, however, seldom found in all the pupils in a class, and then it is advisable to choose the simpler and slower process. The same objection ought not to deter the teacher from treat- ing subtraction in the manner shown before ; the results there being smaller numbers; whereas in multiplication their increase is sudden and great. Teacher. How much is one 1 ? H.OW much is one 2? How much is one 3 ? How much is one 4? How much is one 5 ? &c., &c. How much is one 87 ? Pupils. 1, 2, 3, 4, 5, &c., 87. .70 LESSONS ON NUMBER. T. How much is 3 ones? How mucli is 2 twos? How mucli is 2 threes? How much is 2 fours? P. 2, 4, 6, 8. T. Instead of saying, "how much is 2 ones," we might say, " how much is 1 taken twice," or " how much is twice 1." How much is twice 5? twice 6? twice 7? F. 10, 12, 14. T. How much is twice 8? twice 9? twice 10? P. 16, 18, 20. T. Repeat together twice 1 is 2. twice 2 is 4. twice 3 is 6. &c. twice 10 is 20. P. [Repeat.] T. Do all of you know this readily? — if so, can you tell how much twice 11 is? P. 11 is 10 more 1; hence twice 11 is twice 10 more twice 1, which is 20 more 2, or 22. T. How much is twice 12? P. Twice 10 more twice 2, or 20 more 4, which is 24. T. How much is twice 13? P. Twice 10 more twice 3, or 20 more 6, which is 26. T. How much is twice 14? twice 15? MULTIPLICATION. 71 T. How mucli is twice 16? twice 17? twice 18? twice 19? P. Twice ten more twice 9, or 20 more 18^ which is 38 T. How mucli is twice 2? P. 4. T. How much is twice 2 tens or twice 20? P. 4 tens or 40. T. How much is twice 3? P. 6. T. How much is twice 3 tens or twice 30? P. 6 tens or 60. T. How much is twice 4? P. 8. T. How much is twice 4 tens or twice 40? P. 8 tens or 80. T. How much is twice 5 tens or twice 50? twice 6 tens or twice 60? twice 7 tens or twice 70? tm.ce 8 tens or twice 80? twice 9 tens or twice 90? twice 10 tens or twice 100? Class repeat : twice 20 is 40. twice 30 is 60. twice 40 is 80. &c. twice 100 is 200. 72 LESSONS ON NUMBER. T. Decompose 31 into tens and units. P. 21 is 2 tens more 1. T. How much is twice 21? P. Twice 20 more twice 1, or 40 more 2, which is 42. T. How much is twice 22? P. Twice 20 more twice 2, or 40 more 4, or 44. T. How much is twice 23? twice 24? twice 25j &c twice 31? twice 32? &c. twice 47 ? twice 89 ? P. Twice 80 more twice 9, or 160 more 18, which is 178. T. How much is twice 147? P. Twice 100, more twice 40, more twice 7, or 200 more 80, more 14, which is 294. T. You have now learnt to take any number twice, or as it is usually called to multiply hy 2. If you know how to take 1 twice, or to multiply 1 by 3, which other numbers do you then know how to multiply by 3? P. 10, 11, 100, 101, 111, 1000, &c. T. Why? explain. P. Because twice 1 is 2; therefore twice 1 ten is 2 tens or 20. And because 11 is 10 more 1, twice 11 is twice 10 more twice 1, or 22. Again, because twice 1 is 2, therefore twice 100 is 300, and twice 101 is twice 100 more twice 1, or 303; MULTIPLICATION. 73 and twice 111 is twice 100, more twice 10, more twice 1, or 322. T. If you know how to multiply 2 by 2, what other numbers do you then know how to multiply by 2? P. 20, 21, 22, 200, 201, 202, 221, 222, &c. T. Explain. P. [Give similar reasons as before.] And so on with the other numbers. T. In order then to multiply any number what- ever by 2, it is sufficient to learn to multiply ac- tually which numbers only? P. The numbers, 1, 2, 3, 4, 5, &c., 9. T. What is the result called when two or more numbers are added? P. Their sum. T. In the written exercises what sign did we use to indicate addition ? P. The sign + . T. What is the result called when a lesser number is subtracted from a greater: and what sign did we use to indicate subtraction ? P. Their difference. We used the sign — . T. You have learnt to multiply numbers by 2 ; the result obtained by multiplication is called pro- duct ; and the sign used to indicate multiplication is X . What is the sum, difference, and product of the numbers 2 and 78 ? P. Their sum is 80; their difference 76; and their product 156. 74 LESSONS ON NUMBER. T. "Write this on your slates, using the proper signs for addition, subtraction, and multiplication. P. [Write] 78 + 2=80 sum. 78-2=76 difference. 78x2=156 product. The pupils- are now called upon to give ques- tions to the class on the previous lesson, after which the exercises, Part II., are taken up for further practice. Answers to the Exercises. Lesson 1. Models. 2 X 17=2 X 10+2 X 7=20 + 14=34. 2x89=2x80+2x9=160+18=178. 2x178=2x100 + 2x70 + 2x8=200 + 140 + 16=356. 1. 1. 34 Ans. 7. 158 Ans. ,13. 414 2. 56 8. 176 14. 556 3. 72 9. 192 15. 698 4. 98 10. 246 16. 974 5. 114 11. 274 17. 1190 6. 130 12. 378 18. 1978 Lesson II. To multiply by 3. , Teacher. How much is 2 X 1 ? Pupils. 2. T. How much more is 3 x 1 ? P. 1 more. MULTIPLICATION. 75 T. How much then is 3 x 1 ? P. 3 J 3+1 or 3. T. How much is 2 x 2 ? P. 4. T. How much more is 3 x 3 ? P. 2 more. T. How much then is 3 x 3 ? P. 4 + 3 or 6. T. How much is 2 x 3 ? P. 6. T. How much more is 3x3? P. 3 more. T. How much then is 3 x 3 ? P. 6 + 3 or 9. T. You have learnt to take any number twice or to multiply any number by 3, can you tell what must be done to take any number 3 times, or to multiply it by 3 ? P: First multiply the number by 2, and then add the number to the product. T. Hence you are able to multiply any number by 3 ; multiply 56 by 3. P. 2x56 = 112; therefore 3x56 = 112 + 56 =168. T. Would it not be more convenient to be able to tell at once the product of any number by 3 without first multiplying it by 2 ? The products of which numbei^s is it sufficient to know in order to do this ? [should this question not be answered in a. satisfactory manner] — E 2 76 LESSONS ON NUMBER. T. [continues.] The products of which numbers is it sufficient to know in order to multiply any number by 3 ? P. Of the numbers 1, 3, 3, &c., 9. T. See whether it be sufficient to know the products of these same numbers by 3. P. Yes ; no T. Suppose you know how much 3 x 1=3 is ; can you tell how much 3 x 10 is ? P. Yes, for 3x1=3, therefore 3x1 ten = 3 tens, or 3 X 10=30. T. And if you know how much 3 x 10 is, can you teU how much 3x11 is ? P. Yes, for 11=10+1, therefore 3xll=3x 10+3x1=30 + 3=33. T. You know how much 3 x 3 is, and I think you are able to ascertain from this readily how much 3 X 30 is ? P. Yes, for 3x3=6, therefore . 3 x 2 tens = 6 tens, or 3 X 30=60. T. And now I think, too, you are able to tell how much 3 x 12 is. P. Yes, for 13=10 + 3, therefore 3xl3=3x 10+3x3=30+6=36. T. There are yet at least two other numbers which you are able to multiply by 3, which are they? P. 31 and 33; for 3x31=3x30 + 3x1=60 + 3=63; and 3x32=3x30+3x3=60+6=66. MULTIPLICATION. 11 T. What do you now think of my former ques- tion? P. If we know the products of the numbers 1, 2, 3, 10, by 3, we shall be able to find readily those of any other number. T. Then find the products of these 10 numbers; write them on your slates and learn by heart. When known — Class repeat: 3x 1=8. 3 X 3=6. 3 X 3=9. &c. 3x10=30. T. Apply this to tens. • Class repeat: 3 x 10=30. 3x20=60. 3x30=90. &c. 3x100=300. Here follow promiscuous questions given at first by the teacher, and then by the pupils; every answer, right or wrong, must be subjected to the decision of the class, a practice much recommended as inducing the pupils to take a lively interest, and thereby sustaining their attention. Example. How much is 3 x 89? P. 267; for 3x89 = 3x80+3x9 = 340 + 27 =267. 78 LESSONS ON NUMBER. Answers to the Exercises. Lesson II. Ans.l. Ill Ans.7. 729 Ans .13. 2064 2. 177 8. 807 14. 2391 3. 234 9. 1155 15. 2508 4. 288 10. 1428 16. 2625 5. 339 11. 1779 17. 2061 6. 387 12. 1872 18. 3702 Lesson III. To multiply by 4. According to the plan detailed in the previous lessons, the pupils must be led to ascertain, 1 . That to multiply a number by 4, is to multiply it first by 3, and then to add the number to the product. 2. That it is sufficient to know the products of the nnits by 4, in order to multiply any number by 4. Hence the necessity of actually learning these products- by heart. 3. This done, the results are immediately applied to the multiplication of tens by 4. 4. Questions are given by the teacher and each pupil, in turn. 5. The exercises, Part II., are taken up. Here follovc a few questions, with their solutions, such as the pupils must be able to give, before they begin the next lesson. MULTIPLICATION- 79 Question. How mucli is 4x 39? Answer. 156; for 4x39=4x30+4x9 =120 + 36=156. Q. What is the product of 87 and 4? A. 348; for 4x87=4x80 + 4x7 =320+28=348. Q. Multiply 178 by 4. A. 712; for 4x178=4x100+4x70+4x8 =400+280+32=680+32=712. Answers to the Exercises. Lesson III. Ans. 1 . 188 Ans. 7. 556 Ant !.13. 1468 2. 236 ' 8. 628 14. 1556 3. 296 9. 876 15. 1772 4. 348 10. 992 16. 2348 5. 392 11. 1108 17. 2712 6. 420 12. 1300 18. 3156 Lesson IX. To multiply hy 5. Answers to the Exercises. ns. 1. 430 Ans.S. 1065 J.WS.15. 3390 3. 485 9. 1105 16. 3435 3. 540 10. 1745 17. 3945 4. 685 11. 2290 18. 3990 5. 795 12. 2425 19. 4495 6. 875 13. 2835 20. 4990 7. 920 14. 2880 21. 6170 80 LESSONS ON NUMBER. Lesson V. To multiply hy 6. Answers to the Exercises. Ans.l. 462 Ans.9,. 1302 1 Lms. 15. 3534 2. 534 9. J 404 16. 4068 3. 690 10. 2070 17. 4134 4. 828 11. 2202 18. 4482 5. 894 12. 2736 19. 5256 6. 1122 13. 2868 20. 5923 7. 1194 14. 3402 Lesson VI. To multiply hy 7. Answers to the Exercises. Ans.l. 455 Ans.%. 1512 Ans.lZ. 4032 2. 546 9. 1638 16. . 4368 3. 623 10. 2415 17. 4746 4. 679 11. 2579 18. 5523 5. 728 12. 3381 19. 6379 6. 973 13. 3473 20. 6909 7. 1281 14. 3724 21. 6951 Lesson VII. To multiply hy 8. Answers to the Exercises. Ans.l. 344 Ans.4i. 672 Ans.7. 984 2. 464 5. 768 8. 1512 3. 632 6. 912 9. 1704 MULTIPLICATION. 81 Ans. 10. 1792 Ans. 15. 3032 Ans. 20. 4448 11. 1880 16. 3288 21. 5336 12. 1968 17. 3376 22. 6224 13. 2856 18. 3472 23. 7112 14. 2944 19. 4360 24. 7944 Lesson VIII. To multiply by 9. Answers to the Exercises. Ans. 1. 333 Ans. 9. 1044 Ans. 17. 3942 2. 432 10. 1143 18. 4401 3. 531 11. 2142 19. 4635 4. 549 12. 2241 20. 4887 5. 648 13. 2259 21. 6012 6. 747 14. S258 23. 6975 7. 846 15. 3357 23. 8073 ». 945 16. 2843 24. 8874 Lesson IX. Continued Products. Teacher. How much is 2 x 2 ? Pupils. 4. T. And how much is 2 x 4 ? P. 8. T. How much, then^ is 2 x 2 x 2 ? P. 8. T. How much is 2x3? P. 6. e3 83 LESSONS ON NUMBER. T. Howmuchis2x6? P. 12. T. How much, then, is 3x3x2? P. 13. T. Can you tell what you have done here ? P. We have first multiplied 3 by 3, and again multiplied the product by 3. T. Hence you have connected the numbers 2, 3, 2, by multiplying the first two, and then continued by multiplying the product by the next number; The answer in such a case is called the continued product of the three numbers. What is the continued product of 3, 4, 5 ? P. 60 ; because 3 x 4=12 and 5 x 13=60. T. What is the continued product of 4, 5, 6 ? P. 130; because 4x5=20j and 6x20=130. T. If a number be multiplied by 2, and the product be multiplied by 2, by what number, then, is the first number multiplied ? P. By 4j because it is first taken twice, and the product, then, is taken twice ; hence the first number is taken 4 times, or it is multiplied by 4. T. Instead, then of multiplying 87 by 4, in the manner you have learnt, what might be done ? P. 87 might first be multiplied by 3, and the product again by 2. T. Try it both ways. P. 4x87=4x80 + 4x7=320 + 28=348. Or, 4x87=2x87x2; 2 X 87=174, and 3 x 174=348. MULTIPLICATION. 83 7. What might be done instead of multiplying a number by 6, in the manner you have learnt before ? P First multiply by 2, and the product by 3. The present subject can, at this stage^ be only adverted to, as its thorough application depends on the pupUs being able to decompose numbers into factors, as treated fully in a subsequent chap- ter. This lesson may be omitted, if the class be not properly developed. § 2. MULTIPLICATION BY TENS. Lesson I. To multiply by 10, 20, and 30. Teacher. How many tens are 10 x 1 ? Pupils. 1 ten. ^ , T. How many units is this? P. 10 units. T. How many tens, are 10 x 2 ? P- 2 tens, or 20 units. T. How many tens are 10 x 3 ? P. 3 tens, or 30 units. T. How many tens are 10x4? 10x5? 10x6? &c., &c. 10x11? 84 LESSONS ON NUMBER. P. 4s tens, or 40 units. 5 tens, or 50 units. 6 tens, or 60 units. 11 tens, or 110 units. T. Hence, what may be said in general of a number multiplied by 10 ? P. It is as many tens as there are units in the number. T. How much, then, is 10x56? P. 56 tens, or 560. T. How much is 3 tens x 1 ? P. 2 tens, or 20. T. How much is 2 tens x 2 ? P. 4 tens, or 40. T. How much is 20x3? P. 6 tens, or 60. T. How much is 20x4? 20x5? 20x6?. &c. 20x10? 20x11? P. 8 tens, or 80. 10 tens, or 100. 12 tens, or 120. 20 tens, of 200. 32 tens, or 330. T. If, then, a number is to be multiplied by 3 tens, or by 30, what may be said ? MULTIPLICATION. 85 P. It is multiplying that number by 2, and call- ing the product tens, not units. T. Hence, if you know how to multiply by 2, you are likewise able to multiply P. By two tens, or by 20. T. Multiply 37 by 20. P. 20 X 37=2 X 37 tens=74 tens=740. T. Multiply 89 by 20. P. 20x89=2x89 tens= 178 tens =1780. T. How many tens are 30 x 1 ? P. 3 tens, or 30 units. T. How many tens are 30 x 2 ? P. 6 tens, or 60 units. T. How many tens are 30 x 3 ? 30x4? &e. 30x12? P. 9 tens, or 90 units. 12 tens, or 120 units. 36 tens, or 360 units. T. Hence what may be said of a number multi- plied by 3 tens, or by 30 ? P. It is multiplying that number by 3, and call- ing the product tens. T. If, then, you know how to multiply by 3, you are able to multiply likewise P. By 3 tens, or by 30. T. Multiply 45 by 30. P. 30 x 45 = 3 X 45 tens = 135 tens = 1350 units. 86 LESSONS ON NUMBEE. Answers to the Exercises. Lesson I. Ans. 1. 930 A', m. 7. 640 Ans. 13. 450 2. 870 8. 1130 14. 990 3. 1100 9. 1480 15. 1710 4. 1530 10. 1760 16. 2530 5. 2170 11. 2740 17. 2970 6. 3710 12. 4960 18. 4110 Lesson II. To multipi !w hy 40, 5o; and h «60. In this lesson, the pupils must be led to ascer- tain that, 1. Multiplying a number by 40, is. to multiply it by 4, and to call the product tens. 2. Multiplying a number by 50, is to midtiply it by 5, and call the product tens. 3. Finally, multiplying a number by 60, is to multiply it by 6, and to call the product tens. Answers to the Exercises. 1st, by 40. 2nd, by 50. 3rd, by 60. Ans. 1. 720 Ans. 1. 950 Ans. 1. 660 2. 1360 2. 1250 2. 1320 3. 2320 3. 2350 3. 2100 4. 2680 4. 3400 4! 3360 5. 2920 5. 4100 5. 3840 6. 3800 6. 5050 6. 5700 MULTIPLICATION. 87 Lesson II. To multiply ly 70, 80, and 90. Here, again, the pupils liave to ascertain that multiplying a number by 70, 80, and 90, is to multiply it by 7, 8, and 9 respectively, and to call the products tens. Answers to the Exercises. 1st, by 70. Snd, by 80. 3rd, by 90. Ans. 1. 910 Ans. 1. 1120 Ans. 1. 1350 2. 1330 3. 2960 2. 1890 3. 1890 3. 3280 3. 3150 4. 2450 4. 3840 4. 4140 5. 3430 5. 4320 5. 5130 6. 3710 6. 5040 6. 6120 7. 4690 7. 6160 7. 7110 8. 5880 8. 7120 8. 7560 9. 6440 9. 7840 9. 8910 10. 6790 10. 7600 10 9360 § 3. MULTIPLICATION BY TENS AND UNITS. Lesson I. To multiply by 11, 12 19. In the foregoing lessons, the pupils have learnt to multiply by 1 and by 10; by 2 and by 10; by 3, 4, &c., 9, and by 10 ; they are now to multiply 88 LESSONS ON NUMBEK. by 11, 12, 13, &c.; that is, by 10 + 1, 10+2 10 + 3., &c. The teacher, then will ask: If a number is to be multiplied by 11, what must be done. Pupils. The number must first be multiplied by 10, and then by 1, and the two products must be added. T. Multiply 11 by 11. P. 11 X 11 = 10 >; 11 + 1 X 11 = 110 + 11 = 121. T. Multiply 17 by 11. P. 11 X 17 =10 X 17 + 1 X 17 = 170 + 17 = 187. T. Multiply 89 by 11. P. 11 X 89 = 10 X 89 + 1 X 89 = 890 + 89 = 979. T. If a number is to be multiplied by 12, what must be done? P. First multiply by 10, and then by 2, and add the two products. T. Multiply 12 by 12. P. 12 X 12 = 10 X 12 + 2 X 12^= 120 + 24 = 144. T. Multiply 13 by 12. P. 12 X 13 = 10 X 13 + 2 X 13 = 130 + 26 = 156. T. Multiply 98 by 12. P.. 12 X 98 = 10 X 98 + 2 X 98 = 980 + 196 = 1080 + 96=1176. MULTIPLICATION. 89 Similarly witli the multiplication by 13, 14, 15, &c. T. Multiply 19 by 19. P. 19 + 19 = 10 X 19 + 9 X 19 = 190 + 171 = 290+71=361. T. Multiply 67 by 19. P. 19x67 = 10 X 67 + 9 x 67 = 670 + 603 = 1273. T. K, instead of multiplying 67 by 19, in the manner you have done, I multiplied by 20 ; by how much would the product be more than is required ? P. By once 67. T. "What, then, may be done to obtain 19 X 67? P. Multiply by 20, and from the product sub- tract 67; thus: — 19x67=20x67—1x67. =1840—67=1273. T. I leave it to you to choose either of these two methods. Teacher and pupils give promiscuous questions, after which the exercises. Part II, are taken up. Answers to the Exercises. By 11. Ans. 1. 539 Ans!] 3. 1089 Ans. 5. 1573 2. 957 4. 1331 6. 2387 90 LESSONS ON NUMBEK. By 12. By 13. By 14, Ans. 1, 444 377 266 2. 696 611 490 3. 838 819 644 4. 888 1092 798 5. 996 1235 •952 6. 1164 1387 1106 By 15. By 16. By 17. By 18. Byl! Ans. 1. 255 288 323 288 285 2. 420 464 357 486 494 3. 585 496 544 684 703 4. 615 672 731 882 912 5. 780 848 918 918 1121 6. 945 1024 1105 1116 1159 7. 1110 1200 1292 1314 1368 8. 1275 1376 1479 1512 1577 9. 1440 1552 1666 1710 1786 Lesso» f II. To mut Itipki by 21.22.. 29. Teacher. You have learnt to multiply by 20, and also by 1 ; if, then, you have to multiply a number by 21, what must be done? Pupils. First multiply by 20, then by 1, and add the two products. T. Multiply 37 by 31. P. 31x 37=20x27 + 1x27. =540+37=567. MULTIPLICATION. 91 The pupils must give a similar account as to the multiplication by 23, 23 29. Answers to the Exercises. By 21 By 22. By 23. By 24. Ans. 1. 525 594 644 600 2. 777 836 1035 864 .3. 1008 1078 1334 1128 4. 1176 1254 1472 1392 5. 1407 1606 2001 1656 6. 1869 2156 2208 1800 By 25. By 26. By 27. By 28. By 29. Ans. 1. 625 884 729 1232 841 2. 950 1144 1107 1568 1102 3. 1225 1430 1566 2072 1363 4. 1335 1716 1863 3424 1682 5. 1875 2003 2106 2632 2001 6. 2175 2288 2538 2772 2581 Lesson III. To multiply by 31, 32 39. Teacher. Multiply 31 by 31. Pupils. 31 X 31 = 30 X 31 + 1 X 31 = 930+31 =961. T. Multiply 87 by 32. P. 32 X 87 = 30 x87+2 x 87 = 3610 + 174= 2784. 93 LESSONS ON NUMBER. Answers to the Exercises. By 31. By 32. By 33. By 34. Ans. 1. 961 2784 2541 2278 2. 1302 3136 2904 2652 3. 1333 1024 3267 3026 4. 1674 1408 1122 3094 5. 2015 1760 1485 1122 6. 2356 2112 1848 1564 By 35. By 36. By 37. By 38. By 39. r. 1. 1995 1656 1332 3572 1248 3. 2380 2052 1739 3230 1677 3. 2765 2448 2146 2888 2106 4. 2835 2844 2553 2470 2535 5. 3220 2952 2627 2052 2964 6. 1225 3312 3071 1634 3393 Ans. 1. By 41 By 42. / "3 By 43. M!i7. By 44. 1. 1681 1764 1849 1936 2. 2132 2226 2322 2420 3. 2583 2688 2795 2904 4. 3034 3150 3268 3388 5. 3485 3612 3741 3872 6. 3936 4074 4214 4356 MULTIPLICATION. 93 By 45. By 46. By 47. By 48. By 49. Ans.l. 1710 4370 4183 3648 3087 2. 2205 1656 4277 4176 3626 3. 2295 2162 1504 4704 4165 4. 2790 2668 2021 1872 4704 5. 3285 3082 2538 1968 4263 6. 3780 3588 3055 2496 3724 XJAO Answers to the Exercises. . . . or 70. How much is 16 divided by 2 ? Ans. 8. 16 tens, or 160 . . . 2 ? Ans. 8 tens, or 80. DIVISION. 103 How much is 18 divided by 2 ? Ans. 9. 18 tens, or 180 . . . 2 ? Ans. 9 tens, or 90. How much is 20 divided by 2 ? Ans. 10. 20 tens, or 200 . . . 2 ? Ans. 10 tens, or 100. What remark have you to make ? P. That if it be known how to divide 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, by 2, it is easy to ascertain the quotients of 20, 40, 60, 80, 120, 140, 160, 180, 200; of 200, 400, 600, &c., by 2. T. Which are the intermediate tens to these numbers ? P. 30, 50, 70, 90, &c. T. How much is 20 divided by 2 ? P. 10. T. How much is 10 divided by 2 ? P. 5. T. Then how much is 20+10, or 30, divided by 2? P. 10+5, or 15. T. How much is 50 divided by 2 ? P. 50=40+ 10 j 40 -r- 2= 20; 10 -f- 3 = 5. Hence 50 or 40+10-r2=20+5=25. T. How much is 70, 90, 110, 130, &c., divided by 2? P. 35, 45, 55, 65, &c. T How much is 21 -;-2? 104 LESSONS ON NUMBER. P. 10, with remainder 1. T. How much is 22 -=-2? P. 22 = 20+ 2; 20-r-2 = 10; aiid2-r2=l Hence 22-f-2=10+l = ll. T. How much is 23, 24, 25, &c.,-7-3? P. 11, r. 1; 12; 12, r. 1; &C. T. How much is 31, 32, 33, 34, &c.,-i-2? P. 34-T-2 is 30H-2, and 4-r-2; 30-^2 = 15, and 4h-2=2. Hence 34-4-2=17. T. [Proceeds in a similar manner with the mim- hers 41, 42, 43, &c; 51, 52, &c.] How much is 97-r-3? P. 97H-2 is 90-r-3, and 7^2; 90-i-2=45, and 7-f-2=3, rem. 1. Hence 97-^2=45 + 3, rem. 1=48, rem. 1. T. How much is 159-r-2? 159=100 + 50 + 9, which divided hy 3, =50 + 25+4, rem. 1. Hence 159-h2=79, rem. 1. T. If, then, you know how to divide certain numbers by 2, it is easy to ascertain the quotients of all other numbers by 3. Which are those cer- tain numbers ? P. 3, 3, 4, 5, 6, 7, &c., 30. The pupils are now called upon to give ques- tions to the class, after which the exercises in Part II. are taken up. DIVISION. 105 Answers to the Exercised. 1. The number to be divided is called the divi- dend ; the number which is contained in the divi- dend is called the divisor ; and the number which shows how often the divisor is contained in the dividend is called the quotient. 2. The numbers themselves. 3. Even numbers are such as are exactly divisi- ble by 2 ; odd numbers are such as are not exactly divisible by 3. 4. If odd numbers be divided by 2, the remain- ders are always I. 5. Ans. 4. 6. That if a number be multiplied by 2, and the product be divided by 2, the quotient is always the number itself. 7. That if a number be divided by 2, and the quotient be multiplied by 2, the product is always the number itself. Ans. 8. 94, r 1 ; 1 18, r. 1 ; 175, r. 1 ; 200 r. \; 253, r. 1 ; 300, r. 1 ; 391, r.l; 448,r. 1 j 494; 500, r. 1. 9. 186,7-. 1. Ans. 12. 204. 10. 44. 13. 0, 11. 195, r. 1, F 3 106 LESSONS ON NUMBEB. Lesson II. To divide by B, 4, and 5. The division by 3, 4 ,5, and succeeding numbers, is to be treated precisely in the same way as shown in the preceding lesson. The pupils, accordingly, are to find in which numbers 3. is contained once, twice, 3 times, 4 times 10 times (these are said to be divisible by 3.) Next, to find the quo- tients of all numbers between 3 and 30, and to commit them to memory. Then follows an im- portant exercise: — 3 is contained in 3 once. 3 tensorin 30... 1 ten times. SOtensorin 300 ... 10 ten times, or 100 times. Agaia, 3 is contained in 6 twice. 6 tens, or 60 ... 2 tens, or 20 times. 60 tens, or 600 ... 20 tens, or 200 times. Similarly, 3 is contained in 9 3times. 90 30 times. 900 300times. 12 4time8. 120 40 times. ■■• 1200 400 times. &c. DIVISION. 107 These exercises are important^ because they give the pupils facility in dividing the intermediate tens, 40, 50, 70, 80, 100, &c.; for, since they know that 3 is contained in 30 10 times, the question, how often it is contained in 40, wiU present no new difficultyj they will naturally conclude thajt 3 is contained in 30 + 10, 10 4- 3 times, with remainder 1 ; similarly 3 is contained in 50, that is in 30 + 20, 10 + 6 times, with the remainder 2 ; again, 3 is contained in 70, that is in 60+10, 20 + 3 times, with the remainder 1; and so on with other numbers, which, taught by the above exercise, the pupils will separate each time into two or more convenient numbers: divide each of them and add the quo- tients. For instance, the question, "Divide 87 by 3,^' will be solved thus, by pupils who have gone slowly and regularly through the above exer- cises; viz. 3 is contained in 60 20 times, and in the remainder 27 9 times. Hence, 87-7-3=29. Again, Divide 158 by 3. Solution. 150h-3=50. 8h-3= 2, r. 2. Hence 158-7-3=52, r. 2. And again, Divide 259 by 3. Solution. 240-f-3 = 80 ; i. e. 24 tens + 3 = 8 tens ; and 19-=-3= 6, r. 1. Hence 259-^3=86, r. 1. 108 LESSONS ON NUMBER. Teacher and pupils give questions to tlie class, and the exercises in Part II conclude division by 3. The steps in dividing by 4, 5, 6, are precisely the same. The pupils have to learn by heart the quotients of the numbers 4 to 40, 5 to 50, 6 to 60, by 4, 5, 6, respectively;. — these known, there follow what were called above, important ex- ercises. Questions by the teacher and pupils, and the exercises in Part 11, conclude each lesson. Answers to the Exercises. By 3. Ans. 1. 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, &c 99. 2. Either 1 or 2. 3. No ; for if the remainder be 3, the quo- tient will be 1 more ; the same with 4 or 5. 4. 38, r. 1 ; 84, r. 1 ; 127 ; 135, r. 2 ; 115, r. 2; 229, r. l;233,r. 2; 276, r.l; 328 J 335, r. 2; 780; 1033, r. 2. 5. 152. Ans. 6. 252. 7. 1260. By 4. Ans. 1. 4, 8, 12, 16, 20, 24, 28, 32, 36, &c. 100. DIVISION. 109 Ans. 2. Either 1, or 2, or 3. 3. There cannot he a remainder 4, or more than 4, because 4 would he contained in such remainders. 4. 34; 54, r. 3; 76, r. I; 118, r. 1 ; 147, r. I; 169, r. 1; 197, r. I; 212, r. 3; 227, r. 1; 253; 328, r. 3; 505, r. 1. 5. 104, n 3. Ans. 6. 118, r. 2. 7. 1836. By 5. ^ws. 1. 5, 10, 15, 20, 25, &c. 2. Either 1, 2, 3, or 4. 3. 29, J-. 3 ; 54, r. 1 ; 60, r. 4 ; 96, r. 3 ; 119, r. 3; 123, r.2; 148, r. 1; 171, r. 4; 189, r. 2; 202, r. 1; 429, r. 4 ; 655, r. 3 ; 823, r. 1 ; 1065, r. 4. 4. 107. .4ns. 8. 88. 5. 46, r. 2. 9. 87. 6. 80. 10. 261. 7. 46, r. 1. Lesson III. To divide by 6, 7, 8, 9. 1. The pupils have to commit to memory the quotients of the numbers, 6, 7, 8 ... 60 by 6 ; 7, 8, 9 ... 70 by 7 ; 8, 9, 10 ... 80 by 8; 9, 10, 11 ... 90 by 9. 110 LESSONS ON NUMBER. 2. Then follow important exercises, viz. 6 divided by 6 = 1. 60 ... 6 =1 ten. 600 ... 6 =1 hundred. &c. 12 divided by 6 =2. 120 ... 6 =2 tens. 1200 ... 6 =2 hundreds. &c. Likewise, 7 divided by 7 = 1. 70 ... 7 = 1 ten. 700 ... 7 = 1 hundred. &c. And 8 divided by 8 = 1. 80 ... 8 = 1 ten. 800 ... 8 = 1 hundred. &c. Also, 9 divided by 9 = 1. 90 ... 9 = 1 ten. 900 ... 9 = 1 hundred. &c. 3. Questions by the teacher and pupils to the class. 