BOOK I. (continued 3)

34,86,78,44,01 (50000 34,86,78,44,01(59049

25 00 00 00 00 9000 25

100000 ) 9 86 78 44 01 40 109) 986

9000 9 81 00 00 00 9 981

100000 ) 5 78 44 01 1180 4)57844

18000 4 72 16 00 47216

40 —————— ————

100000 ) 5 78 44 01

18000 4 72 16 00

40 ——————

100000 ) 1 06 28 01 1180 89)1062801

18000 1 06 28 01 1062801

80 —————— —————

9 0 0

163. The rule is as follows: To extract the square root of a number;—

I. Beginning from the right hand, cut off periods of two figures each, until not more than two are left.

II. Find the root of the nearest square number next below the number in the first period. This root is the first figure of the required root; subtract its square from the first period, which gives the first remainder.

III. Annex the second period to the right of the remainder, which gives the first dividend.

IV. Double the first figure of the root; see how often this is contained in the number made by cutting one figure from the right of the first dividend, attending to IX., if necessary; use the quotient as the second figure of the root; annex it to the right of the double of the first figure, and call this the first divisor.

V. Multiply the first divisor by the second figure of the root; if the product be greater than the first dividend, use a lower number for the second figure of the root, and for the last figure of the divisor, until the multiplication just mentioned gives the product less than the first dividend; subtract this from the first dividend, which gives the second remainder.

VI. Annex the third period to the second remainder, which gives the second dividend.

VII. Double the first two figures of the root;[25] see how often the result is contained in the number made by cutting one figure from the right of the second dividend; use the quotient as the third figure of the root; annex it to the right of the double of the first two figures, and call this the second divisor.

VIII. Get a new remainder, as in V., and repeat the process until all the periods are exhausted; if there be then no remainder, the square root is found; if there be a remainder, the proposed number has no square root, and the number found as its square root is the square root of the proposed number diminished by the remainder.

IX. When it happens that the double of the figures of the root is not contained at all in all the dividend except the last figure, or when, being contained once, 1 is found to give more than the dividend, put a cipher in the square root and in the divisor, and bring down the next period; should the same thing still happen, put another cipher in the root and divisor, and bring down another period; and so on.

EXERCISES.

Numbers proposed. Square roots.

73441 271

2992900 1730

6414247921 80089

903687890625 950625

42420747482776576 205962976

13422659310152401 115856201

164. Since the square of a fraction is obtained by squaring the numerator and the denominator, the square root of a fraction is found by taking the square root of both. Thus, the square root of ²⁵/₆₄ is ⅝, since 5 × 5 is 25, and 8 × 8 is 64. If the numerator or denominator, or both, be not square numbers, it does not therefore follow that the fraction has no square root; for it may happen that multiplication or division by the same number may convert both the numerator and denominator into square numbers (108). Thus, ²⁷/₄₈, which appears at first to have no square root, has one in reality, since it is the same as ⁹/₁₆, whose square root is ¾.

165. We now proceed from (158), where it was stated that any number or fraction being given, a second may be found, whose square is as near to the first as we please. Thus, though we cannot solve the problem, “Find a fraction whose square is 2,” we can solve the following, “Find a fraction whose square shall not differ from 2 by so much as ·00000001.” Instead of this last, a still smaller fraction may be substituted; in fact, any one however small: and in this process we are said to approximate to the square root of 2. This can be done to any extent, as follows: Suppose we wish to find the square root of 2 within ¹/₅₇ of the truth; by which I mean, to find a fraction a/b whose square is less than 2, but such that the square of a/b + ¹/₅₇ is greater than 2. Multiply the numerator and denominator of ²/₁ by the square of 57, or 3249, which gives ⁶⁴⁹⁸/₃₂₄₉. On attempting to extract the square root of the numerator, I find (163) that there is a remainder 98, and that the square number next below 6498 is 6400, whose root is 80. Hence, the square of 80 is less than 6498, while that of 81 is greater. The square root of the denominator is of course 57. Hence, the square of ⁸⁰/⁵⁷ is less than ⁶⁴⁹⁸/₃₂₄₉, or 2, while that of ⁸¹/₅₇ is greater, and these two fractions only differ by ¹/₅₇; which was required to be done.

166. In practice, it is usual to find the square root true to a certain number of places of decimals. Thus, 1·4142 is the square root of 2 true to four places of decimals, since the square of 1·4142, or 1·99996164, is less than 2, while an increase of only 1 in the fourth decimal place, giving 1·4143, gives the square 2·00024449, which is greater than 2. To take a more general case: Suppose it required to find the square root of 1·637 true to four places of decimals. The fraction is ¹⁶³⁷/₁₀₀₀, whose square root is to be found within ·0001, or ¹/₁₀₀₀₀. Annex ciphers to the numerator and denominator, until the denominator becomes the square of ¹/₁₀₀₀₀, which gives ¹⁶³⁷⁰⁰⁰⁰⁰/₁₀₀₀₀₀₀₀₀, extract the square root of the numerator, as in (163), which shews that the square number nearest to it is 163700000-13564, whose root is 12794. Hence, ¹²⁷⁹⁴/₁₀₀₀₀, or 1·2794, gives a square less than 1·637, while 1·2795 gives a square greater. In fact, these two squares are 1·63686436 and 1·63712025.

167. The rule, then, for extracting the square root of a number or decimal to any number of places is: Annex ciphers until there are twice as many places following the units’ place as there are to be decimal places in the root; extract the nearest square root of this number, and mark off the given number of decimals. Or, more simply: Divide the number into periods, so that the units’ figure shall be the last of a period; proceed in the usual way; and if, when decimals follow the units’ place, there is one figure on the right, in a period by itself, annex a cipher in bringing down that period, and afterwards let each new period consist of two ciphers. Place the decimal point after that figure in forming which the period containing the units was used.

168. For example, what is the square root of (1⅜) to five places of decimals? This is (145) 1·375, and the process is the first example over leaf. The second example is the extraction of the root of ·081 to seven places, the first period being 08, from which the cipher is omitted as useless.

