Is $2^\sqrt{2}$ irrational? Is it transcendental?
Deciding whether $2^{\sqrt2}$ is irrational/transcendental
7
$\begingroup$
number-theory
transcendental-numbers
rationality-testing
-
5See http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem – 2012-07-22
-
2See [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
-
0@Pink Elephants : Perhaps this should be an answer. – 2012-07-22
-
0@Pink Elephants : Thanks, very interesting, but is there an easy way to prove irrationality without such a theorem? – 2012-07-22
-
6If 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
-
4@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
1 Answers
9
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.