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This question already has an answer here:

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?

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marked as duplicate by MJD, gebruiker, timur, Math1000, Milo Brandt Jun 19 '16 at 16:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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. $\endgroup$ – Emily Jun 19 '16 at 12:39
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    $\begingroup$ I don't think there's a way to do this without using some fairly advanced mathematics. $\endgroup$ – Gerry Myerson Jun 19 '16 at 12:40
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    $\begingroup$ I don't know who'd pose this... See the first answer and the comments here. $\endgroup$ – David Mitra Jun 19 '16 at 12:42
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The irrationality of $\sqrt{2}^{\sqrt{2}}$ is a trivial consequence of the Gelfond-Schneider theorem.

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    $\begingroup$ That is to say, a trivial consequence of a highly nontrivial theorem. $\endgroup$ – Gerry Myerson Jun 19 '16 at 23:42

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