FRACTIONS.] ALGEBRA 531 This remainder is to be divided by 19&, which being done, and the last divisor taken as a dividend as before, the rest of the operation will be as follows : a-b)3a 2 + Z> 3a 2 - Sab from which it appears that the common divisor sought is a b, and remarking that the quantities proposed have also a simple divisor b, the greatest common measure which is required will be b(a b). It will be seen that the examples we have given are not on numbers, but on algebraic quantities. In fact, the axiom and the demonstration founded on it apply, with some restrictions and modifications, to such quantities. The most important of the modifications is this : that the divisor, instead of being a whole number, is an expression of the form x + m, where m is of the nature of a numerical quantity, and does not depend on x. The application of this modified form of the axiom has a wide range in the higher analysis. We offer two addi tional examples for advanced students. Ex. 1. If ax 2 + bx + c, a x 2 + b x + c have a common divisor of the form x + m, prove that (a b - ab ) (b c - be } = (a c - ac f . Multiply the first expression by a, and the second by a, and subtract the products, the difference (a b ab )x + a c ac, is by the axiom divisible by x + m, L , ab ab Again, multiply the first expression by c, and the second by c, and subtract them; the difference (a c-ac )x- + (b c - bc )x is divisible .by x + m, . : x + -- ,isx + m. b c be a c ac Consequently, , . , , , ac-ac ab ab the condition required. Ex. 2. If ax 3 + 3bx 2 + d, bx 3 + 3dx + e, have a common divisor; then ) 2 = 0. Treating this question exactly as the last, viz., multiplying first by b and a, and then by e and d, and subtracting, it appears (if u be written instead of Id ae for brevity) that the two following expressions have a common divisor, 3Z/ 2 *; 2 - 3adx + u and ux 1 - 3bex + 3d 2 , whence, by the last example, the condition is (3beu - Oad 3 ) (3adu - We) = ( 2 - 9i 2 d 2 ) 2 , from which u divides out as a common factor, and the result reduces to that enunciated. 28. PROB. II. To Reduce a Fraction to its Lowest Terms. Rule. Divide both numerator and denominator by their greatest common measure, which may be found by Prob. I. a-x x 3 Ex. 1. Reduce - to its lowest terms. We have already found in the first example of Prob. I. that the greatest common measure of the numerator and denominator is a - x ; and dividing both by this quantity, we have a 3 -2a 2 z+az 2 a 2 -ax In like manner we find the common measure being b(a - b), as was shown in Ex ample 2, Problem I. Ex. 2. Reduce to its lowest terms. a 2 -(6-c) 2 a-b+e Ex. 3. To find the value of - x 2 -2x When * = 2 Here the substitution of ?, in place of x renders the numerator and denominator separately equal to 0. This shows (Art. 20) that x 2 is a divisor of each of them. We get, therefore, (x+l) 2 -3x-3 x+1 a; 2 2x- Ex. 4. Find the value of . > which when x = 2 becomes - x 2 x 3 -4x 2 + 2x + l when x = 1. -4x 3 + 6x 2 -4x+l Dividing numerator and denominator by # 1, the x 2 3x 1 result is -= -. > which, when 1 is written in place x 3 3x- + 3x 1 x 2 of x, becomes - > or infinity. 29. PROB. III. To Reduce a Mixed Quantity to an Improper Fraction. Rule. Multiply the integer by the denominator of the fraction, and to the product add the numerator; and the denominator being placed under this sum, the result will be the improper fraction required. x 2 Ex. 1. Reduce a-x + - to an improper fraction. a+x x 2 (a + x) (a x)+x 2 a 2 a-x + -, Ans. a+x a+x a+x 30. PROB. IV. To Reduce an Improper Fraction to a Whole or Mixed Number. Ride. Divide the numerator by the denominator for the integral part, and place the remainder, if any, over the denominator; it will be the mixed quantity re quired. Ex. 1. Reduce quantities.
First ax + 2x 2 a + x and to whole or mixed x-y x + a + x the answer. And - = x + y a whole quantity, which is the x y 31. PROB. V. To Reduce Fractions having different De nominators to others of the same value, which shall have a common Denominator. Rule. Multiply each numerator separately into all the denominators except its own for the new numerators^ and all the denominators together for the common de nominator. 2 /v2 Ex. 1. Reduce and to fractions of equal value, a-x a+x having a common denominator. ax(a + x) = a^x + ax 2 ) (a 2 - x-} (a-x) = a s - a?x - ax- + x 3 J (a x)(a + x) = a 2 # 2 , the common denominator. Hence ox a 2 x+ox 2 , a 2 x 2 a 3 a 2 x a.^ and a + x
a 2 -x z
