If four magnitudes of the same kind be proportionals, they shall also be proportionals when taken alternately. Given AB:: C: D; to prove that B: D. alternately, A:C; and (Prep.) Take mA, mB any equimultiples of A and B, nC, nD any equimultiples of C and D. (Dem.) Then (V. 15) A:B:: mA: mB; now A: B:: C:D; therefore (V. 11) But C:D :: nC: nD (V. 15); therefore C:D:: mA: mB. mA: mB:: nC: nD (V. 11); wherefore if mAnC, mBn D if mA nC, mB=nD, or if mAnC, mB≤nD; therefore (V. Def. 10) A: C:: B: D. (V. 14); PROPOSITION XVII. THEOREM. If magnitudes, taken jointly, be proportionals, they shall also be proportionals when taken separately; that is, if the first, together with the second, have to the second the same ratio which the third, together with the fourth, has to the fourth, the first shall have to the second the same ratio which the third has to the fourth. Given A+B:B::C+D:D; to prove that A: B :: C: D. by division,' (Prep.) Take mA and nB any multiples of A and B, by the numbers m and n; and first, let mA7nB; to each of them add mB, then mA + mB7mB + nB. But mA + mBm(A+B) (V. 1), and mBnB = (m + n)B (V. 2); therefore m(A + B) 7 (m + n)B. (Dem.) And because A + B : B :: C + D :D, if m(A + B) (m+n).B, m(C + D) 7 (m + n)D, · or mCmDmD +nD; that is, taking mD from both, mCnD. Therefore, when mA is greater than nB, mC is greater than nD. In like manner it is demonstrated that if mA = and if mA < nB, mC ≤ nD; nB, mC = nD; therefore A: B:: C: D (V. Def. 10). PROPOSITION XVIII. THEOREM. : The terms of an analogy are proportional by composition. (Prep.) Take m(A + B) and nB any multiples whatever of A+B and B; and first, let m be greater than n. (Dem.) Then, because A+B is also greater than B, m(A + B) n.B. In this case, there For the same reason, m(C + D) ≈ nD. fore, that is, when m>n, m(A + B) is greater than nB, m(CD) is greater than nD. be proved that when m=n, and And in the same manner it may m(A+B) is greater than nB, and m(C + D) greater than nD. Next, let mn, or nm, than nB, or may be equal to it, m(A+B) be greater than nB; take mB, which is less than nB, mB, or mA (n − m)B (V. 6). > (n − m)D; =nD — mD, then m(A+B) may be greater or may be less; first, let then also, mA + mBnB; from both, and mA7nBBut if mA (n − m)B, mC Now, (n m)D mD, and adding mU that is (V. 1), m(C + D) 7nD. because A: B:: C: D. therefore mCnD to both mC+mDnD, If therefore m (A + B) 7nB, m(C + D) nD. In the same manner it will be proved that if m(A + B) = nB, m(C + D) =nD; and if m(A + B) ≤nB, <nD; therefore A+B:B::C+D: D. m(C + D) PROPOSITION XIX. 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. Given A: B:: C: D, and C less than A; to prove that C:B-D: A: B. A (Dem.) Because A: B :: C:D, alternately (V. 16), A: C:: B:D; and therefore by division (V. 17), A C:C:: B - D Wherefore, again, alternately, A C:B - D::C:D; but A:B:: C:D, therefore (V. 11) A - C: B-D :: :D. A: B. COR. AC: B-D::C: D. PROPOSITION D. THEOREM. The terms of an analogy are proportional by conversion. A:AB:: C: C - D. (Dem.) For since A: B :: C:D, by division (V. 17), A — B B:: C-D: D, and inversely (V. a), B: A-B:: D:C therefore, by composition (V. 18), A: A-B:: C : C – D. If there be three magnitudes, and other three, which, taken two and two in 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. Given three magnitudes, A, B, and C, and other three D, E, and F; such that A: B::D:E; and B: C :: E:F; to prove that if A = DZF. if AC, DF; C, D F; and if AC, A, B, C, D, E, F. (Dem.) First, let AC; then A:B>C:B (V. 8). But A: B:: D: E, therefore also D:EC:B (V. 13). Now BC: E:F, and inversely (V. A), C: B::F:E; shewn that D:EC: B, consequently DF (V. 10). and it has been but A: B there Next, let A = C; then A: B:: C: B (V. 7), :: D: E, therefore, C: B :: D:E; but C: B:: F: E, fore, D: E:: F: E (V. 11), hence D = F (V. 9). Lastly, let and, as was shewn in the first case, C:B:: F:E, and BA:: E:D; therefore, by the first case, if CA, FD; that is, if A 2 C, D ≤ F. If there be three magnitudes, and other three, which have the same ratio taken two and two, but in a cross order; if the first magnitude be greater than the third, the