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How do I prove that there exist some $a\in\mathbb{Q}$ and $b \in \mathbb{R} - \mathbb{Q}$ such that $a^b \in\mathbb{R} - \mathbb{Q}$? I dont need to find what it is, just that it exists. The only numbers I know are irrational for the purposes of this proof are $\sqrt{2}$, $\sqrt[3]{2}$, $\sqrt{3}$, and $log_{2}3$, but I could prove the irrationality of some other number if I needed it to prove this.

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  • $\begingroup$ How about $2^\sqrt{2}$? $\endgroup$
    – molarmass
    Sep 26 '20 at 17:35
  • $\begingroup$ @molarmass how would you prove $2^{\sqrt 2}$ is irrational? $\endgroup$
    – fleablood
    Sep 26 '20 at 17:47
  • $\begingroup$ By the Gelfond–Schneider theorem, $2^\sqrt{2}$ is transcendental and thus it is irrational. In fact, this number is known as the Gelfond–Schneider constant. $\endgroup$
    – molarmass
    Sep 28 '20 at 18:28
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HINT: How big is the set $\left\{2^b:b\in\Bbb R\setminus\Bbb Q\right\}$? How many rationals are there?

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If $a > 0$ and $a \ne 1$ then $a^x$ is injective, one to one. As there are uncountibly meany $x$ there are uncountably many $a^x$. As there are only countably many rationals, not all of the $a^x$ can be rational.

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Given the list of values you know to be irrational, we could simply take $$2^{\log_2 \sqrt 3}=\sqrt 3$$

We remark that $\log_2\sqrt 3 =\frac {\log_2 3}2$ so that too must be irrational (given that $\log_2 3$ is irrational).

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