It is known that $\sqrt{2}^{\sqrt{2}}$ is irrational. Is it true that for any irrational number $a$, $a^a$ must be irrational?
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asked Feb 6, 2014 at 17:23
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11$\begingroup$ Obviously not. There is a number $a > 1$ such that $a^a = 2$. This number cannot be rational. $\endgroup$– Andrej BauerCommented Feb 6, 2014 at 17:39
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1$\begingroup$ @Andrei Bauer my typing the proof took 7 seconds more :) $\endgroup$– Sasha Anan'inCommented Feb 6, 2014 at 17:41
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9$\begingroup$ (1) this is not a race; (2) this question shouldn't have been on MO so it's better not to answer. $\endgroup$– Anthony QuasCommented Feb 6, 2014 at 18:52
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1 Answer
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There is a positive number $a$ such that $a^a=2$. If $a=m/n$ with $m,n\in{\mathbb N}$ coprime, then $m^m=2^nn^m$. As $n\ge1$, we conclude that $m$ is even, sayt $m=2^kl$ with $k\ge1$ and odd $l$. So, $2^{km}l^m=2^nn^m$, implying $2^{km}=2^n$ and $km=n$. A contradiction.
answered Feb 6, 2014 at 17:39