fourth shall be greater than the sixth; and if equal, equal; and if less, less. Given three magnitudes, A, B, C, and other three, D, E, and F, such that A: B :: E:F, and B: C: D: E; to prove that if AC, DF; if A= C, D F, C, D ≤ F. (Dem.) First, let AC. Then A B C : B (V. 8); but A: B:: E: F, therefore, E:FC:B (V. 13). Now, B: C:: D:E, and in versely, C: B:: E:D; therefore, E: F Next, let A= C. Then (V. 7) A:B:: C:B; but A:B:: E:F; therefore, C: B:: EF (V. 11); but B: C:: D: E, and inversely, C: B::E: D; A, B, C, D, E, F. therefore, EF:: E Lastly, let A≤C. Then A: BC: B, therefore, EFC: B, inversely (V. A), C: B:: E:D, but A:B:: E:F, and but B: C:: D: E, therefore DF (V. 10). PROPOSITION XXII. THEOREM. If there be any number of magnitudes, and as many others, which, taken two and two in order, have the same ratio; the first shall have to the last of the first magnitudes the same ratio which the first of the others has to the last. First, let there be given three and other three, D, E, and F, order, have the same ratio; B:C::E:F '; to prove that magnitudes, A, B, and C, which, taken two and two in namely, A: B:: D:E, A:C: D:F. and (Prep.) Take of A and D any equimultiples whatever, mA, mD; of B and E any whatever, nB, nE; and of C and F any whatever, qC, qF. (Dem.) Because A: B::D: E, mA :nB:: mD: nE (V. 4); and for the same A, B, C, reason, nB: qC :: nE: qF. Therefore D, E, mA, nB, qC, qC, equal to it, or less, mD is greater mD, nE, qF. (V. 20), according as mA is greater than than qF, equal to it, or less; but mA, mD are any equimultiples of A and D; and qC, qF are any equimultiples of C and F; therefore (V. Def. 10) A:C:: D:F. Again, let there be given four magnitudes, and other four, which, taken two and two, have the same ratio; namely, A:B::E:F; B:C::F:G; C:D::G:H; to prove that A:D:: E: H. (Dem.) For since A, B, C are three magnitudes, and E, F, G other three, which, taken two and two, have the same ratio, by the foregoing case A:C::E: G. And because also C:D :: G: H, by that same case, A:D :: E: H. In the same manner the demonstration is extended to any number of magnitudes. A, B, C, D, F, G, H. E, PROPOSITION XXIII. THEOREM. If there be any number of magnitudes, and as many others, which, taken two and two in a cross order, have the same ratio; the first shall have to the last of the first magnitudes the same ratio which the first of the others has to the last. First, let there be given three magnitudes, A, B, C, and other three, D, E, and F, which, taken two and two in a cross order, have the same ratio; namely, A: B::E: F, :: D:E; to prove that A:C:: D:F. and B: C (Prep.) Take of A, B, and D any equimultiples mA, mB, mD; and of C, E, F any equimultiples nC, nE, nF. and because also A: B:: therefore, mA: mB:: (Dem.) Because A:B:: E: F, C:: D: E, mB nC :: mD: nE any equimultiples of A and D, if and Next, let there be given four magnitudes, A, B, C, and D, and other four, E, F, G, and H, which, taken two and two in a cross order, have the same ratio; A:B::G: H, C: D:: E: F, For, since A, B, C are three magni B:C::F:G, namely, then A: D:: E: H. A, B, C, D, H. But C: D and tudes, and F, G, H other three, which, taken two and two in a cross order, have the same ratio, by the first case A: C:: F: H. :: E:F; therefore A, C, and D are three magnitudes, E, F, and H other three, which, taken two and two in a cross order, are proportional, hence, by the first case, A: D :: E: H. In the same manner the demonstration may be extended to any number of magnitudes. If the first has to the second the same ratio which the third has to the fourth; and the fifth to the second the same ratio which the sixth has to the fourth; the first and fifth together shall have to the second the same ratio which the third and sixth together have to the fourth. Given A: B:: C: D, and also E:B:: F: D, to prove that, by inversion, B: E:: D: F. But by hypothesis, A:B:: C:D; therefore, by equality (V. 22), A:E::C:F, and by composition (V. 18), A+E:E::C+ Now, again by hypothesis, E: B:: F:D; therefore, by equality, A+E:B::C+F: D. Ratios which are compounded of the same ratios are the same with one another. |