PROPOSITION VII. THEOREM.-Equal magnitudes have the same ratio to the same magnitude; and the same has the same ratio to equal magnitudes. Let A and B be equal magnitudes, and C any other. A and B shall each of them have the same ratio to C: and C shall have the same ratio to each of the magnitudes A and B. DEMONSTRATION. Take of A and B any equimultiples whatever D and E, and of C any multiple whatever F: then, because D is the same multiple of A that E is of B, and that A is equal to B (a): therefore D is equal to E (b) therefore if D be greater than F, E is greater than F; and if equal, equal; if less, less; but D, E, are any equimultiples of A, B, and F is any multiple of C; therefore A is to C as B is to C (c). Likewise C has the same ratio to A, that it has to B, or C is to A as C is to B. For, having made the same construction, D may in like manner be shown to be equal to E; therefore if F be greater than D, it is likewise greater than E; and if equal, equal; if less, less; but F is any multiple whatever of C, and D, E, are any equimultiples whatever of A, B; therefore C is to A as C is to B (c). E (a) Hypoth. SCHOLIUM. This proposition, algebraically expressed, is as follows: THEOREM. If A = B, and C be any third quantity, A: C:: B: C, and CA: C: B. Since A = B, PROPOSITION VIII. THEOREM.-If two magnitudes are unequal, the greater has a greater ratio to any other magnitude than the less has; and the same magnitude has a greater ratio to the less than it has to the greater. Let AB, BC be two unequal magnitudes, of which AB is the greater, and let D be any other magnitude. AB shall have a greater ratio to D than BC has to D: and D shall have a greater ratio to BC than it has to AB. DEMONSTRATION. If the magnitude which is not the greater of the two AC, CB be not less than D, take EF, FG, the doubles of AC, CB (as in Fig. 1). But if that which is not the greater of the two AC,CB be less than D (as in Figs. 2 and 3), this magnitude can be multiplied, so as to become greater than D, whether it be AC or CB. Let it be multiplied until it become greater than D, and let the other be multiplied as often; and let EF be the multiple thus taken of AC, and FG the same multiple of CB: therefore EF and FG are each of them greater than D: and in every one of the cases, take H the double of D, K its triple, and so on, till the F multiple of D be that which first becomes greater than FG: let L be that multiple of D which is first greater than FG, and K the multiple of D which is next less than L. Then, because L is the multiple of D which is the first that becomes greater than FG, the next preceding E G C E Fig. 1. F. Fig. 2. F. C B B L K HD multiple K is not greater than FG; that is, FG is not less than K: and since EF is the same multiple of AC that FG is of CB; therefore FG is the same multiple of CB that EG is of AB (a): that is, EG and FG are equimultiples of AB and CB: and since it was shown that FG is not less than K, and by the construction EF is greater than D; therefore the whole EG is greater than K and D together but K together with D is equal to L; therefore EG is greater than L: but FG is not greater than L: and EG, FG were proved to be equimultiples of AB, BC; and L is a multiple of D; therefore AB has to D a greater ratio than BC has to D (b). Also D shall have to BC a greater ratio than it has to AB. For having made the same construction, it may be shown, in like manner, that L is greater than FG, but that it is not greater than EG; and L is a multiple of D; and FG, and EG were proved to be equimultiples of CB, AB; therefore D has to CB a greater ratio than it has to AB (6). SCHOLIUM. This proposition may be algebraically expressed as follows:THEOREM. If A is B, then A: Cis> B: C, and C: B is > C: A. For if A is B, is> Bi and therefore C B is > C: A. PROPOSITION IX. THEOREM.-If magnitudes have the same ratio to the same magnitude, they are equal to one another: and those to which the same magnitude has the same ratio are equal to one another. Let A, B have each of them the same ratio to C; then A shall be equal to B. DEMONSTRATION. For, if they are not equal, one of them must be greater than the other: let A be the greater: then, by what was shown in the preceding proposition, there are some equimultiples of A and B, and some multiple of C such that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than that of C. Let these multiples be taken; and let D, E be the multiples of A, B, and F the multiple of C, such that D may be greater than F, but E not greater than F: then, because Á is to C as B is to C, and of A, B are taken equimultiples, D, E, and of C is taken a multiple F; and that D is greater than F; therefore E is also greater than F (a): but E is not greater than F; which is impossible: therefore A and B are not unequal; that is, they are equal. Next, let C have the same ratio to each of the magnitudes A and B; then A shall be equal to B. B D (a) V. Def. 5. For, if they are not equal, one of them must be greater than the other let A be the greater: therefore, as was shown in the eighth proposition, there is some multiple F of C, and some equimultiples E and D of B and A, such that F is greater than E, but not greater than D and because C is to B as C is to A, and that F the multiple of the first is greater than E the multiple of the second (a); therefore F the multiple of the third is greater than D the multiple of the fourth: but F is not greater than D; which is impossible. Therefore A is equal to B. SCHOLIUM. The foregoing proposition, algebraically expressed, is as follows: THEOREM. If A: B:: C: B, then A = C; and if B: A :: B: C, then For if A: B:: C: B also A = C. PROPOSITION X. THEOREM.—That magnitude which has a greater ratio than another has to the same magnitude, is the greater of the two; and that magnitude to which the same has a greater ratio than it has to another magnitude, is the lesser of the two. Let A have to C a greater ratio than B has to C; then A shall be greater than B. DEMONSTRATION. For, because A has to C a greater ratio than B has to C, there are some equimultiples of A and B, and some multiple of C, such that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than it (a); let them be taken; and let D, E be the equimultiples of A, B, and F the multiple of C, such that D is greater than F; but E is not greater than F, therefore D is greater than E: and because D and E are equimultiples of A and B, and D is greater than E, therefore A is greater than B (b). Next, let C have a greater ratio to B than it has to A; then B shall be less than A. For there is some multiple F of C, and some equimultiples E and D of B and A, such that F is greater than E, but not greater than D (a): therefore E is less than D: and because E and D are equimultiples of B and A, and that E is less than D, therefore B is less than A (b). SCHOLIUM. This proposition may be algebraically expressed as follows:— therefore CB is > AB; and dividing by B, C is A, or A is > C. |