divide both ab and pq by bq. Then, a = = pq and = q b q ; hence ab a b q (180), or a q :: Ρ : 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 are 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; α = c b d' all the proportions which follow from a d = bc, by the last article, 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 ab :: c d than 1), dividing the first by the second we have or d a+b c+d a-b a+ba-b :: c+d: c-d α and also a+bb-a: c+d: d-c, if is less than 1. = c-d' 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 xmd, or mad mb c while that of the means is mb x-, or But since a: b:: n n' n a n mad cd, by (181) ad = bc, whence, by (180) mad = m bc, and n 187. If the terms of one proportion be multiplied by the terms of a second, the products are proportional; that is, if a bcd, and p: q r s, it follows that ap: bq :: cr a d = bc, and ps = qr, by (180) adps = bc qr, or ap ds = b q × cr, whence (182) ap: bq :: cr: ds. 188. If four numbers are proportional, any similar powers of these numbers are also proportional; that is, if and 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; of the third degree, since in each term there is either a, b, and c, or one of these repeated alone, or with another, so as to make three in all. Thus, 8 a a a b c, 12 abccc, maa a aa, na a b b c, 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,* 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; * A theorem is a general mathematical fact: thus, that every number is divisible by four when its last two figures are divisible by four, is a theorem; that in every proportion the product of the extremes is equal to the product of the means, is another. and if from the last two, two other expressions be formed, in precisely the same manner, the four results will be proportional. For example, if a b c d, and if 2 aaa+ 3 a ab and bbb+abb be chosen, which are both homogeneous with respect to a and b, and both of the third degree; and if the corresponding expressions 2 ccc+3ccd, and ddd+c d d be formed, which are made from c and d precisely in the same manner as the two former ones from a and b, then will 2 aaa+zaab bbb+abb :: 2ccc+3ccd: ddd+cdd C d Then, since α = x, and = b To prove this, let be called a. ī b it follows that = x. But since a divided by b gives x, x multiplied by b will give a, or a = b x. For a similar reason, c = d x. Put bx and da 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 bb b x x x. Hence, 2 aaa+3aab = 2 b x b x b x + z b x b x b Whence (186), b b b ( 2 x x x + 3 x x ): d d d ( 1 + x) :: b b b ( 2 x x x + 3 xx): ddd (1+x), which, when instead of these expressions, their equals just found are substituted, becomes 2 aaa+3aab: bbb + abb: 2ccc+3ccd ddd+cdd. The same reasoning may be * If br be substituted for a in any expression which is homogeneous with respect to a and b, the pupil may easily see that b must occur in every term as often as there are units in the degree of the expression: thus, a a+ab becomes b x b x+b x b or b b (x x+x); aaa+bbb becomes bx b x bx+b bb or b b b (x x x+1); and so on. applied to any other case, and the pupil may in this way prove the following theorems : mab 2 aa+bb :: mcd 2 cc+dd 191. If the two means of a proportion be the same, that is, if a bbc, 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 192. Let a, b, c, d, &c. be in continued proportion; we have then Each term is formed from the preceding, by multiplying it by the same a b b xb; and since = b d b = which is = -; therefore, b' a (which is called the common ratio of |