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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?

Glorfindel
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habs
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    Take $\alpha = \sqrt{2}^{\sqrt{2}}$, and $\beta = \sqrt{2}$, and look at $\alpha^{\beta}$ – DeepSea Jan 12 '16 at 21:36
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    It is known that $\sqrt{2}^\sqrt{2}$ is irrational (transcendental); see this previous Question. – hardmath Jan 12 '16 at 22:07
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    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 $f(z) = |z|.$ – Dave L. Renfro Jan 12 '16 at 22:31
  • @Kf-Sansoo Thanks for that example; I did not think to consider actually testing out $(x,y)=(\sqrt{2},\sqrt{2}),(\sqrt{2}^{\sqrt{2}},\sqrt{2})$. – habs Jan 12 '16 at 23:32
  • @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?

     – habs Jan 12 '16 at 23:37
  • @Harry I'm inclined to think that marking as duplicate would be good since the answers here are equivalent to answers already over there (and any answer here would serve as an answer there) as well as that the answers there do serve this sub-question. On the other hand, it's not an exact duplicate, so it's not like closing as duplicate is an obvious, uncontentious choice (and you're certainly free to vote however you wish). – Milo Brandt Jan 12 '16 at 23:45
  • @hardmath Thanks, that settles the matter. – Did Jan 12 '16 at 23:48

2 Answers2

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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.

André Nicolas
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Let $x = \mathrm{e}$ and $y = \ln(2)$, then $x^y = \mathrm{e}^{\ln(2)} = 2$.

Björn Friedrich
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