1,37,5(1·17260

1

21) 37

21

227) 1650

1589

2342) 6100

4684

23446) 141600

140676

23452) 92400

8,1(·2846049

4

48)410

384

564) 2600

2256

5686) 34400

34116

569204) 2840000

2276816

569208) 56318400

·000002413672221(·001553599

1

25) 141

125

305) 1636

1525

3103) 11172

9309

31065) 186322

155325

310709) 3099710

2796381

30332900

169. When more than half the decimals required have been found, the others may be simply found by dividing the dividend by the divisor, as in (155). The extraction of the square root of 12 to ten places, which will be found in the next page, is an example. It must, however, be observed in this process, as in all others where decimals are obtained by approximation, that the last place cannot always be depended upon: on which account it is advisable to carry the process so far, that one or even two more decimals shall be obtained than are absolutely required to be correct.

A

12(3·46410161513

9

64) 300

256

686) 4400

4116

6924) 28400

27696

69281) 70400

69281

6928201) 11190000

6928201

69282026) 4261799|00

4156921|56

692820321) 104877|4400

69282|0321

6928203225) 35595|407900

34641|016125

69282032301) 954|39177500

692|82032301

692820323023) 261|5714519900

207|8460969069

53|7253550831

B

692820323026)537253550831(77545870549

484974226118

52279324713

48497422611;

3781902102

3464101615

317800487

277128129

40672358

34641016

6031342

5542562

488780

484974

3806

3464

342

277

65

62

3

If from any remainder we cut off the ciphers, and all figures which would come under or on the right of these ciphers, by a vertical line, we find on the left of that line a contracted division, such as those in (155). Thus, after having found the root as far as 3·464101, we have the remainder 4261799, and the divisor 6928202. The figures on the left of the line are nothing more than the contracted division of this remainder by the divisor, with this difference, however, that we have to begin by striking a figure off the divisor, instead of using the whole divisor once, and then striking off the first figure. By this alone we might have doubled our number of decimal places, and got the additional figures 615137, the last 7 being obtained by carrying the contracted division one step further with the remainder 53. We have, then, this rule: When half the number of decimal places have been obtained, instead of annexing two ciphers to the remainder, strike off a figure from what would be the divisor if the process were continued at length, and divide the remainder by this contracted divisor, as in (155).

As an example, let us double the number of decimal places already obtained, which are contained in 3·46410161513. The remainder is 537253550831, the divisor 692820323026, and the process is as in (B). Hence the square root of 12 is,

3·4641016151377545870549;

which is true to the last figure, and a little too great; but the substitution of 8 instead of 9 on the right hand would make it too small.

EXERCISES.

Numbers. Square roots.

·001728 ·0415692194

64·34 8·02122185

8074 89·8554394

10 3·16227766

1·57 1·2529964086141667788495

SECTION VIII. ON THE PROPORTION OF NUMBERS.

170. When two numbers are named in any problem, it is usually necessary, in some way or other, to compare the two; that is, by considering the two together, to establish some connexion between them, which may be useful in future operations. The first method which suggests itself, and the most simple, is to observe which is the greater, and by how much it differs from the other. The connexion thus established between two numbers may also hold good of two other numbers; for example, 8 differs from 19 by 11, and 100 differs from 111 by the same number. In this point of view, 8 stands to 19 in the same situation in which 100 stands to 111, the first of both couples differing in the same degree from the second. The four numbers thus noticed, viz.:

8, 19, 100, 111,

are said to be in arithmetical[26] proportion. When four numbers are thus placed, the first and last are called the extremes, and the second and third the means. It is obvious that 111 + 8 = 100 + 19, that is, the sum of the extremes is equal to the sum of the means. And this is not accidental, arising from the particular numbers we have taken, but must be the case in every arithmetical proportion; for in 111 + 8, by (35), any diminution of 111 will not affect the sum, provided a corresponding increase be given to 8; and, by the definition just given, one mean is as much less than 111 as the other is greater than 8.

171. A set or series of numbers is said to be in continued arithmetical proportion, or in arithmetical progression, when the difference between every two succeeding terms of the series is the same. This is the case in the following series:

1, 2, 3, 4, 5, &c.

3, 6, 9, 12, 15, &c.

1½, 2, 2½, 3, 3½, &c.

The difference between two succeeding terms is called the common difference. In the three series just given, the common differences are, 1, 3, and ½.

172. If a certain number of terms of any arithmetical series be taken, the sum of the first and last terms is the same as that of any other two terms, provided one is as distant from the beginning of the series as the other is from the end. For example, let there be 7 terms, and let them be,

a b c d e f g.

Then, since, by the nature of the series, b is as much above a as f is below g (170), a + g = b + f. Again, since c is as much above b as e is below f (170), b + f = c + e. But a + g = b + f; therefore a + g = c + e, and so on. Again, twice the middle term, or the term equally distant from the beginning and the end (which exists only when the number of terms is odd), is equal to the sum of the first and last terms; for since c is as much below d as e is above it, we have c + e = d + d = 2d. But c + e = a + g; therefore, a + g = 2d. This will give a short rule for finding the sum of any number of terms of an arithmetical series. Let there be 7, viz. those just given. Since a + g, b + f, and c + e, are the same, their sum is three times (a + g), which with d, the middle term, or half a + g, is three times and a half (a + g), or the sum of the first and last terms multiplied by (3½), or ⁷/₂, or half the number of terms. If there had been an even number of terms, for example, six, viz. a, b, c, d, e, and f, we know now that a + f, b + e, and c + d, are the same, whence the sum is three times (a + f), or the sum of the first and last terms multiplied by half the number of terms, as before. The rule, then, is: To sum any number of terms of an arithmetical progression, multiply the sum of the first and last terms by half the number of terms. For example, what are 99 terms of the series 1, 2, 3, &c.? The 99th term is 99, and the sum is

(99 + 1) 99 , or 100 × 99 , or 4950.

2 2

The sum of 50 terms of the series

1 , 2 , 1, 4 , 5 , 2, &c. is ( 1 + 50 ) 50 ,

3 3 3 3 3 3 2

or 17 × 25, or 425.

