Let the two magnitudes AB, CD be equimultiples of the two E, F, and let AG, CH taken from the first two be equimultiples of the same E, F. Then the remainders GB, HD shall be either equal to E, F, or equimultiples of them. A G K с H D First. Let GB be equal to E; HD shall be equal to F. B Demonstration. Because AG is the same multiple of E, that CH is of F (hyp.), and that GB is equal to E, and CK to F; therefore 1. AB is the same multiple of E, that KH is of F; 2. KH is equal to CD (V. Ax. 1) ; take away the common magnitude CH; then but AB, by the hypothesis, is the same multiple of E, that CD is of F; therefore KH is the same multiple of F, that CD is of F; wherefore K F G 3. The remainder KC is equal to the remainder HD; but KC is equal to F (constr.); therefore 4. HD is equal to F. Next. Let GB be a multiple of E. Then HD shall be the same multiple of F. A E B take away CH from both; therefore F с H D Construction. Make CK the same multiple of F, that GB is of E. Demonstration. Because AG is the same multiple of E, that CH is of F (hyp.), and GB the same multiple of E, that CK is of F, therefore E 1. AB is the same multiple of E, that KH is of F (V. 2) ; but AB is the same multiple of E, that CD is of F (hyp.); therefore KH is the same multiple of F, that CD is of F; wherefore 2. KH is equal to CD (V. Ax. 1) ' 3. The remainder KC is equal to the remainder HD ; and because GB is the same multiple of E, that KC is of F (constr.), and that KC is equal to HD; therefore PROPOSITION A.-Theorem. If the first of four magnitudes has the same ratio to the second, which the third has to the fourth; then, if the first be greater than the second, the third is also greater than the fourth; and if equal, equal; if less, less. therefore 4. HD is the same multiple of F, that GB is of E. Take any equimultiples of each of them, as the doubles of each; then if the double of the first be greater than the double of the second, 1. The double of the third is greater than the double of the fourth (V. Def. 5); but if the first be greater than the second, the double of the first is greater than the double of the second; wherefore also 2. The double of the third is greater than the double of the fourth; 3. The third is greater than the fourth; in like manner, if the first be equal to the second, or less than it, 4. The third can be proved to be equal to the fourth, or less than it. Therefore, if the first, &c. Q.E.D. G PROPOSITION B.-Theorem. If four magnitudes are proportionals, they are proportionals also when taken inversely. Let A be to B, as C is to D. Then also inversely, B shall be to A, as D to C. A с D F B H Construction. Take of B and D any equimultiples whatever E and F; and of A and C any equimultiples whatever G and H. Demonstration. First, let E be greater than G, then 1. G is less than E; and because A is to B, as C is to D (hyp.), and of A and C, the first and third, G and H are equimultiples; and of B and D, the second and fourth, E and F are equimultiples; and that G is less than E, therefore A G B D E F 2. H is less than F (V. Def. 5); that is, F is greater than H; с H if, therefore, E be greater than G, A E 3. F is greater than H; in like manner, if E be equal to G, F may be shown to be equal to H; and if less, less; but E, F, are any equimultiples whatever of B and D (constr.), and G, H any whatever of A and C; therefore 4. As B is to A, so is D to C (V. Def. 5). Therefore, if four magnitudes, &c. Q.ED. PROPOSITION C.-Theorem. If the first be the same multiple of the second, or the same part of it, that the third is of the fourth; the first is to the second, as the third is to the fourth. First. Let the first A be the same multiple of the second B, that the third C is of the fourth D. Then A shall be to B as C is to D. B G Construction. Take of A and C any equimultiples whatever E and F; and of B and D any equimultiples whatever G and H. F H Demonstration. Because A is the same multiple of B that C is of D (hyp.); and that E is the same multiple of A, that F is of C (constr.); therefore 1. E is the same multiple of B, that F is of D (V. 3); that is, E and F are equimultiples of B and D ; but G and H are equimultiples of B and D (constr.) ; therefore 2. 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, Fis greater than Н; 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 equimultiples, any whatever, of A, C (constr.); and G, H any equimultiples whatever of B, D; therefore 3. A is to B, as C is to D (V. Def. 5). Next. Let the first A be the same part of the second B, that the third C is of the fourth D. Then A shall be to B, as C is to D. A B D For since A is the same part 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, 1. B is to A, as D is to C; and therefore inversely, 2. A is to B, as C is to D (V. B): Therefore, if the first be the same multiple, &c. Q.E.D. PROPOSITION D.-Theorem. If the first be to the second as the third to the fourth, and if the first be a multiple, or a part of the second; the third is the same multiple, or the same part of the fourth. Let A be to B as C is to D, and A B с D F E Construction. Take E equal to 4, and whatever multiple A or E is of B, make F the same multiple of D. Demonstration. Because A is to B, as C is to D (hyp.), and of B the second, and D the fourth, equimultiples have been taken, E and F, therefore 1. A is to E, as C to F (V. 4, Cor.); but A is equal to E (constr.); therefore 2. C is equal to F (V. A); and Fis the same multiple of D, that A is of B (constr.); therefore 3. C is the same multiple of D, that A is of B. Next. Let A the first be a part of B the second. Then C the third shall be the same part of D the fourth. A B D Demonstration. Because A is to B, as C is to D (hyp.); then, inversely, 1. B is to A, as D to C (V. B); but A is a part of B, therefore B is a multiple of A (hyp.); therefore, by the preceding case, D is the same multiple of C; that is, 2. C is the same part of D, that A is of B. PROPOSITION 7.-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. Then 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. A B C E. F Construction. Take of A and B any equimultiples whatever D and E, and of C any multiple whatever F. D Demonstration. Because D is the same multiple of A, that E is of B (constr.), and that A is equal to B (hyp.), therefore D is equal to E (V. Ax. 1); therefore 1. If D be greater than F, E is greater than F; and if equal, equal; and if less, less; but D, E are any equimultiples of A, B (constr.), and F is any multiple of C; therefore 2. As A is to C, so is B to C (V. Def. 5). Likewise C shall have the same ratio to A, that it has to B. For having made the same construction, D may in like manner be shown to be equal to E; therefore 1. If F be greater than D, it is likewise greater than E; and if equal, equal; and if less, less; but F is any multiple whatever of C; and D, E are any equimultiples whatever of A, B; therefore 2. C is to A, as C is to B (V. Def. 5). Therefore, equal magnitudes, &c. Q.E.D. |