The terms of a proportion, if they be all of the same kind, are proportional by alternation. Let A: B = C :D : it is required to prove A : C = B : D. Take mA, MB any equimultiples of A and B, nC, nD any equimultiples of C and D. and Now, if mA be greater than nC, mB is greater than nD; if mA = nC, mB = nD; and if mA be less than nC, MB is less than nD; :. A: C = B : D. V. 14 V. Def. 5 PROPOSITION 17. THEOREM. The terms of a proportion are proportional by subtraction. Let A + B : B = C + D: D: it is required to prove A : B = C : D. Take mA and nB any multiples of A and B by the numbers m and n; and first let mA be greater than nB. To each of these unequals add mB; then mA + MB is greater than mB + nB. But mA + B = m(A + B), and mB + nB = (m + n)B ; ... m(A + B) is greater than (m + n)В. I. Ax. 4 V. 1, Cor. V. 2 Now because A + B : B = C + D : D, if m(A + B) be greater than (m + n)B, m(C + D) is greater than (m + n) D; V. Def. 5, Cor. or mC + mD is greater than mD + nD; that is, taking mD from both, mc is greater than nD. Hence when mA is greater than nB, mC is greater than nD. Similarly it may be proved that if mA = nB, mC = nD; and if mA be less than nB, mC is less than nD; .'. A : B = C : D. V. Def. 5 COR. The proposition is equivalent to the following: If A : B = C: D, then A – B : B = C − D : D. Hence also, on the same hypothesis, it may be proved that A - B: A = C − D : C; that A: A – B = C : C − D; and that B : A − B = D : C − D. [If it be thought desirable, any one of these changes on the proportion A: B = C : D may be denoted by the word subtraction.] The terms of a proportion are proportional by addition. Let A: B = C :D : it is required to prove A + B : B = C + D : D. Take m(A + B) and nB any multiples of A + B and B. First, let m be greater than n. Because A + B is greater than B; ... m(A + B) is greater than nB. Similarly m(C + D) is greater than nD; ... when m is greater than n, m(A + B) is greater than nB. and m(C + D) is greater than nD. Second, let m = n. In the same manner it may be proved that in this case m(A + B) is greater than nB, and m(C + D) greater than nD. Third, let m be less than n. Then m(A + B) may be greater than nB, or may be equal to it, or may be less than it. First, let m(A + B) be greater than nB; then mA + MB is greater than nВ. Take mB, which is less than nB, from both; V. 1 ... if mA is greater than (n - m)B, mC is greater V. 6 Нур. V. 6 V. 1 If therefore m(A + B) is greater than nB, m(C + D) is greater than nD. In the same manner it may be proved that, if m(A + B) = nB, m(C + D) = nD; if m(A + B) be less than nB, m(C + D) is less than nD. Hence A + B : B = C + D : D. V. Def. 5 COR.-Hence also, on the same hypothesis, it may be proved that A + B: A = C + D :C; that A: A + B = C : C + D; and that B: A + B = D : C + D. [If it be thought desirable, any one of these changes on the proportion A: B = C : D may be denoted by the word addition. The words addition and subtraction, as being more significant of the operations performed on the terms of the proportion, have been substituted for composition (componendo) and division (dividendo), which are the translations of the words (σύνθεσις, διαίρεσις) used by the Greek geometers.] PROPOSITION 19. THEOREM. If a whole magnitude be to a whole as a magnitude taken from the first is to a magnitude taken from the other, the remainder shall be to the remainder as the whole to the whole. Let A: B = C: D, and let C be less than A: it is required to prove A – C: B − D = A : B. .. .. .. ... Because A: B = C :D ; A : C = B : D, by alternation; A - C:C = B – D: D, by subtraction; Нур. V. 16 V. 17 V. 16 V. 11 If there be three magnitudes, and other three, which, taken two and two in direct order, have the same ratio; if the first be greater than the third, the fourth shall be greater than the sixth; and if equal, equal; and if less, less. Let A, B, C be three magnitudes, and D, E, F other three, such that A : B = D : E, and B : C = E : F: it is required to prove that if A be greater than C, D will be greater than F; if A = C, D will = F ; if A be less than C, D will be less than F. First, let A be greater than C; then A: B is greater than C: B. But A: B = D : E; V. 8 Нур. V. 13 then C is greater than A; and, as was shown in case first, C: B = F: E, and B : A = E : D. ... by case first, if C be greater than A, F is greater than D, that is, if A be less than C, D is less than F. If there be three magnitudes, and other three, which, taken two and two in transverse order, have the same ratio; if the first be greater than the third, the fourth shall be greater than the sixth ; and if equal, equal; and if less, less. Let A, B, C be three magnitudes, and D, E, F other three, such that A : B = E : F, and B : C = D : E : it is required to prove that if A be greater than C, D will be greater than F; if A = C, D will = F; if A be less than C, D will be less than F. First, let A be greater than C; then A: Bis greater than C: B. V. 8 |