4. The exercises in Part II., conclude each lesson. DIVISION. Ill Answers to the Exercises. By 6. Ans. 1. 6, 12, 18, 24, 36, 42, 48, 54, 60, 66, &c. Ans. 2. Either 1, 2, 3, 4, or 5. 3. 23; 34, r. 1 ; 64, r. 5; 78, r. 5; 97, r. 4; 111, r. 1; 149, r. 5; 148; 156, r. 4; 167,_r. 1; 352, r. 2; 535; 720, r. 1. Ans. 4. 28. ^jis. 6. 192, r. 5. 5. 28. 7. 43, r. 1. By 7. Ans. 1. 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, &c. 2. Either, 1, 2, 3, 4, 5, or 6. 3. 16, r. 2; 32, r. 1; 48; 65, r. 1; 81; 96, r. 6; 127; 141; 176,/-. 2; 335; 493, ;-. 5. J.ns. 4. 29, 7-. 1. Ans. 6. 3x4x5x6=360. 5. 55, r. 4, 7. 1016. 112 LESSONS ON NUMBER. By 8. Ans. 1. 8, 16, 24, 32, 40, 48, 56, 64, &c. 2. Either 1, 2, 3, 4, 5, 6, or 7. 3. 19, r. 1; 32, r. 4; 46, r. 3; 60, r. 2; 74, r. 1; 85, r. 4; 95, r. 3; 109, r. 7; 123, r. 3; 140, r. 2 ; 293, r. 1. ^n«. 4. 33, r. 2. ^ws. 6. 7x9x10=630. 5. 61, r. I. 7. 1380. By 9. Ans. 1. 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, &c. 2. Either 0, 1, 2, 3, 4, 5, 6, 7, or 8. 3. 15, r. 1 ; 27, r. 4 ; 39, r. 7 ; 52, r. I; 63, ?•. 3; 75, r. 6; 88; 99, r. 2; 109, r. 8; 108, r. 6; 146. .4ms. 4. 8, r. 6. Ans. 6. 8x10x11=880. 5. 58, r. 6. 7. 1507. Lesson IV. To divide by 10, 11, 12. Answer to the Exercises. By 10. Ans. 1. 10, 20, 30, 40, &c. DIVISION. 113 .2. 73 is not divisible by 10, because 3 re- mains after division. 130 is divisible by 10, because remains after divi- sion. All numbers ending in 0, are divisible by 10. 3. All numbers of ■which the unit figure is either 0, 2, 4, 6, or 8^ are divisible by 2. All numbers of which the unit figure is either or 5, are divisible by 5. 4. 31, r. 7; 58, r. 6; 61; 79, r. 3; 89, r. 7; 99, r. 9; 100; 101; 220; 307. By 11. Ans.l. 11, 22, 33, 44, &c. 2. 9, r. 5; 19, r. 8; 29, r. 7; 50, r. 5; 57, r. 6; 67, r. 7; 77, r. 8; 90, ?•. 9; 91 r. 10. 3. 13, r. 8. 4. How mnch is the sum of 58 and 93, divided by 11? 5. 36.— Divide 66 by 11, and multiply the quotient by 6. 6. 40, r. 4. Divide the difference of 731 and 287 by 11. 7. 280. 114 LESSONS ON NUMBEE. By 12. Ans.l. 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180, 192. 2. 12, r. 9; 23, r. 2; 29, r. 11; 37, r. 4; 47, r. 3; 64, r. 5; 74, r. 6; 76, r. 10; 83, r. 4, 3. 17, r. 9. Divide the sum of 59, 67, and 87, by 12. 4. 63, r. 10. Divide the difference between 1139 and 373,.by 12. 5. 172, r. 6. Divide the product of 345 and 6, by 12. 6. 64.— To the quotient of 180 by 12, add 49. 7. 12x12 = 144. — Divide the continued product of 12, 12, and 12, by 12. N.B. The method exhibited in these lessons applies to the division by the numbers, 13, 14, 15, &c. The limit to these exercises is that where the pupils cease to conceive with clearness, and calculate with facility; yet it is advisable to re- quire the pupils to divide all numbers up to 100 or 200, by any number not exceeding these. Another useful exercise is the following: — a number is given, which it is required to divide by 2, 3, 4, 5, &c., successively. The pupils DIVISION. 115 may exhibit a similar question on their slates^ thus: — Quest. Divide 750 by 2, 3, 4, 5 9 Dividend / Divisors, 2. 3. 4. 5. 750. I Quotients, 375 250 187, n 2 150 Dividend j Divis. 6. 7. 8. 9. 750. IQuot. 125 107,/-. 1 93,r.6 83,r.3 116 LESSONS ON NUMBER. CHAPTER V. MULTIPLES, AND DIVISOKS OF NUMBERS PRIME NUMBERS SQUARE NUMBERS. Lesson I. Multiples and Divisors. The subjects of this chapter are important, as they are a preparation for fractions. The opera- tions to be performed are easy; the object is to gain clear ideas, and these obtained, to express them in proper terms. Teacher. Multiply 1 by 1, by 2, by 3, by 4, &c., by 10. Pupils. 1, 2, 3, 4, &c., 10. T. The numbers 1, 2, 3, 4, &c., 10, are called multiples of 1. Which is the first multiple of 1 ? P. One. T. Which is the second multiple of 1 ? P. Two. T. Which is the seventh multiple of 1 ? P. Seven. T. Tell me a number which is not a multiple of 1. P. Every number is some multiple of 1. DIVISION. 117 T. Multiply 2 by 1, by 3, by 3, &c., by 10. P. 2, 4, 6, . . . 20. T. The numbers 2, 4 6, . . . 20, are called mul- tiples of 2. Whicb is tbe first multiple of 2 ? — and which is the ninth multiple of 2 ? P. Two and eighteen. ' T. Is every number some multiple of 2 ? P. No ; the numbers 3, 5, 7, &c., are not mul- tiples of 2. T. Write down the multiples of 3, 4, 5, 6, 7j &c. P^ The multiples of 3 are 3, 6, 9, 12, 15, 18, &c. of 4 are 4, 8, 12, 16, 20, 24, &c. of 5 are 5, 10, 15, 20, 25, 30, &c. of 6 are 6, 12, 18, 24, 30, 36, &c. of 7 are 7, 14, 21, 28, 35, 42, &c. T. Try to express in general the meaning of the word multiple. P. A number which contains another number a certain number of times exactly, is said to be a multiple of the latter. T. Name two numbers, of which the one is not a multiple of the other, and state the reason why it is not. P. 18 is not a multiple of 7, because 18 does not contain 7 exactly. T. Pind 6 numbers which ate multiples of 9, 7, 10, 13, and 15. 118 LESSONS ON NrMBER. P. Multiples of 9 are 18, 27, 36, 45, &c. of 7 are 14, 21, 28, 35, &c. of 10 are 20, 30, 40, &c. of 13 are 26, 39, 52, &c. of 15 are 30, 45, 60, &c. T. Find all numbers of wMch 100 is a multiple. P. 100 is a multiple of 1, 2,4, 5, 10, 20, 25, and 50. T. In order to answer this last question, what were you obliged to do ? P. Find all numbers which are exactly con- tained in 100, or by which 100 is divisible. T. It is for this reason these numbers are said to be divisors of 100. Name a number which is a divisor of 12. P. 1, or 2, or 3, or 4, or 6, or 12. T. Why? P. Because each of these numbers is contained exactly in 12. T. Name two numbers, of which the one is not a divisor of the other; and state the reason why it is not. P. 5 is not a divisor of 12, because it is not contained in 12 exactly. T. Find some numbers which are divisors of 15, 20, 40, 90, 3, and 7. P. Divisors of 15, are 1, 3, 5, and 15. 20, ... 1,2, 4, 5, 10, and 20. ... ■ 40, ...l,2,4,5,8,10,20and40. DIVISION. 119 Divisors of 90, are 1, 2^ 3, 5, 6, 9, 10, 18, 30, 45, and 90. 3, ... 1 and 3. 7, ... 1 and 7. Answers to the Eceereises. Ans. 1. A number which contains another num- ber a certain number of times exactly^ is said to be a multiple of the latter. Ans. 2. A number which does not contain ano- ther number a certain number of times exactly, is said to be not a multiple of the latter. Ans. 3. AU numbers are multiples of 1. Ans. 4. Multiples of 13, are 26, 39, 52, &c. 27, ... 54, 81, 108, &c. 59,... 118,177, 236, &c. 87,... 174,261,348, &c. Ans.5. 108. Ans. 6. 104 is a multiple of 1, 2, 4, 8, 13, 26, 52, 104. Ans. 7. A number which is contained in another number exactly, is said to be a divisor of the latter. Ans. 8. A number which is not contained in another number exactly, is said not to be a divisor of the latter. Ans. 9. 1 is a divisor of every number. 120 LESSONS ON NUMBEE. Ans. 10. Divisors of 28, are 1/2, 4, 7, 14, 28. ... , 39, ... 1, 3, 13, 39. 48, ... 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. 17, ...1,17. 29, ...1,29. 70, ... 1, 2, 5, 7, 10, 14, 35, 70. 50, ... 1, 2,5, 10, 25, 50. 80, ... 1, 2, 4, 5, 8, 10, 16, 20, 40, 80. 66, ... 1, 2, 3, 6, 11, 22, 33, 66. 112, .. V2, 4, 7,8, 14,16, 28,56, 112. 120, .... 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. 200, 1,2,4,5,8,10,20, 25, 40, 50, 100, 200. Lesson II. Common Multiples and Common Divisors. Teacher. Find the numbers of which 12 is a multiple. . Pupils. 12 is a multiple of 1, 2, 3, 4, and 6. ' T. We will now reverse the question. Find a ' number which is of multiple of 1, 2, 3, 4, and 6. DIVISION. 12L p. 12 is a multiple of them. T. For this reason 12 is said to be a common multiple of 1, 2, 3, 4, and 6. Find a common mul- tiple of 3 and 5. P. 15 is a common multiple of 3 and 5. T. Why? P. Because 15 is a multiple of 3, and also a multiple of 5; therefore 15 is a multiple of 3 and of 5 at the same time, or in common. T. When is a number said to be a common multiple of two or more numbers? P. When it contains each of these numbers exactly. T. Find a common multiple of 5 and 4. P. 20. T. Find other numbers which are likewise com- mon multiples of 5 and 4. P. 40, 80, 100, &c. T. But what is the least number which is a common multiple of 5 and 4 ? P. 20. T. Hence what may be said of 20? P. 20 is the least common multiple of 4 and 5. T. Find the least common multiple of 2 and 3 Ans.6. 2 and 4 4. 2 and 5 10. 2 and 6 6. 2 and 7 , . . . . 14. 2 and 8 8. G 122 LESSONS ON NUMBER. 3 and 4 Ans.l2. 3 and 5 15. 3 and 7 21. Sands 24. 3 and 9 9. 4 and 5 20. 4 and 6 12. 4 and 7 28. 4 and 8 8. 4 and 9 36. 4 and 10 20. 4 and 11 44. 4 and 12 12. 2, 3, and 4 12. Pound thus: — least common multiple of 2 and 3 is evidently 6; and the least common multiple of 6 and 4 is 12. Hence the least common multiple of 2, 8, and 4 is 12. Similarly, the least common multiple of 2, 3, 5 ^ms.30. 2,3, 6 6. 2, 3, 7 42. Of 2, 3, 4, 5, 6. Found thus;:— Least common multiple of 2 and 3 is 6. 6 and 4 is 12. 12 and 5 is 60. 60 and 6 is 60. Hence least common multiple of 2,3,4,5, 6, is 60, DIVISION. 123 N.B. Before proceeding, the pupils must be able to ascertain with readiness the least common multiple ofj at least, any combinations of 2, 3, 4, or 5 simple numbers. Teacher. Name a number of which 2 is a divisor? Pupils. 2 is a divisor of 4. T. Name another number of which 2 is likewise a divisor. P 6. T. Hence, what may be said of the number 2? P. 2 is a divisor of 4 and of 6 ; 2 is a common divisor of 4 and 6. T. See whether the numbers 4 and 8 have a common divisor. P. 2 is a common divisor of 4 and 8. T. These numbers have another common divisor, besides 2; what is it? P. 4. T. Hence the common divisors of 4 and 8 are P 2 and 4. T. Find the common divisors of 6 and 12. P. 1, 2, 3, and 6. T. And what may be said of 6 ? P- 6 is the greatest divisor of 6 and 12. T. Find the common divisors of 12 and 20, and see which of them is the greatest. P. The common divisors of 12 and 20 a.re, 1, 2, and4; — the greatest is 4. g2 124 LESSONS ON NUMBER. T. Find the greatest common divisor Of 15 and 20 J.ws.5. 6 and 24 6. 12 and 30 6. 4, 8, and 16, found thus : — The greatest common divisor of 8 and 16 is 8, and of 8 and 4 is 4. Hence the greatest common divisor of 4, 8, and 16, is 4. Similarly, of 5, 10, and 15 Ans.5. 9, 12, and 15 3. 36, 18, and 12 6. 4, and 7 1. 6, 7, and 8 1. Answers to the Exercises. 1. A number is said to be a common multiple of two or more numbers, when it contains each of them exactly. 2. The least number which contains each of them exactly, is said to be the least com- mon multiple. 3. The least common multiple Of 2, 3,4, 5 is 60. 2, 3,4, 6 is 12. 2, 3,4, 7 is 84. 2, 3,4, 8 is 24. 2, 3,4, 9 is 36. DIVISION^ 125 Of 2, 3, 4, 10 is 60. 3,4, 5, 6 is 60. 3, 4, 5, 1 is 420. 3, 4, 5, 8 is 120. 3,4,5, 9 is 180. 3.4, 5, 10 is 60. 4, 5, 6, 7 is 420. 4.5, 6, 8 is 120. 4,5, 6, 9 is 180. 4, 5, 6, 10.. is 60. 2,4, 6, 8, 10 is 120. 2, 6,8, 10 is 120, 3, 7, 9,5 is 315. 5, 6, 7, 8, 9 is 2520. 2, 3, 4, 5, 6, 7, 8, 9, 10 is 2520. 4. A number which is contained in two or more nmnbers exactly, is said to be their com- mon divisor. 5. And the greatest number which is so con- tained in two or more numbers, is said to be their greatest common divisor. 6. The greatest common divisor Of 18 and 48 is 6. 36 and 24 : is 13. 80 and 48 is 16. 60 and 84 is 12. 100 and 25 is 25. 144 and 96 is 48. 126 LESSONS ON NUMBER. Of 91 and 104 is 13. .84 and 126 is 42. 153 and 187 is 17. 95 and 133 is 19. Lesson III. Prime Numbers. Teacher. Pind the divisors of the numbers 1 1 , 12, and 13. Pupils. The divisors of 11 are 1 and 11. 12 are 1, 2, 3, 4, 6, and 12, 13 are 1 and 13. T. "What is there to he remarked as to the number of divisors? P. Some numbers have more divisors than others. T. What is the least number of divisors a number can have ? P. Two divisors ; 1 and the number itself. T. Give some instances of this. P. The divisors of 11 are 1 and 11. 13 are 1 and 13. 7 are 1 and 7. 5 are 1 and 5. T. Such numbers are called prime numbers. Find 10 numbers which are prime numbers. P. 1, 2, 3, 5, 7, 11, 13, 17, 19, 23. T. Is 14 a prime number? p. No; because its divisors are 1, 2, 7, and 14. DIVISION. 127 T. Can a prime number be divided into two or more numbers, which, multiplied together, produce that number? P. No ; for 5 cannot be divided into two num- bers, which, multiplied together, produce 5. T. Can a number which is not prime be so divided? P. Yes; for instance, 10=2x5; 12=3x6. T. Name two numbers which, multiplied toge- ther, produce 20. P. 4 and 5. T. Are these numbers prime numbers? P. 5 is prime, but 4 is not. T. Since 4 is not prime, find two numbers which, multiplied together, produce 4. P. 2 and 2. T. [Writing on the slate.] Hence, 20=4x5=2x2x5; is this correct? What are the numbers 2, 2, 5 ? P. They are prime numbers. T. Let us take another number which is not prime.: — Choose. p. 24. T. Resolve 24 into two numbers. P. 4x6. T. Are these prime numbers? Resolve each into two numbers. P. 4=2x2, and 6=2x3. 128 LESSONS ON NUMBEK. T. Hence, 24=4 x 6=2 x 2 x 2 x 3. Resolve, in a similar manner, 20, 30, 40, into prime numbers. P. 20=4x5=2x2x5. 30=6x5=2x3x5. 40=8x5=4x2x5=2x2x2x5. T. Hence, every number which is not prime, may be resolved P. Into other numbers which are prime. T. You said before, 40=8 x 5. Now, consider- ing the numbers 8 and 5 as producing the number 40, when multiplied, they are called factors of 40 — Name two other factors of 40. P. 2 and 20, or 4 and 10. T. And if these factors be prime numbers, the number is said to be resolved into its prime factors. Resolve 27 into prime factors. P. 27=3 xa=3x 3x3. T. Resolve 28 into prime factors. ' P. 28=4x7=2x2x7. T. Resolve 30, 32, 36, 40, into prime factors. P. 30=3x10=3x2x5. 32=4x8=4x2x4=2x2x2x2x2. 36=4x9=2x2x3x3. 40=4x10=2x2x2x5. Answers to the Exercises. 1 . A number having no other, divisors besides unity and the number itself, is called a prime number. DIVISION. 1 2. The prime numbers are- - 1. 17. 43. 73. 101. 2. 19. 47. 79. 103. 3. 23. 53. 83. 107. 5. 29. 59. 89. 109. 7. 31. 61. 97. 113. 11. 13. 37. 41. 67. 71. 119. 129 3. Numbers whicb, multiplied together^ pro- duce a given number, are said to be factors of that number. 4. Prime numbers which, multiplied together, produce a given number, are said to be prime. factors of that number. 5. 60=2x3x3x5. 66=2x3x11. 70=2x5x7. 74=2x37. 80=2x2x2x2x5. 85=5x17. 89 is a prime number. 90=2x3x3x5. 100=2x2x5x5. 102=2x3x17. 115=5x23. 200=2x2x2x5x5. 300=2x2x3x5x5. 400=2x2x2x2x5x5. 500=2x2x5x5x5. g3 130 LESSONS ON NUMBER. Lesson IV. Square Numbers. This lesson supposes the pupils possessed of some elementary notions of form. Their attention being directed to a comparison between squares constructed upon multiple lines, they are thence to deduce the truths which follow :— 1 I 2 3 a i JYote. These elementary notions are conveyed in " Lessons on Solids;" a little work whieh has long been in use in Clieam School, and is now preparing for publication. Pupils who have not received this instruction, may easily be led to the Icjiowledge that a square is a surface having four equal sides, and its angles right angles. DIVISION. 131 If fig. a represent a square, of which one side be 1 iach long, each of the other sides is likewise 1 inch long, and the surface a is called 1 square inch; and if one. side be 1 foot, 1 yard, or 1 mile long, the surface a is called 1 square foot, 1 square yard, or 1 square mile. Again, if fig. h represent a square, of which one side is 2 inches long, each side is 2 inches long; hence, dividing each side into 2 equal parts, and drawing lines joining the sections, the surface 6 is a square containing 4 squares, each equal to fig. a; that is 4 square inches, 4 square feet, 4 square yards, or 4 square miles, according as fig. o is 1 square inch, foot, yard, or mile. Likewise, if fig. c represent a square, of which one side is 3 inches long, each side . is 3 inches long; and hence, dividing each side into 3 equal parts, and drawing lines joining the sections, the surface c is a square containing 9 squares, each equal to fig. a; that is, 9 square inches. By a similar process, it may be ascer- tained, that if fig. d represent a square, of which one side is 4 inches long, the surface cZ is a square containing 16 square inches; and if squares be constructed upon lines 5, 6, 7, 8, 9, 10 inches long, the surfaces will be found to contain 25, 36, 49, 64, 81, and 100 square inches respectively. Hence it may be said in general : — The square of. 1 inch length is 1 square inch. 2 inches ... is 4 ... inches. 3 is 9 ]32 LESSONS ON NTJMBEB. The square of 4 inches length 5 ... is 16 is 25 square inches. 6 is 36 7 is 49 8 is 64 9 is 81 10 is 100 And if instead of "inch length/' we have foot, yard or mile length, we should have square feet, square yards, square miles. Now, omitting the idea of inch, foot, yard, mile, it may be said generally: — The square of 1 is 1. 3 is 4. 3 is 9. 4 is 16. 5 is 25. The square of 6 is 36, 7 is 49. 8 is 64. 9 is 81. 10 is 100. Originally, the numbers 1, 3, 3, 10, represent the length of the sides; And the numbers 1, 4, 9, 100, the squares constructed upon them. In the abstract the numbers 1, 2, 3, 10, are called roots; And the numbers 1, 4, 9, ... . 100, their squares respectively; and for this reason 1, 4, 9, 36, 49, &c., are called square numbers; and 2, 3, 5, 6, 7, &c., are not square numbers. To arrive at these results, the teacher may pro- ceed thus : — Teacher. Tell me what you know of a square. DIVISION. 133 Pupils. A square is a surface having 4 equal sides, and 4 right angles. T. [Describes a square upon the slate.J If fig. a is a square of which one side is 1 inch long, what else is known. P. Each of the other sides is likewise 1 inch long. T. And what may the surface a be called ? P. One square inch. T. And if one side of fig. a represent 1 foot, 1 yard, or one mile, what may the surface a be called each time. P. One square foot, one square yard, or one square mile. T. Let one of you come here, and describe a square, of which one side is two inches long. P. [Describes upon the school slate fig. 6.] T. Compare this square with the former, which is the greater surface? P. The second square, the square b. 134 LESSONS ON NUMBER. T. What must be done to ascertain how many- times it is greater? P. Divide each of the sides into two equal parts, and draw lines j oining the sections. T. Do it. P. [At the slate.J P. It is 4 times as great. T. Hence if fig. h represent a sqnare, of which a side is 2 inches long, the whole surface eon- tains ? P. 4 square inches. T. And if a side be 2 feet, 2 yards, or 2 miles long — P. The whole surface contains 4 square feet, 4 square yards^ or 4 square miles. T. Describe, each on your slates, a square of which a side is 3 inches long ; proceed, as we have done just now, and tell me how many square inches it contains. P. 9 square inches. T. Proceed in a ,similar way with a square, of which a side is 4 inches long. P. It contains 16 square inches. T. Continue in a similar manner describing DIVISION. 135 squares whose sides are 5, 6, 7, 8, 9, and 10 inches long. P. We have not room enough. T. How can you remedy this ? P. We must describe smaller squares, and sup- pose the sides to be so many inches long. T. What have you found? P. A square, whose side is 5 inches long, con- tains 25 square inches ; if 6 inches long, it contains 36 square inches; if? ... 49 &c., &c., &c., T. If a square contain 4 square inches, how many square inches lie on one side ? P. Two. T. And how many such rows of two squares each does the whole contain ? P. Two rows. T. If a square contain 9 square inches, how many squares lie on one side of it ? 136 LESSONS ON NUMBBK. P. Three squares. T. And how many such rows of 3 squares each does the whole contain ? P. 3 rows. T. If a square contain 100 squ.are inches, how many squares lie on one side of it? P. 10 squares. T. And how many such rows of 10 squares each does the whole contain ? P. 10 rows. T. Now suppose a square, on one side of which lie 1 1 square inches, how many such rows of 1 1 squares each must the whole contain ? P. 11 rows. T. How many square inches then must a square contain of which a side is 1 1 inches long ? P. 11 times 11, that is, 121 square inches. T. And how many square inches does a square contain of which the length of a side is 12 inches.? P. On 1 side lie 1 2 squares, and there must he 12 rows each of 12 squares; it contains, therefore, 12 times 12, that is, 144 square inches. T. How many square inches are in a square of which a side is 13, 14, 15, 16, 17, &c., inches long? P. 13 X 13 or 169. 14 X 14 or 196. 15 X 15 or 225. } Square inches. 16 X 16 or 256. 17 X 17 or 289. DIVISION. 137 T. Endeavour to express, in words, how tHe content (area) of a square is found, if the length of its side be known. P. Multiply the number corresponding to the; length of the side by itself. T. Let us repeat — If the side of a square be 1 inch, 1 foot, 1 yard, 1 mile long, what is the surface called each time? P. 1 square inch, 1 square foot, 1 square yard, 1 square mile. T. In general, then, if the side of a square be called one, the surface must likewise be called — P. One. T. Or, in other words, the square of 1 is P. 1. T. In a similarly way, the square of 2 is P. 4. T. The squares of 3, 4, 5, &c.;, are? P. 9, 16, 25, &c. T. Speaking of numbers, then, without reference to length, the numbers 1, 4, 9, 16, 35, &c., are called square numbers; and the numbers 1, 2, 3, 4, 5, &c., which represent the lengths of the sides of the squares, are then called, not sides, but square roots. "What is the square of 8? P. 64. T. And what is the square root of 64? P. 8. T. What is the square root of 1 ? P. 1. 138 LESSONS ON NUMBER. T. What is the next square number, and what is its square root? F. The next square number is 4, and its root is 2. T. Is the square root of 3 more or less than 1 ? P. It must be more than 1, but less than 2, because the square root of 4 is 2. T. And what may be said of the square root of 3? P. It must be more than the square root of 2, but still less than 2. T. The next number is 4, and its square root is exactly P. 2. T. The next nuitiber is 5, and its square root is P. More than 2 but less than 3, because the square root of 9 is 3. T. The next numbers are 6, 7, 8, and their square roots are P. All more than 2, but less than 3. T. What are the square roots of 10, 11, 12, 13, 14, 15? P All more than 3, but less than 4. T. What is the square root of 50? P. More than 7, but less than 8. T. Now, if a number is more than 7, but less than 8, how much must it be? P. A little more than 7; — 7 and, a part of 1. T. You do not know what part of 1 to call this. DIVISION. 139 In our next lesson we will speak of the parts of 1, 01 fractions, as they are called. The ideas of the division of unity-j and of the necessity of this division, are thus strongly awakened in the minds of the pupils; and the 'moment has now arrived to acquaint them with the parts of unityj or what are called fractions. Answers to the Exercises. Ans. 1. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196. 2. 100, 400, 900, 1600, 2500, 3600, 4900, 6400, 8100, 10000. 3. The square root of a number represents the side of which the number repre- sents the square; or the square root of a number is such as, being multiplied by itself, produces the given number. 4. More than 9, but less than 10. 140 CHAPTER VL— FRACTIONS. § 1. DENOMINATIONS OF HALVES, THIRDS, FOURTHS, &C. Lesson I. Halves. The first notions of fractions are obtained^ like the first ideas of number, from the senses. As before, a line furnishes sufficient means to effect that purpose. To obtain halves, then, a line is divided into 2 equal parts; to obtain the idea of thirds, it is divided into 3 equal parts; to obtain fourths, fifths, sixths, &c., it is divided into 4, 5, 6, &c., equal parts successively, as will be seen in detail from what follows. Teacher. Draw a straight line on your slate, and divide it into two,equal parts. Pupils. 1 . T. What part is one of these divisions of the whole line? P. One-half. T. What is the other part of the whole line? P. Likewise one-half. FRACTIONS; 141 T. What are tlie two parts taken together? P. The whole line, T. How many halves, then, are in one line? P. Two halves. T. By what means is the half of anything ob- tained? H. By dividing it into two equal parts, and taking one of these parts. T. Imagine the number 1 divided into 2 equal parts ; what will yon caU each of these parts? P- One-half. T. One half is written thus, \. How many halves together make 1 ? P. 2 halves. T. If we wished to express this shortly in writing, how could it be done? P. Thus, i + i=l. T. Two halves is written thus, |; hence 1 is the same as . Write the answer on your slates. T. Since in 1 there are f , how many halves are there in 2 ? P- Twice f ; that is, 4 halves. T. How will you write 4 halves? P. Thus, |. T. Hence 2 is the same as |-; — express this shortly in writing. P. i=2. T. And how much is 4? 143 LESSONS ON NUMBER. P One and one half. T. Write this on your slates. T. How many halves are there in 3^ 4, 5^ 6^ 7, &c.? P. 6, 8, 10, 12, 14, &c., halves. T. You say 14 halves are 7 ; tell me why ? P- Because in 1 there are f ,^nd, therefore, in 7 there must be 7 times f, that is y. T. How many halves are there in 17? P. In 1 there are |, and in 17 there are 17 times "I, which are ^. T. Let us write what we have learnt upon the slate. [Writing.] 1 =^ h 6= V. 2 = f. 7= V- 3=f. 8 = -■ 4 = f. 9 = iJ. 5 = LD. 10 = if- &c. this. P. [Eead] 1 is equal to 2 halves 2 are equal to 4 halves. 3 are equal to 6 halves. &c., &c. T. We will now reverse the questions. How much are |, ^, J-, |, &c. P. 1, 3, 3, 4, &c. T. How much are ^ ? P. 24. FRACTIONS. 