173. The first term being given, and also the common difference and number of terms, the last term may be found by adding to the first term the common difference multiplied by one less than the number of terms. For it is evident that the second term differs from the first by the common difference, the third term by twice, the fourth term by three times the common difference; and so on. Or, the passage from the first to the nth term is made by n-1 steps, at each of which the common difference is added.

EXERCISES.

Given. To find.

Series. No. of terms. Last term. Sum.

4, 6½, 9, &c. 33 84 1452

1, 3, 5, &c. 28 55 784

2, 20, 38, &c. 100,000 1799984 89999300000

174. The sum being given, the number of terms, and the first term, we can thence find the common difference. Suppose, for example, the first term of a series to be one, the number of terms 100, and the sum 10,000. Since 10,000 was made by multiplying the sum of the first and last terms by ¹⁰⁰/₂, if we divide by this, we shall recover the sum of the first and last terms. Now, ¹⁰,⁰⁰⁰/₁ divided by ¹⁰⁰/₂ is (122) 200, and the first term being 1, the last term is 199. We have then to pass from 1 to 199, or through 198, by 99 equal steps. Each step is, therefore, ¹⁹⁸/⁹⁹, or 2, which is the common difference; or the series is 1, 3, 5, &c., up to 199.

Given. To find.

Sum. No. of terms. First term. Last term. Common diff.

1809025 1345 1 2689 2

44 10 3 29 14

5 45

7075600 1330 4 10636 8

175. We now return to (170), in which we compared two numbers together by their difference. This, however, is not the method of comparison which we employ in common life, as any single familiar instance will shew. For example, we say of A, who has 10 thousand pounds, that he is much richer than B, who has only 3 thousand; but we do not say that C, who has 107 thousand pounds, is much richer than D, who has 100 thousand, though the difference of fortune is the same in both cases, viz. 7 thousand pounds. In comparing numbers we take into our reckoning not only the differences, but the numbers themselves. Thus, if B and D both received 7 thousand pounds, B would receive 233 pounds and a third for every 100 pounds which he had before, while D for every 100 pounds would receive only 7 pounds. And though, in the view taken in (170), 3 is as near to 10 as 100 is to 107, yet, in the light in which we now regard them, 3 is not so near to 10 as 100 is to 107, for 3 differs from 10 by more than twice itself, while 100 does not differ from 107 by so much as one-fifth of itself. This is expressed in mathematical language by saying, that the ratio or proportion of 10 to 3 is greater than the ratio or proportion of 107 to 100. We proceed to define these terms more accurately.

176. When we use the term part of a number or fraction in the remainder of this section, we mean, one of the various sets of equal parts into which it may be divided, either the half, the third, the fourth, &c.: the term multiple has been already explained (102). By the term multiple-part of a number we mean, the abbreviation of the words multiple of a part. Thus, 1, 2, 3, 4, and 6, are parts of 12; ½ is also a part of 12, being contained in it 24 times; 12, 24, 36, &c., are multiples of 12; and 8, 9, ⁵/₂, &c. are multiple parts of 12, being multiples of some of its parts. And when multiple parts generally are spoken of, the parts themselves are supposed to be included, on the same principle that 12 is counted among the multiples of 12, the multiplier being 1. The multiples themselves are also included in this term; for 24 is also 48 halves, and is therefore among the multiple parts of 12. Each part is also in various ways a multiple-part; for one-fourth is two-eighths, and three-twelfths, &c.

177. Every number or fraction is a multiple-part of every other number or fraction. If, for example, we ask what part 12 is of 7, we see that on dividing 7 into 7 parts, and repeating one of these parts 12 times, we obtain 12; or, on dividing 7 into 14 parts, each of which is one-half, and repeating one of these parts 24 times, we obtain 24 halves, or 12. Hence, 12 is ¹²/₇, or ²⁴/₁₄, or ³⁶/₂₁ of 7; and so on. Generally, when a and b are two whole numbers, a/b expresses the multiple-part which a is of b, and b/a that which b is of a. Again, suppose it required to determine what multiple-part (2⅐) is of (3⅕), or ¹⁵/₇ of ¹⁶/₅. These fractions, reduced to a common denominator, are ⁷⁵/₃₅ and ¹¹²/₃₅, of which the second, divided into 112 parts, gives ¹/₃₅, which repeated 75 times gives ⁷⁵/₃₅, the first. Hence, the multiple-part which the first is of the second is ⁷⁵/₁₁₂, which being obtained by the rule given in (121), shews that a/b, or a divided by b, according to the notion of division there given, expresses the multiple-part which a is of b in every case.

178. When the first of four numbers is the same multiple-part of the second which the third is of the fourth, the four are said to be geometrically[27] proportional, or simply proportional. This is a word in common use; and it remains to shew that our mathematical definition of it, just given, is, in fact, the common notion attached to it. For example, suppose a picture is copied on a smaller scale, so that a line of two inches long in the original is represented by a line of one inch and a half in the copy; we say that the copy is not correct unless all the parts of the original are reduced in the same proportion, namely, that of 2 to (1½). Since, on dividing two inches into 4 parts, and taking 3 of them, we get (1½), the same must be done with all the lines in the original, that is, the length of any line in the copy must be three parts out of four of its length in the original. Again, interest being at 5 per cent, that is, £5 being given for the use of £100, a similar proportion of every other sum would be given; the interest of £70, for example, would be just such a part of £70 as £5 is of £100.

Since, then, the part which a is of b is expressed by the fraction a/b, or any other fraction which is equivalent to it, and that which c is of d by c/d, it follows, that when a, b, c, and d, are proportional, a/b = c/d. This equation will be the foundation of all our reasoning on proportional quantities; and in considering proportionals, it is necessary to observe not only the quantities themselves, but also the order in which they come. Thus, a, b, c, and d, being proportionals, that is, a being the same multiple-part of b which c is of d, it does not follow that a, d, b, and c are proportionals, that is, that a is the same multiple-part of d which b is of c. It is plain that a is greater than, equal to, or less than b, according as c is greater than, equal to, or less than d.

