tiple of B that F is of D (b); that is, E and F are equimultiples of B and D; but G and H are equimultiples of B and D; therefore, if E be a greater multiple of B than G is of B, F is a greater multiple of D than H is of D; that is, if E be greater than G, F is greater than H; in like manner, if E be equal to G, or less than it, F may be shown to be equal to H, or less than it; but E, F are any equimultiples whatever of A, C; and G, H any equimultiples whatever of B, D; therefore A is to B as C is to D (c). Next, let the first A be the same submultiple of the second B that the third C is of the fourth D; A shall be to B as C is to D. For since A is the same submultiple of B that C is of D, therefore B is the same multiple of A that D is of C; wherefore, by the preceding case, B is to A as D is to C; and therefore inversely, A is to B as C is to D (d). SCHOLIUM. The foregoing proposition, expressed algebraically, is as follows: THEOREM.-If A = m a, and B = m. b, or if b and B = then A: a :: B: b. m For, in the first case, PROPOSITION D. THEOREM.-If the first be to the second as the third to the fourth, and if the first be a multiple or submultiple of the second, the third is the same multiple or submultiple of the fourth. Let A be to B as C is to D; and first let A be a multiple of B, then C shall be the same multiple of D. DEMONSTRATION. Take E equal to A, and whatever multiple A or E is of B, make F the same multiple of D; then, because A is to B as C is to D (a); and of B the second, and D the fourth, equimultiples have been taken, E and F; therefore A is to E as C is to F (b); but A is equal to E, therefore C is equal to F (c); and Fis the same multiple of D that A is of B: therefore C is the same multiple of D that A is of B. Next, let A be a submultiple of B; then C shall be the same submultiple of D. Because A is to B as C is to D (a); then inversely, B is to A as D is to C (d); but A is a submultiple of B, that is, B is a multiple of A; therefore, by the preceding case, D is the same multiple of C; that is, C is the same submultiple of D that A is of B. SCHOLIUM. This proposition is the inverse of the preceding; it may be algebraically expressed as follows: THEOREM. If A: a :: B: b, and A = either m.a a b Or then B = m. b, or -. m For m =A 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 Ĉ is to B (c). E Hypoth. (b) V. Ax. 1. V. Def. 5. 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. which is not the greater of the two AC, Then, because L is the multiple of D which is the first that becomes greater than FG, the next preceding E F. G C E F. G A C B Fig. 1. LKH D (a) V. 1. E Fig. 3. Fig. 2. F. C 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, 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 |