143 T. Tell me how you found this. P. Because f = 1, as often as f are contained in ^, so many ones we shall have. Now f are contained in. ^8 34 times; therefore ^ = 24. T. Instead of saying ones, it would be better to say— P. Units. T. Hence ^ are equal to — P. 24 units. T. How many units are there in ^ , V, ^, ¥? P. H, 181 39, 481. N.B. The pupils must be able to state how they have obtained these answers. T. There is another sort of question connected with the idea of halves; when you are able to answer them, we shall proceed. It is this ■ — What is 1 of 1? P i T. What is i of 2? P. I or 1. T. What is i of 3? P. f , or li. T. What is i of 4, of 5, of 6, of 1, &c. ? P. 2, 2i 3, 31 &c. T. What is half of 17? P. Because ^ of 1 is J. ^ of 17 must be 17 x i ; that is V ^^ H- 144 LESSONS ON NUMBER. T. What are f of a number? P. The whole number. T. What are I of 59? P. 59. T. And if one half of a number be taken 3 times, what must the answer be? P. More than the number ; the number more one half of the number. T. What then are f of 12? P. 18 ; because ^ of 12 = 6, therefore f of 12 = 3x6=18. or because f of 12 = 12, and ^ of 12 = 6, therefore | of 12=12 + 6=18. T. And if one-half of a number be taken 4 times, what must be the answer? P. Twice the number; because f of a number are equal to the number, and |^ of a number must therefore be equal to twice the number. T. How much is I of 8 ? P. 16 ; because J of 8 =4, therefore, ^ of 8 = 4x4=16; or, because | of 8=8; therefore f of 8=2 x 8=16. T. What are f of 20? P. 50. T. What are I of 4? P. 3x4 or 12. &c., T. Before we proceed, I wish you to tell me FRACTIONS. 145 what sort of questions we have had since we began fractions. P. 1. The first sort was. How many halves are there in 1, 2, 3, 4, 5, &c. 2. The second sort was, How many units are there in t' 'S> ¥' °''^v 2 > °''^- 3. The third sort was. How much is f of 14, | of 26, &c. The teacher and pupils are to give questions to the class, after which the exercises in Part II. are to he taken up. Answers to the Exercises. Ans. 1. If a number be divided iato two equal parts, each of those parts is called one half of the number. 2. 1A4. 190. 2^6. 2 ' 2 J 2 ' 6|0. 15J8. lyg. 3. 38i; 47i; 56i; 172i; 3941; 493. 4. f of 49=73i; V of 12= 66; fof 58=145; V of 5= 32i; fof 89=311J; V of 3= 21; f of 100=400; V of 20=150; I of 43=1931-; ^ on5=U2i; Lesson II. Thirds. Teacher. [Draws a straight line upon the slate.J 146 LESSONS ON NUMBER. a r 1 6. Let us divide the straight line c d a b into 3 equal parts, and put the letters c d s,t the points of section. What part of the whole line is a c? P. One-third. T. And cd,db? P. Each one-third. T, How many thirds then make up the whole line? P. 3 thirds. T. What part of the whole line ia a d? P. 2 thirds. T. By what means is a third of anything ob- tained? P. By dividing it into three equal parts, and taking one of them. T. And by what means are two-thirds of any thing obtained? P. By dividing it into 3 equal parts, and taking 2 of them. T. If now we imagine the number 1 divided into 3 equal parts, what will you call each of the parts? P. Oner-third. T. One-third is written thus |. Let us examine this mode of writing one-third. What does the number 3 express? P. It shows that 1 has been divided into 3 equal parts. FRACTIONS. 147 T. And what does the 1 express? P. That one of these 3 equal parts has been taken. T. How are two-thirds of one obtained? P. By supposing 1 divided into 3 equal partsj and taking two of them. T. Try, now, to express two-thirds in writing. T. You have written 2 above 3, and separated both numbers by a small line drawn between them Some of you have put 3 above 2. Did I proceed thus in writing -t? — Which number have I put below the line? — which above? Remember, then, the number which shows into how many parts 1 has been divided, is put below that which shows how many of the parts are taken. Hence, what does f mean? P. That 1 has been divided into 2 equal parts, and that 3 of such parts have been taken. It means ^ taken three times. T. Now express 3 thirds. P i T. And what are f equal to? P. f =1. T. In our last lesson on halves, we hadf=l; what were the questions after that? P. How many halves are in 3, 3, 4, &c. T. Let us, then, ask ourselves how many thirds there are in 2, 3, 4, 5, &c. P A SL 13 L5 fee 148 LESSONS ON NUMBER. T. How many thirds are there in 31 ? P. ^ ; because in 1 there are f; in 21, there- fore, there must he 21 times |, that is ^. T. What was the next sort of questions in our lessons on halves? P. Just the reverse. How many units are there m^,if ^r. &c. T. How many units are there in ^ ? P. 16^; because -1=1; therefore, as often as f are contained in ^, so many units there will be. Now f are contained in *-^.\Q times and -^ more; therefore « = 16i. T. And what was the third sort of questions in halves. P. How much is f of 18; f of 29, &c. T. How much, then, is ^ of 2? P. f ; because -^ of 1=-^; therefore ^ of 2 must be 2 times -^, that is -|. T. How much is ^ of 4, 5, 6, &c., 100? P. iof4=li; iof5=l|; 4- of 6=3; &c. and 1 of 100=331. T. How much is f of 3? P. 1^; because ^ of 2=f ; therefore f of 2 must be twice f ; that is -f, or 1^. T. How much is f of 7, 14, 20, 50. P. Because ^ of 7=|-=3i; therefore f of 7=2 x 2|=44 . FRACTIONS. 149 Again, Because -^ of 14= ^ =4f ; therefore f of 14=3 x 4f =9i. Also, Because i of 20=2J> = 6|; therefore -| of 20 = 2 x 6f =13i And, Because i of 50=^.=16f ; therefore f of 50=2 x 16|=33i. Similar solutions are required for each of the questions given by the teacher and the pupils, after which the exercises in Part II. are taken up. Answers to the Exercises. Ans.l. f means that 1 has been divided into 2 equal parts, and that one of these parts has been taken 5 times. And -f means that 1 has been divided into 3 equal parts, and that one of • these parts has been taken 5 times. 3. 7i = y^ ; 19f = SJ. a_8 s = 9h and V = 14. V = 15, and V = 22^. 11' = 39, and 11'= 58i. ifi = 47, and i|-i= 70-1-. 2|i = 67, and 2^1= 100^. 150 LESSONS ON NUMBER. AnsA. fof 19= =13f; fof 59= = 78| fof 53= =35i; fof 48= = 80. fof 80= =531; fof 50= =100. fof 86= =57i; iof 81= =189. f of 100= =66f; fof 82= =318|. f of 126= =84; f of 100= =300. • Lesson III. Fourths. Teacher. Draw a straight line on your slates and divide it into 4 equal parts. Pupils. 1 1 1 . T. What part of the whole line is one of these parts? P. 1 fourth.. T. And what are 2 parts? P. 2 fourths. T. What are 3 parts? P. 3 fourths. T. And what are the 4 parts taken together? P. 4 fourths, or the whole line. T. If, now, we imagine the number 1 to- be divided into 4 equal parts, what is each of the parts called? P. One-fourth. T. Write in figures one-fourth. P J T. What are 2 parts called? P. |. FRACTIONS. 151 T. What are 3 parts called? T. What does the number 4 indicate? P. That 1 has been divided into 4 equal parts. T. It is called the denominator. Thus, in the fraction f , what is the name of the parts into which 1 has been divided? P. Thirds. T. Hence the number 3 names or denominates the parts into which 1 has been divided. And m. the fraction f, what does 3 indicate? p. That 3 of the 4 equal parts have been taken. T. That is, it shows the number of parts taken, and it is thence called the numerator. In f , Which is the denominator — which the numerator? P. 3 is the denominator, and 2 the numerator. T. How many fourths are there in 1, 2, 3, 4, 5, &c 100? P A A JLS 1J5 2J> 4ilO T. Our second sort of questions are the reverse of the above; that is, how many units are there ■ini,f,i,f.f, &c ^r? P. li, 1|, If, 2, 2i . . . . 25. T. And our third sort of questions are, what isiof2, 3, 4 17? P S. i I AX T. What is f of 37? P. iof37=V=9i; and f of37=3x9i=18f. 152 LESSONS ON NUMBER. T. How mucli is | of 33? P. 4of23=sJ=5|; and I of 33=3x51 =17 J. Answers to the Exercises. Ans.\. The denominator of a fraction indicates how many equal parts 1 has been divided into; and the numerator shows how many of the equal parts have been taken. 3. 5i = Sj3: 71= V; 19i=¥; 37S = iF; 113^ = *f'; 2393 = H'; 27U = 1(L8 5 4 ■ i\ IT = Ri- U = 5-S-' 3J — 41 ¥=i9i; ¥=17-1; ¥= .91. ^6=43.; S«=28f; ^e = 21f. i|-s=76i; H'=51; H'= 38i. H'=72i; H'=54J; «F=195ij «f6=146|; 6|6=322; 6|6 = 166f. 4. fofl7=lli; -«- of,33=49f. I of 17=121 J I of 25= 43i. fof82=54f; f of 136=272. I of 82=611; f of 116=261. i of 96=72; V of 44=110. f of 71=881 ; V of 29= 79i. FRACTIONS. 153 Lesson IV. Fifths, Sixths, Sevenths. If a straight line be divided into 5, 6, 7 equal parts, the pupils will arrive at the idea of 5ths, 6thSj 7ths. A process quite analagous to that pursued in the previous lessons, is then applied to the number 1 . The following are instances of these sorts of questions, referred to in Lessons I, II, III. 1. Reduce 1, 2, 3, 4 98, 99, 100 to 5ths. 1,3,3,4 99, 100 to 6ths. ... 1,3,3,4 99, 100 to 7ths. 2. Reduce to units f, h t\ ¥; '#"• i A i 0-9 IS.O •• • •• • 6' 6> 6 6 > 6 • S. ± XS> 0-9 lilO ■ • ■ • • • 7' 1> 7 7 > r • 3. How much is f of 27? f of 39; f of 63, &c.? f of 31; I of 48; f of 55, &c.?. f of 17; f of 30; A of 43; f of 100, &c. ? Answers to the Exercises. Ans. 1. H'r-r-> f';'r>"^°i''^'- 3. 17f;10f 93j23A;33f,43f;68|; 67i;l 31;135|. h5 154 LESSONS ON NUMBER. Ans. 3. A of 53 = 42f ; f of 316 = 129|- ; f of 89 = 58|-; f of 331 = 132| j I- of 1 17 = 46|- ; f of 459 = 367^- ; fo{ 189 = 1511; A SA4 . 5AZ • 6iO . 1QJ4 . 1 U 8 . 2 T_5 4 ^' 6»aJ6> 6^ 6^ 6' 5. 15i; 2^^, 30|; 36^; 57^; 76; 96; 113; 148f. 6. i of 86 = 14f ; -f of 187 = 155|; I of 93 = 31; iof 44= 51-|; |-ofll5=57f; I of 80 = 106f; -I of 139 = 93|. 7 6At . 819. 13-D2- 1C98. 2Lg8. 2 9_D 5 '• T'7' T> T' 7> 7" 8. 7|; 12f; 13f; 21f; 27; 31; 35f; 45|; 59|; 77; 97f 9. iof 86= 12f; fof200 = 171f; f of 94= 26f; f of 319 = 2734 ; I of 112 = 48 ; 4 of 480 = 342f ; A of 186 = 106f ; A of 590 = 337f ; f of 190 = 1354 ; I- of 640 = 274f ; Lesson V. Eights, Ninths, Tenths. Answers to the Exercises. Ann 1 4X3. 63.8. 7X6. 1048. 14J.6. ltJ4. JJ.nS. ±. 8'8»8J 8 > S > 8 > 25_52 8 • 2. 11|; 18; 29.i; 39|; 56|; 64f; 84; 97i; 106; 112^. FRACTIONS. 155 iof 49= 61; I of 156= 97|; I of 73 = 18f; f of 200 = 150; I of 100 = 37|; I of 346 = 302f^ I of 115 = 57|; 4 7JL4 . 9 AS. 13J7. 1044. 34l29 9 > 9 ' 9 » 9 > 9 ■ 5. 4i; 12|; 20f ; 24^; 39f 6. I of 86= 9|; | of 134= 74|; •f of 144 = 32; |of242 = 161|; I of 63 = 21; |of313 = 248|; A of 115 = 511; |of516 = 458f 7 86D. 14.40. 2iZ0. 3^2.0 '• 10' 10 r 10 > 10" 8 Q-3-- 14-S-- 18-i-- 25-5-- 37-i- "• ^10' ^10> -"^"lOJ ^"10> "'lO' Ans.9. Jg-of 87= S-J^; foOfll5= 23; tVofl86= 55^^; 3^ Of 217:- 863^; j^ of 316 = 158; ^ of 419 =2515^; 3Lof528 = 369-ifio; 3^ of 674 = 539/0; ^ of 749 = 674-^0 • 156 LESSONS ON NUMBEB. § 3. OF THE DIFFERENT ilXPB-ESSIONS FOR THE SAME FRACTIONS. Lesson I. Different Expressions for Halves. If a straight line be divided- into 3 equal parts, and each of tliese parts be further subdivided into 2, 3j 4, 5, &c., equal parts, the whole line wiQ thus be divided into 4, 6, 8, 10, &c., parts, of which the half contains 2, 3, 4, 5, &c. , I I I in~ I I I I I I I I I I I I ] i 1 I That is, one half of the line is the same as 3 fourths, as 3 sixths, as 4 eighths, as 5 tenths, &c. If the same kind of division be afterwards applied to the number 1, we shall have i=A = i=A = -£- = -fi- fro 2 — 4 — 6 — 8 — 10 — y-i) °^^' By a similar process it will be found That J-=-2.=^-=-i-=-5- &n And so on as will be shown in the following lessons ; — Teacher. Draw a straight line on your slates, which divide into 3 equal parts. What is each of the parts called? Pupils. One-half T. Subdivide each of these halves again into 3 equal parts, into how many parts is the whole line then divided? FKACTIONS. 157 P. Into 4 equal pai-ts. T. What then is each of them called ? P. One-fourth. T. And how many of these fourths are there in half the line? P. Two. T. Hence it may be said that one half is the same as P. One half is the same as two-fourths. T. Instead of dividing each half into 2 equal parts, divide it into 3 equal parts, and state what you observe. P. One half is the same as 3 sixths. T. Continue to divide each half into 4, 5, 6, &c., equal parts, and state each time what you observe. P. One half is the same as 4 eighths, or as 5 tenths, or as 6 twelftjis, &c. T. Let us now apply what we have learnt to the number 1. Suppose 1 to be divided into 3 equal parts, each of them then is called T. And if each of these halves be divided into 2 equal parts, the number 1 is then divided into P. Fourths. T. And ^ contains how many of these fourths? P. Two. T. Come here and write this on the slate. P i=2 -^ • 3 — 4- 158 LESSONS ON NUMBER. T. And if each ^ is divided into 3 equal parts, the number 1 is then divided into P. 6 sixths. T. And I contains how many of these 6 sixths? P. Three. T. Hence i=f =-|. In how many different ways have we expressed -^P P. In two different ways. T. Try to find 20 different expressions for |. -"■ • 2 — 4 — 6^8 — 10 — 12 — 14) '^^■ T. Express I in 14ths. P. 1=1^; and | therefore = ^ of ^, which isJj. T. In which numbers can \ be expressed? P. In 4ths, 6ths, 8ths, lOths, 12ths, &c. T. Generally speaking, in which numbers? P. In all numbers which are multiples of 3. T. Can \ be expressed in 3rds, 5ths, or 7ths? P. No, because 3, 5, 7, are not multiples of 2. Lesson II. Different Expressions for Thirds, Fourths, Fifths, Sfc. Teacher. Divide a line into 3 equal parts, what is each of the parts called? Pupils. One-third. T. Subdivide each of these 3rds into 2 equal parts, and state what you observe. P. One-third is the same as 2-sixths. FRACTIONS. 159 T. If \ve wish, to find other dififerent expressions for \, what must be done? P. Each third must be divided successively into 3, 4, 5j 6, &c., equal parts. T. Apply the same way of reasoning to the number 1, and find 6 different expressions for ^. -^ • S — 6 — 9 — 12 15 — 18* T. In what numbers can ^ be expressed? P. In 6ths, 9ths, 12ths^ &c. ; in general, in all numbers which are multiples of 3. T. Instead of saying, " express ^ in 27ths," it is usual to say, " reduce ^ to 27ths;" that is, how many 27ths are equal to ^? P 4 — -2- T. Why? P. Because 1 = ^, t therefore must be -^ of %\, which are -^. T. And how many 27ths are equal to f, or reduce f to 27ths? P. 1^ = -!^; because ^ =-jV» f nmst be twice ^, which are ^. T. Reduce | to 36ths. P. 1 = -If j ^ therefore must be \ of ff, which is -^ ; and |- must be twice ^, which ^° 3 6" T. Hence if you know in what numbers 4ths, 5ths, 6ths, &c., can be expressed, you will be able to find different expressions for f , ^, f , &c. In what numbers can 4ths be expressed ? 160 LESSONS ON NUMBER. P. In 8ths, ISths, 16tlis, &c. ; in all numbers which are multiples of 4. T. Reduce \ to 20ths. P. 1 = ■^■^, because 1 = f g ; therefore i must be J of 1^, which are ^V T. Eeduce | to 38ths. P. I = |i, because 1 = ff ; therefore J = | of f| = /g-, and T. Find 6 different expressions for |. P 3=£. = .9__i5._Jj3_X6.— Al ■^ • 5 8 12 — 16 — 20 24 — 28 T. In what numbers can 5ths be expressed? P. In lOths, 15ths, 20ths &c. ; in all numbers which are multiples of 5. T. Reduce i to 25ths. ^- i = ^. because l=f|; therefore ^ = ;J oiU = -h- T. Reduce f to 60ths. -P- f = If, because 1= fa; therefore -i- = -i- of -ff = ^, and 6 — ^^ 60 — 60- T. Find 6 different expressions for f. P A fi_ — _a_ — JLS. — JL5. := li = it -'• S — 10 — 15 — 20 — 26 — 30 35* A similar mode of proceeding is to be followed as to the reduction of 6ths, 7ths, 8ths, &c., to other expressions. The teacher will have observed that the use of the line has been abandoned for a sort of systematic reasoning ; yet, whenever the idea is FKACTIONS. 161 not quite clear in the pupil's mind, it is recom- mended to have recourse to the senses, by requir- ing a line to be divided, according as the question may require. Answers to the Exercises. Halves. JXnS. 1. 2 — 12 — aO — 64 — 86 — 17 2* 2 i _ jsj. _ .5J1 _ i_o 5 _ ijta _ 34X *• 2 — IS — 38 — 70 — 94 — 238' Thirds. "• 315 — 39 57 87 147 A X—3-6.— A3.— JU..6. = 1 72 ^- 3 — 64 — 78 — 174 25 8" Fourths. 5 i — -L6- — ia = -SA = _3-L., "• 4 64 72 96 12 4* fi A — Ai ^ -lA. = J.aA =; 3JtA "• 4 — 68 — 100 — 248 300" Fifths. '• b — 55 — 96 — 135 300* Q i — AA — AiX = 408. "• 6 — 90 — 485 670" A jj. = _a = Aifi. 6 — 90 485 670* Sixths. Q X == liL ^ -JL2_ ^ -SLL. "'• 6 78 114 282 A = iL5. = _2J iL3.A T — 78 — 114 — 282- • Sevenths. J-'J' 7~91 — 12616827 3* ■S- = -2-6- = -S-0- = -^;r = ■^^. 7 — 91 126 168 27 3' 16162 LESSONS ON NUMBER. A A9. ^ _5_i! 3L2_ Hi 7 — 91 — 126 — 168 — 273* A — A3. l^^ = _2.6_ = -1-S.&. 7 — 91 — 126 — 168 273" i = A.5_ 9 Q_ = J-2JL = X2J- 5 — 91 — 126 — 168 273* A =r ia = jjLa — JL JL*. = sua. 7 91 126 168 273- Eights Ana n i = X2. ^ 1 5 — - 1 7 — 43 ■""*• •'••'■• 8 — 96 — 120 — 136 — 3 3 6- i — 3A = 4/; — _£J X^fi. 8 — 96 — 120 136 — 3o6" A = AQ. = 7s =:_a-5_ = .axa. 8 96 120 136 336' i =.aA — JLiLS. =^ n 9 = 894 8 96 — 120 136 336' Ninths. 12 J- = -1-3- = -J^- = -i?4t. '-^- 9 — 117 162 306* X 2-fi- 3-Q_ =: _6_8_. 9 — 117 -"162 — 306- A — jS_S_ __3L2_ =: lA^. 9 — 117 — 162 30 6 A 6.4- = -SJL = J44. 9 — 117 — 162 30 6' X — 91 1 26 JL3_a 9 — 117 — 162 — 306" 5. = 1 OA = iA4 = ii|.. 9 — 117 — 162 30 6* Miscellaneous. 13 -i = JL — 81 10 — 300' L4 1 Sfi IS — 135' a_ — j_3 11 — 121" lA—AA 16 — 48- U as la — ft- JUL _.S_3. 17 — 34- A = if- jLl_£_a 18 72* i-|=li. ^f = lf- FKACTIONS. 163 Lesson III. Reduction of Fractions to the same Denominator, Teacher. Whicli are tlie several expressions for halves ? Pupils. 4ths, 6ths, 8ths, lOths, 12ths, &c. P. And wtich are the several expressions for thirds ? P. 6ths, 9ths, 12ths, &c. T. If, then, it be required to express -J and ^. both by other fractions of the same denomination, which will you chose ? P. 6ths, or 12ths, or 18ths, or 24ths, &c. T. Then reduce '- and ^ to fractions having the same denominator. P 1 = A and JL = -2- . or i = A, andi=j%; or i = -h> and i = A; or, &c. y. How will you proceed in order to reduce •^ and \ to other fractions having the same de- nominator ? P. First find the several expressions for -^ and \, and then choose those which are the same. T. Keduce -J- and \ to fractions having the same denominator, or, as it is usual to say, to a common denominator. 164 LESSONS ON NUMBER. P 1 — A = i_A — _S 6_ fee 4 — 812 — 16! ^^' Hence the same denominations for \ and \ are, — 4ths, Sths, 12tlis, 16ths, &c. andi= f = A = A.&c- T. What are the numbers 4, 8, 12, 16, &c., of the denominators 3 and 4 ? P. They are common multiples of 2 and 4. T. Hence, if two fractions of diflferent denomi- nators are to be reduced to a common denominator, what must be done ? P. We must find the common multiples of the different denominators,* and reduce the fractions to these denominations. T. Reduce \ and f to a common denominator. '■ P. The common multiples of 2 and 3 are, — 6, 12; 18, 24, &c. Hence i=-3. fi_ = _a_ — jjj. &c r. Reduce ^ and ^ to a common denominator. P. The common multiples of 2 and 5 are, — 10, 20, 30, 40, &c. Hence i=-5-=J-Q- = -LS=.2Ji &e AiCUl^c 2 — 10 — 20 — 30 — 401 ^*^' and i=:-2; 1 fi_=_a_ &/. See Chap. V., Lesson IT. FRACTIONS. 165 T. Reduce ^ and -| to a common denomi- nator. P. The common multiples of 3 and 4 are, — 12, 24, 36, 48, 60, &c. TTpTIPP -2- — -8 ifi. — SA. := 3Ji -^ A£i fen xaeiioe 3 — y^ — 24 — 35 — 43 — ggj °''^" clllU 4 — j^ — 24 — 3a — 48 — 6 0' ^^• T. Which of the several common denominators is the least ? P. 12. T. Hence, if 2 fractions are to be reduced to the least common denominator, what must be done ? P. Find the least common multiple of their denominators, and reduce the fractions to that denomination. T. Eeduce f and ^ to. the least common denomi- nator. P. The least common multiple of 8 and 5 is 40 j lipncp 3. == 1-S. and -t ^ 3i T. Reduce ^, \, and ^ to the least common denominator. P. The least common multiple of 2, 3, and 4 is 12; hence i = A; i = A; and i = ^. Pupils and teacher give questions to the class, after which follow the exercises in Part II. 4 — 36 > i = ^i- 5 ^ 40 5' i- = U- i f = lf. 7 = T6> 1 = ff- 166 LESSONS ON NtTMBER. Answers to the Exercises. 1. 2. 3. 4. •'• 9 90>10 90" "• 10 60>13 6 0' '• 9 — »6> 12 — 36- 8 -B. = JLfi.. i = ii "• 9 — 18 18 18* Q 3 = _2_. i = .ajL *'■ 24 > 6 2 4" ■■-"■ 12 60' IS — 60* 13 AS. AO AlS. i.ti AM. A2. AS A^ AS. ■■-"■ 6 0^ 6 01 6 OJ 6 J 6 0^ 6 0^ 6TJJ 6 0' 6 0" J."*. 2 45 2 4J 2 4> 2 4? 2 4J 2 4" IK 380 3 IK 3 3 6 36 •*■"• 420J 420» 420J 420" 16 _&i)_ 9 0.. 10 0. XQJ. 1Q8 •'^'-'' 120> 120J 1205 120> 120' 1 7 2X0 . 270 280 ^12. 155 315' 3155 316* 1« 1 3 5 1 44 1 5 1 60 ^"' 18 05 18 05 18 05 18 0' § 3. ADDITION or FRACTIONS. Lesson I. Addition of Fractions which have the same Denominator. Teacher. Tell me what we have been learning concerning fractions. ruACTioNS. 167 Pupils. We have learnt. 1st. What halves, thirds, fourths, &c., are; 2nd. To reduce units to fractions; 3rd. To reduce fractions to units; 4th. To reduce a fraction to other denomina- tions; 5th. To reduce several fractions to th« least common denominator. It is of great use to put, from time to time, questions similar to the above to the pupils, and require them to illustrate each case hy examples. Teacher. All that now remains to be learned respecting .fractions is, how to add them; to subtract one from another; to multiply and to divide them. We will begin with addition of fractions. Give a question which you think very easy to answer. Pupils. Add ^ and ^. T. How much is ^ + ^ ? P. f , or 1. T. Give some other easy question. P. Add i and f . T. How much is i+f? P. f , or IJ. T. Add I and |. P -2- + -2. = L* = 7. T. Add ^ and i- P ^ 168 LESSONS ON NUMBER. T. Add I and I . P. I or IJ. T. Add I and |. P. |+A=i = 2. T. "What kind of fractions have we been add- ing? P. Tractions having the same denominator. T. Add f and f . Similar questions are to be given in the addition of 6ths, 7ths, 8ths, &c. Lesson II. Addition of Fractions having different Denominators. Teacher. If now we have to add fractions of different denominators, suppose ^ + \, what will you do? Pupils. First reduce -^ and -J- to a common deno- minator, and then add them ; thus : — ■1-1- J- = A4.^ = i Teacher. Add \ and -f. T. Add 1 and J. i'. i+J =1+4=1. T. Add ^ and i. P. j+S = i+|=| = i|. T. Add i and |-. P i+A = _5_+_a_— j_a. — i_i_ J^- gT^fiiflilT— io~ ■■■ To- FRACTIONS. 169 T. Add 1 and |. ■* • 2 '6 — 6~6 — 6 — •'■a- T. There is something to be remarked here The answer is If, do you know of another expres- sion for -f? P. Yes ^, or -f^, or ^, &c. T. True; but do you know of another expres- sion for -f below 12ths? P. Yes, i. T. Now, it is usual to express a fraction by the least expression possible; or, as it is called, to reduce it to its lowest terms. Hence f reduced to its lowest terms is P +. T. Add I and f. a 1^ 7 1 4, ~ 1 4 — 14" T. Can \\- be reduced to lower terms ? P. No. T. Add i- and 3%. T. Can If be reduced? p Yes JL2-=-5. T. In our next lesson we shall learn what frac- tions may be reduced to lower terms, and what may not. Teacher and pupils, as usual, give similar ques- tions to the class, after which the exercises in Part II. are taken up. I 170 LESSONS ON NUMBBK. Answers to the Exercises, Ans. 1, lA. /4«s.l3. I|-I=lii' 2. lii- - 14. Iff- 3. IJ-a 15. in- 4. Ifl- 16. Ill- 5. Ifi- 17. lil- 6. Iff- 18. 1-U- 7. l|i- 19. Ifl- 8. 1 B9 ■^no- 20. ii*=iR- 9. il 09 -■■1 32- 21. ■^80- 10. 3 3- 22. ifB-=lff. 11. i|i. 23. ■^7 ■^3S' 12. ift- Lesson III. Teacher. State the different expressions for ^. T. Since -1^=^, what must be done to bring ■j^ back again to its lowest denomination, -|? P. Divide its nunierator and its denominator by 6. T. State the different expressions for ^. -^ • 3 — 6 — 9 — 12 — ISJ"'''^' T. Since i=-j^, what must be done to reduce -j'^g- to its lowest term, -J-? P. Divide its numerator and its denominator by 5. FRACTIONS. 171 T. Let us now consider the fraction \^, for instance, and see by what number its numerator and denominator must be divided, in order to reduce ^ to its lowest terms. P. They must be divided by 2, and then \^ will be reduced to \. T. Can you reduce ^ to lower terms? P. No, because there is no number by which both 7 and 9 are divisible. T. Reduce -If to its lowest terms. P. 18 and 24 are both divisible by 2, therefore, -^=35^; and 9 and 12 are divisible by 3, there- fore, -^i=\ ; and hence if =i-. T. In general, then, when can a fraction be re- duced to lower terms, and when not? P. It can be reduced when both numerator and denominator are divisible by some one number ; and it cannot be reduced when numerator and denominator are not both divisible by the same number. T. Eecollect then to reduce the answers you obtain by adding, subtracting, multiplying, or dividing fractions, always to their lowest terms. Can you tell some reason why it is usual to do so? P. Because in the fraction ^, which, when re- duced, is |- ; it is easier to imagine the number 1 divided into 4 equal parts, and 3 of these parts taken, than to- imagine 1 divided into 12 equal parts, and 9 of these taken. I 2 172 LESSONS ON NUMBER. T. We will now add 3 or more fractions, [writing upon tlie school-slate.] Find tlie sum of \, \, and \. P. The least common denominator is 12, and T. Add i f , and \. T. Add i, i, and f . -*^- 2T^4T5 — 20 I 20 '20 — 20 — *20- T. Add i, i, and ^i^. JT. 2~4T 12 12 ' 12~12 12 6' T. Add li, 2i and S^-. P. li + 2i+3i=l3 4J 5' "''^' T. If, then, a number be multiplied by ^, or by \, I, &c., the product must in each case be FRACTIONS. 179 P. Less tlian the number. T. What is the meaning of multiplying by 1 ? P. Taking a number once. T. Hence what is the meaning of multiplying a number by ^ ? P. Taking it J times, or to take ^ of it. T. What is the meaning of multiplying 1, 2, 3,4, &c.,byi? P. Taking J of 1, 2, 3,. 4, &c. T. How much, then, is 1, 2, 3, 4, &c., multi- plied by i ? T. How much is 17 multiplied by ^ ? P. V, or 8i. T. What is the njeaning of multiplying by f ? P. Taking ^ of a number twice. T. How much is 18 multiplied by f ? P. ^ of 18 taken twice; ^ of 18 = 9, which taken twice, = 18. T. What have you to remark, if a number is to be multiplied by | ? P. It is the same as multiplying it by 1, since T. What is the meaning of multiplying by ^ ? P. Taking a number ^ times, or to take ^ of it. T. And what is the meaning of multiplying byf? p. Taking J- of a number twice. T. Multiply 17 by f . 