# Deciding whether $2^{\sqrt2}$ is irrational/transcendental [duplicate]

This question already has an answer here:

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

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## marked as duplicate by Najib Idrissi, Servaes, Tom-Tom, Claude Leibovici, hardmathSep 16 '15 at 12:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

– Julian Rosen Jul 22 '12 at 8:44
@Pink Elephants : Perhaps this should be an answer. – Patrick Da Silva Jul 22 '12 at 8:50
If it were so easy, it wouldn't have been on the list of Hilbert's problems, would it? – J. M. Jul 22 '12 at 8:58
@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. – Marco Disce Jul 22 '12 at 11:57

## 1 Answer

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.

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