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:

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

The sum of 50 terms of the series

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.

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.

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,

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

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,

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

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

184. From (114) it follows that

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

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

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

dividing the first by the second we have

or  a + b : a - b ∷ c + d : c - d

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

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.

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:

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

while that of the means is

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

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

For, if we write the proportion twice, thus,

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

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

Which are in continued proportion, since

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

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

b = ar c = br = arr d = cr = arrr

and so on; whence the series

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

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:

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,

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

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

Similarly, the second is

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

EXAMPLES.

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

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

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

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

though always less than

may be brought as near to

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

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

as was shewn in the last article.

EXERCISES.