180 LESSONS ON NUMBEE. P. 17 X f is^of 17 X 2; •| of 17 = V, which taken twice, = 1.4 = 111 T. What is the meaning of multiplying by \, -L JL i ^-p ? 5J ei 1) "''^' ■ P. Taking \i^,\, ^, &c. of a number. T. And what is the meaning of multiplying by I? P. Taking J of a number 3 times. T. Multiply 9 by J. P. 9 X I is i of 9 X 3; i of 9 = 1, which X 3 = V = 6|. T. What is the meaning of multiplying by ^ ? P. Taking ^^ of a number 6 times. T. Are you able to multiply a whole number by a fraction? P. Yes; we have learnt it in our first lessons on fractions. T. What kind of questions were those? P. To take %, %, •*, &c., of a number; and this is the same as to multiply a number by \, or I, A &c. Lesson III. Fractions by Fractions. Teacher. We must now learn to multiply a fraction by a fraction; and we will begin with ascertaining what is the meaning of multiplying^, for instance, by ^. You know the meaning and the result if ^ be multiplied by 1. FRACTIONS. 181 Pupils. YeSj it means to take ^ once, which gives }. T. What J then, is the meaning of multiplying Jbyi? P. Taking i of L. T. How much is that? One or two of the pupils, perhaps, will answer this question correctly, the majority not. Re- course must then be had to ocular demonstration. T. Draw a straight line ; divide it into halves, each half again into halves; now tell me what part of the whole line one of these second halves is. P. One-fourth of the line. T. Apply the same reasoning to the number ^, and tell me what | of ^ is? P ^ T. Hence how much is ^ x i ? P. J of i, or i. T. What is the meaning of multiplying ^ by |^? P. Taking i of i T. You may ascertain this by drawing a line; how will you proceed? P. Divide a line first into thirds, each third then into halves, and see what part ^ of J is of the whole line; it is \ of it. T. How much then is ^ x ^? ^ P. lofi, ori. T. And how much is ^ x i? P. 1 of I, or i. 182 LESSONS ON NUMBER. T. What is the meaning of multiplying \, ^, i, \, &c., by i? P. Taking 1 of \, of ^, of ^ of i, &c. T. Hence if you wish to learn how to multiply a fraction by J, you must be able to ascertain readily how much \ of \, of ^, &c., is. Need you always take a line and actually divide it? P. No, we can imagine it. T. Wellj then, ascertain either by drawing a line and dividing itj or by supposing it divided, how much ^ is of ^ of \, of f , \, |, I, f, f, h &c. The pupils must be able to draw up the foL lowing results : — i of i = J. 3 — 6' iL = 1 3 3" 4- = i 4 — 8' 2^ = J- 4 4- S. = A 4 8* jT. a little reasoning wiU save you a great deal of trouble. For instance, how much is |- of ^? T. How much, then, is i of f , |-, f , ^? P. 2x-L, Sxffl, 4x-Jiy, Tx-^, or ^, j^,, 10* 10' T. And if you know how much ^ of ^ is, can you tell me how much f, |^, |, of ^ is? iofi — 14' 3. — _3_ T — 14- i = -^ 8 15- JL — J_ 12 24* _a_ — _a- 10 20* ijt — xa. 31 — 31" FRACTIONS. 183 P. Yes,foriofi = ^; f ofi = 3x Vo=f6. |ofi = 5 xf^=A = ^ T. Hence, how much is | multiplied by f ? iof|=3 xfo=3^; and |off = 4x^=^^ = f=l|. T. How much is -f x f ? P J- of 1 = -L.. -t . 2 "■■• 9 — 18' f of 1 = 8x^5 = 3^=1; and I of f = 5 X I = V = 3|; therefore .a. X -S. = 2-2- A sufficient number of questions relating to the multiplication of fractions by halves, ought to be given before proceeding further ; and it must be remarked, that most children will -soon discover the rule, viz. to multiply numerator by numerator, and denominator by denominator; but since it is not the object of this treatise to enter upon rules, but merely to prepare for them, the teacher ought frequently to require of his pupils to explain how they have obtained the_ result. From the above, the mode of proceeding as to the multiplication by ^, \, ^, &c., may be antici- pated, and a short outline will be sufficient, 184 LESSONS ON NUMBER. The pupils must ascertain that 1 ofi ^ i 3 = i 1. 4 = 1 ii i &c. 1 5 i oii — -^ i = 15 i = 20" X = 26 -k = ■3V J of i = X 8 J, = 1"2 X = iV i &c. 1 '20 ioi ■4 = 1 2 X 3 = 18 i = 2 4 i = 3"0 6 &c Is &c. This done and committed to memory, is all that is necessary. Teacher. What is the meaning of multiplying by i? _ Pupils. Taking I of a number. T. What is the meaning of multiplying ^ by ^? P. Taking i of 1 T. How much is that? P J- T. How much is f x ^ ? p -X of A- JL of JL=-i-- and ^ of f =4 X -ix=^i; therefore Ax 1-=-*- T -^ 3 — 2 !• T. How much is | x f ? FllACTlONS. 185 P. i of |. Now J of > = ,^ 2 T ' and i of I = 8x^ =-^- therefore | of f = 2 x /^ = if . Hence | x -| = if . T. How much is -^ x | ; P. Sof^. Now I of I =J^; therefore J of 1 = 7 xJ^ = ^; and I of I = 3 X J^ = f i. ^»m. ^wsz«er« to the Exercises. 1. Fractions by Integers. Ans. 1. 3. Ans. 11. 27f. 2. 5. 12. 41|. 3. 7. 13. 73. 4. 12. 14. 78. 5. 317. 15. 29. 6. 4i. 16. 52. 7. 13|. 17. 91. 8. 25i. 18. 112. 9. 40|. 19. 142. 10. 32|. 20. 311. 2. Integers by Fractions. 311. Ans. 6. 343^, 2. 64f. 7. 78^. 3. 92i. 8. 249f. 4. 88|. 9. 24. 5. 270. 10. 91f. 186 LESSONS ON NTJMBEK. 3. Fractions by Fractions. 1. -^ Ans. 9. l^ 2. a lo'. if 3. U 11- ^ 4. i-f 12. i^, 5. A 13. ^T, 400 _1 100 6. A 14- '• 28 ^''- TOO « _fi_ IR 81 4. Integers and Fractions by the same. Teacher. What is tlie meaning of multiplying n by h Pupils. Taking J of 1|. T. How much is that? P. iof l=iandAof i = ij therefore i of 1^ = i + ^ = f + i = |. T. A little reflection wUl assist you in this mul- tiplication. ^ How much is 1^? T. Hence IJ multiplied by ^ is the same as- P. 4 X 2 J which is |. T. How much is 1^ x ^. P livi = 4vi=:A -*• ■'■s'^a — i ^ 2 — 3" T. How much is If x |? P. l|x|=|x |=||=l^ = li. FRACTIONS. 187 T. How much is 3| x f. P. 3|xf= VxA = Ai = 2_'L = 2i? T. And how much is 1^ x 1^. P. The same as | x f, which is f, or 2|. T. How much is l^xlf? P. lixH = |xf = V =2f- r. What is the sum, the difference, and the product of I and f? i-|-=i^— 2#=^> difference. And I X f = 2^, product. Questions like the above may be written on the school slate, and the pupils be desired to answer them, performing each operation mentally. Answers to the Exercises. Ans. Ans'. 1- A- -4ws, . 7. 2. 2. li. 8. 2i-i. 3. if. 9. 3s%. 4. 5|. 10. 6ii. 5. 7^. 11. 4^3- 6. 3|i. 12. 5|i. Sum. Djiference. Prodi 13. IH- n- n- 14 H- i- 3i. 15. 51i. lA- 81. 16. lOA. 3i^. 254-. 188 LESSONS ON NUMBEK. Sum. Difference. Produ Ans. 17. m laV 2M. 18. ^U H m- 19. m i* ^■ 20. ^th f^ 9^. § 6. DIVrSION OF FRACTIONS. The parts of this subject may be presented in the order observed in the four following sections:— 1. To divide fractions by integers; 2. To divide integers by fractions ; 3. To divide fractions by fractions; 4. To divide integers and fractions (mixed num- bers) by the same. The object of division, generally, is to ascertain " how often one number is contained in another number ; " and the teacher is strongly recommended to abide by this definition in all cases, although a great effort must be made on the part of the pupils to solve the questions belonging to the first of the four sections. 1. Fractions by Integers, Teacher. What is the meaning of division ? Pupils. Ascertaining how often one number is contained in another number. 71 How often is 1 contained in 1 ? P. Once. FRACTIONS. 189 T. Hence, how much is 1 -r- 1 ? P. 1. T. Now if a number be greater than 1, can it be contained as often? P. No; less times. T. How often then is 2 contained in 1 ? P. Less than once ; only half as often as 1 is contained. T. How much is that ? P. ^ times. T. How much then is l-;-3? P A T. How often is 3 contained in 1 ? P. \ times. T. Hence, how much is l-i-3? P. i. T. How often is 4, 5, 6, 7, &c., contained in 1 ? P- i, h h h &c., times. T. Hence, how much is 1 -;- 4, 1 -f- 5, 1 -r- 6, 1 H- 7, &c. ? T. Now, since 1 is contained in 1 once, will 1 be contained as often in a number which is less than 1 ? P. Noj less times. T. Hence, how often is 1 contained in ^ ? P. Only ^ times. T. How much then is J-f-l? P 1 T. And how often is 2 contained in k ? 190 LESSONS ON NUMBER. P- Only ^ times of what 1 is contained in J, that is I times, T. How much then is |-t-2? T. And how much is I divided by 3, 4, 5, 6, 7, &c. P. i. i _L _1_ _L 6' 8> 10' 12' 1 4- T. How often is 1 contained in -^^ ? P. Only \ times. T. How much then is ^-f- 1 ? P ^ T. How often is 2 x 1, or 2 contained in 1 ? P. |- of -^ times, -^ times. T. Hence i-r-3 is? P -L T. How much is ^ divided by 4, 5, 6, &c. ? ■'"^' 12' 15J 18' ^f'-' r. And how much is f divided by 1 ? P. f. y. Why? P. Because ^-f-l=^, therefore •§-7-1=2 x-L=f- T. How much is 1-7-2 ? P. -!■; because f-f-l=f, therefore ^ -r- 3 = ^ of -i = JL 3 — 3- T. How much is l-^\l P. i. r. How much is |-^2? P. f; because |-t-1=|, therefore J -^ 3 = -J of FRACTIONS. 191 The pupils must continue to give similar solu- tions for each question which the teacher may give. / These questions may be solved in another way. Dividing by 2 signifies to take ^ of a number. 3 i ... 4 J &c., &c. Hence, |-i-2=J of J=|. i— 3=J of i=J &C. i_:_K 1 nf 1 — -L 3— 'J — 5 01 -J— 15- i— Q=-i- of i=J &c. Still the first mode of considering division would be preferable. 2. Integers by Fractions. Teacher. How often is 1 contained in 1, 2, 3, 4, &c.? Pupils. Once, twice, 3 times, 4 times, &c. T. How often then is J contained in 1, 2, 3, 4, &c.? P. J is contained in 1, twice; 2, 4 times; 3j 6 times; 4, 8 times; &c. 192 LESSONS O^ NUMBER. T. How much then is 1-r-i, 2-r-h, 3-^^, 4~|? P. 2, 4, 6, 8. r. How often is -^ contained in 1,2, 3, 4, &c. ? P. ■J- is contained in I, 3 times ; '. . •■• 2, 6 times ; ... 3, 9 times; ••• ... 4, 12 times; &c. r. How much then is 1- ~i, 2-r^, 3--i, 4-i? p. 3, 6, 9, 12. T. Since -^ is contained in 1, 3 times, how often must f be contained in 1 ? P. Only ^ as often as ^ is contained in 1, that is, ^ of 3, or f times. T. Then how much is 1 -f-^ ? P. i or li. T. How much is 2-r-f ? P. 3, because 2-=-^ = 6, therefore 2-r-f = | of 6=3. T. Howmuchis3-T-f ? P. f or 4J ; because 8 -j-^ = 9 ; therefore 3-=--! = i of 9 = 4i. T. How much is 1, 2, 3, 4, &c., divided by J. P. 1-ri = 4; 2-ri = 8; 3^^ = 12; 4^i = 16, &c. T. How much is 1 H- J ? P. |- or 1-|-; because l-=-| =4, therefore l-r-J =^ of 4 = 1 = H. FRACTIONS. 193 T. How much is 7-f-|? P. "-f or 9^; because 7 -=- 4 = 28, therefore 7-|=iof28=8^=9i. Similar solutions are required for each of the questions which the teacher and pupils may give to the class. The principle upon which these solutions depend can be made very obvious to the senses; for it is clear that the smaller the mea- sure the greater the result obtained by applying it to an object. Suppose a foot measure to be applied to the length of a table 10 times; it is not necessary actually to apply a \ foot measure, or ii fj ij fj &c., of a foot measure; for since a foot measures the object 10 times, a \ foot must evi- dently measure it 2x 10, that is 20 times; ^ of a foot 3 X 10, that is 30 times; but a -| foot measure only -^ of 30, that is 15 times; a \ foot measure 4 X 10, that is 40 times; but a | foot measure only ^ of 40, or ^gO, or 13^ times. The intelligence of children of 8 or 9 years old readily seizes these truths, and little effort is then required to apply a similar train of reasoning to abstract numbers. Whenever a child seems perplexed by a question of the kind, he will surely arrive at the true result, if accustomed thus to reason. Suppose the question; divide 3 by -J, the child begins from what he knows to be true, viz. : — 3h-1=3; therefore 3^-^=3x9=27; and hence 3^J-= V of 27= V=3f. K 194 LESSONS ON NUMBEK. Again, divide 7 by f . Solution. Because l-f-^=9; therefore 7-r-i=7x9=63; and therefore 7-=-f =i of 63= ^ =7^. These solutions the pupils are required to per-' form mentally; but the written exercises should be solved as shown above. Answers to the Exercises. I. Fractions by Integers. Ans.l. J. 4* A: ns. 9. A- 2. ^- 10. 120- 3. 6 4' 11. J. 4. ^^ 12. 81' 5. -is- 13. 49* 6. f- s 14. 1 01" 7.. 3 3- 15. 1ST 12 0' 8. 19 10 0' 2. Integers hy Fractions. Ans.l. 8. Ans.l. 35f. 2. »i- 8. 4f. 3. 18|. 9. Vi- 4. 10*. 10. 8. 5. 36f. 11. 7. 6. 161- 12. 7. FRACTIONS. ns. 13. 8. Ans .17. 13. 14. 10. 18. 4. 15. 11. 19. 5. 16. 12. 30. 9. 195 3. Fractions by Fractions. The solutions of questions of this section are only an extension of the principles laid down in the former section to fractional numbers. For, since it is known that ^ is contained in 1 twice, it necessarily follows that ^ is contained in J of 1 only ^ of twice, that is, once; or, because 1 -^ | = 2 ; therefore i-^^=j of 2 = 1. Again, because it is known that ^ is contained in 1, 3 times; it follows, that ^ is contained in -l only half of 3 times, that is f times; in other words, because 1 -r--^ = 3 ; therefore i-Hf =i of 3=f =1|. And so on with other questions of the kind. Teacher. How often is | contained in 1 ? Pupils. Twice. T. Hence, how often must ^ be contained in ^ of 1, or in ^? p. Only ^ the number of times it is contained in 1, that is once. k2 196 LESSONS ON NUMBER. T. How much then is ^-i-^? P. 1. T. Agaiiij how often is \ contained in 1 ? P. Twice. T. Then how often must \ be contained in the 3rd part of 1, or in ^? P. Only the 3rd part of the number of times it is contained in 1, that is f times. T. How much then is i-r-i? P. f; because 1-7-^=3; therefore i-=-i=i of 2 = 2. T. Also, because l-r-i=2; therefore -^-f-| is P. i of 3, or \, or J. T. And, because 1-^^=2 therefore ^H-i= is P. iof2=f. r. And, because 1-^^=2, therefore -^-7-^ is P J- of 2=^=i T. How much is i, 1, Jg^, &c.,-7-i. P. Because l-=-^=2; therefore i-i-i=i of 2=f =i. Again, because 1-4-^=2; therefore -^-7-^=1 of 2 =|-; and because l-h^=2; therefore A^i=A of ^=fo=h T. How much is -f-r-i? FRACTIONS. 197 P. Because l-i-i=2; therefore ^-r-i=\ of 2=|; aud therefore f -i-i=2 x f =f =1 l, T. Divide I, f, -I, f, &c., by i. P. Because l-7-i=3; therefore i-~^=i of 3=f =J ; and therefore | -7- 1 =3 x 1=^=11, Also, because 1-t-^=2; therefore i-=-i=i of 2=f ; anaf-i-i=4x|-=f=lf Again, because 1-t-^=2; therefore i-T-i=i of 2=|=i; andiH-i=5xJr=4=lf. Also, because l-f-|=3; therefore i-f-i=| of 2=^; andf-f-i=6x|-=V'=lf T. How much is J-i-i? P. Because l-=-|^=3; therefore |-i-i=i of 3=f=l^. T. And how much is J-r-f ? P. Because l-r-^=3; therefore i-f-^=i of 3=|j and therefore J-r-f =5 of f =|. 06s. In this solution, the teacher has to bring forcibly before the minds of his pupils the prin- ciple, that, "if a number be contained in another 198 LESSONS ON NUMBER. number 6 times, twice the former can be contained in the latter only ^ of 6 times, that is 3 times." And 3 times the former, only ^ of 6 times, that is, twice. 4 times the former, only J of 6 times, that is, -| = | times. &c. And hence, because it has been ascertained that i-i-'k=%; therefore i--2xi=Jof|=i. Teacher. How much is ^-i-i ? Pupils. Because l-i-^=3; therefore i-i-i=i of 3=1 ; and therefore ^-f-f =i of 1=|. T. How much is 4 H-f? P. Because 1-7-^=3} therefore i-i-i=i of 3=|; and4-i-|=iof|=|. T. How much is |-=-|^? P. Because 1-7-^=3; therefore l^\=l of 3=|; andi-ri=3x|=|; and therefore i-Hf=^ of ^=1 = 1^. The pupils must be able to give a similar solution for each question. FKACTIONS. 199 The principle remains unaltered^ whatever the question may be; for let it be required to divide Because l^i=9; therefore ■|-7-i=| of 9=^; andf^i=3x-a= and therefore ■f-=-|=i of V=f6- Again, to divide f by f . Because l-r-^=5; therefore ■^-r-J=^ of 5=|; hence |-f-i=5xf=V; and therefore f-r-f =i of ^ =f|=] J-^, Answers to the Exercises. ns.l. H- /4ns. 13. 60 7T 2. IVa- 14. If 3. n- 15. 3ii 4. If 16. as 5. i- 17. 2if. 6. l^- 18. I. 7. A3. 64,- 19. |. 8. IJLI ■•■iO" 20. |. 9. 40- 21. H- 10. 44- 22. S- 11. lU- 23. 1- 12. Hf 24. |. 200 LESSONS ON NUMBER. 4. Integers and Fractions by the same. Teacher. "What are the kinds of questions yon have now learnt to perform in division of frac- tions ? Pupils. ] . To divide fractions by integers ; 2. To divide integers by fractions; 3. To divide fractions by fractions. T. What do you think now remains to be learnt ? P. To divide integers and fractions by integers and fractions. T. Give a question of the kind, and let us see if you are able to answer it. P. Divide l^ by l-^. T. Which is the greater of these numbers? jr. ij. T. Can you tell whether the, answer (quotient) is to be less than 1, or more than 1? P. It must be more than 1, because \\ is less ihan 1^ ; it must, therefore, be contained more than jnce in IJ. ' T. Well, then, how much is 1^-rli? P. \\-—\\ is the same as f -r-| ; and because 1-^^=3, therefore i-H|=iof3=f, and f-H- ^==3 X f=f; and therefore Henceli-r-lJ=l^. FRACTIONS. 201 T. Divide 1^^ by \\. Is tlie quotient to be more or less than 1 ? P. Less than 1, because \\ is more than 1^, and cannot, therefore^ be contained once in 1^. Now li--lJ=A^i which is -|. T. Hence you can always teU beforehand, whe- ther the quotient is to be less or more than 1 ; and I advise you to ascertain that before you begin finding the answer, as it wUl show you whether your answer approaches the truth or not. — Divide 31 by 41. P. The 1 quotient must be less than 1. 3|H r41=V-HV- Now 1 H r i=5; 4h ^ i=l; 1-5 _: 4 • - i=¥; therefore VH-V=84=f3- ^«*- The pupils must give a similar account for each of the questions given to them; and in the exer- cises of Part II., write out, as shown above, the process by which they have obtained these re- sults. Answers to the Exercises. 1. if-. 8. A- 9. 5|. 10. 7|. 11. 4f. 12. ^V k3 ns.l. h 2. 1-5- 3. ••■20 4. 2Jl 21' 5. ^i' 6. li- 202 LESSONS ON NUMBER. Answers to the Promiscuous Questions. Ans.6. 17. 7. ^. 10. -aVo- ns.l. 8' 2. 3ii. 3. H- 4. 1-0-8- 5. SAA. 203 CHAPTER VII. PROPORTIONS & PROGRESSIONS. § 1. ARITHMETICAL PKOPORTION^ OR BQUI-DIFFERENCE. Teacher. What is the difference between two equal numbers? Pupils. Nothing. T. Name two numbers whose difference is 1. P. 1 and 2. T. Whose difference is 2 P. 1 and 3. T. Whose difference is 3 P. 1 and 4. T. Which of these is the greater; the first or the second? P. The second, namely 4. T. Name two other numbers whose difference is also 3 ; the second being the greater of the two. P. 5 and 8. 204 LESSONS ON NtlMBER. T. Hence, of the four numberSj 1, 4, 5, 8, it may be said that the difference between the 1st and 2nd is the same as P. The difference between the 3rd and 4th. T. I will write this upon the slate. [Writing.] 1~4=5~8. Besides this, there is something else to be re- marked concerning these 4 numbers. What is it? P. That the second is by as much greater than the 1st, as the 4th is greater than the 3rd. T. Can these 4 numbers be so placed, that the 1st is by as much greater than the 2nd, as the 3rd is greater than the 4th. P Yes, thus: 4~1=8~5. T. Find 4 other numbers of which the same may be said as of these 4. P. 2~5=7~10, or 5~2=10~7. T. What is the difference between the first and second pair of these numbers ? P. 3. T. Hence it may be said that their difference, 3, is common to each pair, or that they have a com- mon difference. Find two pairs of numbers whose common dif- ference is 4. P. 6, 10, and 11, 15. T. How am I to place these 4 numbers, so that the order before mentioned may be observed? PROPORTIONS AND PROGRESSIONS. 205 P. 6~10=11~15, or 10~ 6=15~11. T. Can they be placed so that the difference be- tween each pair shall still be 4, and this order not be observed? P. Yes; 6~10=15~11, or 10~ 6=11~15. T. Now the 1st of these is less than the 2ndj but the 3rd is P. Greater than the 4th. T, Or, the 1st is greater than the 2nd, but the 3rd is P. Less than the 4th. T. Now I wish you to find two pair of numbers whose common difference is 5, and to place them so as that the order before mentioned may be ob- served. P. 5~10=11~16, or 10- 5 = 16-11. T. Find 2 pair whose common difference is 6. P. 7~13=15~21. 13- 7=21-15. T. Two pair of numbers, similarly related, and placed in such an order, are said to form an arith- metical proportion, or equi-difference. Now state how 2 pair of numbers must be re- lated, and how they must be placed, in order to form an arithmetical proportion. p. The difference between each pair must be 206 LESSONS ON NUMBER. the same; and they must be so placed that accord- ing as the 1st is greater or less than the 2nd, the 3rd is likewise greater or less than the 4th. T. Find 6 arithmetical proportions. P. 1~ 3= 9~ 5; 13-17=25-29; 18-12=20-26; 33-54=17-38; 49-15=86-52; 100-80=30-10. T. Do 17-15 = 39-35 form an arithmetical proportion? P. No; because they have not a common dif- ference. T. Do 18-5 = 10-33 form an arithmetical proportion? P. No; because though they have a common difPerencej yet the 1st is greater than the 2nd; but the 3rd is less than the 4th, which must not be. T. If, now, the first 3 numbers, or terms, as they are usually called, be known, you will be able, I think, to find the 4th term. For instance, the difierence between 3 and 7 is the same as the dif- ference between 9 and what other number? P. 5 or 18. T. Trui; but recollect that the four numbers must form an arithmetical proportion^ and that I mentioned the lesser first. P. It must be 13, and not 5. PROPORTIONS AND PROGRESSIONS. 207 T. Hence 3~7=9~13; but had I said the difference between 7 and 3=the difference between 9 and what other number? your answer would have been P. 5. T. Let us try another question. [Writing on the slate.] 13~9=19~ Is the 4th term to be less or greater than 19? P. Less than 19^ because the 1st term, 13, is greater than the 2nd term, 9; hence the 3rd term, 1 9, must be greater than the 4th term, or the 4th term must be less than 19. T. And by how much must it be less than 19? P. By as much as 9 is less than 13; that is, by 4. T. How much, then, is the 4th term ? P. 19—4, or 15. T. Hence 13~9=19~15. What, then, must be inquired iato, in order to find the 4th term? P. First, whether it is to be greater or less than the 3rd; and then, by how much it is to be greater or less. T. Find a fourth number which shall form an equi-difference with 5, 11, 17. P. The difference between 5 and 11=6; and since 5 is less than 11, the 3rd term, 17, must be 208 LESSONS ON NUMBER. less than the 4th; that is, it must be less than the 4th by 6; the number, therefore, is 17+6=23. Hence 5-11=17-23. T. Find a 4th number which shall form an equi- difference with the numbers ^, ^, \. P. The difference between and since | is greater than -^ by |, therefore ^ must be greater than the 4th term by -J. Hence i— i=i%— A=iVj t^^ 4th term; and therefore i~i=i~-iV- A similar solution is required for each question; and since the mind has to dwell some time upon each question, the teacher is advised to write the 3 numbers each time upon the school-slate, or let the pupils write the question upon their own slates. Answers to the Exercises. AnsA. 74. Ans. 10. 48. 5. 100. 11. 1 2* 6. 33. 12. ii- 7. 51. 13. -U- 8. 61. 14. ^.■ 9. 29. 15. 80/,. PROPORTIONS AND PROGRESSIONS. 209 § 2. ARITHMETICAL PROGRESSION. Teacher. Name 4 numbers whose successive dif- fereaces are equal. Pupils. 1, 2, 3, 4. T. What is the common difference? P. 1. T. Name 4 other numbers whose successive dif- ferences are 2. P. I, 3, 5, 7. T. What difference is there between these four numbers, and 4 numbers forming an equi-differ- ence? P. The numbers 1^ 3, 5, 7, form an equi-differ- euce; but the difference, 2, is not only common to the 1st and 2nd pairs, but also is the difference between the 2nd and 3rd terms. T. Name 4 other numbers whose successive differences are 3. P. \, 4, 1, 10. T. Can there be more than 4 numbers having this property. P. Yes, as many as you please; thus: — 1, 4, 7, 10, 13, 16, 19, 22, &c. T. A series of numbers, whose successive differ- ences are equal, is called an arithmetical progres- sion. Form an arithmetical progression, the num- 210 LESSONS ON NUMBER. bers, or terms, as they are calledj having a com- mon difference, 4. P. I, 5, 9, 13, 17, 21, 25, &c. T. You made 1 the first term ; could the series begin with 2? P. Yes; 2, 6, 10, 14, 18, 33, &c. T. With 3? P. Yes; 3, 7, 11, 15, &c. T. In short, it might begin with any number. If, then, I tell you to form a progression, begin- ning with 5, and the difference of whose terms is 7, can you do it ? P. Yes; 5, 12, 19, 36, 33, &c. T. IS the 1st term of a progression be 3, and the 3nd term 7; what is the difference of the terms? P. 5. T. And what is the progression? P. 2, 7, 13, 17, 33, 27, &c. r. The first term is 9; the 2nd term 16; what is the difference, and what the following terms ? P. The difference is 7, and the terms are, — 9, 16, 33, 30, 37, &c. T. If the 1st term be 100; the 2nd, 97; what is the difference, and what are the terms of the progression ? P. The difference is 3, and the progression is — 100, 97, 94, 91, 88, 85, 82, &c. T. What is there to be remarked concerning the PROPORTIONS AND PROGRESSIONS., 211 terms of this series, when compared with those of the former? P. The terms in this series are decreasing ; in the former increasing. T. Hence, an arithmetical progression may he P. Either increasing or decreasing. T. The 1st term of a series is 85 ; the 2nd 73. Is the series decreasing or increasing? P. Decreasing, because the 3nd term is less than the 1st. T. What are the terms of the series ? P. The difference of the terms is 12 ; hence the terms are 85, 73, 61; 49, 37, &c. T. If the 1st term be 1, and the common differ- ence be likewise 1, what is the series ? P. 1, 3, 3, 4, 5, &c. T. What is the 10th term of this series ? P. 10. T. If the 1st term be one, the common difference 2, what is the tenth term ? P. 19. T. How have you found this ? P. We have written 10 terms of this progres- sion, thus : — 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. T. And had I asked you to find the 100th term, what would you have done ? 212 LESSONS ON NUMBER. P. We should have drawn up 100 terms of the progression. T. This would have been rather troublesome; now, if you will pay attention and reflect a little, we shall be able to find any term whatever, with- out first writing down all the previous terms. [Writing upon the slate :] — 1. 2. 3. 4. 5. 6. 7. 