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

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

I crossposted to MO: https://mathoverflow.net/questions/138247

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mathlove
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    Cannot be done. – Will Jagy Jul 31 '13 at 06:39
  • Meanwhile, perhaps this is what you are thinking about: http://math.stackexchange.com/questions/104119/can-an-irrational-number-raised-to-an-irrational-power-be-rational – Will Jagy Jul 31 '13 at 06:54
  • @Will Is there some intuition for why no irrationality proof likely exists (short of proving transcendence)? – user7530 Jul 31 '13 at 07:24
  • One way could be using the proof of Gel'fond-Schneider's theorem, as opposed to the Gel'fond-Schneider's theorem. – OR. Jul 31 '13 at 07:30
  • @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 07:57
  • I think what Will means is that no one knows how to prove only the irrationality of $\sqrt{2}^{\sqrt{2}}$; all known arguments establish the transcendence, and in fact can be made to work in much more generality (namely that of the Gel'fond-Schneider Theorem). As far as I know it's certainly conceivable that a more elementary proof exists. I don't even know how to adequately formalize a statement to the contrary, let alone prove it. (That might make for a good question, by the way: are there any theorems in mathematics of the form "There is no elementary proof of..."?) – Pete L. Clark Jul 31 '13 at 08:50
  • @RGB: Unless we're working in some specific formal proof system, I can't see why the distinction you're making is of any significance. – Pete L. Clark Jul 31 '13 at 08:53
  • It has the same meaning as your first comment. The proof can be applied to the particular case of $(\sqrt{2})^x$, and then, perhaps, simplified. Do you see the significance of your own comment? – OR. Jul 31 '13 at 09:01
  • RGB: "Do you see the significance of your own comment?" Is that a sincere question? I'm really trying to understand what you mean. Do me a favor and help me out with that. In particular, part of what I'm saying (and I think what Will was saying) is that it seems that the proof of Gel'fond-Schneider does not simplify when one concentrates on this particular example. If you would like to weigh in on this point, please do so. – Pete L. Clark Jul 31 '13 at 09:05
  • It means what you said in your first comment. One could take the proof of the theorem and try to simplify it for the particular case of the function $(\sqrt{2})^x$. We can also use for the simplification that the aim is to prove that $\sqrt{2}^{\sqrt{2}}$ doesn't satisfy a linear polynomial equation with integer coefficients, instead of a any polynomial equation of higher degree. – OR. Jul 31 '13 at 09:09
  • @RGB: Thanks for the clarification. Yes, of course we can try. But it is my understanding that many people have tried and no one has succeeded. – Pete L. Clark Jul 31 '13 at 09:12
  • In the vague memory I have of the proof I remember lemmas bounding solutions to polynomials in exponential functions. The fact that we only care about polynomials of degree $1$ should make many of those lemmas trivial. I don't see why dealing $\sqrt{2}$ specifically should give any simplification, but trying to prove irrationality, instead of transcendency should cut down the proof. – OR. Jul 31 '13 at 09:28

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I'm posting an answer just to inform that the question has received an answer by Mark Sapir on MO.

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

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