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?

Martin Sleziak
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    See [this](http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem). – 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*](http://books.google.com/books/about/Irrational_Numbers.html?id=ov-IlIEo47cC). – 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.) – Andrés E. Caicedo Jul 18 '13 at 15:48
  • @Avatar I haven't noticed that feature before! Thanks. – Potato Jul 18 '13 at 15:52
  • @Potato: It's a brand new feature. – Asaf Karagila Jul 18 '13 at 17:39
  • @AsafKaragila Wonderful. – Potato Jul 18 '13 at 17:58
  • Since the answers so far seem to miss the difficulty, let me make here some general remarks: A constructive proof of the Gelfond-Schneider result for $x=\sqrt 2^\sqrt2$ *cannot* begin by assuming that $x$ is algebraic, hoping to derive a contradiction. If we tried to follow this outline, we would have to begin with the assumption that $x$ is *not transcendental*, which without excluded middle is not (necessarily) the same as saying that it is algebraic. But then the proof vanishes, as we cannot use the putative polynomial of $x$ to derive estimates (we do not have a polynomial anymore)! – Andrés E. Caicedo Jul 19 '13 at 05:00
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    @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 http://www.cs.nyu.edu/pipermail/fom/2005-March/008846.html – Carl Mummert Jul 19 '13 at 11:44
  • Yes, Carl, but without excluded middle, "not not algebraic" does not give you "algebraic", does it? (And the OP seems to be requiring to avoid the use of excluded middle.) Anyway, you know this better than I do. Maybe I'm overlooking something. (If this is not an issue, then yes, the bounds in these arguments are obtained effectively.) – Andrés E. Caicedo Jul 19 '13 at 14:49
  • Mark Sapir's [answer](http://mathoverflow.net/a/138291/6085) on MathOverflow. – Andrés E. Caicedo Apr 26 '14 at 02:06
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    How could a proof of $\textit{ir}$rationality be constructive? – Jack M May 29 '14 at 16:43
  • possible duplicate of [Deciding whether $2^{\sqrt2}$ is irrational/transcendental](http://math.stackexchange.com/questions/173804/deciding-whether-2-sqrt2-is-irrational-transcendental) – Alex M. Sep 16 '15 at 10:55
  • @JackM, it can't be. [Discussion in chat about it.](https://chat.stackexchange.com/transcript/message/39379412#39379412) – Wildcard Aug 12 '17 at 05:19

1 Answers1


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.

Shuhao Cao
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    Not a constructive argument (in the sense of mathematical logic), though. – Andrés E. Caicedo Jul 19 '13 at 05:00
  • @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 05:07
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    I don't know if there is one. It seems a difficult question. – Andrés E. Caicedo Jul 19 '13 at 05:11
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    To all the voters out there: the OP asks whether the irrationality of ${\sqrt 2} ^{\sqrt 2}$ can be proved CONSTRUCTIVELY. I fail to see how this upvoted answer addresses the question. – Alex M. Jun 05 '17 at 20:50
  • @AlexM. This question shows up as the answerable one in a "duplicate of..." chain for a number of questions that were not asking for this requirement. Some of them weren't even asking about $\sqrt{2}^\sqrt{2}$. – GPhys Jun 19 '17 at 06:35
  • Why should increasing the number of interpolation points makes the error goes to zero? This is not a general fact of interpolation, right? – Wirdspan Sep 10 '20 at 14:55