179. Four numbers, a, b, c, and d, being proportional in the order written, a and d are called the extremes, and b and c the means, of the proportion. For convenience, we will call the two extremes, or the two means, similar terms, and an extreme and a mean, dissimilar terms. Thus, a and d are similar, and so are b and c; while a and b, a and c, d and b, d and c, are dissimilar. It is customary to express the proportion by placing dots between the numbers, thus:

a : b ∷ c : d

180. Equal numbers will still remain equal when they have been increased, diminished, multiplied, or divided, by equal quantities. This amounts to saying that if

a = b and p = q,

a + p = b + q,

a - p = b - q,

ap = bq,

a b

and — = —.

p q

It is also evident, that a + p-p, a -p + p, ap/p, and a/p × p, are all equal to a.

181. The product of the extremes is equal to the product of the means. Let a/b = c/d, and multiply these equal numbers by the product bd. Then,

a × bd = abd (116) = ad,

b b

and c × bd = cbd = cb:

d d

hence (180), ad = bc.

Thus, 6, 8, 21, and 28, are proportional, since

6 = 3 = 3 × 7 = 21 (180);

8 4 4 × 7 28

and it appears that 6 × 28 = 8 × 21, since both products are 168.

182. If the product of two numbers be equal to the product of two others, these numbers are proportional in any order whatever, provided the numbers in the same product are so placed as to be similar terms; that is, if ab = pq, we have the following proportions:—

a : p ∷ q : b

a : q ∷ p : b

b : p ∷ q : a

b : q ∷ p : a

p : a ∷ b : q

p : b ∷ a : q

q : a ∷ b : p

q : b ∷ a : p

To prove any one of these, divide both ab and pq by the product of its second and fourth terms; for example, to shew the truth of a: q ∷ p: b, divide both ab and pq by bq. Then,

ab = a , and pq = p ; hence (180),

bq q bq b

a = p , or a : q ∷ p : b .

q b

The pupil should not fail to prove every one of the eight cases, and to verify them by some simple examples, such as 1 × 6 = 2 × 3, which gives 1: 2 ∷ 3: 6, 3: 1 ∷ 6: 2, &c.

183. Hence, if four numbers be proportional, they are also proportional in any other order, provided it be such that similar terms still remain similar. For since, when

a = c ,

b d

it follows (181) that ad = bc, all the proportions which follow from ad = bc, by the last article, follow also from

a = c .

b d

184. From (114) it follows that

1 + a = b + a ,

b b

and if a be less than 1,

b

1 - a = b - a ,

b b

while if a be greater than 1,

b

a - 1 = a - b .

b b

Also (122), if a + b be divided by a - b

b b

the result is a + b

a - b

Hence, a, b, c, and d, being proportionals, we may obtain other proportions, thus:

Let a = c

b d

Then (114) 1 + a = 1 + c

b d

or a + b = c + d

b d

or a + b: b ∷ c + d: d

That is, the sum of the first and second is to the second as the sum of the third and fourth is to the fourth. For brevity, we shall not state in words any more of these proportions, since the pupil will easily supply what is wanting.

Resuming the proportion a: b ∷ c: d

or a = c

b d

1 - a = 1 - c , if a be less than 1,

b d b

or b - a = d - c

b d

that is, b-a: b ∷ d-c: d or, a-b: b ∷ c-d: d,

if a be greater than 1.

b

Again, since a + b = c + d

b d

and a - b = c - d ( a being greater than 1)

b d b

dividing the first by the second we have

a + b = c + d

a - b c - d

or a + b : a - b ∷ c + d : c - d

and also a + b : b - a ∷ c + d : d - c,

if a be less than 1.

b

185. Many other proportions might be obtained in the same manner. We will, however, content ourselves with writing down a few which can be obtained by combining the preceding articles.

a + b : a ∷ c + d : c

a : a - b ∷ c : c - d

a + c : a - c ∷ b + d : b - d.

In these and all others it must be observed, that when such expressions as a-b and c-d occur, it is supposed that a is greater than b, and c greater than d.

186. If four numbers be proportional, and any two dissimilar terms be both multiplied, or both divided by the same quantity, the results are proportional. Thus, if a: b ∷ c: d, and m and n be any two numbers, we have also the following:

ma : b ∷ mc : d

a : mb ∷ c : md

a : mb ∷ c : md

n n

ma : nb ∷ mc : nd

a : b ∷ c : d

m m m m

a : b ∷ c : d

m m n n

and various others. To prove any one of these, recollect that nothing more is necessary to make four numbers proportional except that the product of the extremes should be equal to that of the means. Take the third of those just given; the product of its extremes is

a × md, or mbc ,

n n

while that of the means is

mb × c , or mad .

n n

But since a : b ∷ c : d, by (181) ad = bc,

whence, by (180), mad = mbc, and mad = mbc .

n n

Hence a , c , and md, are proportionals.

n n

187. If the terms of one proportion be multiplied by the terms of a second, the products are proportional; that is, if a: b ∷ c: d, and p: q ∷ r: s, it follows that ap: bq ∷ cr: ds. For, since ad = bc, and ps = qr, by (180) adps = bcqr, or ap × ds = bq × cr, whence (182) ap: bq ∷ cr: ds.

188. If four numbers be proportional, any similar powers of these numbers are also proportional; that is, if

Then a : b ∷ c : d

aa : bb ∷ cc : dd

aaa : bbb ∷ ccc : ddd

&c. &c.

For, if we write the proportion twice, thus,

a : b ∷ c : d

a : b ∷ c : d

by (187) aa : bb ∷ cc : dd

But a : b ∷ c : d

Whence (187) aaa : bbb ∷ ccc : ddd and so on.

189. An expression is said to be homogeneous with respect to any two or more letters, for instance, a, b, and c, when every term of it contains the same number of letters, counting a, b, and c only. Thus, maab + nabc + rccc is homogeneous with respect to a, b, and c; and of the third degree, since in each term there is either a, b, and c, or one of these repeated alone, or with another, so as to make three in all. Thus, 8aaabc, 12abccc, maaaaa, naabbc, are all homogeneous, and of the fifth degree, with respect to a, b, and c only; and any expression made by adding or subtracting these from one another, will be homogeneous and of the fifth degree. Again ma + mnb is homogeneous with respect to a and b, and of the first degree; but it is not homogeneous with respect to m and n, though it is so with respect to a and n. This being premised, we proceed to a theorem,[28] which will contain all the results of (184), (185), and (188).

