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Prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using the Gel'fond-Schneider's theorem.

I'm interested in this problem because I knew that ${\sqrt2}^{\sqrt2}$ is a transcendental number by the Gel'fond-Schneider's theorem. I've tried to prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using the Gel'fond-Schneider's theorem, but I'm facing difficulty. I need your help.

I crossposted to MO. http://mathoverflow.net/questions/138247/prove-that-sqrt2-sqrt2-is-an-irrational-number-without-using-a-theorem

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Cannot be done. –  Will Jagy Jul 31 '13 at 6:39
    
Meanwhile, perhaps this is what you are thinking about: math.stackexchange.com/questions/104119/… –  Will Jagy Jul 31 '13 at 6:54
    
@Will Is there some intuition for why no irrationality proof likely exists (short of proving transcendence)? –  user7530 Jul 31 '13 at 7:24
    
One way could be using the proof of Gel'fond-Schneider's theorem, as opposed to the Gel'fond-Schneider's theorem. –  ABC Jul 31 '13 at 7:30
    
@Will Jagy: It is not what I'm thinking about. I've already known its answer. I'm looking for a 'simple' proof. –  mathlove Jul 31 '13 at 7:57
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I'm posting an answer just to inform that the question has received an answer by Mark Sapir on MO.

http://mathoverflow.net/questions/138247/prove-that-sqrt2-sqrt2-is-an-irrational-number-without-using-a-theorem

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