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In solving the following problem: Prove or disprove that there is a rational number $x$ and an irrational number $y$ such that $x^y$ is irrational. I let $x=2$ and $y = \sqrt 2$, so that $x^y = 2^\sqrt 2$.

Why is this not enough? How come I have to go through the case whether $x^y = 2^\sqrt 2$ is rational?

If $2^\sqrt 2$ is rational than let $x = 2^\sqrt 2$, and $y = \sqrt 2 / 4$

$x^y = (2^\sqrt 2)^{\sqrt 2 /4} = 2^{(\sqrt 2*\sqrt 2) /4} = 2^{2/4} = 2^{1/2} = \sqrt 2$ (previous value for y that was established as irrational.

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    $\begingroup$ And how do you know that $2^{\sqrt{2}}$ is irrational? $\endgroup$ – Zeekless Jan 20 at 21:50
  • $\begingroup$ Elliott and @Zeekless: Oh... it isn't rational. $\endgroup$ – David G. Stork Jan 20 at 21:50
  • $\begingroup$ If it were rational (which it isn't, but you don't know that for sure), then you'd have to go with $\left(2^\sqrt2\right)^\sqrt2$ $\endgroup$ – Ivan Neretin Jan 20 at 21:54
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    $\begingroup$ @David G. Stork, you misunderstand the question. Your last edit of the title is wrong. $\endgroup$ – Zeekless Jan 20 at 21:55
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    $\begingroup$ The more usual question is to show that there are irrational $a,b$ such that $a^b$ is rational. Something being irrational is common but often hard to prove, like for $2^{\sqrt 2}$. It is irrational, but I don't know an easy proof. You have tried to display an example for your question, but have not supplied any proof that it is irrational. The Gelfond-Schneider theorem says it is transcendental but I have never seen the proof. $\endgroup$ – Ross Millikan Jan 20 at 22:03
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Let $\mathbb{I}$ denote the set of irrational numbers.

Function $f : \mathbb{I} \rightarrow \mathbb{R} : x \mapsto 2^x$ is an injection.

$\mathbb{I}$ is uncountable $\Rightarrow$ image of $f$ is uncountable.

The set of rational numbers is countable $\Rightarrow$ image of $f$ contains something more than rationals.

There exist such irrational $x$ that $2^x$ is irrational.

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If you want to prove the statement consider first that for positive rational $a$ and real $b$ we have that $a^b$ is a continuous function of $b$ and given a positive real number $x$ we have $x=a^b$ when $\log x = b \log a$ or $b =\frac {\log x}{\log a}$ where we can take logarithms to any sensible base. (Note that $a=1$ is a special case here, so we avoid choosing $a=1$ when we have a choice).

Now the expressions $a^b$ where $a\in \mathbb Q^+; b\in \mathbb Q$ are countable and we can choose $x\in \mathbb R^+$ which is neither in the set of such expressions nor in $\mathbb Q$ (the union of two countable sets is countable). $x$ is irrational and is not expressible in the form $a^b$ with both $a$ and $b$ rational - so we choose a positive rational $a\neq 1$ and the corresponding $b$ must be irrational.

This is a non-constructive proof, which shows that there will be lots of examples.

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