1 ८ Coroll. Hereby it is manifest that a space commenfurable to a medial-space, is also medial. Lemma. Plate IV. Fig. 21. To find out two medial right-lines A, B, commenfurable in length and also two A C commenfurable only in power. (a) Let A be any μ, (b) take BIA, and (c) C A, (d) and 'tis evident the thing is is done. PROP. XXV. Fig. 29. A rectangle DB contained under DC, CB mcdial rightlines commenfurable in length is medial. Upon DC describe the square DA. Because AC: (DC) CB (a) : : DA: DB, and DCCB; (6) shall DADB. (c) therefore DB is μν. Which was to be demonstrated. PROP. XXVI. Fig. 30. A rectangle AC comprehended under medial right-lines AB, BC commenfurable only in power is either rational or medial. Upon the lines AB, BC, (a) defcribe the squares AD, CE; and upon FG (b) make the rectangles FH = AD, (b) and IK AC (b) and LM=CE. The squares AD, CE, that is, the rectangles FH, LM, (c) are μα and 1.. therefore GH, KM, having the fame proportion (d) are ș, (e) and (f) therefore GHX KM is fr. But because AD, AC, CE, that is FH, IK, LM, (g) are; (b) and so GH, HK, KM also; (k) thence HKq GH x KM. (1) therefore HK is 6, or, or IH (GF ;) if (m) then the rectangle IK or AC is fr, but if (n) then AC is μν. Which was to be dem. If A and E are Lemma. Fig. 31. only, Then first, shall Aq, Eq, Aq + Eq, Aq-Eq (a). And secondly Aq, Eq, Aq + Eq, Aq-EqAE and 2 AE. For A: E (6): : Aq. AE (6): : ΑΕ: Eq. therefore seeing A (c) LE, (d) shall Aq AE, (e) and 2 AE. Alfo Eq (d) and 2 AE. wherefore because Aq + Eq and Aq-Eq Aq and Eq; (f) therefore Eq, (f) and Aq-Eq be AE, and 2 AE. Hence AE, (e) shall Aq+ a lem. 21. 10 and 13. 6. b 2. lem. 10. 10. c 3. lem 10. 10. d conft. and 24. 10. C 24. 10. a 46. 1. 24. 10. d 23. 10. e 10. 10. f 20. 10. gsch.22.6. h 1.6. m 20.10. η 22. 1Ο. a byp. and 16. 10. b 1. 2. c hyp. d 10. 10. e 14. 10. f 14.10. g 14. 10. and 7. 10. h cor. 7. 2. a cor. 16.6. b hyp. € 3. 10. € 1. 10. f 13. 10. g lem. 26. ΙΟ. h fch. 12. 10. a byp. 10. c fch. 12. 10. a fch. 12. 10. 6 16. 10. cfch. 12. 10. a lem. 21.10. b 13.6. C 12. 6. d 22. 10. e conftr. f 10. 10. g 24. 10. h 17.6. h fch. 12. 10. Hence also thirdly, Aq, Eq, Aq + Eq, Aq - Eq, 2 PROP. XXVII. Plate IV. Fig. 32. A medial rectangle AB exceedeth not a medial rectangle AC by a rational rectangle DB. EF, Upon EF (a) make EG-AB, (a) and EH=AC. The rectangles AB, AC, i, e. EG, EH, (6) are μα; (c) therefore FG and FH are EF. Whence, if KG, (d) i. e. DB be pr, (e) then shall HG be HK; (f) wherefore HGFH. (g) and confequently FGqL FHq. But FH is f. (6) therefore is FG p. but FG was FG. f before. Which is contradictory. Schol. Fig. 29, 33 1. A rational rectangle A E exceeds a rational rectangle AD by a rational rectangle CE. For ΑΕ (α) τι AD, (6) therefore AEL CE (c) wherefore CE is pv. Which was to be demonstrated. 2. A rational rectangle AD joined with a rational rectangle CF makes a rational rectangle AF. For AD (a) CF, (6) wherefore AFL AD and CF; (c) and so AF is pv. Which was to be demonftrated. PROP. XXVIII. Fig. zz. To find out medial-lines (C and D) which contain a rational rectangle CD. (a) Take A and B (6) make A:C::C: B. (c) and A:B::C: D I say the thing required is done. For AB (Cq) (d) is ur, (d) whence C is u. But because ABCD (F) therefore C☑ D; (g) and confequently D is u. Moreover by permutation A:C:: B: 1). i. e. C: B::B: D; (b) therefore Bq=CD. But Bq is pr; (b) therefore CD is pv. Which was to be done. In numbers, let A be 23 and B 6. therefore C is v✓ 12. make √ 2: √6::γ 12 : D. or v4 : v√ 36 : : 21/12: D; then shall D be v√ 108; but v√ 12 X √ 108 = √ 1296 =√ 36 = 6. therefore CD is 6, likewife C: D::1:√ 3; wherefore CD. PROP. 1 PROP. XXIX. Plate IV. Fig. 34. To find out medial right-lines commenfurable in power only, D and E, containing a medial rectangle DE. (a) Take A, B, C, make A : D (6) :: D: B. (c) and B:C::D:E I lay the thing desired is performed. For AB (d) = Dq; and AB (e) is ur, therefore D is μ; and B (f) C, (g) whence D Ε. (6) E is μ. Moreover B: Cf): : D: E, and by permutation B :D ::: E. i. e. D:A: C:E; (1) therefore DE =AC. But AC (m) is pr; therefore DE is ur. Which was to be done. In numbers, let A be 20, and B, Therefore D is +80000; and E 200, and C, v 80. fore DE =√14 1024000000 / 32000. and D:E:: 10: 2. wherefore D E. alem21.13. e 22. 10. f conftr. g 10. 