190. If any four numbers be proportional, and if from the first two, a and b, any two homogeneous expressions of the same degree be formed; and if from the last two, two other expressions be formed, in precisely the same manner, the four results will be proportional. For example, if a: b ∷ c: d, and if 2aaa + 3aab and bbb + abb be chosen, which are both homogeneous with respect to a and b, and both of the third degree; and if the corresponding expressions 2ccc + 3ccd and ddd + cdd be formed, which are made from c and d precisely in the same manner as the two former ones from a and b, then will

2aaa + 3aab : bbb + abb ∷ 2ccc + 3ccd : ddd + cdd

To prove this, let a be called x.

b

Then, since a = x, and a = c ,

b b d

it follows that c = x.

d

But since a divided by b gives x, x multiplied by b will give a, or a = bx. For a similar reason, c = dx. Put bx and dx instead of a and c in the four expressions just given, recollecting that when quantities are multiplied together, the result is the same in whatever order the multiplications are made; that, for example, bxbxbx is the same as bbbxxx.

Hence, 2aaa + 3aab = 2bxbxbx + 3bxbxb = 2bbbxxx + 3bbbxx

which is bbb multiplied by 2xxx + 3xx

or bbb (2xxx + 3xx)[29]

Similarly, 2ccc + 3ccd = ddd (2xxx + 3xx)

Also, bbb + abb = bbb + bxbb

= bbb multiplied by 1 + x

or bbb(1 + x)

Similarly, ddd + cdd = ddd (1 + x)

Now, bbb : bbb ∷ ddd : ddd

Whence (186), bbb(2xxx + 3xx): bbb(1 + x) ∷ ddd(2xxx + 3xx): ddd(1 + x), which, when instead of these expressions their equals just found are substituted, becomes 2aaa + 3aab: bbb + abb ∷ 2ccc + 3ccd: ddd + cdd.

The same reasoning may be applied to any other case, and the pupil may in this way prove the following theorems:

If

a : b ∷ c : d

2a + 3b : b ∷ 2c + 3d : d

aa + bb : aa - bb ∷ cc + dd : cc - dd

mab : 2aa + bb ∷ mcd : 2cc + dd

191. If the two means of a proportion be the same, that is, if a : b ∷ b: c, the three numbers, a, b, and c, are said to be in continued proportion, or in geometrical progression. The same terms are applied to a series of numbers, of which any three that follow one another are in continued proportion, such as

1 2 4 8 16 32 64 &c.

2 2 2 2 2 2 2 &c.

3 9 27 81 243 729

Which are in continued proportion, since

1 : 2 ∷ 2 : 4 2 : 2 ∷ 2 : 2

3 3 9

2 : 4 ∷ 4 : 8 2 : 2 ∷ 2 : 2

3 9 9 27

&c. &c.

192. Let a, b, c, d, e be in continued proportion; we have then

a : b ∷ b : c or a = b or ac = bb

b c

b : c ∷ c : d b = c bd = cc

c d

c : d ∷ d : e c = d ce = dd

d e

Each term is formed from the preceding, by multiplying it by the same number. Thus,

b = b × a (180); c = c × b;

a b

and since a = b , b = c

b c a b

or c = b × b.

a

Again, d = d × c ,

c

but d = c , which is = b ;

c b a

therefore, d = b × c, and so on

a

If, then, b (which is called the common ratio of the series)

a

be denoted by r, we have

b = ar c = br = arr d = cr = arrr

and so on; whence the series

a b c d &c.

is a ar arr arrr &c.

Hence a : c ∷ a : arr

(186) ∷ aa : aarr

∷ aa : bb

because, b being ar, bb is arar or aarr. Again,

a : d ∷ a : arrr

(186) ∷ aaa : aaarrr

∷ aaa : bbb

a : e ∷ aaaa : bbbb , and so on;

that is, the first bears to the nᵗʰ term from the first the same proportion as the nᵗʰ power of the first to the nᵗʰ power of the second.

193. A short rule may be found for adding together any number of terms of a continued proportion. Let it be first required to add together the terms 1, r, rr, &c. where r is greater than unity. It is evident that we do not alter any expression by adding or subtracting any numbers, provided we afterwards subtract or add the same. For example,

p = p - q + q - r + r - s + s

Let us take four terms of the series, 1, r, rr, &c. or,

1 + r + rr + rrr

It is plain that

rrrr - 1 = rrrr - rrr + rrr - rr + rr - r + r - 1

Now (54), rr-r = r(r-1), rrr -rr = rr(r-1), rrrr-rrr = rrr(r-1), and the above equation becomes rrrr -1 = rrr(r-1) + rr (r-1) + r (r-1) + r-1; which is (54) rrr + rr + r + 1 taken r-1 times. Hence, rrrr-1 divided by r-1 will give 1 + r + rr + rrr, the sum of the terms required. In this way may be proved the following series of equations:

1 + r = rr - 1

r - 1

1 + r + rr = rrr - 1

r - 1

1 + r + rr + rrr = rrrr - 1

r - 1

1 + r + rr + rrr + rrrr = rrrrr - 1

r - 1

If r be less than unity, in order to find 1 + r + rr + rrr, observe that

1 - rrrr = 1 - r + r - rr + rr - rrr + rrr - rrrr = 1 - r + r(1 - r) + rr(1 - r) + rrr(1 - r);

whence, by similar reasoning, 1 + r + rr + rrr is found by dividing 1-rrrr by 1-r; and equations similar to these just given may be found, which are,

1 + r = 1 - rr

1 - r

1 + r + rr = 1 - rrr

1 - r

1 + r + rr + rrr = 1 - rrrr

1 - r

1 + r + rr + rrr + rrrr = 1 - rrrrr

1 - r

The rule is: To find the sum of n terms of the series, 1 + r + rr + &c., divide the difference between 1 and the (n + 1)ᵗʰ term by the difference between 1 and r.

