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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closed as offtopic by Emil Jeřábek, Andrej Bauer, Gerald Edgar, Ian Agol, Stefan Kohl Feb 6 at 17:44This question appears to be offtopic. The users who voted to close gave this specific reason:



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. 

