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3. Prove that any Integens . Proof (by contradicton): ,9 t (a?-3)

3. Prove that any Integens . Proof (by contradicton): ,9 t (a?-3)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 55RE
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I did the proof and my teacher said I was wrong so can u please help me prove this ? Let a^2 = 9q + 3 in the proof of contradiction
23. Prove that any Integens 9 t a^-3) Proof (by contradicton): a E Z and 91(a²-3) So 3la See hit 9 a?-3 - 99 hy definitim of dwisibility. for Some integer k.by definition of the quotient - remainder Suppose not . That is suppose or a = 3K +2 theovem. case 1. a = gu +2 .Then a²3 = 99 Gu +1)°-3- 9g ay substitutun. qk'tok -2=99 by algebra 3(34? +2K-3Q) = ? by algebra by al gebra het l = 3K*+2K-39 lez since 3,2,K° &,q E Z the closure of integers under additrons and mulhplications, But l = 5 € 2 therefore this leads to the contradictión 9 |la?-3). Case 2: a= 3k +2 Thin that a-3 = 99 (3Kt2)-3 =99 ay subshitution - 9k?+ 12K +1 = 3(3K²+ HKB-39)= 3k?+4K -39 = 3 by algebra . Let d= 3k2+4K -39. d is integers since 3 4, K²,k 9 E z bey the closure of mtegers under additionsand multiplications. But -1 d = that 9 la?3). 2 there fore this leads to the contradiutron
Transcribed Image Text:23. Prove that any Integens 9 t a^-3) Proof (by contradicton): a E Z and 91(a²-3) So 3la See hit 9 a?-3 - 99 hy definitim of dwisibility. for Some integer k.by definition of the quotient - remainder Suppose not . That is suppose or a = 3K +2 theovem. case 1. a = gu +2 .Then a²3 = 99 Gu +1)°-3- 9g ay substitutun. qk'tok -2=99 by algebra 3(34? +2K-3Q) = ? by algebra by al gebra het l = 3K*+2K-39 lez since 3,2,K° &,q E Z the closure of integers under additrons and mulhplications, But l = 5 € 2 therefore this leads to the contradictión 9 |la?-3). Case 2: a= 3k +2 Thin that a-3 = 99 (3Kt2)-3 =99 ay subshitution - 9k?+ 12K +1 = 3(3K²+ HKB-39)= 3k?+4K -39 = 3 by algebra . Let d= 3k2+4K -39. d is integers since 3 4, K²,k 9 E z bey the closure of mtegers under additionsand multiplications. But -1 d = that 9 la?3). 2 there fore this leads to the contradiutron
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