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I am learning proofs and a question was posed which asked us to prove that $\sqrt{2}^{\sqrt{2}}$ is irrational. They mentioned this - Hint: try using the log10 function...

I tried my hand at the proof by contradiction. Assuming $\sqrt{2}^{\sqrt{2}}$ is rational. Hence,

$\sqrt{2}^{\sqrt{2}} = \frac{p}{q} \implies \big({\frac{p}{q}}\big)^{\sqrt{2}} = 2$

Can I now say, that we know that $\sqrt{2}$ is irrational, but $\frac{p}{q}$ and $2$ are rational;

If $a$ is rational and $b$ is irrational, $a^b$ must be irrational. *(1)

In our case $a^b = 2$ which is rational: a contradiction to our original assumption.

This is just tomfoolery on my part. I don't know if *(1) is even true or not; how can I approach such a problem?

akad3b
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  • Your statement is incorrect about irrational powers of rational numbers: let $a=1$. In fact, it is possible to take an irrational power of an irrational number and recover a rational number. – Emily Jun 19 '16 at 12:39
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    I don't think there's a way to do this without using some fairly advanced mathematics. – Gerry Myerson Jun 19 '16 at 12:40
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    I don't know who'd pose this... See the first answer and the comments here. – David Mitra Jun 19 '16 at 12:42

1 Answers1

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The irrationality of $\sqrt{2}^{\sqrt{2}}$ is a trivial consequence of the Gelfond-Schneider theorem.

Jack D'Aurizio
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