6. If GK be greater than HX, then LN is greater than MP; and if equal, equal; and if less, less (V. Def. 5) ; · but if GH be greater than KX, then, by adding the common part HK to both, GK is greater than HX (I. Ax. 4); wherefore also 7. LN is greater than MP; and by taking away MN from both, LM is greater than NP (I. Ax. 5) ; therefore 8. If GH be greater than KX, LM is greater than NP. In like manner it may be demonstrated, that if GH be equal to KX, LM is equal to NP; and if less, less; but GH, LM are any equimultiples whatever of AE, CF (constr.), and KX, NP are any whatever of EB, FD; therefore 9. As AE is to EB, so is CF to FD (V. Def. 5). PROPOSITION 18.-Theorem. If magnitudes, taken separately, be proportionals, they shall also be proportionals when taken jointly; that is, if the first be to the second, as the third to the fourth, the first and second together shall be to the second, as the third and fourth together to the fourth. Let AE, EB, CF, FD be proportionals; that is, as AE to EB, so let CF be to FD. A Then they shall also be proportionals when taken jointly; that is, as AB to BE, so shall CD be to DF. G L E B N P M K H CFD Construction. Take of AB, BE, CD, DF any equimultiples whatever GH, HK, LM, MN; and again, of BE, DF take any equimultiples whatever KO, NP. Demonstration. Because KO, NP are equimultiples of BE, DF; and that KH, NM are likewise equimultiples of BE, DF; therefore if KO, the multiple of BE, be greater than KH, which is a multiple of the same BE, then G A L 1. NP, the multiple of DF, is also greater than NM, the multiple of the same DF; CFD and if KO be equal to KH, NP is equal to NM; and if less, less. First. Let KO be not greater than KH; therefore A 2. NP is not greater than NM; and because GH, HK, are equimultiples of AB, BE, and that AB is greater than BE, therefore 3. GH is greater than HK (V. Ax. 3); but KO is not greater than KH; therefore L 4. GH is greater than KO. In like manner it may be shown, that LM is greater than NP. Therefore, if KO be not greater than KH, then 5. GH, the multiple of AB, is always greater than KO, the multiple of BE; and likewise LM, the multiple of CD, is greater than NP, the multiple of DF. Next. Let KO be greater than KH; therefore, as has been shown, 6. NP is greater than NM. G E T K N K T B H H CFD And because the whole GH is the same multiple of the whole AB, that HK is of BE, therefore 7. The remainder GK is the same multiple of the remainder AE, that GH is of AB (V. 5), which is the same that LM is of CD. In like manner, because LM is the same multiple of CD, that MN is of DE, therefore 8. The remainder LN is the same multiple of the remainder CF, that the whole LM is of the whole CD (V. 5) ; but it was shown that LM is the same multiple of CD, that GK is of AE; therefore GK is the same multiple of AE, that LN is of CF; that is, 9. GK, LN are equimultiples of AE, CF. And because KO, NP are equimultiples of BE, DF, therefore if from KO, NP there be taken KÈ, NM, which are likewise equimultiples of BE, DF, 10. The remainders HO, MP are either equal to BE, DF, or equimultiples of them (V. 6). First, let HO, MP be equal to BE, DF; then because AE is to EB, as CF to FD (hyp.), and that GK, LN are equimultiples of AE, CF; therefore 11. GK is to EB, as LN to FD (V. 4, Cor.) ; but HO is equal to EB, and MP to FD; wherefore GK is to HO, as LN to MP; therefore 12. If GK be greater than HO, LN is greater than MP (V. A); and if equal, equal; and if less, less. A But let HO, MP be equimultiples of EB, FD. Then, because AE is to EB, as CF to FD (hyp.), G K T E B M H P and that of AE, CF are taken equimultiples GK, LN; and of EB, FD, the equimultiples HO, MP; 13. If GK be greater than HO, LN is greater than MP; and if equal, equal; and if less, less (▼. Def. 5) ; which was likewise shown in the preceding case. But if GH be greater than KO, taking KH from both, GK is greater than HO; (I. Ax. 5); wherefore also 14. LN is greater than MP; and consequently adding NM to both, LM is greater than NP (L. Ax. 4); therefore 15. If GH be greater than KO, LM is greater than NP. In like manner it may be shown, that if GH be equal to KO, LM is equal to NP; and if less, less. And in the case in which KO is not greater than KH, it has been shown that GH is always greater than KO, and likewise LM greater than NP; but GH, LM are any equimultiples whatever of AB, CD (constr.), and KO, NP are any whatever of BE, DF; therefore 16. As AB is to BE, so is CD to DF (V. Def. 5). 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 the whole AB be to the whole CD, as AE a magnitude taken from AB, is to CF a magnitude taken from CD. Then the remainder EB shall be to the remainder FD, as the whole AB to the whole CD. A с EF B Demonstration. Because AB is to CD, as AE to CF; therefore, alternately, 1. BA is to AE, as DC to CF (V. 16); and because if magnitudes taken jointly be proportionals, they are also proportionals when taken separately (V. 17); therefore, as BE is to EA, so is DF to FC; and alternately, 2. As BE is to DF, so is EA to FC; but, as AE to CF, so, by the hypothesis, is AB to CD; therefore also 3. BE the remainder is to the remainder DF, as the whole AB to the whole CD (V. 11). Wherefore, if the whole, &c. Q.E.D. Cor. If the whole be to the whole, as a magnitude taken from the first is to a magnitude taken from the other; the remainder shall likewise be to the remainder, as the magnitude taken from the first to that taken from the other. The demonstration is contained in the preceding. PROPOSITION E.-Theorem. If four magnitudes be proportionals, they are also proportionals by conversion; that is, the first is to its excess above the second, as the third to its excess above the fourth. Let AB be to BE, as CD to DF. Then BA shall be to AE, as DC to CF. A E с F B A Demonstration. Because AB is to BE, as CD to DF, therefore by division 1. AE is to EB, as CF to FD (V. 17) ; and by inversion, BE is to EA, as DF is to CF (V. B); wherefore, by composition, 2. BA is to AE, as DC is to CF (V. 18). PROPOSITION 20.-Theorem. If there be three magnitudes, and other three, which, taken two and two, have the same ratio; then 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, which taken two and two have the same ratio, viz., as A is to B, so is D to E; and as B to C, so is E to F. Then if A be greater than C, D shall be greater than F; and if A be equal to C, D shall be equal to F; and if A be less than C, D shall be less than F. B с D E F First. Let A be greater than C. Then D shall be greater than F. Demonstration. Because A is greater than C, and B is any other magnitude, and that the greater has to the same magnitude a greater ratio than the less has to it (V. 8); therefore 1. A has to B a greater ratio than C has to B ; but as D is to E, so is A to B (hyp.); therefore 2. D has to E a greater ratio than C to B (V. 13) and because B is to C, as E to F, by inversion, 3. C is to B, as F is to E (V. B); and D was shown to have to Ę a greater ratio than C to B; therefore |