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Theorem 4.8.1 Irrationality of V2 V2 is irrational.

Theorem 4.8.1 Irrationality of V2 V2 is irrational.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter7: Real And Complex Numbers
Section7.1: The Field Of Real Numbers
Problem 26E: Prove that if and are real numbers such that , then there exist a rational number such that ....
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Suggest you refer to Theorem 4.8.1. Please help me prove this using Theorem 4.8.1
Theorem 4.8.1 Irrationality of V2 V2 is irrational. Proof (by contradiction): [We take the negation and suppose it to be true.] Suppose not. That is, suppose V2 is rational. Then there are integers m and n with no com- mon factors such that m 4.8.1 n [by dividing m and n by any common factors if necessary]. [We must derive a contradic- tion.] Squaring both sides of equation (4.8.1) gives bonsh m2 2 = Or, equivalently, m2 = 2n². 4.8.2 Note that equation (4.8.2) implies that m is even (by definition of even). It follows that m is even (by Proposition 4.7.4). We file this fact away for future reference and also deduce (by definition of even) that 2k for some integer k. 4.8.3 m = Substituting equation (4.8.3) into equation (4.8.2), we see that ² = (2k) = 4k = 2n². m Dividing both sides of the right-most equation by 2 gives n? = 2k?. Consequently, n is even, and so n is even (by Proposition 4.7.4). But we also know that m is even. [This is the fact we filed away.] Hence both m andn have a common factor of 2. But this contradicts the supposition that m and n have no common fac- tors. [Hence the supposition is false and so the theorem is true.]
Transcribed Image Text:Theorem 4.8.1 Irrationality of V2 V2 is irrational. Proof (by contradiction): [We take the negation and suppose it to be true.] Suppose not. That is, suppose V2 is rational. Then there are integers m and n with no com- mon factors such that m 4.8.1 n [by dividing m and n by any common factors if necessary]. [We must derive a contradic- tion.] Squaring both sides of equation (4.8.1) gives bonsh m2 2 = Or, equivalently, m2 = 2n². 4.8.2 Note that equation (4.8.2) implies that m is even (by definition of even). It follows that m is even (by Proposition 4.7.4). We file this fact away for future reference and also deduce (by definition of even) that 2k for some integer k. 4.8.3 m = Substituting equation (4.8.3) into equation (4.8.2), we see that ² = (2k) = 4k = 2n². m Dividing both sides of the right-most equation by 2 gives n? = 2k?. Consequently, n is even, and so n is even (by Proposition 4.7.4). But we also know that m is even. [This is the fact we filed away.] Hence both m andn have a common factor of 2. But this contradicts the supposition that m and n have no common fac- tors. [Hence the supposition is false and so the theorem is true.]
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