194. This may be applied to finding the sum of any number of terms of a continued proportion. Let a, b, c, &c. be the terms of which it is required to sum four, that is, to find a + b + c + d, or (192) a + ar + arr + arrr, or (54) a(1 + r + rr + rrr), which (193) is

rrrr - 1 × a, or 1 - rrrr × a,

r - 1 1 - r

according as r is greater or less than unity. The first fraction is

arrrr - a , or (192) e - a .

r - 1 r - 1

Similarly, the second is

a - e .

1 - r

The rule, therefore, is: To sum n terms of a continued proportion, divide the difference of the (n + 1)ᵗʰ and first terms by the difference between unity and the common measure. For example, the sum of 10 terms of the series 1 + 3 + 9 + 27 + &c. is required. The eleventh term is 59049, and ⁽⁵⁹⁰⁴⁹ ⁻ ¹⁾/₍₃₋₁₎ is 29524. Again, the sum of 18 terms of the series 2 + 1 + ½ + ½ + &c. of which the nineteenth term is ¹/₁₃₁₀₇₂, is

1

2 - ———

131072 131070

————— = 3 ——— .

1 - ½ 131072

EXAMPLES.

9 terms of 1 + 4 + 16 + &c. are 87381

10 ...... 3 + 6 + 12 + &c. ... 847422675

7 49 201768035

20 ...... 1 + 1 + 1 + &c. ... 1048575

2 4 8 1048576

195. The powers of a number or fraction greater than unity increase; for since 2½ is greater than 1, 2½ × 2½ is 2½ taken more than once, that is, is greater than 2½, and so on. This increase goes on without limit; that is, there is no quantity so great but that some power of 2½ is greater. To prove this, observe that every power of 2½ is made by multiplying the preceding power by 2½, or by 1 + 1½, that is, by adding to the former power that power itself and its half. There will, therefore, be more added to the 10th power to form the 11th, than was added to the 9th power to form the 10th. But it is evident that if any given quantity, however small, be continually added to 2½, the result will come in time to exceed any other quantity that was also given, however great; much more, then, will it do so if the quantity added to 2½ be increased at each step, which is the case when the successive powers of 2½ are formed. It is evident, also, that the powers of 1 never increase, being always 1; thus, 1 × 1 = 1, &c. Also, if a be greater than m times b, the square of a is greater than mm times the square of b. Thus, if a = 2b + c, where a is greater than 2b, the square of a, or aa, which is (68) 4bb + 4bc + cc is greater than 4bb, and so on.

196. The powers of a fraction less than unity continually decrease; thus, the square of ⅖, or ⅖ × ⅖, is less than ⅖, being only two-fifths of it. This decrease continues without limit; that is, there is no quantity so small but that some power of ⅖ is less. For if

5 = x,

2

2 = 1 ,

5 x

and the powers of ⅖ are 1 , 1 ,

xx xxx

and so on. Since x is greater than 1 (195), some power of x may be found which shall be greater than a given quantity. Let this be called m; then 1/m is the corresponding power of ⅖; and a fraction whose denominator can be made as great as we please, can itself be made as small as we please (112).

197. We have, then, in the series

1 r rr rrr rrrr &c.

I. A series of increasing terms, if r be greater than 1. II. Of terms having the same value, if r be equal to 1. III. A series of decreasing terms, if r be less than 1. In the first two cases, the sum

1 + r + rr + rrr + &c.

may evidently be made as great as we please, by sufficiently increasing the number of terms. But in the third this may or may not be the case; for though something is added at each step, yet, as that augmentation diminishes at every step, we may not certainly say that we can, by any number of such augmentations, make the result as great as we please. To shew the contrary in a simple instance, consider the series,

1 + ½ + ¼ + ⅛ + ¹/₁₆ + &c.

Carry this series to what extent we may, it will always be necessary to add the last term in order to make as much as 2. Thus,

(1 + ½ + ¼) + ¼ = 1 + ½ + ½ = 1 + 1 = 2 (1 + ½ + ¼ + ⅛) + ⅛ = 2. (1 + ½ + ¼ + ⅛ + ¹/₁₆) + ¹/₁₆ = 2, &c.

But in the series, every term is only the half of the preceding; consequently no number of terms, however great, can be made as great as 2 by adding one more. The sum, therefore, of 1, ½, ¼, ⅛ &c. continually approaches to 2, diminishing its distance from 2 at every step, but never reaching it. Hence, 2 is celled the limit of 1 + ½ + ¼ + &c. We are not, therefore, to conclude that every series of decreasing terms has a limit. The contrary may be shewn in the very simple series, 1 + ½ + ⅓ + ¼ + &c. which may be written thus:

1 + ½ + (⅓ + ¼) + (⅕ + ... up to ⅛) + (⅑ + ... up to ¹/₁₆) + (¹/₁₇ + ... up to ¹/₃₂) + &c.

We have thus divided all the series, except the first two terms, into lots, each containing half as many terms as there are units in the denominator of its last term. Thus, the fourth lot contains 16 or ³²/₂2 terms. Each of these lots may be shewn to be greater than ½. Take the third, for example, consisting of ⅑, ¹/₁₀, ¹/₁₁, ¹/₁₂, ¹/₁₃, ¹/₁₄, ¹/₁₅, and ¹/₁₆. All except ¹/₁₆, the last, are greater than ¹/₁₆; consequently, by substituting ¹/₁₆ for each of them, the amount of the whole lot would be lessened; and as it would then become ⁸/₁₆, or ½, the lot itself is greater than ½. Now, if to 1 + ½, ½ be continually added, the result will in time exceed any given number. Still more will this be the case if, instead of ½, the several lots written above be added one after the other. But it is thus that the series 1 + ½ + ⅓, &c. is composed, which proves what was said, that this series has no limit.

198. The series 1 + r + rr + rrr + &c. always has a limit when r is less than 1. To prove this, let the term succeeding that at which we stop be a, whence (194) the sum is

1 - a , or (112) 1 - a .