1. 3. 5. 7. 9. 11. 13. 1. 4. 7. 10. 13. 16. 19. 2. 7. 12. 17. 22. 27. 32. What have I written here upon the slate ? P. 7 terms of each of 4. progressions. T. The first of these may serve to show us the number of terms in each of the others ; we will begin with the 2nd series. Its fitrst term is 1 ; what is the difference of the terms? P. 2. T. Now, tell me how the 2nd term 3 is obtained? P. By adding the difference, 2, to the 1st term, 1. T. Is this true for each of the other two pro- gressions ? P. Yes ; for, in the third, the first term is 1, and the difference 3 ; 2nd term is 1+3=4; again, in the last progression, the first is 2, difference 5; and the second term is 2 + 5=7. PROPORTIONS AND PROGRESSIONS. 213 T. Hence the 2iid term of every progression is equal to p. To the first term, more the difference. T. And how is the third term obtained from the 2nd? P. By adding the common difference 2 to it ; thus the 2nd term is 3, difference 2 ; hence the 3rd term is 3+2=5. T. Now, recollect^ we ascertained before that the 2nd term is the same as the 1st term, + the difference ; hence the third term must be P. The same as the 1st term, more the dif- ference ; and more the difference again, that is, more twice the difference. T. Repeat what you have ascertained concern- ing the 2nd and 3rd term. P. The 2nd term = 1st term + the difference ; 3rd term = 1st term + twice the diff. T, See if this be true for each of the other 2 progressions. P. Yes, it is. T. Now observe how the 4th term is obtained from the 3rd. P. By adding the common difference to it. T. And how is it obtained from the 1st term? T. By adding 3 times the difference to the 1st term. T. Hence the 4th term is equal to 214 LESSONS ON NUMBER. P. The 4tli term is equal to the 1st term + 3 times the common difference. T. I think you will now be able to find out how the following terms are obtained from the 1st term. The pupils will thus find, that The 2nd term = 1st term, + once the common difference. The 3rd term = 1st term^ + twice the common difference. The 4th term = 1st term, + 3 x the common difference. The 5th term = 1st term, + 4 x the common difference. The 6th term = 1st term, + 5 x the common difference. &c. The 10th term = 1st term, + 9 x the common difference. &c., &c. if the series be increasing; but if decreasing, the difference must be subtracted from, instead of added to, the 1st term. T. If the 1st term be 1, and the common difference 3, what will be the 10th term? P. The 10th term= 1st term, + 9 x the difference; now the 1st term = 1, difference = 3; hence the 10th term = 1 + 9 x 3 = 28. PROPORTIONS AND PROGRESSIONS. 215 T. By what means will you ascertain if this be true ? P. By writing 10 terms of the progressions 1^ 4, 7, &c. The pupils should be called upon to verify their answers for each question. Answers to the Exercises. Ans. 3. Ij 7 j 16; 19; 25; 31. 4. 100; 91; 82; 73; 64; 55. 5- h f ; ifo; If; IjV; 3- 6. 20; 19i, 18|; 18; 171; 16f. 7. See § 2. Arithmetical Progression. 8. 43. Ans. 12. 27^. y. 97. 13. 80. 10. 51 14. 293. 11. 6. 15. If. § 3. GEOMETRICAL PROPORTION. Teacher. What is meant by comparing one thing with another? Pupils. Ascertaining in what respect the 2 things are alike, or in what respect they diflFer from each other. T. Hence, if we had to compare two numbers with each other, what would you do? 216 LESSONS ON NUMBER. P. Ascertain whether they are equal to each other, or not. T. Compare the numbers 2 and 6. P. They are not equal to each other. T. And if 2 numbers are not equal to each other, what may be said of them respectively? P. That the one is greater than the other; or that the one is less than the other. T. By what operation do you ascertain by how much 6 is greater than 2 ? P. By subtracting the lesser number 2 from the greater number 6. T. And what is the result of this operation called? P. The difFerence between 6 and 2, which is 4. T. Is there any other way of compa^ng these two numbers with each other ? P. Yes, by ascertaining how often the one is contained in or contains the other. T. By what operation do you ascertain how often 2 is contained in 6, or 6 in 2 ? P. By dividing 6 by 2, or 2 by 6. T. And what is the result of this operation called? P. The quotient of 6-;- 2, which is 3, or of 2-r-6; which is \ or \. T. Name 2 numbers, of which the difference is the same as that between 2 and 6. P. 12 and 16. PROPORTIONS AND PROGRESSIONS. 217 T. Do you recollect what was said of 4 numbers similarly related as the numbers 2, Q, \2, 16? P. Yes ; they form an equi-difference, or arith- metical proportion, T. And can you find more pairs of numbers, whose difference is the same as that between 2 and 6? P. Yes, as many as you please ; 14 and 18 ; 20 and 24, &c. T. No; find 2 numbers, of which the quotient is the same as that of 6 divided by 2. P. 12 divided by 4; or 18-f-6; 30-j-lO, &c. T. Hence what may be said of the 4 numbers 6, 2, 12, 4. P. That the 1st 6 divided by the 2nd 2, is the same as the 3rd 12 divided by the 4th 4. T. In short, that 6-7-2 = 12-r-4. Now find 2 numbers whose quotient is the same as that of lH-2. P. 3-7-6, or 4-i-8, 5-f-lO, &c. T. How did you find these numbers? P. Since 1-7-2=1, any two numbers of which the one is ^ of the other must, when the less is divided by the greater, have a quotient = \ ; such numbers are 3-r-6, 4-7-8, 5 -=-10. T. In short, then, l-7-2=3-=-6. Find two numbers whose quotient is the same as that of 2-rl. P. Because 2-7-1=2; any two numbers, of which the one is twice as much as the other, must, L 218 LESSONS ON NUMBER. \rlien the greater is divided by the less have a quotient = 2 such are 6-f-3, 8-i-4, 10-7-5. T. We have found before that l-7-2=3-=-6; and noWj that 2-f-l=6-i-3. — What conclusion can you make? P. That of 4 numbers, if the 1st divided by the 2nd, is equal to the 3rd divided by the 4thj then also is the 2nd divided by the 1st, equal to the 4th divided by the 3rd. T. Any 4 numbers having this property are said to be proportional numbers. So that it may be said, 1 has the same proportion to 2, as 3 has ... ... to 6, or, as 4 has ... ... to 8. &c. And that, 2 has the same proportion to 1, as 6 has ... ... to 3, or, as 8 has ... ... to 4. &c. And this is shortly written thus : — 1 2 : : 3 6. 1 2 : : 4 : 8. 2 1 : : 6 3. And 2 1 : : 8 . 4. PROPORTIONS AND PROGRESSIONS. 219 Which is read thus : — as 1 is to 2, SO is 3 to 6; as 2 is to 1, SO is 6 to 3 ; &c. Now, find 2 numbers which have the same pro- portion that 1 has to 3. P. 2 to 6; 3 to 9; 4 to 12, &c.; or any num- bers of which the first is \ of the other. T. How have you ascertained this? P. By dividing 1 by 3, which is ^, and any two numbers of which the first is \ of the other, are in the proportion of 1 to 3. T. Find two numbers in the proportion of 3 to 1. P. 6 to 2j 9 to 3; 12 to 4, &c.; or any two numbers of which the first is 3 times as much as the other. T. Hence, to ascertain two numbers, which have the same proportion to each other as two other numbers, what must be done? P. Divide the one by the other; and find two numbers, which, when divided, have the same quotieht. T. Speaking of proportional numbers, this quo- tient is called the ratio, which the two numbers have to each other. What is the ratio of 6 to 12 ? p _6_ or i T. And what is the ratio of 12 to 6? P. V or 2, 230 LESSONS ON NUMBER. T. And of 4 numbers -whicli form a proportion according as tte 1st is greater or less than the 2nd, the 3rd must be P. Greater or less than the 4th. T. Hence, as 1 : 4 : : 5 to what other number? —Is it to be less or greater than 5 ? P. It must be greater than 5, since the first term^ 1 , is less than 4, the 2nd. T. "What is the number? P. 20 ; since the ratio of 1 to 4 is J, hence the number of which 5 is J is the number required; which, therefore, is 4 x 5, or 20. T. This is called finding the 4th proportional to the 3 numb r I, 4, 5. Find the 4th proportional to 4, 1, 5. P. Since f-=4, the ratio of 5 to the number must be 4, that is, 5 must be 4 times as much as the number, which is, therefore, \ of 5, or 1^; hence, 4:1: : 5 : 1^. T. As 2 is to 3, so is 4 to what 4th number? P. Because 3 is f of 3, 4 must be f of the number; hence \ of the number = ^ of 4; and V therefore the number must be ^ of 4 taken 3 times, which is 6 ; therefore, 3 : 3 : : 4 : 6. N.B. The pupils will not find it diflicult to ascertain the 4th proportional, whien the 1st term is either a imultiple of the 2nd, or the 3nd a mul- tiple of the first. When, however, the ratio of the PROPORTIONS AND PROGRESSIONS. 221 first two terms is f, |, f, f, f, ^, &c., that is, fractional, they will naturally experience greater diflSculty. Before entering then upon similar questions, it will be advisable to go through the following exercises. 6 is the half of what number? Ans. Of 2x6; that is, of 12. 6 is the third part of what number ? Ans. Of 3 X 6 ; that is, of 18. 6 is -f of what number? Ans. Of the half of 6 taken 3 times;, that is of 9. 6 is 5 of what number? Ans. Of 4x6; that is, of 24. 6 is I of what number ? Ans, Of -^ of 6 taken 4 times; that is, of 8. 6 is ^ of what number ? Ans. Of 5x6; that is, of 30. 6 is I" of what number? Ans. Of J of 6 taken 5 times; that is, of 15. 6 is 4 of what number ? Ans. Of ^ of 6 taken 5 times ; that is, of 10. 6 is |- of what number ? Ans. Of i of 6 taken 5 times ; that is 232 LESSONS ON NUMBER. 6 is -^ of what number? Ans. Of ^ of 6 taken 7 times ; that is, 11 is f of what number? Ans. Of -^ of 11 taken 8 times; that is, &c., &c. These and similar questions having been duly practised^ no difficulty will be met with in the following : — Teacher. As 7 is to 1, so is 13 to Pupils, l-f-. Because the 1st term is 7 times as much as the 2nd, the 3rd must be 7 times as much as the 4th. Now 13 is 7 times \^, which is If. T. As 7 is to 2, so is 8 to P. 2\; because 7 is i of 2, therefore 8 must be I of the number. Now 8 is \ of ^ of 8 taken twice ; that is, fx2, orV=2|. T. If your answer, 2f , be correct, what must be P. 1; divided by 2, must give the same quotient as 8, divided by 2f : — 7-i-2=|, and 8-f-2f=8-j-L6=|. Hence the answer is correct. PROPORTIONS AND PROGRESSIONS. 223 Answers to the Exercises. N.B. The answers to the first 7 questions are contained in § 3, Geometrical Proportions. Am. 8. Eatio.ofSto 7 = =^. 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(From " The Museum of Science and Art. ) 1 vol., with 89 Illustrations. 2s. * Lardner. The Electric Telegraph, Popularised. With 100 Illustrations. (From " The Museum of Science and Art.") 12mo., 250 pages. 2s., cloth lettered. * A Guide to the Stars, in Twelve Planispheres, shewing the Aspect of the Heavens for eveiy Kight in the Year. With an Introduction. 8vo. 6s. 6d., cloth. LOGIC. De Morgan's Formal Logic ; or, The Calculus of Inference, Necessary and Probable. 8vo. 6s. 6d. Boole's Investigation of the Laws of Thought, on which are founded the Mathematical Theories of Logic and Probabilities. 8vo. Us. * Neil's Art of Reasoning : a Popular Exposition of the PrmoipleB of Logic, Inductive and Deductive; with an Introductory Outline of the History of Logic, and an Appendix on recent Logical Developments, with Notes. Crown 8vo. 4b. 6d., cloth. LESSONS ON FORM; OR, AN INTRODUCTION TO GEOMETRY.' AS GIVEN IN A PESTALOZZIAN SCHOOL, CHEAM, SURREY. BY C. REINER, TEACHER OF MATHEMATICS IN CHEAM SCHOOL. LONDON: PRINTED FOR TAYLOR AND WALTON, SOOKSELLERS AMD FVBLISIIEIIS TO THE UNIVERSITY OF LOKBON, UPPER GOWER STREET. 1837. LONDON : PRINTED BY SAMUEL BENTLEY, Doiset Street, Fleet Street. PREFACE. Bacon has made an observation to this effect — that a man really possesses only that knowledge, which he in some sort creates for himself. To apply to in- tellectual instruction the principle implied in these words was the aim of Pestalozzi. It is a principle admitting of various degrees, as well as modes of application, in the different branches of human know- ledge ; but in no one can it be more extensively ap- plied than in Geometry. That science is peculiarly the creation of the human mind, in whiph, independ- ent of external nature, and complete in its own re- sources, it builds up the solid but airy fabric of its abstractions. It needs no laboratory to test its con- clusions, no observatory to obtain data for its cal- culations; rendering ; m'*d to other sciences, it asks none for itself. Hence, that teacher will act most fn conformity IV PREFACE. with the genuine character of the science, and con- sequently will render the study of it the most inter- esting and the most improving, who invites and trains his pupils to create the largest portion of it for them- selves. In Geometry, the master must not dogmatise, either in his own person or through the medium of his book ; but, he must lead his pupils to observe, to determine, to demonstrate for themselves. In order to accomplish this, he must study the intellectual process in the acquisition of original mathematical knowledge ; and having ascertained what are the con- ditions of successful investigation, he must so arrange his plan of instruction as that these conditions may be' perfectly supplied. He cannot fail to perceive that the leading requisites are a clear apprehension of the subject matter, and well-formed habits of mathemati- cal reasoning. To these must of course be added a familiar acquaintance with the science as far as it has been elaborated. The master, led by these considerations, will, in directing the first labours of his pupils, consider it as his especial aim, to enable them to form clear apprehensions at" the subject matter of Geometry, and then to develope the power of mathe- matical reasoning. Aware that clearness of appre- PREFACE. V hension can take place only when the idea to be formed is proximate to some idea already clearly formed — ^when the step, which the mind is required to take, is really the next in succession to the step already taken, he will commence his instruction exactly at that point where his pupils already are, and in that manner which best accords with the mea- sure of their development. As his pupils are unaccus- tomed to pure abstractions, he will not commence with abstract definitions. But supposing them, through the medium of ' Lessons on Objects' to have had their attention directed to the forms which matter assumes, he will present in his first lessons a transition from the promiscuous assemblage of forms to a particular group of them, consisting of the sphere, the cone, the pyramids, the prisms, and the five regular bodies. In conformity with the plan pursued in ' Lessons on Ob- jects,' the pupils will examine these solids, state what they perceive at the first glance, then by more close and attentive examination, directed by the master, discover and supply the deficiencies in their first perception, and afford him an occasion for con- necting their new ideas with adequate technical expressions. a3 VI PREFACE. The master's next aim must be to cultivate the power of abstract mathematical reasoning. With a view to this end, he maiy advantageously avail himself of the knowledge, obtained by the pupils from the solids> in the manner above described. Here, then, he will lead them to deduce the necessary consequences from the facts which they know to be true, and then invite them to examine the object and see whether their reasoning has led to a correct result. Thus, if a ' child has ascertained and knows that two sides of different planes are requisite to form an edge, and that a certain solid (an octahedron) is bounded by eight triangular planes, he will be required to deter- mine from these data the number of .edges which that solid has. He will reason thus : — Eight triangular faces have twenty-four sides ; two sides form one edge : therefore, as many times as there are two sides in these twenty-four sides, so many edges that body must have, — that is, twelve edges. This result being obtained, the object is presented to him for examination, and he perceives by actual observation the truth of that conclusion at which he had arrived by abstract reasoning. These lessons form the basis of the Introduction to PREFACE. Vll Geometry, and their results are, correct ideas of the subject matter of the subsequent lessons, adequate expressions for these ideas, and sound knowledge of the definitfons, which form the connecting link be- tween physical and abstract truths. In the former part of this work, a mode of ac- complishing these points is set forth : in the se- cond part, the further development of the power of abstract reasoning is connected with a direct pre- paration for the study of Euclid's Elements. That work exhibits a series 'of mathematical reasonings and deductions, arranged in the most perfect lo- gical order, so that the truths demonstrated rest, in necessary sequence, on the smallest possible num- ber of axioms and postulates. But, admirable as it may be in itself, viewed simply in relation ■ to the science, it is not, viewed paedagogically, an elemen- tary work. It is fitted for the matured, and not for the opening mind. The judicious teacher will desire to present to his pupils the subject matter of Euclid in such a mode and in such order as that in studying it the higher faculties of their minds may be most effectually exercised and improved. For thus only can the intellectnal food be assimilated to the Vlll PREFACE. intellect - itself — be received, as it were, into its sub- stance, and nourish, and strengthen, and expand its powers. These Lessons on Form present a mode in which these principles are applied: other modes, perhaps better ones, may be arranged ; — we but say with Horace, Si quid novisti rectius istis Candidas imperti ; si non, his utere mecum. It has been found in the actual use of these Les- sons for a considerable period, that a larger average number of pupils are brought to study the Mathe- matics with decided success, and that all pursue them in a superior manner. There is much less of mere mechanical committing to memory, of mere otiose admission and comprehension of demonstrations ready- made, and proportionably more of independent judg- ment and original reasoning. They not only learn Mathematics, but they become Mathematicians. Hence, when Euclid's Elements and the higher branches of Mathematics are to be read, the pupils are found competent to demonstrate for themselves the greater part of the propositions, and have re- course to books only for occasional correction or im- PREFACE. IX provement of their processes, and for fixing more firmly in their memory the results. These advantages arise from the application of a principle generally neglected in early education, but' deserving of attentive consideration and universal adoption ; namely, that " Every course of scientific instruction should be preceded by a preparatory course, arranged on psychological principles." First FORM THE MIND, THEN fURNISH IX. C. Mayo. CONTENTS. Page Lesson I. — Introduction . . . ; 1 II.— Ditto ...... 5 III. — Description of the five regular Solids . 9 IV.— Ditto . . . . . .11 v.— Ditto . . . . . 13 VI. — The Rhomboidal Dodecahedron . ■ 18 VII. — The Bipyramidal Dodecahedron . 20 The Trapezohedron . . . .21 VIII.— The Pyramid .... 22 IX.— The Prism . . . . ., 25 X. — Solids bounded by Curved Faces. The Sphere. 26 XI The Cylinder .... 28 -XII.— The Cone .... 31 SURFACES. CHAPTER I. — Straight Lines : Angles. Section I. — One and Two Straight Lines . . .33 II. — Three Straight Lines ... 48 III.— One Triangle . . . .63 IV.— Two Triangles— their Equality . . 74 V Equality of Triangles . . .88 VI. — Quadrilateral Figures ... 99 Xll CONTENTS. Page Section VII. — Equality of Squares, Rectangles, and Parallelo- grams , . . . .113 VIII. — Proportional Triangles . . 139 IX — Polygons 149 CHAPTER II. Section I, — One Circle — One and Two Straight Lines in a Circle 161 II. — Three and more Straight lines in a Circle . 178 III. — Lines without a Circle — Tangents 191 IV.— Two Circles .... 206 Straight Lines .... 210 Angles ..... 211 Triangles . 211 Quadrilateral figures 212 Polygons ..... 212 Circles ..... 213 LESSONS ON FORM, BEING AN INTRODUCTION TO GEOMETRY. LESSON I. The master places before his pupils a variety of objects, among which there should be (the following solids, viz.) the five regular solids, viz. several of the prisms and pyramids, the cylinder, the cone, and the sphere. Master. — I have set these objects before you, that you may find out some properties common to them all. Endeavour to discover them. (The answer of each pupil should be subjected to the consideration of the class, and be tried if in reality it equally applies to all objects.) Pupils. — These and all other objects occupy a space. M. — In how many directions does each extend ? P. — In three directions : in length, in breadth, and in depth. One of the pupils said, " and in thickness." M. (holding up a book.) — Which would you call the length of this book ? — Which is its breadth ? Does B 2 LESSONS ON FORM, BEING it extend in another direction? — By what word will "you describe it ? P. — Thickness. M. — Name an object of which it would be proper to say depth instead of thickness, P. — A well extends in length, in breadth, and in depth : so does the sea, a pond, a lake, a river. M. — Objects considered with reference to these three dimensions only are called solids. What other * property have all solids in common ? P. — They are all bounded by a surface, M. (holding up a sphere and a prism) — In what does the surface of one of these objects chiefly differ from the surface of the other ? P. — The one is composed of several surfaces, and the other is bounded only by one curved surface. M. — In what does a surface ^consist? P, — In extension of length and breadth : a surface is the boundary of anything, M. — What happens if a surface be removed from an object? P. — A part of the object is likewise removed by removing a surface. M. — Is the object, by doing so, increased' or de- creased ? P, — It is decreased. M.-^\n how many directions is it decreased ? P. — It is decreased either in length, or in breadth, or in thickness. M, — Can a surface exist without the object of AN INTRODUCTION TO GEOMETRY. 6 which it is a surface? Can you hold a surface in your hand without holding the object itself? JP.— No. On this question being asked, one of the pupils said, a shadow is a surface existing without a concomitant solid. The master will of course convince his pupils of the error, should a similar answer be given. M. (holding: up a prism.') — What is meant by the surface of this object ? P. — The assemblage of the several surfaces which bound it. M. — If we wish to distinguish one of these several surfaces from the total number of surfaces, it is usual to call it one of its faces. Now state what may be said in general of the number of faces by which all objects are bounded. P. — All objects are bounded either by one face only, or by several faces. M. — Now examine more minutely the faces of these objects, and class those together which you think to have similar faces. The master should allow the pupils some time for arranging the objects before him into groups, until they have perceived that they may be classed pro- perly into three distinct groups; — the one compre- hending those which are bounded by plane faces ; the next, those that are bounded by plane and curved faces; and lastly, those that have only one curved surface. P. — All these objects are either bounded hy pltine b2 4 LESSONS ON FORM, BEING fanes, or by plane and curved faces, or by only one curved surface. 'M. — Now examine the boundaries of the faces of the first group you have mentioned. What do you observe ? P. — They are all straight lines. M. — And the boundaries of the other group ? P. — Curved lines and straight lines, or only curved lines. The substance of the lesson is then written on the school-slate by the master, and the pupils are required to commit it to memory. Thus : — 1. All objects discernible by the senses are extend- ed in three dimensions : namely, in htigth, in breadth, and in depth or thickness. 2. Objects considered with reference to these three dimensions are called solids. 3. The surface of a solid is its length and breadth considered without reference to its depth. 4. Every solid is bounded either by one surface only, or by several _/«!ces. 5. Solids are either bounded hy plane faces, or by both plane and curved faces, or by only curved faces. 6. The boundaries of faces are either straight lines, or both straight and curved lines, or only curved lines. When the above is committed to memory, it is effaced from the slate, and the pupils are required to write it from memory and in the same order. AN INTRODUCTION TO GEOMETRY. LESSON II. At the beginning of this and every following lesson, the pupils ought to be required to recapitulate the pre- ceding lesson, first viva voce, and then by writing it out on their slates. M. — We will now first examine those solids which are bounded by plane faces only. See in what respect their faces differ. P. — In size, in shape, and in the number of straight lines which bound them. M. — Speaking of the boundaries of faces, it is usual to call them sides, instead of lines. I have brought here a considerable number of solids which are bound- ed by plane faces. Arrange them according to the number of sides by which some of their faces ar.e bounded, beginning with the least. What is the least number of sides by which some of the faces are bounded ? P. — By three straight lines — by three sides. M. — And by what word will you express the space which three straight lines inclose ? P,—r-A three-sided face — a triangle. M. — Imitate a triangle on your slates. How many lines are necessary to inclose a space ? Try one, two, three. If a space is inclosed by two lines, what sort of lines must these be ? b LESSONS ON FORM, BEING P. — Either a straight and 'a curved line, or two curved lines. M. — Which face have you placed next in succession to the triangle ? P.