I have to prove or disprove this statement. I am stuck on which method to use to solve this. I can chose from direct, contra positive and contradiction
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2How about $\sqrt 2$? – J. W. Tanner Oct 06 '20 at 02:21
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direct proof is best I think. – Asinomás Oct 06 '20 at 02:22
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1A better and well known question is whether there is an irrational number $a$ such that $a^{\sqrt{2}}$ is rational. – markvs Oct 06 '20 at 02:24
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@JCAA $a=\sqrt{2}^{\sqrt{2}}$. This can be proven by exhaustion/cases. – Sage Stark Oct 06 '20 at 02:24
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@Pendronator: I know the answer to that problem for about $50$ years. As I said, it is a well known problem. – markvs Oct 06 '20 at 02:27
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@JCAA do you have a reference for a proof for this statement? A quick google search as not clear for me. – IntegrateThis Oct 06 '20 at 02:29
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2@IntegrateThis: Search for Gelfond-Schneider theorem. – markvs Oct 06 '20 at 02:31
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Cf. https://math.stackexchange.com/questions/104119/can-an-irrational-number-raised-to-an-irrational-power-be-rational – J. W. Tanner Oct 06 '20 at 02:42
2 Answers
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If $p$ is a prime then $\sqrt{p}$ is irrational, and moreover $(\sqrt{p})^2 = p$ is an integer.
IntegrateThis
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