1 - r 1 - r 1 - r

The terms decrease without limit (196), whence we may take a term so far distant from the beginning, that a, and therefore

1 - a ,

1 - r

shall be as small as we please. But it is evident that in this case

1 - a

1 - r 1 - r

though always less than

1

1 - r

may be brought as near to

1

1 - r

as we please; that is, the series 1 + r + rr + &c. continually approaches to the limit

1 .

1 - r

Thus 1 + ½ + ¼ + ⅛ + &c. where r = ½, continually approaches to

1 or 2,

1 - ½

as was shewn in the last article.

EXERCISES.

The limit of 2 + 2 + 2 + &c.

3 9

or 2 ( 1 + 1 + 1 + &c. ) is 3

3 9

... 1 + 9 + 81 + &c. ... 10

10 100

... 5 + 15 + 45 + &c. ... 8¾

7 49

199. When the fraction a/b is not equal to c/d, but greater, a is said to have to b a greater ratio than c has to d; and when a/b is less than c/d, a is said to have to b a less ratio than c has to d. We propose the following questions as exercises, since they follow very simply from this definition.

I. If a be greater than b, and c less than or equal to d, a will have a greater ratio to b than c has to d.

II. If a be less than b, and c greater than or equal to d, a has a less ratio to b than c has to d.

III. If a be to b as c is to d, and if a have a greater ratio to b than c has to x, d is less than x; and if a have a less ratio to b than c to x, d is greater than x.

IV. a has to b a greater ratio than ax to bx + y, and a less ratio than ax to bx- y.

200. If a have to b a greater ratio than c has to d, a + c has to b + d a less ratio than a has to b, but a greater ratio than c has to d; or, in other words, if a/b be the greater of the two fractions a/b and c/d,

a + c

b + d

will be greater than c/d, but less than a/b. To shew this, observe that (mx + ny)/(m + n) must lie between x and y, if x and y be unequal: for if x be the less of the two, it is certainly greater than

mx + nx

m + n

or than x; and if y be the greater of the two, it is certainly less than

my + ny

m + n

or than y. It therefore lies between x and y. Now let a/b be x, and let c/d be y: then a = bx, c = dy. Now

bx + dy

b + d

is something between x and y, as was just proved; therefore

a + c

b + d

is something between a/b and c/d. Again, since a/b and c/d are respectively equal to ap/bp and cq/dq, and since, as has just been proved,

ap + cq

bp + dq

lies between the two last, it also lies between the two first; that is, if p and q be any numbers or fractions whatsoever,

ap + cq

bp + dq

lies between a/b and c/d.

201. By the last article we may often form some notion of the value of an expression too complicated to be easily calculated. Thus,

1 + x lies between 1 and x , or 1 and 1 ;

1 + xx 1 xx x

ax + by lies between ax and by ,

axx + bbyy axx bbyy

that is, between 1/x and 1/by. And it has been shewn that (a + b)/2 lies between a and b, the denominator being considered as 1 + 1.

202. It may also be proved that a fraction such as

a + b + c + d

p + q + r + s

always lies among

a , b , c , and d ,

p q r s

that is, is less than the greatest of them, and greater than the least. Let these fractions be arranged in order of magnitude; that is, let a/p be greater than b/q, b/q be greater than c/r, and c/r greater than d/s. Then by (200)

is less than and greater than

a + b a b and c

p + q p q r

a + b + c a + b and a c and d

p + q + r p + q p r s

a + b + c + d a + b + c and a d

p + q + r + s p + q + r p s

whence the proposition is evident.

203. It is usual to signify “a is greater than b” by a > b and “a is less than b” by a < b; the opening of V being turned towards the greater quantity. The pupil is recommended to make himself familiar with these signs.

SECTION IX. ON PERMUTATIONS AND COMBINATIONS.

204. If a number of counters, distinguished by different letters, be placed on the table, and any number of them, say four, be taken away, the question is, to determine in how many different ways this can be done. Each way of doing it gives what is called a combination of four, but which might with more propriety be called a selection of four. Two combinations or selections are called different, which differ in any way whatever; thus, abcd and abce are different, d being in one and e in the other, the remaining parts being the same. Let there be six counters, a, b, c, d, e, and f; the combinations of three which can be made out of them are twenty in number, as follow:

abc ace bcd bef

abd acf bce cde

abe ade bcf cdf

abf adf bde cef

acd aef bdf def

The combinations of four are fifteen in number, namely,

abcd abde acde adef bcef

abce abdf acdf bcde bdcf

abcf abef acef bcdf cdef

and so on.

205. Each of these combinations may be written in several different orders; thus, abcd may be disposed in any of the following ways:

abcd acbd acdb abdc adbc adcb

bacd cabd cadb badc dabc dacb

bcad cbad cdab bdac dbac dcab

bcda cbda cdba bdca dbca dcba

of which no two are entirely in the same order. Each of these is said to be a distinct permutation of abcd. Considered as a combination, they are all the same, as each contains a, b, c, and d.

206. We now proceed to find how many permutations, each containing one given number, can be made from the counters in another given number, six, for example. If we knew how to find all the permutations containing four counters, we might make those which contain five thus: Take any one which contains four, for example, abcf in which d and e are omitted; write d and e successively at the end, which gives abcfd, abcfe, and repeat the same process with every other permutation of four; thus, dabc gives dabce and dabcf. No permutation of five can escape us if we proceed in this manner, provided only we know those of four; for any given permutation of five, as dbfea, will arise in the course of the process from dbfe, which, according to our rule, furnishes dbfea. Neither will any permutation be repeated twice, for dbfea, if the rule be followed, can only arise from the permutation dbfe. If we begin in this way to find the permutations of two out of the six,

a b c d e f

each of these gives five; thus,

a gives ab ac ad ae af

b ... ba bc bd be bf

and the whole number is 6 × 5, or 30.

Again,

ab gives abc abd abe abf

ac ... acb acd ace acf

and here are 30, or 6 × 5 permutations of 2, each of which gives 4 permutations of 3; the whole number of the last is therefore 6 × 5 × 4, or 120.

