Powers changed from one Term to the other of a Fraction. Now a fraction is multiplied either by multiplying its numerator or by dividing its denominator; and it is divided either by dividing its numerator or by multiplying its denominator. Hence, It has the same effect to multiply one of the terms of a fraction by a quantity, which it has to multiply the other term by the reciprocal of the quantity. 82. Corollary. If either term of a fraction is multiplied by the power of a quantity, this factor may be suppressed, and introduced as a factor into the other term with the sign of the power reversed. By this means, a fraction can be freed from negative exponents. x8+3x3 nents. Ans. x4+1 Product of Means equals that of Extremes. 5. Free the fraction 1+x-2+y-2 from negative ex ponents. 1+x-2-y-2 84. The preceding rules for fractions may all be applied to ratios by substituting the term antecedent for numerator, and consequent for denominator. SECTION IV. 85. A proportion is the equation formed of two equal ratios. Thus, if the two ratios A: B and C: Dare equal, the equation A: B = C:D is a proportion; and it may also be written The first and last terms of a proportion are called its extremes; and the second and third its means. Thus, A and D are the extremes of this proportion, and B and C its means. 86. If the ratios of the preceding proportion are reduced to a common consequent, in the same way in which fractions are, by art. 67, reduced to a common denominator, we have AxB:BxD = B x C : BXD; Product of Means equals that of Extremes. that is, A × D and BXC have the same ratio to B x D, and are consequently equal, that is, AXD=BX C, or the product of the means of a proportion is equal to the product of its extremes. This proposition is called the test of proportions, that is, if four quantities are such that the product of the first and last of them is equal to the product of the second and third, these four quantities form a proportion. Demonstration. Let A, B, C, D be four quantities such that AxD = B x C. We have, by dividing B × D, AxD:BxD = B x C : B × D, or, by reducing these ratios to lower terms, as in art. 40, A: B = C:D; that is, A, B, C, D form a proportion. 87. Corollary. If A, B, C, D form a proportion, we obtain from the preceding test A:CB: D B: A = D:C B:DA:C D: C = B : A, &c.; that is, the terms of a proportion may be transposed in any way which is consistent with the application of the test. To find the Fourth Term of a Proportion. 88. Problem. Given three terms of a proportion, to find the fourth. Solution. The following solution is immediately obtained from the test. When the required term is an extreme, divide the product of the means by the given extreme, and the quotient is the required extreme. When the required term is a mean, divide the product of the extremes by the given mean, and the quotient is the required mean. 89. EXAMPLES. 1. Given the three first terms of a proportion respectively A, B, C; find the fourth. BC Ans. A 2. Given the three first terms of a proportion respectively 2ab2, 3a2b, 663; find the fourth. Ans. 9ab2. 3. Given the three first terms of a proportion respectively am, a", ar; find the fourth. Ans. an+r-m. 4. Given the first term of a proportion a3 62, the second 3 as b3, the fourth 7 ab; find the third. Ans. 7a-4. 5. Given the first term of a proportion 6 am-26, the third 15 a3 65, the fourth 40 a-(m-1); find the second. Ans. 16 a-4-4. 6. Given the three last terms of a proportion respectively a2-b2, 2 (a+b), a2+2ab+b2; find the first. Ans. 2 (a-b). 90. When both the means of a proportion are the same quantity, this common mean is called the mean proportional between the extremes. Mean Proportional. Continued Proportion. Thus, when A: B = B : C, B is a mean proportional between A and C. 91. If the test is applied to the preceding proportion it that is, the mean proportional between two quantities is the square root of their product. 92. A succession of several equal ratios is called a continued proportion. Thus, A:BC: D E: F, &c. is a continued proportion. 93. Theorem. The sum of any number of antecedents in a continued proportion is to the sum of the corresponding consequents, as one antecedent is to its consequent. Demonstration. Denote the value of each of the ratios in the continued proportion of the preceding article by M, and we have whence M = A : B = C : D = E : F, &c.; A=BxM C=DxM E = F × M, &c.; and the sum of these equations is A+ C + E + &c. = (B + D + F + &c.) × M; |