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It's possible to prove nonconstructively that there exists irrational numbers $x$ and $y$ such that $x^y$ is rational, but that proof only proves that such numbers exist and does not specify what they are.

What is a constructive proof that there are two irrational numbers $x$, $y$ such that $x^y$ is rational, i.e. what are those numbers?

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    $\begingroup$ Take $\alpha = \sqrt{2}^{\sqrt{2}}$, and $\beta = \sqrt{2}$, and look at $\alpha^{\beta}$ $\endgroup$
    – DeepSea
    Jan 12, 2016 at 21:36
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    $\begingroup$ It is known that $\sqrt{2}^\sqrt{2}$ is irrational (transcendental); see this previous Question. $\endgroup$
    – hardmath
    Jan 12, 2016 at 22:07
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    $\begingroup$ If we don't have to use real numbers, then $e^{\pi i} = -1$ is an easy construction (assuming we don't have to prove the irrationality of $e$ and $\pi).$ Reminds me of how easy it is to get a continuous nowhere differentiable function when you get to use complex numbers --- just use $f(z) = |z|.$ $\endgroup$
    – Dave L. Renfro
    Jan 12, 2016 at 22:31
  • $\begingroup$ @Kf-Sansoo Thanks for that example; I did not think to consider actually testing out $(x,y)=(\sqrt{2},\sqrt{2}),(\sqrt{2}^{\sqrt{2}},\sqrt{2})$. $\endgroup$
    – habs
    Jan 12, 2016 at 23:32
  • $\begingroup$ @MiloBrandt Thanks for pointing that out; I would not consider this question a duplicate of that question, but I do admit that the answer to this question is contained within the answers to that question (though notably, this question is not answered in the accepted answer to that question). Because of this, I'm not sure whether or not to mark it as a duplicate. What do you think? $\endgroup$
    – habs
    Jan 12, 2016 at 23:37

2 Answers 2

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Let $x=3^{1/2}$ and $y=\log_{3}(4)$. Then $x^y=2$.

The proof that $x$ is irrational is familiar. For $y$, suppose $y=p/q$ where $p$ and $q$ are positive integers. Then $3^{p/q}=4$, so $3^p=4^q$. This is impossible, since $4^q$ is even and $3^p$ is odd.

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  • $\begingroup$ Thank you. I chose this as the accepted answer because it explains why $x$ and $y$ are irrational. $\endgroup$
    – habs
    Jan 12, 2016 at 23:28
  • $\begingroup$ You are welcome. $\endgroup$
    – André Nicolas
    Jan 13, 2016 at 3:31
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Let $x = \mathrm{e}$ and $y = \ln(2)$, then $x^y = \mathrm{e}^{\ln(2)} = 2$.

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    $\begingroup$ It would improve your Answer (which I already upvoted) if you added some details about why $e, \ln 2$ are known to be irrational. $\endgroup$
    – hardmath
    Jan 12, 2016 at 22:09

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