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Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number s such that s2 = 6.

Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number s such that s2 = 6.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The Algebraic And Order Properties of Real Number

I already provide theorem 2.1.4

Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number s such that s2 = 6.
Transcribed Image Text:Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number s such that s2 = 6.
2.1.4 Theorem There does not exist a rational number r such that r = 2. Proof. Suppose, on the contrary, that p and q are integers such that (p/q)? = 2. We may assume that p and q are positive and have no common integer factors other than 1. (Why?) Since p? = 2q, we see that p² is even. This implies that p is also even (because if p = 2n – 1 is odd, then its square p = 2(2n² – 2n + 1) – 1 is also odd). Therefore, since p and q do not have 2 as a common factor, then q must be an odd natural number. Since p is even, thenp = 2mfor some m e N, and hence 4m? = 2q², so that 2m? = q². Therefore, q? is even, and it follows that q is an even natural number. Since the hypothesis that (p/q) = 2 leads to the contradictory conclusion that q is both even and odd, it must be false. Q.E.D.
Transcribed Image Text:2.1.4 Theorem There does not exist a rational number r such that r = 2. Proof. Suppose, on the contrary, that p and q are integers such that (p/q)? = 2. We may assume that p and q are positive and have no common integer factors other than 1. (Why?) Since p? = 2q, we see that p² is even. This implies that p is also even (because if p = 2n – 1 is odd, then its square p = 2(2n² – 2n + 1) – 1 is also odd). Therefore, since p and q do not have 2 as a common factor, then q must be an odd natural number. Since p is even, thenp = 2mfor some m e N, and hence 4m? = 2q², so that 2m? = q². Therefore, q? is even, and it follows that q is an even natural number. Since the hypothesis that (p/q) = 2 leads to the contradictory conclusion that q is both even and odd, it must be false. Q.E.D.
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