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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
@DavidMitra: many thanks, I was not aware of that theorem. – Ron Gordon Jul 18 '13 at 15:43
One can find this result, and its proof, in Ivan Niven's book Irrational numbers. – David Mitra Jul 18 '13 at 15:46
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.) – Andrés E. Caicedo Jul 18 '13 at 15:48
@Andres: isn't the definition of "transcendental" just "not algebraic"? Then a constructive proof that a number is transcendental is precisely a proof of a contradiction from the assumption that number is algebraic. I am not very familiar with the actual proof so I can't say much about the details, but see – Carl Mummert Jul 19 '13 at 11:44

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. – Andrés E. 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
I don't know if there is one. It seems a difficult question. – Andrés E. Caicedo Jul 19 '13 at 5:11

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 '14 at 19:45

The so-called Gelfond–Schneider theorem says that, if $a$ and $b$ are algebraic numbers such that $a\neq 0, 1$ and $b$ is irrational, then $a^b$ is transcendental.


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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

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