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From this question I thought: the algebraic numbers are closed under the operations $+,\div,\times,-,\sqrt[n]{\cdot}$ for arbitrary $n\in\Bbb N$.

However, as far as I know, they are not closed under exponentiation. Then my questions:

  1. It is known if $\pi$ can be represented by $a^b$ for some algebraic numbers $a$ and $b$?

  2. If the above question is positive, it is known if is possible to represent $\pi$ with the form $\left(\sqrt[n]{m}\right)^{\sqrt[p]{q}}$, for integers $n,m,p,q$?

Thomas Andrews
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Masacroso
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    I am guessing that this is unknown, but I'll just note that the answer is probably no, as there are only countably infinite many numbers of the form $a^b$ for $a$, $b$ algebraic. – idok Jan 18 '18 at 19:02
  • As to your question, why do you think the algebraic numbers aren't closed under exponentiation? – Allawonder Jan 18 '18 at 19:08
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    Probably nobody can answer this question for sure (and if some body can, s/he won't do it for free), but you can bet for the answer 'no'. – ajotatxe Jan 18 '18 at 19:08
  • IOW, do you have a counterexample? If so, you might want to include it in your question. This would make it better. – Allawonder Jan 18 '18 at 19:09
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    @Allawonder $i^i=e^{-\pi/2}$ is known to be transcendental, I believe. – Thomas Andrews Jan 18 '18 at 19:10
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    @Allawonder Gelfond-Schneider's theorem. – ajotatxe Jan 18 '18 at 19:10
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    Yep: https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem says that if $a\neq 0,1$ and $b$ irrational with both algebraic, then $a^b$ is transcendental. – Thomas Andrews Jan 18 '18 at 19:12

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