— One which is bounded by four sides. M. — Imitate it on your slates. Which of the faces come next ? P. — The five-sided face ; then the six, seven, and eight-sided face. M. — Imitate all these faces on your slates. Ex- amine the three-sided figure on your slates : in how many points do its three sides meet ? P, — In three points. M. — {Draws a triangle upon the school-slate). I will put the letters a, b, c at the three points, in order that we may be able to distinguish one side from the others. By what word will you express the position of the line a b, to the line b c 9 P. — The line a b is inclined to the line b c. M, — And how many inclinations have the three sides to each other ? , P. — Three inclinations. M. — The inclination which one line has to another AN INTRODUCTION TO GEOMETRY. 7 line is called an angle. How many angles are in a three-sided figure ? P. — Three angles. M. — See how many angles there are in each of the figures on your slates. P. — A four-sided figure has four angles ; a five-sided, five ; a six-sided, six ; a seven-sided, seven ; and an eight-sided has eight angles. TlSf.— Can you imagine a figure having nine, ten, eleven, etc. sides? Describe them on your slates, and observe how many angles each figure has. P. — Every figure has as many angles as it has sides. M. — You have mentioned another word instead of three-sided figure. P. — Yes, a triangle. M. — From what circumstance do you think it is called thus ? P. — From its having three angles. M. — The names of these several faces are derived sometimes from the number of their angles, and sometimes from the number of their sides. Thus, a three-sided face is sometimes called a trilateral figure (from the Latin tres, three, and latus, a side), or a tri- angle ; a four-sided face is called a quadrilateral figure (from the Latin quatuor, four, and latus, a side) ; a five-sided face, a pentagon (from the Greek irivrs, five, and yavia, angle) ; a six-sided face, a hexagon, (from the Greek t^, six, and yavia, angle) ; a seven- sided face, a heptagon (from the Greek iVra, seven. 8 LESSONS ON rOJlM, BEING ^ and yavLa, angle); an eight-sided face, an octagon (from the Greek oktu, eight, and ymna, angle). And if this mode of expression be extended to faces which are bounded by many sides, they are csD^eA. polygons, (from the Greek iro\i/c many, and yavia, angle). As before, the pupils are called upon to reproduce the lesson on their slates ; the substance of which is then arranged into sentences, and written by the master on the large school-slate, the pupils commit- ting them to memory. 1 Solids bounded by plane faces diflPer in shape and in the number of their faces. 2. — Their faces diflfer in the number of their sides. 3. — A face bounded by three sides is called trila- teral; by four sides, quadrilateral ; by five sides, a pentagon ; by six sides, a hexagon ; by sev - 1 sides, a heptagon ; by eight sides, an octagon ; by nine or more sides, a polygon. 4. — An angle is the inclination of two lines to one another which meet in a point. 5. — A trilateral face has three angles, it is therefore called a triangle ;- a quadrilateral has four angles ; a pentagon has five, a hexagon six, a heptagon seven, an octagon eight; a polygon has as many angles as it has sides.' AN INTRODUCTION TO GEOMETRY. 9 LESSON III. M. — What other parts do you discover on these solids ? P. — Corners and edges. M. — How are the corners formed ? P. — By several angles of different planes meeting in one point ; or by several edges meeting in one point. M. — How many edges or angles of different faces are at least required to form a corner or solid angk ? Try, one — two — three. P. — Three at least. M. — Instead of "corners," say solid angles; how are the edges formed ? P. — By the meeting of two faces. DESCRIPTION OF THE FIVE REGULAR SOLIDS.* M. — Which of these five solids is bounded by the least number of faces? By how many faces is it bounded ? This solid is therefore called Tetrahedron (from the Greek rtrpa, four, and £^,001, seats). M. — What are the four faces ? P. — Four triangles. M. — How many sides have four triangles ? P. — Twelve. M. — How many of these sides are there to each of the edges ? P. — Two sides. • Tetrahedron, Hexahedron Octahedron, Dodecahedron, Icosa- hedron, B 5 10 LESSONS ON FORM, BEING M. — How many edges therefore must this solid have ? P. — Six edges; because there are six twos in twelve. jf..^Now take the solid, examine it, and see whe- ther it is so. — How many angles are there about each corner or solid angle ? It is important that the pupils be convinced by actual examination of the solid, that the calculation which they have made is strictly true. P. — Three angles. M. — These angles are called plane anffles ; can you tell why ? P. — Because they are the angles of the plane faces. M. — How many plane angles are there in the four triangular faces ? P. — Twelve plane angles. iff. — How many solid angles must the tetrahedron have? P. — Four solid angles; because about every solid angle there . arfe three plane angles, and there are four threes in twelve. M. — See whether it is so. SUBSTANCE OF THE LESSON. L' — Two faces meeting laterally form an edge. 2. — Three or more edges meeting in one point form a solid angle. 3 The tetrahedron is a solid bounded by four tri- angular faces : it has six edges, and four solid angles. AN INTRODUCTION TO GEOMETRY. 11 LESSON IV. M. — Compare the sides of the faces of the tetra- hedron. What do you observe ? P. — They are of the same length ; they are equal. M. — How will you call a triangle which has three equal sides ? P. — An equal-sided triangle. M. — Call it an equi-lateral triangle (from the Latin cBquus, equal, and lotus, side). Describe an equi-late- ral triangle on your slates ; put letters at the angles. — Are all triangles necessarily equi-lateral ? P. — No ; for two sides of a triangle may be equal to each other, and the third unequal ; or the three sides may he unequal. The master desires the pupils to draw such trian- gles upon their slates ; after which, he may describe an equilateral, an isosceles, and a scalene triangle upon the school-slate, and, pointing to them, continue. M. — A triangle having only two of its sides equal to each other is called an isosceles (from the Greek laoi, equal, and axiXot, a leg) ; and the unequal side is called its base. And a triangle having none of its sides equal to each other is called a scalene (from the Greek aKali,a, to limp,, and P. — Three , -, ., . • i-.T tt * " — " Oi e i straight Imes may be equal. ' 2nd. Two 7 J ., „ a c a e f of them may be equal, and the third unequal. 3rd. -All , 3 , , ft ^6 e ie. „/ Because the parallel lines a b and c d intersect ef, the exterior /^emb ■= interior and opposite L dne. But /, e >w 6 = its vertical /_ a mf; AN INTRODUCTION TO GEOMETRY. 53 therefore, /_ a mfr=. /_dne. Again, Because the parallels a h and c d intersect the straight line ef, the ext. /_ am e ^=. int. opp. L e n e : But, z « »» e = its vertical Z. hmf; therefore, L h mf ■=. /_ cue. M. — What axiom did you make use of in this de- monstration ? P. — Angles or things which are equal to the same angle or thing, are equal to each other. M. — Let the parallel lines a b and c d meet the line ef in the points b and d. Which / / 'S of these angles would you call i j interior angles ? -P. — The angles abd and c d b are interior angles. M. — What angle is equal to the angle ab d9 P. — The angle c df is equal to the angle a b d; be- cause Z c df is the exterior, and Lah d, the in- terior and opposite angle. M. — Hence ^ a b d + /_c d b is equal to P.— Z. cdf + ^cdb. M. — And what do you know respecting the angles cdf+cdb? P.— A cdf + Z.C db = 2 rt. z_s. inf.— -What do you thence conclude ? P. — That the two interior angles, ab d + cdb =, 2 rt. z_s, likewise. M. — Express, now, in words the truth we have just discovered. 54 LESSONS ON FOKM, BEING P. — If two parallel lines meet another straight line, they make the two interior angles, together, equal to two right angles. M. — Demonstrate this truth. P. — Let the parallel lines ab,c rf meet the line ef'm the points b d, the two interior angles a 6 (Z and crffi shall, to- ^ b i- f gether, be equal to two right angles. Because the lines a b and c d are parallel, the ext. ^ c df zz. int. opp. ^ ab d. To each of these angles add /_ c db ; then, ^ c df + ^c db ■=! ^ab d + /_c d b. But Z c ^/ + Z.C db = 2rt. ^s; therefore, /_ abd -{■ /_c db -^ likewise, 2 rt. /.s- M. — What axioms did you use in this demonstra- tion? P.—" If equals be added to equals, the sums are equal ;" and, " things which are equal to the same thing are equal to each other." M. — Demonstrate this truth from the equality of the alternate angles. In doing, so, you will, of course, use intersected lines. P. — Because a b and c d are "" parallel lines, ^b e/= /, cfe, /f each being an alternate angle. To each of these angles add z a ef; then, Z *c/+ Z«e/= Acfe+/:. aef But, Z*e/-t- Zae/= 2 rt. /_&; therefore, the 2 int. ^s, c f e + a ef= likewise, 2 rt. Zs. AN INTRODUCTION TO GEOMETRY. 55 M. — Since the 2 int. /.s, a ef + cfe = 2 rt. ^s, what must the 2 exterior /.s, c/m + a e m be ? -P. — The 2 €xt. /,s, cy w + a e m, must be equal to 2 rt. /_s : because the 2 interior angles, eye + a ef, together with the 2 exterior angles, cfn + ae m = A rt. angles, and it is known that the 2 interior ^s, cfe + aef-=. 2 rt. ^s ; therefore, the 2 ext ./.s, c/m + a em, must be equal to 2 rt. ^s. ■ Jf. — Demonstrate this on your slates. P .T— Because ^s cfn + cfe =. 2 rt. /.s, and, also, ^s, « e ?» + a e/ = 2 rt. Z.^ ; therefore, ^s, c/« + cfe + aem + a ef= 4 rt. ^s. But, ^s c/e + aef-=. 2rt. ^s; because a 6 and c/ are parallel : therefore, the 2 ext. Z.^,cfn -\- a e m ■=. likewise, 2 rt. ^s. M. — If it be known that, the , g, exterioranglecrfeiswoi equal to the interior and opposite angle f a S rf, what must be concluded ? J g P. — That the lines ab and d c are not parallel. M. — Again, if itbeknown that the angles a e/ and dfe are not equal, what must be concluded ? /a c V / f /l i /a c V / if / I 56 LESSONS ON FORM, BEING P. — That the lines a h and c d are not parallel. M. — Again, if itbeknown that the 2 interior angles a ef ■{■ cfe are not equal to 2 rt. /.s, what must be concluded ? P.— That the lines a b and e d are not parallel. M. — If the ^s, aef -\- cfe, are not equal to 2 rt. ^s, what may they be, together? P. — They may be together either more or less than 2 rt. /_&. M. — If the angles a ef+ cfe be kss than 2 rt. /,s, what will happen if the lines ab, c d, be produced both ways ? P. — ^The lines a b and c «? will meet, toward a and c. M. — If the angles bef + dfe be more than 2 rt. ^s, what will happen if the lines a b and c dbe pro- duced both ways ? P. — The lines a b and erf will rwt meet toward the points b, d. M. — Hence, if any two straight lines intersect an- other straight line, on which side of this line will the two lines meet, if produced far enough ? P. — The lines will meet on that side of the in- tersecting line where the two interior angles are toge- ther less than two right angles. M. — If, then, we wished to ascertain • „ whether the lines a b and c d are parallel or not, what must be done ? " AN INTRODUCTION TO GEOMETRY. 57 -P. — A third line must be drawn so as to intersect a h and c d ; and, then, we must ascertain whether the exterior angle is equal to the interior and opposite, or not ; or, whether the alternate angles are co-equal or not ; or, whether the two interior angles are, toge- ther, equal to, less than, or more than, two right angles. M. — In the 6 for the words " is or' are greater than ;" also, the sign /, for the word " therefore," and the sign '.* for the word " because.'' Express Nos. 4 and 5 in words. P. — If one side of a triangle be produced, the exterior angle is greater than either of the interior and opposite angles. M. — Write the demonstration of this truth on your slate, and introduce the signs yist recommended. 72 LESSONS ON FORM, BEING P. '.' /_%, acd+ acb=:2 rt. Z,s, and /.s, abc + acb<^2 It, /_s ; •*• Z.S, acd + acb'^ /_s, a b c + a c b. From each of these unequals take away /_ a c b — Then, the remaining ext. /_ac d'^ the int. opp. /, ab c. In the same way, it may be shown that the ext. ^a c d is greater than the int. opp. /_b a c. M. — Compare the exterior angle of a triangle with the sum of the two interior opposite angles. P. — The exterior angle a c d is equal to the sum of the two inferior opposite angles a b c + b a c: for, V Z.^, a c d + a cb ^ 2 rt, /_s, and, also, /Is, a c b + abc-\-bac'=.2xt. /_& ; .'. Z.S) a c d + a c b ^ /_s, a cb + a b c -i- ba c. Take away /_a c b from each of these equals — .". the remaining ext. ^ a c d^ remaining int. opp. ^s, a b c + b a c. M. — Produce two of the sides of a triangle. What is the sum of the two exterior angles ? P. /_s, acd + acb =. 2rt. Z.S, and /,s, ebc + abc-=2 rt. Z.S ; .*. ZSj acd + acb + ebc,+ abc = 4rt. zs; . e/ and, .: Z.s,acd + acb + ebc + abc + bacy 4 rt. /_s. From these unequals take Z.s, acb+abc + bac, which = 2 rt. ^s; Then, the remaining ext. Zs, a c rf + ei c > 2 rt-^s. AN INTRODUCTION TO GEOMETRY. 73 Hence, two exterior angles of a triangle are together greater than two right angles. M. — Produce each of the sides of a triangle. What is the sum of the three exterior angles ? P. — Each exterior angle to- gether with its adjacent angle =: two rt. ^s; ,•. the three ext. 2ls together with the angles of the triangle ■=■ six rt. ^s. But the ^s of the triangle = 2 rt. /.s ; /. the three exterior /.s == 6 rt. ^s — 2 rt. /.s = 4 rt. ^s. Hence, if each of the sides of a triangle be produced, the sum of the three exterior angles =: four rt. ^s. M. — Produce each of the sides of a triangle both ways. What is the sum of the twelve angles ; and, what is the sum of the nine exterior angles ? P. — The sum of the twelve angles = 12 rt. /J,; and .•, the sum of the 9 ext. /.s = 10 rt. ^s. SUBSTANCE OF SECTION III. 1. Any two angles of a triangle are together less than two right angles. 2. The interior angles of every triangle are together equal to two right angles, 3. A right-angled triangle is that which has a right angle. 4. An obtuse-angled triangle is that which has an ohtuse angle. 74 LESSONS ON FORM, BEING 5. An acute-angled triangle is that which has three acute angles. 6. If one side of a triangle be produced, tjie ex- terior angle is greater than either of the interior and opposite angles. 7. . If one side of a triangle be produced, the exterior angle is equal to the two interior and opposite angles. 8. If each of the sides of a triangle be produced, the three exterior angles are, together, equal to four right angles. SECTION IV. TWO TRIANGLES — THEIR EftUALITV. M. — What may be said, on comparing the angles' of two triangles ? P. — ]. The angles of one triangle are, together, equal to the angles of an^ other triangle ; because, their sum, in each, is equal to two right angles. 2. One angle of the one may be equal to an angle of the other. 3. Two angles in the one may be equal to two An- gles in the other, each to each. 4. The three angles of the one may be equal to the three angles of the other, each to each. 5. The three angles of the one may be wwequal to the three angles of the other, each to each. M. — If an angle of one triangle be equal to an angle of another triangle, what may be said of the other two angles, in each ? P. — The sum of the other two angles of the one triangle must be equal to the sum of thie ■ remaining AN INTRODUCTION TO GEOMETRY. 75 two angles of the other ; because, the sum of the three angles of the one is equal to the sum of the three angles of the other, and if the equal angles be sub- tracted from these equals, the remaining angles must be equal. M. — If two triangles have two angles of the one equal to two angles of the other,, each to each, what may be said of the remaining third angles ? P. — They must be equal, — for the reason alleged in the former case. M. — ^What may be said, then, of the angles of these triangles ? P. — The angles of the one are equal to the angles of the other, each to each. M. — Draw two triangles having one angle of the one equal to one angle of the other. P. — Let the angle b a c be equal to the angl6 e df. i M. — If the angle bac be equal to the angle e df, what happens, if fy the triangle e d f be placed upon the triangle a be, so, that the point d may be upon the point a, and the angle e df upon the angle bac? P- — The side e d must fall upon the side a h, and the side df upon the side ac. M. — And, where may the ppuits e and/ fall? e2 76 LESSONS ON FORM, BEING P. — Either somewhere on the sides a 6 and a c, — when de and df are, each, less than a b and ac ; e I Or, one of them may fall upon a b and the other beyond the point c, — when d e is less than a h, and • ^ a 6 c. But ^b a c"^ int. opp. ^c db ; much more .•. /_h ac'y' /_ab c. Hence, in every triaUvgle the greater angle is opposite to the greater side. M. — If, then,- it be known that one angle of a tri- angle is greater than another angle, what must be concluded with respect to the sides subtending [op- posite to] these angles ? P. — The side subtending the greater angle must be greater than the side subtending the less. M. — Which, then, of the sides of a right-angled triangle is the greatest ? -P. — The side subtending the right angle. M. — And in an obtuse-angled triangle, which is the greatest side ? P. — The side subtending the obtuse angle. M. — In an equilateral triangle, compare the sum of any two sides with the remaining side. P. — Any two sides of an equilateral triangle must together be greater than the rema!ining [third] side) because all the sides are equal. , M. — But, in any triangle, are two sides together greater or less than the remaining third side ? P. — The sidesi b a + a c must be greater than b c, because 6 c is the shortest distance between the points ^i_ . — -^c b and c. For same reason, ab + bc'^ ac, and ac-|-Jc>a6. AN INTRODUCTION TO GEOMETRY. 85 M. — This truth may be demonstrated by converting two sides into one : endeavour to do so. P, — Produce the side a'b at the point a, and make a d, the part produced, equal to a c, and join dc: then '.• ad:z: ac, h^ Z. adc^ j/_acd ; .". /_h cd'y^ /_adc. And ■ .■ to the greater angle the greater side is opposite; .•, hd~^ dc, but bd zzba + ac : .'. b a + ac~^bc. Hence, any two sides of a triangle are together greater than the remaining side. S M. — In the triangle ab c, let the side J c be greater than ab ; how would you discover the ex - c cess of 6 c over a 6 — i.e. by how much 6 c is greater than ab? P. — By cutting off from the greater, 6 c, a part equal to ab; the remaining part must be the difference, in length, between b c and a b. M. — Compare this difference between two unequal sides of a triangle with the remaining [third] side. P, — Let 6 c be greater than ab . o, frofti b c cut oS bd=- ab : i ^^^ ^\ dc\& the difference between b c and a b. Then, because any two sides of a triangle are to- gether greater than the third side^ ba + ac"^ be; but ba = bd; 86^ LESSONS ON FORM, BEING .',bd -{■ ac'^ b c. From these unequals take-away 6 rf, which is common to both — there remains ac^ dc. Hence, the difference between any two sides of a triangle is less than the third side. M. — The same truth can be demonstrated by means of the angles : try this method. P. — Let dc he the difference a between b c and ab ; join at?.- then •: ba = bd, ^^^ I \:c ^ bad= j/_ bda; but the ext. ^ a «? c > int. opp. Z,bad ; .'. /_ adc > ^ bda. Also, the ext. /. bda > int. opp. /_dac ; much more .•. ^adc"^ /_dac ; but to the greater angle the greater side is opposite — .•. ac'^d c — that is, the difference, d c, between 6 c and a 6, i&kss than the third side, a c. M. — Compare the three sides of a triangle with the double of any one side. P. — The three sides of any triangle are together greater than double the length of any one side ; for, ab + ac being > 6 c, a, add be io each of these unequals; ^^ ■ \ then ab + ac + bc'^bc + bc. a .^ Xc SUBSTANCE OF SECTION IV. 1. If two triangles have one angle of the one equal to one angle of the other, the sum of the remaining AN INTRODUCTION TO GEOMETRY. 87 two angles of the one is equal to the sura of the re- maining two angles of the other. 2. If two triangles have two angles of the one equal to two angles of the other, each to each, the third angle of 'the one is equal to the third angle of the other ; that is, the triangles are equiangular. 3. If two triangles have two sides of the one equal to two sides of the other, each to each, and have like- wise the angles contained by these sides equal, their third sides are equal, the triangles are equal, and the remaining angles of the one are equal to the remaining angles of the other, each to each, namely, those to which the equal sides are opposite. 4. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides, equal to them, of the other, the base of that triangle which has the greater angle is greater than the base of the other. 5. The angks at the base of an isosceles triangle are equal. 6. In an isosceles triangle, the straight line which bisects the vertical angle bisects the base. 7. In an isosceles triangle, the straight line which bisects the vertical angle stands at right angles to [is perpendicular to,] the base. 8. In an isosceles triangle^ if the base be bisected, the straight line joining the vertical angle and the point of bisection bisects the vertical angle and stands at right angles [is perpendicular to,] to the base. bo LESSONS ON FORM, BEING 9. If a triangle be equilateral, it is likewise equi- angular. 10. If two angles of a triangle be equal to one an- other, the sides opposite to them are likewise equal ,' that is, the triangle is isosceles. 11. If a triangle be equiangular, it is likewise equi- lateral. 12. In every triangle the greater side is opposite to the greater angle. 13. In every triangle the greater angle is subtended by the greater side. 14. Any two sides of a triangle are together greater than the third side. 15. The diflFerence between any two sides of a triangle is less than the third side. 16. The three sides of every triangle are together greater than double the length of any one side. SECTION IV. EQUALITY OF TRIANGLES. M. — Describe a triangle ; in it take any point, and, from the extremities of any one side of the triangle, draw lines to that point: what is the result ? a P. — Two triangles, ah c and hdc. \ M. — ^What have these two triangles in common 9 P.— The base b c. AN INTRODUCTION TO GEOMETRY. 89 M. — Compare the sum of the sides h d and d c with the sum of the sides h a and a c. Obs As it is important that the pupils should find-out a method of demonstration for themselves, the master ought, in this and every similar instance, to withhold assistance as long as he perceives the ma- jority of the class actively engaged in the investiga- tion of the question. If, ultimately, the pupils should not succeed in discovering a demonstration, he may direct their attention to the main points in the ques- tion : thus, with respect to the preceding — M. — ^What are you required to do ? P. — To compare h d + d c with b a -\- a c. M. — What does that mean ? P. — To try whether b d + d c is equal to, or greater, or less than b a + a c. M. — ^When the sides of triangles are to be compared with each other, which of the preceding truths will guide you ? P. — " The greater side subtends the greater angle ;" or, " any two sides of a triangle are together greater- than the third." M. — If you adopt the former of these truths, how must you draw a line so as to find a relation between b a and bd? P. — We must join the points a and d. M. — ^Do this, and see if it will assist you. The pupils will find that it cannot assist them, be- cause the point d is not determined. M. — ^And, if you wish to use the other truth, you have mentioned, what must be done ? 90 LESSONS ON FOKM, BEING Here, the master should leave the pupils to their own resources : they will, doubtless, find that either b d or c d must be produced, — if the preceding les- sons have been thoroughly understood. When any of the pupils have succeeded, let the master describe a triangle on the larffe school-slate, and the successful pupil submit his demonstration to the class, the master writing it down as the pupil proceeds. ThUs : (Pupil dictating, and the master writing.') Produce bd to e ; then, ba + ae being "^ be, add e c to each of these uneqiials — .'.ba + ac"^ be + ec. Also, de + ec being "^dc, add bd to each of these unequals — .", be + ec "^ bd + dc. But, it has been shown that ba + ac"^ be + ec; much more .•, is b a + a c "^ b d + e c. Hence, if a point be taken in a triangle, the straight lines drawn to it from the extremities of any one side are, together, less than the other two sides of the triangle. The master may, now, let the rest of the class read [not aloud] what is written on the slate ; and, there- AN INTRODUCTION TO GEOMETRY. 91 after, the demonstration being rubbed-out, let each pupil, in turn, be called-upon to give an oral demon- stration of the problem. ' ^ If the class consists of many pupils, the recapitula- tion of the solution by each pupil is apt to occasion inattention, on the part of those who fully know it. It is, then, advisable to let each pupil take a share of the demonstration — ^not in regular rotation, but at the call of the master, in order, the more effectually, to sustain universal attention. This exercise having been gone-through, let the pupils be called-upon to re-produce the whole in writing, on their own slates — the master carefully revising what they have written. Should there be, after all, some, weaker, pupils in- capable of doing so, the master may assist them by writing, on the school-slate, the main points of the problem, thus : a Show that 1. b a + ac'^ be + ec. 2. be + ecybd + dc. 3. Draw the necessary inference. The preceding process has been found service- able even to those who fiilly understand the demon- stration ; as, they are thereby led to resolve the prob- lem into its component parts. 