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Is $2^\sqrt{2}$ irrational? Is it transcendental?

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    See http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem2012-07-22
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    See [Gelfond–Schneider constant](http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_constant) at Wikipedia. Somewhat related are also this question [Real Numbers to Irrational Powers](http://math.stackexchange.com/questions/2574/real-numbers-to-irrational-powers) and this MO question [About the proof of the proposition “there exists irrational numbers a, b such that a^b is rational”](http://mathoverflow.net/questions/56930/about-the-proof-of-the-proposition-there-exists-irrational-numbers-a-b-such-tha)2012-07-22
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    @Pink Elephants : Perhaps this should be an answer.2012-07-22
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    @Pink Elephants : Thanks, very interesting, but is there an easy way to prove irrationality without such a theorem?2012-07-22
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    If it were so easy, it wouldn't have been on the list of [Hilbert's problems](http://en.wikipedia.org/wiki/Hilbert_problems), would it?2012-07-22
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    @J.M.: As far as I understand it the Hilber's problem is to decide wheter it is trascendental, not to decide whether it is irrational.2012-07-22

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According to Gel'fond's theorem, if $\alpha$ and $\beta$ are algebraic numbers (which $2$ and $\sqrt 2$ are) and $\beta$ is irrational, then $\alpha^\beta$ is transcendental, except in the trivial cases when $\alpha$ is 0 or 1.

Wikipedia's article about the constant $2^{\sqrt 2}$ says that it was first proved to be transcendental in 1930, by Kuzmin.