Again,

abc gives abcd abce abcf

abd ... abdc abde abdf

and here are 120, or 6 × 5 × 4, permutations of three, each of which gives 3 permutations of four; the whole number of the last is therefore 6 × 5 × 4 × 3, or 360.

In the same way, the number of permutations of 5 is 6 × 5 × 4 × 3 × 2, and the number of permutations of six, or the number of different ways in which the whole six can be arranged, is 6 × 5 × 4 × 3 × 2 × 1. The last two results are the same, which must be; for since a permutation of five only omits one, it can only furnish one permutation of six. If instead of six we choose any other number, x, the number of permutations of two will be x(x-1), that of three will be x(x-1)(x-2), that of four x(x -1)(x-2)(x-3), the rule being: Multiply the whole number of counters by the next less number, and the result by the next less, and so on, until as many numbers have been multiplied together as there are to be counters in each permutation: the product will be the whole number of permutations of the sort required. Thus, out of 12 counters, permutations of four may be made to the number of 12 × 11 × 10 × 9, or 11880.

EXERCISES.

207. In how many different ways can eight persons be arranged on eight seats?

Answer, 40320.

In how many ways can eight persons be seated at a round table, so that all shall not have the same neighbours in any two arrangements?[30]

Answer, 5040.

If the hundredth part of a farthing be given for every different arrangement which can be made of fifteen persons, to how much will the whole amount?

Answer, £13621608.

Out of seventeen consonants and five vowels, how many words can be made, having two consonants and one vowel in each?

Answer, 4080.

208. If two or more of the counters have the same letter upon them, the number of distinct permutations is less than that given by the last rule. Let there be a, a, a, b, c, d, and, for a moment, let us distinguish between the three as thus, a, a′, a″. Then, abca′a″d, and a″bcaa′d are reckoned as distinct permutations in the rule, whereas they would not have been so, had it not been for the accents. To compute the number of distinct permutations, let us make one with b, c, and d, leaving places for the as, thus, ( ) bc ( ) ( ) d. If the as had been distinguished as a, a′, a″, we might have made 3 × 2 × 1 distinct permutations, by filling up the vacant places in the above, all which six are the same when the as are not distinguished. Hence, to deduce the number of permutations of a, a, a, b, c, d, from that of aa′a″bcd, we must divide the latter by 3 × 2 × 1, or 6, which gives

6 × 5 × 4 × 3 × 2 × 1

3 × 2 × 1

or 120. Similarly, the number of permutations of aaaabbbcc is

9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 .

4 × 3 × 2 × 1 × 3 × 2 × 1 × 2 × 1

EXERCISE.

How many variations can be made of the order of the letters in the word antitrinitarian?

Answer, 126126000.

209. From the number of permutations we can easily deduce the number of combinations. But, in order to form these combinations independently, we will shew a method similar to that in (206). If we know the combinations of two which can be made out of a, b, c, d, e, we can find the combinations of three, by writing successively at the end of each combination of two, the letters which come after the last contained in it. Thus, ab gives abc, abd, abe; ad gives ade only. No combination of three can escape us if we proceed in this manner, provided only we know the combinations of two; for any given combination of three, as acd, will arise in the course of the process from ac, which, according to our rule, furnishes acd. Neither will any combination be repeated twice, for acd, if the rule be followed, can only arise from ac, since neither ad nor cd furnishes it. If we begin in this way to find the combinations of the five,

a b c d e

a gives ab ac ad ae

b ···· bc bd be

c ···· cd ce

d ···· de

Of these, ab gives abc abd abe

ac ···· acd ace

ad ···· ade

bc ···· bcd bce

bd ···· bde

cd ···· cde

ae be ce and de give none.

Of these, abc gives abcd abce

abd ···· abde

acd ···· acde

bcd ···· bcde

Those which contain e give none, as before.

Of the last, abcd gives abcde, and the others none, which is evidently true, since only one selection of five can be made out of five things.

210. The rule for calculating the number of combinations is derived directly from that for the number of permutations. Take 7 counters; then, since the number of permutations of two is 7 × 6, and since two permutations, ba and ab, are in any combination ab, the number of combinations is half that of the permutations, or (7 × 6)/2. Since the number of permutations of three is 7 × 6 × 5, and as each combination abc has 3 × 2 × 1 permutations, the number of combinations of three is

7 × 6 × 5 .

1 × 2 × 3

Also, since any combination of four, abcd, contains 4 × 3 × 2 × 1 permutations, the number of combinations of four is

7 × 6 × 5 × 4 ,

1 × 2 × 3 × 4

and so on. The rule is: To find the number of combinations, each containing n counters, divide the corresponding number of permutations by the product of 1, 2, 3, &c. up to n. If x be the whole number, the number of combinations of two is

x(x - 1) ;

1 × 2

that of three is

x(x - 1)(x - 2) ;

1 × 2 × 3

that of four is

x(x - 1)(x - 2)(x - 3) ;

1 × 2 × 3 × 4

211. The rule may in half the cases be simplified, as follows. Out of ten counters, for every distinct selection of seven which is taken, a distinct combination of 3 is left. Hence, the number of combinations of seven is as many as that of three. We may, therefore, find the combinations of three instead of those of seven; and we must moreover expect, and may even assert, that the two formulæ for finding these two numbers of combinations are the same in result, though different in form. And so it proves; for the number of combinations of seven out of ten is

10 × 9 × 8 × 7 × 6 × 5 × 4 ,

1 × 2 × 3 × 4 × 5 × 6 × 7

in which the product 7 × 6 × 5 × 4 occurs in both terms, and therefore may be removed from both (108), leaving

10 × 9 × 8 ,

1 × 2 × 3

which is the number of combinations of three out of ten. The same may be shewn in other cases.

EXERCISES.

How many combinations of four can be made out of twelve things?

Answer, 495.

What number of 6 out of 8 Answer, 28

combinations 4 11 330

can be made of 26 28 378

6 15 5005

How many combinations can be made of 13 out of 52; or how many different hands may a person hold at the game of whist?

Answer, 635013559600.