10. h 24. 10. k conftr. and cor 4.5. 1 16.6. m 22. 6. To find out two square numbers (DEq and CDq) so that the number composed of them (CEq) be square also. Take AD, DB like plane numbers (of which let both be even, or both odd) viz. AD, 24. and DB 6. The total of these (AB) is 30; the difference (FD) 18. Half of which (CD)is 9. (a) Now the like plane numbers AD, DB, have one mean number proportional, namely DE; therefore it is evident that every of those numbers CE, 218.8. CD, DE, are rational, and by consequence CEq (6) b 47.1. (CDq+ DEq) is the square number required. Whereby it will be easy to find out two square numbers, the excess of which is a square or not a square number, namely by the fame construction (c) shall CEq C 3. ax. 1. CDq be = DEq. But : : a 24. 8, But if AD, DB be plane numbers unlike, the mean proportional line (DE) shall not be a rational number, and so neither shall the excess (DEq) of the square numbers, CEq, CDq, be a square number. Lemma 2. 4 2. To find out two fuch square numbers B, C, as the number compounded of them D is not Square, Also to divide a square number A into two numbers B, C, not Squares. A, 3. B, 9. C, 36. D, 45. 1. Take any square number B, and let C be=4 B, and D = B + C. I say the thing is done. For Bis Q, by the constr. likewise because B: C:: 1 :4::Q:Q; (a) therefore C also shall be a square number. But because B + C (D): C:: 5:4:: not Q: Q. (b) therefore shall not D be a square number. Which was to be done. А, 36. B, 24. C, 12. D, 3. E, 2. F, 1. 2. Let A be some square number. Take D, E, F, plane numbers unlike, and let D be = E + F. make D:E:: A:B. and D: F:: A: C. I say the thing required is done. For because D: E + F:: A:B + C, and D = E+ F, (a) therefore shall A = B + C. Now suppose B to a 14. 5. b 21. def. 7. be square, (b) then A and B, (c) and confequently Dand E are like plane numbers. Which is contrary to the Hyp € 26.8. The fame absurdity will follow if C be supposed a square number, Therefore, &c. PROP. XXX. Fig. 36. To find out two rational right-lines AB, AF, commenJurable only in power, so that the greater AB shall be in power, more than the less AF, by the Square of a right-line BF, commensurable to it felf in length. Let A B be f. (a) Take the iquare numbers CD, CE, so that CD - CE (ED) be not Q. (t) and make CD: ED:: ABq: AFq. In a circle described upon the diameter AB (c) fit AF, and draw BF. Then I fay AB, AF, are the lines required. 1 For For ABq: AFq (d): : CD: ED; (e) therefore ABq AFq. But AB is ; (f) therefore AF is also ῥ. But because CD is Q: and ED not Q: (g) therefore shall AB be A. Moreover by reason of the (b) rightangle AFB, is ABq (k) = AFq+ BPq; therefore feeing ABq: Arq:: CD: ED, by converfion of proportion shall ABq: BFq:: CD: CE:: Q: Q. (1) Therefore AB BF. Which was to be done. In numbers, let there be AB, 6; CD, 9; CE, 4; wherefore ED, 5. Make 9:5:: 36: (Q: 6.) AFq. then AFq shall be 20; and confequently AF 2c. Therefore BFq = 36-20=16. Wherefore BF is 4. PROP. XXXI. Plate IV. Fig. 36. To find out two rational lines AB, AF commenfurable only in power, so that the greater AB shall be in power more than the less AF by the square of a right-line BF incommenfurable to it felf in length. d conftr. e 6. 10. f fch. 12. 10. g 9. 10. h 31. 3. k 47. I. 19. 10. Let AB be p. (a) Take the square numbers CE, ED, so az lem. 29. that CD = CE + ED be not Q, and in the rest follow the conftruction of the preced. prop. I say then the thing required is done. 10. For, as above, AB, AF, are alfo ABq: BFq:: CD: ED. therefore since CD is not Q: AB, BF (6) shall b 9. 10. be. Which was to be done. In numbers, let there be AB, 5. CD, 45. CE = 36. ED=9. Make 45:9:: 25 (ABq.): 5 (AFq); therefore AF=√5. confequently BFq=45-25-20, Wherefore BF=20. PROP. XXXII. Fig. 37. To find out two medial-lines C, D, commenfurable only in power, comprehending a rational rectangle CD, so that the greater C be more in power than the leffer D by the square of a right-line commensurable in length to the greater. ; (a) Take A and B so as Aq - BqTA (6) and make A:C:: C: B. (c) and A: B:: C: D. I say the thing is done. For because A and (d) B are (e) therefore shall C. (f) (√ AB) be μ. (g) and thence alfo CD (16) there- In a 30. 10. b 13. 6. C 12.6. d conftr. e 22. 1Ο. f 17.6. g 10. 10. h 24. 10. k 17. 6. 115. 10. |