534 reducing the second equality will be identical with, that of reducing the first, when -,T, -, are written in place of a, a b c I, c. Now the former result is a 3 + 1 5 + c 3 + ale = 0, .*. the latter is + = + + = 0. a 6 b* 6 ALGEBRA prove that [SUKDS. abc fix. 4. Prove that and equal to 1 if m = n. ins: (x a 1 )(cc-a 2 We have + &c., is equal to if m>n, This is easily proved by resolv- -. into partial fractions (Art. 33). + . . . (x-a n ) x-a^ x-fi 2 &c., &c. . (1), whence, writing a lt a.,, &c., successively for x, we get A i; A 2 , &c.
- The given quantity is A^A.,-1- ... + A B , and the
equation marked (1), gives, by equating coefficients of like powers of x, the result required. Ex. 5. If =-, then q s P~1 r ~ s -r. P i r T P + ( l r + S For - + ! = - + ! gives 7 -= , , p 1 r and - - 1 = - 1 gives p-q Ex. G. If -+ a b Divide the former by the latter. L 1 a + b + c , then a~ = U 1 = <? n + b ab a b , . either c c(a+b + i I 11 or = - -r- - ab ac be In the latter case, -r = -rH c- ab ac J_ be which is not changed by interchanging c and b or c and a, JL_JL_JL c 2 b z a? so that on either alternative the proposition is true. ad be ac bd Ex. 7. Given that and a not a b c+d ab+cd equal to I, nor c equal to d ; to prove that a + 1 = c + d ; and that cither of the fractions equals Write the equality thus, ac bd _ a-b + (c-d) ad be a-b (c d) Apply Example 5, and there results, ac bd+ad bc a-b ^.e. ac bd ad+bc c d (a-b)(c+d) a-b .
- 7 (; , 7
- = ; , whence a + b = c + d
(c-d)(a + b) c-d If now a c or d b be written by a single symbol a*, the first fraction becomes (c + x)(b + x)-cb b + c + x a+b V6 + VC-- 1 (y+s x+w) (z + x y+w) (x+y (x+7/ + s-rt) 3 (5 + c-a) (c + a ?))( + & c) Deal with the reciprocals of x, y, z, u; thus, y + z - x + u = y + z (x - it) = 4 fjayzxu(c + b a) . Hence, by symmetry, the numerator of the left hand frac tion becomes 64 ijabc y^z z x^w (b + c - a) (c + a - b) (a + b - c) . /I IN /I 1 Also, a; 4- V + 2 - u xy{ - + - + zid --- y x/ M / Hence the result. SECT. IV. SURDS. 3G. It has been already observed (Art. 23), that the root of any proposed quantity is found by dividing the exponent of the quantity by the index of the root ; and the rule has been illustrated by examples, in all of which, however, the quotient expressing the exponent of the result is a whole number; but there may be cases in which the quotient is a fraction. Thus, if the cube root of a 2 were required, it might be expressed, agreeably to the method of notation 2. already explained, either thus, /a 2 , or thus, a 3 . Quantities which have fractional exponents are called surds, or imperfect powers, and are said to be irrational, in opposition to others with integral exponents, which are called rational. Surds may be denoted by means of the radical sign, but it will be often more convenient to use the notation of fractional exponents. The following examples will show how they may be expressed either way. 1 i . 3 2. Jab The operations concerning surds depend on the following principles : 1. If the numerator and denominator of a fractional exponent be either both multiplied or both divided by the same quantity, the value of the power is the same. Thus, a a . 2. The product of like powers (integral or fractional) is the same power of the product. Thus, a a 6- = (a&)*. 37. I. Reduction of a Rational Quantity to the form oj a Surd of any given denomination. Ilule. Ileduce the exponent of the quantity to the form of a fraction of the same denomination as the given surd. Ex. Ileduce a 2 to the form of the cube root.
Here the exponent 2 must be reduced to the form of a
