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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 off-topic by Emil Jeřábek, Andrej Bauer, Gerald Edgar, Ian Agol, Stefan Kohl Feb 6 at 17:44

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9  
Obviously not. There is a number $a > 1$ such that $a^a = 2$. This number cannot be rational. –  Andrej Bauer Feb 6 at 17:39
    
@Andrei Bauer my typing the proof took 7 seconds more :) –  Sasha Anan'in Feb 6 at 17:41
4  
(1) this is not a race; (2) this question shouldn't have been on MO so it's better not to answer. –  Anthony Quas Feb 6 at 18:52

1 Answer 1

up vote 7 down vote accepted

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.

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