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The fact that there exists irrational number $a,b$ such that $a^b$ is rational is proved by the law of excluded middle, but I read somewhere that irrationality of $\sqrt{2}^{\sqrt{2}}$ is proved constructively. Do you know the proof?

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See this. –  David Mitra Jul 18 '13 at 15:32
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@DavidMitra: many thanks, I was not aware of that theorem. –  Ron Gordon Jul 18 '13 at 15:43
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One can find this result, and its proof, in Ivan Niven's book Irrational numbers. –  David Mitra Jul 18 '13 at 15:46
    
@Potato: you can retract your vote. –  Aang Jul 18 '13 at 15:47
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Interesting question! Just about all proofs of transcendence I'm aware of essentially assume the target is algebraic and conclude that there is an integer strictly between $0$ and $1$. Not sure whether anything like this can be turned into a constructively (or intuitionistically) valid argument. I am also not aware of any proof of irrationality of $\sqrt2^{\sqrt2}$ other than the transcendence proof of Gelfond-Schneider. (Of course, I'm probably just overlooking something.) –  Andres Caicedo Jul 18 '13 at 15:48

5 Answers 5

Since this is a well-established result, this is a community wiki post.

Relevant question: Deciding whether $2^{\sqrt2}$ is irrational/transcendental

Kuzmin proved the following claim in 1930:

Theorem: If $\alpha\neq 0,1$ is algebraic, $\beta$ is positive and rational, not a perfect square, then $\alpha^{\sqrt{\beta}}$ is transcendental.

Unfortunately the paper is in Russian and I failed to find an English translation. A corollary of this is that $2^{\sqrt{2}}$ is transcendental, and so is its square root $\sqrt{2}^{\sqrt{2}}$.

The outlines of both Gelfond and Kuzmin's constructive proof can be found here.

As David Mitra pointed out the comments, Niven's book had a section dedicated to this. I love Niven's book so much. The technique is similar to the adapted proof I posted here, proof by contradiction.

Rough idea about the construction: First assuming $\alpha^{\sqrt{\beta}}$ is algebraic. Then using sufficient large degree Lagrange interpolation polynomial to approximate $e^{(\ln \alpha)x}$ at points $\{a+ b\sqrt{2}\}$ for $a,b\in \mathbb{Z}$. Let the number of points go to infinity the error will go to zero, this shows a transcendental function $\alpha^x$ can be interpolate using countably many algebraic points. Contradiction.

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Not a constructive argument (in the sense of mathematical logic), though. –  Andres Caicedo Jul 19 '13 at 5:00
    
@AndresCaicedo Thanks for the heads up, I didn't realize there could be a constructive proof in the sense of mathematical logic. I appreciate if you write one, I want to learn it as well. :) –  Shuhao Cao Jul 19 '13 at 5:07
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I don't know if there is one. It seems a difficult question. –  Andres Caicedo Jul 19 '13 at 5:11

The so-called Gelfond–Schneider theorem says that if a and b are algebraic numbers such that a,b neq 0,1 and if b is not a rational number, then a^b is transcendental...

Source: http://en.wikipedia.org/wiki/Gelfond–Schneider_theorem

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This was mentioned in the comments, but does it help to answer the question about proving constructively that $\sqrt 2^\sqrt 2$ is irrational? –  Jonas Meyer Jul 18 '13 at 20:29

concerning the $a^b$ stuff

$e$ and and natural $\log$ of a rational number also proves that irrational number to and and irrational number can be rational and that is a lot easier

$e^{\ln2} = 2$ for example.

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I think the Question here is not merely whether one irrational number raised to another can be rational $a^b$, but how to show this constructively for $a=b=\sqrt{2}$. So unless you see a way to prove that, I think this would be better as a Comment than as an Answer. If you participate in the Community and earn 50 reputation points, you will be able to Comment on the posts of others. –  hardmath Dec 5 at 19:45

The proof is shown here. If $\sqrt2^\sqrt2$ is irrational, it can be raised to the $\sqrt2$ power again and become rational ($(\sqrt2^\sqrt2)^\sqrt2=2$.

This proves that one irrational number can be raised to an irrational power and become rational.

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Nice proof, but it doesn't answer this question: proving $\sqrt{2}^{\sqrt{2}}$ is irrational. –  Douglas S. Stones Jul 18 '13 at 17:42
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@GottfriedHelms: Yes, I mention that in my comment, but not at first submission, and it might have been edited in after you started writing yours. Applicability of GS is the reason for the first comment on the question, by David Mitra, and was explicitly mentioned in Andres Caicedo's comment subsequently. (When I wrote "answer to the question" I meant proof that the number is irrational, but I don't know about constructiveness.) –  Jonas Meyer Jul 18 '13 at 18:08
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So, nothing to do with the question, right? –  Andres Caicedo Jul 18 '13 at 18:52
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@GottfriedHelms " does that GS-theorem actually satisfy ...?" Not as far as I can tell. As I mentioned in the comment Jonas referred to above, all proofs of transcendence use a format which does not seem amenable to constructive mathematics. –  Andres Caicedo Jul 18 '13 at 18:58
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-1. This is not an answer to the question. –  Qiaochu Yuan Jul 18 '13 at 20:26

I'm not sure about $\sqrt2^{\sqrt2}$, but you can prove there exist $\text{irrational}^{\text{irrational}}$ which are rational using $\sqrt 2^{log_2 9}$.

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$\log_2(4) = 2$ –  EuYu Jul 18 '13 at 17:39
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This could be fixed by instead considering $\sqrt 2^{\log_2 9}$, but it would still not answer the question. –  Jonas Meyer Jul 18 '13 at 18:18

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