92 LESSONS ON FORM, BEING Moreover, it is not an unnecessary practice to in- vert diagrams in all possible ways, — and, in this in- stance, to produce, now he, and, at other times, de. h M. (continues}. — Where must the point rf be taken so that bd + do shall be less than any other two lines drawn to another point in the triangle ? P. The point d must be taken very near the side b c, — it must be taken in the side be; thus : M. — And where must the point d be taken so that bd + dc may be greater than any other two lines drawn to another point in the triangle ? P. — The point d must be taken very near the ver- tex, a, of the triangle, thus: a M. — And, what is to be observed, if the point rf is taken in the vertex a ? P. — The lines b d and c d, then, coincide with the sides, b a and ca, of the triangle. M. — ^^And, what may, then, be said of the two tri- angles abc and bdc? AN INTRODUCTION TO GEOMETRY. 93 P. — That the sides of the triangle b dc coincide with the sides of the triangle a be, each with each ; and consequently, that the triangles are equal to each other. M. — You are already acquainted with one in- stance of the equality of triangles'; what is it? P. — Two triangles are equal when they have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal. M. — Now, I think, from what you have before re- marked respecting the triangles abc and bdc, you will be able to find out another instance of the equality of triangles. What is it ? P. — Two triangles are equal when the three sides of the one are equal to the three sides of the other, each to leach. M. — Right; and what angles are equal in two such triangles ? P. — Those angles which are opposite to the equal sides. M. — Illustrate what you have said, by drawing two triangles having these requisites. In the triangles abc and def, \{ab r: de,ac-=. df, and be ^ ef; then, /_acb— z. dfe, ^abc— ^ def, and /ibac = ^e df, — and A abe^z. A def. 94' LESSONS ON FORM, BEING M. — If, then, in the triangles ahc and d ef, ab ■=. de and ac ■=. df, but the base ef is greater than the base be ; what must be concluded with respect to the angle e df, which is opposite to the greater base ? P.' — The angle e df must be greater than the angle bac, M. — State this proposition fully, in words. P. — If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other, the angle contained by the sides of that triangle which has the greater base is greater than the angle contained by the sides, equal to them, of the other. M. — There is, yet, something to be remarked with regard to the angle which the lines bd, cd contain — {^describing the figure on the slate), a Compare the angle bdc with the angle h a c. P. — The angle b dc '\s greater than the angle bac ; for ".' exterior /_ bee o{ A bae '^ int. opp. ^ bac, and the exterior /_ bdc of K dec "^ int. opp. /_bec; much more .•. is ^ 6 dc > ^ J a c. AN INTRODUCTION TO GEOMETRY. 95 The pupils repeat the demonstration and, then, write it on their own slates ; the master, as before, writes upon the school-slate the following : Show that 1. ^hec~^ /_hac. 2. ^bdcy- Z.bec. 3. Thence draw the necessary consequence. M. — Hence, the angle formed by two lines drawn from the extremity of any side of a triangle to a point within it is .greater than P. — The angle contained by the other two sides of .the triangle. M. — Compare the angles d b c and deb with the angles abc and acb. P. — The angles dbc and deb are, evidently, less than the angles abc and a c fi, — because they are only parts of the latter angles. M. — Where must the point d be taken, so that the angles dbc and deb may become equal to the angles abc and acb? P. — The point d must be taken in coincidence with the point a. M. — And what may, then, be said of the triangles abc and dbc? P. — They are equal to each other. M. — After supposing, then, the angles dbc and deb equal to the angles abc and acb, each to each, there are two oilier parts in the triangles abc and dbc, which are equal to each other, or which these triangles have in common, if we consider them separated from each other. What are they ? 96 LESSONS ON FORM, BEING P.— The side b c. M. —And how is this side situated with respect to the angles ? P. — It is adjacent to them. M. — Describe two ,triangles having the following re-^ quisites : two angles, and the side adjacent to them, of the one, equal to two angles, and the side adjacent to them, of the other. Let /_bac ■=. /_ def, ^ ach ■=■ ^ dfe, and ac ■=. ef. M. — If one of these triangles, we suppose to be ap- plied to the other triangle, so that the point e may be upon the point a, and the side ef upon the side a c, what must happen ? P. — The point/ must fall upon the point c, -because ef-=. ac ; and rfy must coincide with b c, because Z, dfe =. /_acb ; and e d must coincide with a b, because /^defzz /_bac ; and, therefore, the point rf must fall upon the point i, — and the triangle d ef must coincide with the triangle a be, and be equal to it. M. — Here, then, is a third instance of equality in triangles : what is it ? P. — Two triangles are equal, when they have two AN INTRODUCTION TO GEOMETRY. 97 angles of the one equal to two angles of the other, each to each, and have likewise the sides adjacent to the equal angles" equal to each other. M, — Repeat, now, all you have learnt from the in- vestigation of the triangles ahc and hdc. The pupils repeating, the master writes their state- ments on the slate. Thus : — 1. hd ■\- dc'xi less than h a -^ ac. 2. b d and dc are least, when the point d is taken in the side c. S. bd and d c are equal to 6 a and a c, when the point d is taken in the vertex a. 4. Hence, two triangles are equal when three sides of the one are equal to three sides of the other, each to each. 5. Again : if two, triangles halve two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other, the angle contained by the sides of that which has the greater base is greater than the angle contain- ed by the sides, equal to them, of the other. 6. The angle b dc is greater than the angle b ac. 7. The angles dbc and deb are less than the angles abc and acb, 8. When the angles dbc and deb are equal to the 98 LESSONS ON FORM, BEING angles a Jc and acS, each to each, the point «? coin- cides with the vertex a, 9. Hence, two triangles are equal when they have two angles of the one equal to two angles of the other, each to each, and have likewise the sides adjacent to the equal angles equal to each other. M. — And, if the angles hdc anddbc be equal to the angles bac and a be, each to each, what must follow ? P. — The angle deb must be equal to the angle a c b; and the triangle hde must coincide with the triangle ab e, and be equal to it. M. — Hence, if two triangles have two angles of the one equal to two angles of the other, each to each, and have likewise one side equal to one side, how must these sides be situated in order that the triangles may be equal ? P. — They must either be adjacent to the equal angles, or they must be opposite to the equal angles. M. — This, then, is a fourth instance of equality in -triangles : state it. P. — Two triangles are equal when they have two angles of the one equal to two angles of the other, each to each, and have likewise one side equal to one side, — namely, those opposite to the equal angles. This truth the master, now, writes on the slate, in addition to the others already there. The whole is, then, committed to memory by the pupils. A recapitulation of this paragraph is not given, — its restdt having just been written on the school- slate. AN INTRODUCTION TO GEOMETRY. 99 SECTION VI. QUADRILATERAL FIGURES. M. — We have, now, become acquainted with the most general truths respecting triangles. Write them - on your slates, that I may see whether you remem- ber them all. This having been done satisfactorily, let the pupils be called-upon to demonstrate any one of the pro- blems : or rather, let each pupil, in turn, assign a problem, for solution, to the class. After this useful exercise, the master may proceed thus : M. — What, do you think, should be our next step, after the investigation of trilateral figures ? P. — To investigate quadrilateral figures. M. — State all you know of quadrilateral figures. (Lesson V. Introduction.) Let the pupils repeat what they remember respect- ing them. M. — Into what two groups may all quadrilateral figures be classed ? P. — Into parallelograms and trapeziums. M. — We shall begin with parallelograms, and, first, consider the manner in which a parallelogram is con- structed. Draw a parallelogram, and give a definition of it. 7, r2 100 LESSONS ON FORMj BEING P. — If a 6 is parallel to c d, and a c is parallel to b d, — the figure ab dci&& parallelogram. M. — Hence, a parallelogram is P. — A four-sided figure whose opposite sides are parallel. M. — Well, one relation between the sides of paral- lelograms being known, you may be able to discover another : try. P. — The opposite sides must be equal. M. — ^Demonstrate this. ^^ [, P. — Let a J d c be a parallelogram, its opposite sides shall be equal ; namely, a b'=. cd, and ac-=.bd. Join cb: '.' a 6 is parallel to c d, .'. Z_abc-=. alternate /_dcb ; and • . • a c is parallel to b d, .•. i^acb ■=. alternate /_dbc : now, c 6 is common to the triangles acb and dbc; y . A ach ■=. A dbc, ab = cd, and ac := bd. M. — Why are the triangles acb and bdc equal to each other ? P. — Because they have two angles, and the side adjacent to them, of the one, egtial to two angles, and the side adjacent to them, of the other, each to each. t AN INTRODUCTION TO GEOMETRY. 101 M. — And why, then, is a b =: c d, and a c = bd? P. — Because these are the sides which are opposite to the equal angles in the triangles. M. — Then, conversely, if a four-sided figure has its opposite sides equal, what will you conclude as to the figure ? P. — That the figure is a parallelogram. M, — Demonstrate this. a -^ P. — Let abdc be a four-sided figure whose op- posite sides are equal : the figure is a parallelo- gram. Join cb: then, • .• ab ■=. c d, and ac-=.bd, and cbh common to the triangles acb,bdc, .•. A. acb'=. A bdc, . and /_abc'=. ^dc d. But, these are alternate angles ; -• . ab is parallel to c d, also L acb'=. Ldbc. Again, these are alternate angles ; .". ac is parallel to bd: ■ and, hence, the figure ab dcis a. parallelogram. M. — Why are the triangles abc and dbc equal to each other ? P. — Because they 'have three sides of the one equal to three sides of the other, each to each. 102 LESSONS ON FORMj BEING M. — And, why is the angle abc equal to the ^ngle dbc, and the angle acb equal to the angle dbc? P. — Because these are opposite to the equal sides in the equal triangles; M. — Again, if, in the four- sided figure abed, you knew that a 6 is equal and parallel to the opposite side c d, what would you conclude the figure to be ? P. — A parallelogram. M. — Demonstrate this. P. — Let ab dche a, four-sided figure of which the opposite sides ab, cd are equal and parallel : the figure shall be a parallelogram. Join c d : then • . • a 6 is parallel to c d, .*. ^abc ■=. alternate ^dcb, and ab ■=. cd: and c6 is common to the triangles abc, deb — .•. AflsSci: Adcb, and /_acbz:Adbc. Now, these are alternate angles ; .-. a c is parallel to db, — and . ■ . the figure abdc is a, parallelogram. M. — Why is the triangle abc equal to the triangle deb? AN INTRODUCTION TO GEOMETRY. 103 P. — Because these triangles have two sides of the one, a b and b c, equal to two sides of the other, c d and b c, each to each, and have likewise the angles abc and deb, contained by them, equal to each other. il/.^-What else is known of the lines a e and b d, besides their parallelism ? P. — a c is equal to b d, M. — Well, how have a c and b d been drawn ? P. — Joining the extremities a, c, and b, d, of the parallel and equal straight lines ab, cd. M. — Hence, the two straight lines which join the extremities of two equal and parallel straight lines Finish the sentence. P. — Are equal and parallel. M. — That is not quite correct. Repeat what has been said, and see if it be true in every case. P.— No: for, in ^t.^^^ —^ the annexed figure, ^>°^~^ -t a e? is not equal to c h, and yet these two lines have been drawn so as to join the extremities of two equal and parallel straight lines. M. — Now, alter the preceding statement in con- formity with \kA% finding. P. — The two straight lines which join the extre- mities of two equal and parallel straight lines in the same direction, or toward the same parts, are equal and parallel. M. — What, then, is the general truth respecting the sides of parallelograms ? 104 LESSONS ON FORM, BEING P. — The opposite sides of parallelograms are equal. M. — Investigate the angles ^ 6 of a parallelogram. / / P. — 1. The sum of the interior angles is equal to four right angles. 2. The opposite angles are equal. M. — ^Demonstrate these positions. p. — Let abdche a parallelogram ; the sum of its inferior angles shall be equal to four right angles. Join be: then, •.• ^ s of A a 6 c = 2 rt. Z s, . and, also, Z s of a «?c 6 = 2 rt. /,s ; .•. /,s of the AS a b c + d c b z= 4 rt. /_s. But, /.s of AS abc ■{■ deb make up the /Is of the parallelogram abdc ; .•. the sum of the ^s of a parallelogram = 4 rt. ^s. Their opposite angles are, also, equal to each other : ' ■,• it has been demonstrated that A«6c= A deb, and ^abc-=. ^deb, and ^acbzz. /_dbe ; ,•. the whole ^abd =: the whole /. dca; also, Z cab=. l_ cdb : .*. the opposite angles of a parallelogram^ are equal to each other. AN INTRODUCTION TO GEOMETRY. 105 Let the pupils, now, be called-upon to demonstrate the last truth without the auxiliary line c h. M. — The line c b joining the vertices of the opposite angles is called a diagonal. What may be said of it ? P. — A diagonal divides a parallelogram into two eqval parts. M. — Demonstrate this. a ]> ••• ah'=. cd, and ac-=.hd, and I. cah ■=. /_ cdb s .'. £^ cab :=. A cdb, and .•. the diagonal c b divides the parallelogram into two equal parts. M, — How many diagonals may be drawn in a pa- rallelogram ? P.— Two. M. — What may be said of them ? P. — The diagonals of a pa- rallelogram bisect each other. For, :' a b =: c d, and Z. abe=z L dee, and L baezz Lcd^;. .: A abe^ a cde, and be ■=Lce, and « e =z 6 are parallelograms ; p b and bo,) and, •.• /. a /rf is a rt. Z. , ■'• Ls lad + lda = a.Tt. Z-, andj .". ^slad + lda-= z. dac. But, Z Ida = alternate £ pad; .: remaining /_ lad = Lpac, and rt. L ald=.x\.. L ape, and ad = ac; .". A ald^ A ai/>c, and a I == ap = b c ; also ld = pc = ab. And, in the same way, it may be shown, that A. a be = Acne, and ab = en =: Id; also, be = en = al. But, V lb = mn, and Im ■=b n, and Id^ en, and al = en ; .: be=-md, and ab = m e, — and .'. Aa6c^ A dm e, — and .•.lm=:lb = bn=n m; and Z. ^Jn is a rt. Z ; .-. fig. /w is a square. Again, :• pc =: ab^ ha, and eg =:bc = ap, .■.pg=ph=ho=og; and Z jo^'o is a rt. Z ; AN INTRODUCTION TO GEOMETRY. 131 ••• fig A^ is a square. But, hk=zab, and ap = a I; .: hp = bl, .: square kg = square In : also, ■.• kb, bf=ab,bc, each to each, and rt. /. kbf= rt. Z. ab c, ■•• A kbf=: A abc. But, A kbf=k of, and a abc= a ape ; .'. Asabc + one + emd + dia = = AS abc + ape + kbf+ kof. From the square / n, take the triangles abc, cue, e m d, and dla ; and from the square hg, take the triangles abc, ape, h bf, and k of: .: remaining square dc = remaining square A S + sq. S^. The several parts of the preceding demonstration may be thus separated : show that 1. The triangles abc, aid, ene, and emd, are equal to each other. 2. That, therefore, In is a. square. 3. That hg IS & square. 4. That A^ is equal to In. 3. That the triangles kbg, ab c, k of, and ap c, are equal to each other, and to the triangles abc, aid, d me, and ene. 6. And, therefore, that the square de is equal to the square hb + sq. bg. M. — Express fully, in words, the truth you have demonstrated. P. — In a right-angled triangle, the square described 132 LESSONS ON FORM, BEING upon the side opposite to the right angle, is equal to the sum of the squares described upon the sides con- taining the right angle. M. — If required to construct a square which shall be equal to two given squares, in what manner would you proceed ? P.— Draw two straight lines at right angles to each other, and equal to the bases of the given triangles respectively ; the square upon the straight line joining their extremities shall, then, be equal to the two given squares. M. — In an obtuse-angled triangle, squares being de- scribed upon the sides, it is required to compare the square described upon the side subtending the ob- tuse angle with the squares on the sides containing that angle. The pupils should be left to find-out, unaided, the excess of this square above the sum of the other squares. Should, however, the master have occasion to assist them, he may proceed thus: tt c M. — If A a c J is obtuse-angled, at c, what must be done in order to make it a right-angled triangle ? P. — Produce b c, and, from a, draw a rf at right angles to b d. AN INTRODUCTION TO GEOMETRY. 133 M. — Now, find whether the squares o£ bd and a d are greater or less than the squares of 6 c and a e. P. — The square of 6 rf is greater than the square of 6 c, hy the square of cdf and twice the rectangle con- tained by 6 c, cd ; and the sqiiare of a d is less than the square of a c, hy the square ofcd: therefore, the squares of b d and a d are, together, greater than the squares of 6 c and ac, by twice the rectangle contained hy be, c d. M. — And, to what square are the squares o(bd and a d, together, equal ? , P. — To the square of a b. M. — Draw, thence, the necessary inference. P' — The square of a 6 is, therefore, greater than the squares of b c and a c by twice the rectangle con- tained by 6 c, c d. Exhibit this demonstration on your slates ; andy in- stead of writing " the square of a b," write a V, P. bd' "^ b c^, by c d^ and twice the rectangle b c, cd, and ad' < ac', by cd' (because ac' =:ac' + cd'); ,\ b d' + a d^ y>- b c^ + a c', by twice the rectangle b c, cd: hatbtP + ad' = ab',^ .•, ab''p-be'' + ac% by twice the rectangle b c, nd. M. — Instead of producing b c, what else might have been done in order to make the obtuse-angled triangle, a be, a right- angled triangle ? 134 LESSONS ON FORM, BEING P.— The side ac might have been produced, and from b a perpendicular been drawn to it. M.—-Try whether, in this case, the demonstration would be the same. [The pupils will find it would.] ^' — Express, in words, the truth you have demon- strated. P. — In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the oppo- site side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rect- angle contained by the side upon which, when pro- duced, the perpendicular falls, and, the straight line intercepted, without the triangle, between the perpen- dicular and the obtuse angle. Obs, — While it is, of course, not to be expected, that the pupils can express the result of a demonstra- tion in' terms so accurate and precise as the preceding — yet, the teacher is earnestly counselled to require them, in every case, to state, in their otim words, the truths which they have discovered, — altering and amending their expressions till they assume the form of a strict and regular proposition: for, they will, thus, be led to perceive clearly the necessary con- ditions and exact limits of the results of their in- vestigations, and to acquire a readiness in embodying them in correct and suitable phraseology. Perhaps, indeed, of all the various exercises arising out of geometrical studj', few tend more than this toward AN INTRODUCTION TO GEOMETRY. 135 the general improvement of the mental faculties, — which, be it recollected, is a principal object of this Treatise. M. — Draw any acute-angled triangle, and compare the square described upon any one of its sides with the sum of the squares described upon the other two sides. If the master should find it necessary to assist his pupils, he may proceed thus : ^- — ^Let abdhea triangle right-angled at d; what must be done in order to make it an acute-angled tri-, angle ? P- — Produce bd to any point, c, — and draw a c. M. — Now, compare b (? with b d^. P. be- is greater than bd^ by dc' and twice the rectangle bd, dc. M. — Also, compare a M. — Produce each of the sides of a quadrilateral figure, — a pentagon, — a hexagon — of any polygon; and find the sum of the exterior angles. The pupils will find this sum to be, in every case, equal to four right angles. Hence, if each of the sides of any rectilineal figure be produced, the sum of the exterior angles is equal to four right angles. AN INTRODUCTION TO GEOMETRY. 161 CHAPTER II. SECTION I. ONE CIRCLE — ONE AND TWO STRAIGHT LINES IN A CIRCLE. M. — ^State all you know of a circle. Let the pupils state what they remember of Lesson XI. [Introd.] ; and^ to facilitate their recollection^ let the sphere, cylinder, and cone, be presented to them. M. — By what means may a correct circle be de-. scribed ? P. — By a pair of compasses, or by means of a string. The master haSj here, a fit opportunity of making several observations on the use of the Mathematician's Compass in describing a circle, and of instructing the pupils to draw circles without its assistance. M. — What is the fixed point called ? P. — The centre of the circle. M, — And, what is the curved line called ? p. — The circumference of the circle. M. — Andj what is the space bounded by the cir- cumference called ? P. — The area of the circle, or the circle. , M. — What do you know respecting the position of the centre of a circle with regard to the circumference ? ,162 LESSONS ON FORM, BEING P. — The centre of a circle is such, that all straight lines drawn from it to the circumference are eqiiaJ to each other. M. — ^What are such lines called ? F. — Radii of the circle ; and one is called a Radius. ' M. — Can a circle have more than one centre ? P. — No ; for, then, there would be radii not equal to one another. M. — And, what is a straight line passing through the centre and terminated both ways by the circum- ference called ? P. — A diameter of the circle. M. — Compare the portions of a circle obtained by drawing a diameter. P. — They are equal to one another, and are called «emj-circles [^Z^circles] . M. — Compare, likewise, the portions of the circum- ference cut-ofF by a diameter. P. — They are, likewise, equal to one another. Jf.— These and other portions of the circumference are called arcs. When is an arc a sewt'-circumfer- ence? P. — When the straight line joining its extremities is a diameter. M. — Describe a circle, and draw, in it, any straight line terminated both ways by the circumference. In how many diiferent ways can such a line be drawn ? P. — Either passing through the centre, or not. , M. — In a circle, a straight line, which is terminated AN INTRODUCTION TO GEOMETRY. 163 both ways by the circumference, and which is not a dianaeter, is called a Chord. Compare a chord with a diameter. P. — A chord is always less than a diameter. For, join c d and ce ; then •.• any two sides of a triangle are together greater than the third side, .■. c c? + c e > of e. But cd + ce ■=. ab, the diameter of the circle ; .•. afi > de. M. — From the centre erect a perpendicular on a chord ; and, then, compare the segments of the chord. P. — The chord a J is bisected by the perpendicular cd. For, join a c, c 6 ; .". a acb is an isosceles -A ; and, c d being perpendicular to a b, ad-rz bd. 'M. — Hence, if, from the centre, erf be drawn, bi- 164 LESSONS ON FORM, BKING seating the chord ab, what are the angles which it makes with ab? P. — Right angles: for, then, ca — cb, and ad=db, and c rf is common ; ,'. a acd^z a cdb, and /, ' Z. cda ■=. ^ cdb. M. — I think, from the truth you have here demon- strated, you will be able to devise some method of finding the centre of a circle. P. — Yes ; for, it must be in the perpendicular which is drawn bisecting a chord ; thus: ^ draw any chord abm& circle, and let c d bisect a b, and be perpendicular to it ; bisect c d vaf: then/ must be the centre of the circle. M. — Find the different ways in which two straight lines may be drawn in a circle. [The pupils ascertain this.] The most important of the several cases are, 1. When the two straight lines are diameters. M. — If two straight lines in a circle are diameters, where is their point of intersection ? P. — In the centre of the circle. M. — How may two diameters intersect each other ? AN INTRODUCTION TO GEOMETRY. 165 P. — Either at right angles, or not. M. — Compare the arcs intercepted by these diame- ters. ' P. — They are equal to one another ; and each is one-fourth part of the circumference. M. — Compare the areas intercepted by the diame- ters and arcs. P. — They are equal to one another; and each is one-fourth part of the circle. M. — They are^ on this accoAint, called Quadrants. [fr. quadrans, Lat.] If, then, an angle at the cen- tre of a circle is a right angle, what part of the cir- cumference is the arc which subtends it ? P. — One-fourth. M. — And, if the arc is not a fourth part of the cir- cumference, what must the angle which it subtends at the centre be ? P. — Either an obtuse or an acuie angle, according as the arc, which it subtends, is more or less than one- fourth of the circumference. M. — Compare the arcs and the areas intercepted by two diameters which do not intersect each other at right angles. P. — The opposite arcs and areas are equal to each other ; the smaller arcs subtend the equal acute angles,' and the two greater arcs subtend the equal obtuse angles at the centre. M. — And, if the acute angles be made greater or smaller, what will, consequently, be the case with the subtending arcs ? 166 LESSONS ON FORM, BEING P. — They will, consequently, become greater or smaller also. M. — Hence, what relation exists between the arcs of a circle and the angles which they subtend at the centre ?