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Question Number 190480 by aba last updated on 03/Apr/23

$$\mathrm{Proof}\mathrm{that}{\left(\sqrt{2}\right)}^{\sqrt{2}}\in \mathbb{R}\mathrm{\setminus }\mathrm{Q}$$

Answered by mehdee42 last updated on 04/Apr/23

$$lem:ifp\notin Q\Rightarrow \sqrt{p}\notin Q$$$$clim:2^{\sqrt{2}}\notin Q$$$$proof:if{2}^{\sqrt{2}}=\frac{p}{q}\in Q\Rightarrow p=2^{\sqrt{2}}q$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\Rightarrow \sqrt{\left(2^{\sqrt{2}}\right)}={\left(\sqrt{2}\right)}^{\sqrt{2}}\notin Q$$$$$$

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