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Introduction:

If we take $a=2^\sqrt[3]{2}$ which is transcendental by Gelfond-Schneider Theorem, and $b=\sqrt[3]{4}$ which is algebraic irrational because it is root of monic-irreducible polynomial over $\mathbb{Q}$ of degree $3$ which is $x^{3}-4$.(The polynomial is irreducible by the theorem: If polynomial in $F[x]$ where $F$ is field has degree $2$ or $3$ and doesn't have roots in $F$, then it is irreducible over $F$ ).

So, there exist transcendental $a$ and algebraic irrational $b$ such that $a^{b}$ is rational.

My question: Can the opposite happen, i.e, there exist algebraic irrational $a$ and transcendental b, such that $a^{b}$ is rational?

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Yes. Let $a=\sqrt{2}$. By the intermediate value theorem, there is some $0<x<1$, for which $\sqrt{2}^x$ is rational. $x$ can not be rational, and, by Gelfond-Schneider, it can not be algebraic either so it must be transcendental. (As in the comments you can solve for $x$ if you want, but you don’t need to.)

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  • $\begingroup$ Your solution is smart and beneficial, thanks! $\endgroup$ Commented Jun 19 at 19:31
  • $\begingroup$ The following is just discussion, so irrational to irrational is rational in the following cases: 1) Transcendental to algebraic irrational, one of the examples is the one I mentioned in the body of my question. 2)Algebraic irrational to transcendental, one of the examples is in your answer, one specific example is the solution to $(\sqrt{2})^{x}=\frac{7}{5}$, $x=2*\frac{ln(\frac{7}{5})}{ln(2)}$. 3) Transcendental to Transcendental, one of the examples is $e^{ln(2)}$. Note, Algebraic irrational to Algebraic irrational is not included by Gelfond-Schneider Theorem. $\endgroup$ Commented Jun 19 at 19:41
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    $\begingroup$ You have not worked out why $x$ cannot be rational , and it is not at all obvious. $\endgroup$
    – Peter
    Commented Jun 19 at 20:04
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    $\begingroup$ @Peter Consider a rational in reduced form, then $\sqrt{2}^{p/q}$ being rational must mean that $2^p$ is a $2q$-th perfect power, so $q$ divides $p$, meaning that $p$ is an integer…. $\endgroup$ Commented Jun 19 at 20:11

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