> P. — They increase or decrease according as the angles at the centre increase or decrease. M. — It is from this circumstance, that arcs are said to be the measures of the angles which they subtend at the centre. 2. When two straight lines in a circle are not dia- meters. M. — When two straight lines in a circle are not to be diameters, in what two different ways can they be drawn ? / P. — Intersecting one another, or, not intersecting one another. M. — If they do not intersect each other, when are they equal to each other ? P. — When they are equi-distant from the centre. M. — By what means would you measure the distance of a straight line from a point ? P. — By a perpendicular drawn from the point to the line. M. — Why & perpendicular ? Could any other straight line serve the same purpose ? p, — No : for, from the point, any number oiuneqwal straight lines maybe drawn to a straight line, but only one perpendicular. M. — Then, demonstrate that chords which are equi- distant from the centre are equal to each other. AN INTRODUCTION TO GEOMETRY. 167 P. — From the centre c, draw the perpendiculars cf, eg ; and join cb, cd. Then •.• ab, de, are equi-distant from the centre, ef=cg; and cb ■=. cd, as they are radii ; and, bf IS one-half of a 6 ; also, g di& one-half of d e. But, ••• Z cfb is a rt. Z. ; .:cV = cr + bf; similarly, c d' zz gd' + cgK But, cb^ ■=. ccP (because cb := cd); .■.cf + br- = gd^ + cs^. ■ But, c/2 zz c g' (because cf=. c g) ; .-.bf-gd, andj .•. bf'=.gd. But bf is one-half of a b, andgd is one-half of de ; .: a Jzz d e. M. — And, when it is known that chords in a circle are equal to each other, what inference must be drawn ■respecting their distances from the centre ? 168 LESSONS ON FORM, BEING P. — Their distances must be equal : for, :• ah-=.de, and c/, eg are perpendiculars, drawn to them from the centre, •■• bf—gd, and cb'zz. cd ; .: cp + bf = cg'+gd^ (as before). But J/' = gd^; .: cp = cf, and ••• c/= eg; ^ that is, the perpendiculars are equal ; and, therefore, the chords ab, de, are equi-distantfrom the centre. M. — And, when are two chords not equal to each other ? P. — When they are at unequal distances from the centre. For, ifrfe be "^ ab, dg is > If. But, cf<' + bf=cg^ + dg\ and dg' > bf ; .-.cf^ycg'; that is, cf^cg. Hence, the chords are at unequal distances from the centre : and the lesser chord, a b, is farther from the centre than the greater chord, d e. AN INTRODUCTION TO GEOMETRY. 169 3. When two straight lines in a circle intersect each other. M, — If two chords in a circle intersect each other, where may their point of intersection be ? P. — Either in the circumference of the circle, or ivithin the circle. M. — In the former case, the angle made by two chords meeting in the circumference is called the angle at the circumference — a [in contra-distinction to the angle bdc at the centre]. What have the angles bac and hdc in common? P. — The arc b c subtending them both. M. — Hence, these angles are said to stand upon the same arc. Compare them. P. — The angle bdcat the centre is greater than the angle bac at the circumference ; — it is doubh the angle at the circumference. For, join a d and produce it to e ; then '.• da=.db, Z. dab ■= /_ dba ; and .'. the ext. Z edb \s double L dab. For a similar reason, L edc is double /_ da 6} and .•. the whole Lbde.'\% double L hac. 170 LESSONS ON FORM, BEING > as previously. M. — In this figure, the angle had\% an angle at the circumference, and the angle bed, a.t the centre; and they stand upon the same arc b d. Is Z. bed, likewise, double L bad? P. — Yes. Join a c and produce it to e ; Z. eedis double /_ cad; also, Z. ecb is double Z. cab; ,: remaining Z. bed is double the remaining Z. b ad. M. — Hence, when an angle at the circumference is one-half of a right angle, what is the angle at the centre, upon the same arc ? P, — Double one-half of a right angle,'^that is, one right angle. M. — And what part, therefore, of the whole cir- cumference is the arc upon which they stand ? P. — One-fourth of the eircun\ference. M. — And, when an an^le at the circumference is a right angle, what is the corresponding angle at the centre ? P. — There can, in this case, be no angle at the centre ; because, no angle can be equal to double a right angle,- or, to two right angles. AN INTRODUCTIOK TO GEOMETRY. 171 M. — Nowj what part of the circumference is the arc which subtends a right angle at the circumference ? P. — It must be a semi-circumference. M. — If, then, ah angle at the circumference stands upon a semi-circumference, what must that angle be? P. — A right angle. M. — Demonstrate that — When an angle at the cir- cumference stands upon a semi-circumference, it is a right angle. P. — If the arc ae b is a semi-circumference, a b passes through the centre c. Join c d: then •.' cd ^ ca, and cd=: cb, /_cda-=. L cad, and /. cdb^z l_ cbd; and .•• whole Ladh-=. /_%cad ■\- cbd. NoWj if one of the angles of a triangle is equal to the sum of the other two, it is a right angle ; .'. Z arffi is a rt. Z- M. — And, if it be known that an arc is less than a semi-circumference, what must be concluded with re- spect to the angle which it subtends at the circum- ference ? i2 172 LESSONS ON FORM, BEING P. — It must be kss than a right angle ; it must be an cuiute angle. M. — Demonstrate this. , P. — "•• the arc adbi& a semi-circumference, .•. the arc 6? 6 is less than a semi-circumference. But, in A adb, /_ a db is a. rt. ^ ; .-. L dab'xs less than a rt. /.. M. — And, if the arc be greater than a semi-circum- ference — ? P. — The angle which it subtends at the circum- ference must be greater than a right angle ; it must be obtuse. For, •.• the arc a bis a. semi-circumference, .-. the arc bae is greater than a semi-circumference. But Z, adb is art. /_ ; .: L edb yrt. L, i.e. Z. erffi is an obtuse angle. M. — There is an important truth dependent on the relation of angles at the centre and angles at the circumference, upon the same arc. At the centre, can there be several angles which stand upon the same arc ? P.— No. M. — :But, can the same arc subtend several angles at the circumference ? P. — Yes, — an indefinite number. M. — And, what may be concluded respecting them? P. — They must all be equal. AN INTRODUCTION TO GEOMETRY. 173 Join cb, cd ; .•. Z. bed, at the centre c, is double of each of Z.s bad, bed, and bfd ; and /. Z. bad= /. bed^ bfd. M. — Look at the figure you have adopted, in de- monstrating this truth. What part of the circum- ference is the arc b d? — what kind ofangles are bad, bed, bfd ? — Now, show that angles which stand on arcs greater than the semi-circumference are, like- wise, equal to each other. P. :• arc ad"^ serai-circutnfererice, .-. arc ab ed not. If the angles at the centre are equal, the arcs which AN INTRODUCTION TO GEOMETRY. 181 subtend them must be equals (since arcs are the mea- sures of angles at the centre,) and the straight lines which join the extremities of the radii are equal. For, A cab evidently = a cbd; and .: a b = b d. M. — Hence, chords which cut-ofF equal arcs of a circle must be ? P. — Equal to each other. M. — And, equal chords in a circle must cut-ofF — — ? P. — Equal arcs. But, if the angles at the centre are not equal to each other, the corresponding arcs are not equal to each other ; nor are the lines which join their extremities equal to each other and the greater line is that which subtends the greater angle. 2, Of three straight lines in a circle, two may be radii, and the third a diameter. 182 LESSONS ON FORM, BEING If Z.acd= /_dce=. j!_ech, each of them = |s of a rt. L ; .•• /_scad -^ cda = ^ rt. /,. But /_c ad= ^cda; .: each of z.s cad, cda = f rt. /. : hence, a acd\s equiangular, and .'., also, equilateral, and ca^cd=ad. But A acd= A dee =: Ae cb ; .'. ad'=.de = eh. 3. Of three straight lines in a circle, two may be diameters, and the third a radius. No particular truth arises from this case. 4. They may be chords. These, if equal to each other, are equi-distant from the centre. And, if equi-distant from the centre, they are equal to each other. Two of them may be equal to each otheij ; the third, then, is not equi-distant, with them, from the centre. They may be all parallel, or two only may be pa- rallel with each other. No particular truth arises from this case, unless a perpendicular be drawn, from the centre, upon one of them ; it is, likewise, perpendicular to each of them, and bisects each of them. 5. Three straight lines in a circle may intersect each other. Their point of intersection may be within the circle' or in the circumference. AN INTRODUCTION TO GEOMETRY. 183 If it is within the circle, and the lines are equal to each other, the point of intersection is the centre. (This has already been shown.) If the lines are not equal to each other, the rect- angles contained by the segments of each are equal to each other. For, the rectangle ag,gb = rectangle eg, gd ; also, rectangle ag,gb = rectangle eg, gf: .; rectangle ag, gh c= rectangle eg, gd-= rectangle eg, gf- If they are in the circumference, and the angles which they make are equal to each other, the arcs which subtend them' are, likewise, equal to each other. a For, let Z 6a«?= Z daet 18A LESSONS ON FOKM, BEING •'' /L b c d ■= ^ dee,— AS ^s at the centre ; and .'. arc b d =l arc de. 6. Tliey may intersect each other in two points. Nothing determinate can be said of them, in this case, unless more is known. 7. They may intersect in three points, and form a triangle. That triangle may be right-angled, obtuse-angled, or acute-angled. M. — Where are the angular extremities of such a triangle ? P. — In the circumference. M. — A triangle, or any other figure, of which all the angles are in the circumference of a circle, is said to be inscribed in the circle. Where is the centre of the circle, when the inscribed triangle is right-angled ? P. — In the intermediate or middle point of the side which subtends the right angle ; because every right angle at the circumference is. subtended by a semi- circumference, and the line which joins the extremities of a semi-circumference is a diameter. M. — Endeavour to demonstrate this, by means of auxiliary lines. AN INTRODUCTION TO GEOMETRY. 185 P. — Let /. 6 ffl c be a rt. ^ ; bisect 6 c in rf: the point d shall be the centre of the circle. Draw de, dfaX rt. ^s to a &, ac, respectively, and join da: then •,■ de is at rt. /_& to a b, .: ae =:b e, and de is common to as deb, dea; .•. db ^da. In the same way, it may be shown that da = dc ; ,: db = da=zdc; and .". d is the centre of the circle. M. — But, when the inscribed triangle is obtuse or acute, where is the centre ? P. — When it is. obtuse, the centre of the circle must be loithout the triangle, beyond the side which sub- tends the obtuse angle ; because an obtuse angle is sub- tended by an arc which is greater than a semi-cir- cumference. If it is an acute-angled trianglcj the centre of the circle is within the triangle. M. — Yes : and, when the inscribed triangle is equi- lateral, where is the centre of the circle ? P. — In the centre of the triangle; because, since the sides are equal, they are equi-distant from the centre. The pupils may, now, be left to draw four straight lines in a circle. The truths, arising from their vari- 186 LESSONS ON FORM, BEING ous combinations, will afford the means of recapitulat- ing most of those positions which have before been established, and which need, here, no repetition. Those which are of chief importance, however^ are the following. M.—l(four chords be drawn so as to join the ex- tremities of two diameters, what may be said of them. P. — The opposite chords are equal to each other ; and, therefore, they form a parallelogram. For, c a, c b = cd, ce, each to each, and /_ach =. ^dce ; .•.ab = de; similarly, ad =i be. But, if the opposite sides of a quadrilateral figure are equal to each other, the figure is a parallelogram.;. ,: abed is a parallelogram. M. — And, when will the inscribed quadrilateral figure be a square ? P. — When the diameters cut each other at right angles. AN INTRODUCTION TO GEOMETRY. 187 a For, then ab =.bd ; and .-.abedis equilateral. But ^ 6 a af is a rt. ^ [because it is subtended by the semi-circumference Serf] ; .•. abde is rectangular, and .•. aeb dis a. square. M. — Draw any four chords in a circle, so as to form a quadrilateral figure, and determine what may be said of the sum of the opposite angles. P. — The sum of the opposite angles of any qua- drilateral figure, inscribed in a circle, is equal to two right angles. 188 LESSONS ON FORM, BEING Join a c and bd: then V ^s cad, cbd are upon the same arc cd, ^cad = JlcJt?; for a similar reason, ^hac= ^bdc ; .: whole ^ bad^ Z.^ ebd + bdc. To each of these equals add ^bcd ; .: /.s bad + bcd= ^s cbd + bdc + bcd. But /_s cbd + bdc + bed = 2 n. ^s; .: ^sb ad + bcd = 2 rt. ^s; and, ■•• all the angles of the figure abed =z 4 rt. /_s, •'• /.s abe + a dc. = 2 rt. /is. iHf. — Then, what are the onli/ quadrilateral figures which may be inscribed in a circle ? P. — A square ; a rectangle ; and a trapezium, whose opposite angles are together equal to two right angles. The pupils are, now, required to draw^ve, six, seven, Sfc. chords in a circle. These combinations, however, offer no important results which can be adequately ap- preciated by them, at this stage of their geometrical knowledge. From this general remark may be ex- cepted, however, the particular case of an inscribed equilateral and equiangular hexagon, — each side of which is equal to the radius of the circle ; whence the truth, that, tfie radius of a circle may be applied six times, exactly, to the circumference. — The demonstration of this being easy, will be readily discovered by the pupils. AN INTRODUCTION TO GEOMETRY. 189 SUBSTANCE OF SECTION II. 1. If, from a point in a circle, there can be drawn more than two equal straight lines to the circum- ference, that point is the centre of the circle. 2. Of all straight lines that can be drawn from a point in a circle to the circumference, the greatest is that which passes through the centre. 3. And, of the others, that which is nearer to the one in which the centre is, is greater than any one more remote. 4. And, from the same point, only pairs of equal straight lines can be drawn to the circumference. 5. Chords which cut-ofF equal arcs are equal to each other. 6. If chords are parallel, a perpendicular drawn from the centre to one of them, when produced to their cir- cumference, is perpendicular to each of them, and it bisects them. 7. When angles at the circumference are equal to each other, the arcs upon which they stand are, like- wise, equal to each other. 8. A figure is said to be inscribed in a circle, when all its angular extremities are in the circumference of the circle. 9. If a right-angled triangle be inscribed in a circle, the intermediate or middle point of the side which subtends the right angle is the centre of the circle. 10. If an obtuse-angled triangle be inscribed in a circle, the centre of the circle is without the 190 LESSONS ON FOKM, BEING triangle, beyond the side which subtends the obtuse angle. 11. When an acute-angled triangl.e is inscribed in a circle, the centre of the circle is within the tri- angle. 12. When an equilateral triangle is inscribed in a circle, the centre of the circle is, likewise, the centre of the triangle. 13. The four straight lines which join the extrem- ities of two diameters, in a circle, form a rectangular parallelogram. 14. If two diameters intersect each other at right angles, the four straight lines joining their extrem- ities form a square. 15. The opposite angles of any quadrilateral figure inscribed in a circle are together equal to two right angles. 16. A square, — a rectangle, — and a trapezium of which the opposite angles are together equal to two right angles, — are the only quadrilateral figures which can be inscribed in a circle. 17. If an equilateral and equiangular hexagon is inscribed in a circle, each of its sides is equal to the radius of the circle. 18. Hence, the radius of any circle may be applied six times to the circumference. AN INTRODUCTION TO GEOMETRY. 191 SECTION III. LINES without A CIRCLE TANGENTS. The object of this section is the investigation of such truths as arise from the relations of straight lines without a circle. The method of proceeding is here, likewise, closely similar to that which has been ob- served in the preceding sections. A point is, first, assumed without the circle, from which straight lines are drawn so as to intersect the circle. Next, — one tangent is drawn, and the angles, which a diameter, drawn to the point of contact, makes with it are investigated. Two tangents, three tangents, &c. are, then, drawn in succession, and the nature of the circumscribing figures is investigated. M. — When is a point withoitt a circle ? P. — When its distance from the centre is greater than that measured by a radius. M. — If, from a point without a circle, it be requir- ed to draw a straight line to the circle, — find in how many different ways this can be done. P. — A straight line, so drawn, may cict the circle ; or, it may only meet the circle ; or, it may touch the circle ; or, it may neither cut it, meet it, nor touch it. M. — If a straight line is drawn so as to cut a circle, — determine in what manner it may cut it. 192 LESSONS ON FORM, BEING P. — It may either pass through the centre, or not pass through the centre. M. — ^If, from the same point, two lines be drawn so as to cut the circle, in what manner may they cut it? Pt — One of them may pass through the centre, and the other may not pass through it; or, neither of them may pass through it. Jf.— Compare two suph lines, in .each case. P. — If one of them passes through the centre, it is greater than the one which does not pass through the centre. Let a b pass through the centre c, and a d cut the circle ; a b shall be greater than a d. Join dc ; .-, dc + ca'^ a,d: but dc^ cb ; .: aby^ad. And, if two straight lines be drawn from the point a, they may be equal to eaph other, or they may not be equal to each other ; a d is equal to a e, when ^dcb= /.ecb; AN INTRODUCTION TO GEOMETRY. 193 ••• then, /_dc a = ^eca, and dc, ca-=ec, c a, each to each ; .'. ad-= ae. M. — Hence, of all straight lines which can be drawn from the point a so as to cut the circle, what will be the greatest ? P. — The one which passes through the centre ; and of the others, that which is nearer to the one in which is the centre, will be greater than one more remoter Also, only two equal straight lines can be drawn from the same point so as to cut the circle, one on each side of the greatest line. Obs. — The demonstration of these positions being similar to the analogous case in the preceding section, it is unnecessary to repeat it here. The pupils, how- ever, must demonstrate what they have stated above, — and, indeed, should always be required to demon- strate whatever admits of demonstration. M. — When do you consider a straight line as meet- ing, and when as touching, a circle ? P. — A straight line touches a circle when it meets the circumference, and, being produced, does not cut t. M. Compare straight lines drawn, from the same point, to meet the circle. P. — That is the least line which, when produced, passes through the centre of the circle ; and of the others, that which is farther from the least line is always greater than one more contiguous to it. 194 LESSONS ON FORM, BEING Of the straight lines ag, a e, let a g, produced, pass through the centre c, and join cd, ed: then •.• c t? + d a > c a, and cd = cgj .: remainder ag ^^a d; that is, the one which, when produced, passes through the centre, is less than one which, when produced, does not pass through the centre : .•. a^.is the least. Again, •■■ <^ is a point in a a ec, .: ae + ec"^ a d + dc ; but ec ^ dc ; .•. ae'^ ad ; that is, a line which is farther from the least line is greater than one nearer to it. A\so,\S Adca = Z.fea, a d =: af; '.• A. adc ■= A. afc. Hence, only a pair of equal straight lines can be drawn from the same point so as to meet the circle, one on side of the least line. AT* UJTRODnCTION TO GEOMETRY. l95 M. — When a straight line touches a circle, it is called a tangent ; and the point in which it touches the circumference is called the point of contact. From the point of contact draw a straight line so as to cut the circle, and determine the angles it makes with a tangent. P. — A straight line from the point of contact either passes through the centre of the circle, or it does not pass through the centre. If it passes through the centre, it is at right angles to the tangent. lid For, if a is the point of contact, any straight lines, ch, cd, from the centre are always greater than c a : .: c a is the least line which can be drawn from a point to a straight line ; .-. c a is at rt. ^s to ab. Jf.— But, could not another point, b for instance; be likewise in the circumference of the circle ? P.— No : for, . then, a b would cut the circle, and, therefore, not be a tangent. M. — Can you show, that a straight line joining any two points in the circumference of a circle is within the circle ? k2 196 LESSONS ON FORM, BEING Let a, b, be any two points in the circumference : the straight line joining them is vnthin the circle. From c, the centre, draw c d at rt. Z. s to a 6, and join cb ; .•.cby>- cd (because it subtends a rf . Z )> and .•. the point d is within tlie circle : but the point d is in the straight line a b; .-. the points a, b, are in the same line with d, and .•. a ^ is within the circle. M. — Hence, what may be said of a tangent? P. — A tangent touches a circle in one point only ; and the straight line from the centre to the point of contact is at right angles to the tangent. M. — If, then, the angle which a straight line from the centre makes with another line meeting the cir- cle is not a right angle, what must be concluded ? P. — That the straight line which meets the circle is not a tangent, and that, being produced, it must, therefore, cut the circle. M. — Now, investigate the angles which a straight line from the point of contact makes with the tangent, when it does not pass through the centre. AN INTRODUCTION TO GEOMETRY. 197 f V '^ c / e ^--- ^^ Let a 6 be a tangent, d the point of contact, and rfe a straight line, not passing through the centre c: ^edb :=. /_ upon the arc dg e, and /_ eda-^L L upon the arc dfe. Draw dcf, and joinye; in the arc d e take any point, g, and join dg, g e ; •'• Afdh is a rt. /_• But /_ d efis a rt. /_, it being /_ upon the semi-cir- cuinference/rf; .:/_fdh- /_def: and /I def-=L ^s e/if + erf/; .-. ^fdh- Z.sefd + edf, .: Z-^fde + edb-=. ^sefd + e c?/ and .•, /_edhr=. /. e/rf, upon the arc af^ e. Again,. •.• opposite /_% efd + dg e =z 2 rt. /.s, .: /_&ed a + edb-= ^sefd + dge: but, i_edh— Ze/rf; .•. ^eda-=. Z. dge, upon the arc dfe. 198 LESSONS ON FORM, BEING i!f.— Express this truth in words. P — If) from the point of contact, a straight line be drawn so as to cut the circle, the angles which this line makes with the tangent are equal to the angles upon the adjacent arcs. , M. — You said that, a straight line from the point of contact either passes through the centre or it does not ; but, cannot a straight line be drawn between the tangent and the circumference ? P.— No. For, /. c a 6 is a rt. /. ; .•. /. c a«? is not a rt. /, ; and .'. ad\% not a tangent. Hence, a d must cut the circumference. M. — Consequently, how many tangents can be drawn from the same point without a circle ? P. — Only two tangents, one upon each side of the straight line which passes, from the point, through the centre. M. — Draw these, and compare them. P — They are equal to each other. AN INTRODUCTION TO GEOMETRY. 199 Join cb, cd : .*. /_scba, cda are rt. ^s, sa&.:ca^=ch' + ab<'= cd'' + ad'; .*. ab' =:: a d-, and .'.ab ■=. ad, M. — There is another interesting truth respecting a tangent, connected with those you have already ascertained. I will assist you in discovering it. Let a 6 be a tangent, and a da. straight line cutting the circle in e ; join d b and e b. Compare the as ad b, a eb. 200 LESSONS ON FORM, BEING P. — A a db is similar to a a c J. For, •■• he cuts the circle, .: ^ahe=. /_adh, upon the arc he; and j/_ had IS common to the two as : ••. /_bea ■= /_ahd ; •'. A aeb is similar to a ab d. M. — And, what are the homologous sides, in these triangles ? P. — a e, a b, and e b, are homologous to a 6, a d, and d b, respectively. i M. — What proportion results from the first pair of these homologous sides ? P. ae : ab '.: ab : ad. M. — And what equality, therefore, exists between the reptangles contained by the extremes and means of their proportion? P. — The rectangle ae, ad-=. rectangle ab,a d, — that is, rectangle ae, ad^abK M. — Express this equality in words. P. — If, from a point without a circle, two straight lines be, drawn, whereof one is a tangent and the other cuts the circle, the rectangle contained by the line which cuts the circle and by the part of it without the circle is equal to the square of the tangent. M. — If, then, from the same point without a circle, several straight lines be drawn cutting the circle, what may be said of the rectangles contained by each of them and by its corresponding segment without the circle ? P. — They are equal to each other. AN INTRODUCTION TO GEOMETRY. 201 Draw « e, a tangent : .-. rectangle a b, af-= a e* ; and rectangle ac, ag ^ ae^ ; and rectangle ad, ah = ae': .: these rectangles are equal to each other. M. — If two tangents be drawn, not from the same point without a circle, ascertain the manner in which they may be drawn. P. — They may be drawn either parallel to each other, or not. If they are parallel, the straight line which joins their points of contact passes through the centre of the circle. /J 202 LESSONS ON FORM, BEING Let a 6 be parallel to