# Prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using a theorem

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

We know 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.

This question has been asked previously on math.SE without receiving any answers.

• This is the relevant MSE thread. Jul 31 '13 at 14:40
• You want something a bit more precise. For example, I expect you do not want to deduce this from Kuzmin's result preceding Gel'fond-Schneider. Jul 31 '13 at 14:42
• @Andres Caicedo:Thank you very much for good information. As you wrote, the answer on your page is not what I want. Jul 31 '13 at 14:55
• What leads you to expect that this should be possible? There's only one context in which I've seen a discussion of proving something about $\sqrt 2^{\sqrt 2}$ without using Gelfond-Schneider; see math.hmc.edu/funfacts/ffiles/30002.3-5.shtml for example. Jul 31 '13 at 15:28
• $\left(\sqrt2^\sqrt2\right)^2=2^\sqrt2$ appears to be irrational, and it looks like an easier thing to prove... Jul 31 '13 at 15:43

You do not need to use Gelfond-Schneider theorem, you can just repeat one of the well known easy proofs of that theorem. For example, a much stronger theorem is proved in an Appendix to Lang's "Algebra", the proof is only 4 pages long. The proof uses only elementary linear algebra, some calculus and first notions of Galois theory. That proof can be significantly shortened if you want to prove irrationality only.

Edit I have written a more or less complete proof here . Galois theory is not needed there (except for the fact that the norm of an integral element is an integer), but one needs the maximal modulus theorem for analytic functions. I am sure that can be avoided also. From calculus, one needs the Taylor formula (no integration is required).

• That theorem also uses a little bit of complex analysis --- the statement of the theorem mentions poles of meromorphic functions. It also mentions transcendence degree, though perhaps you count that among the "first notions of Galois theory." You also need to know about integral closures, free modules, dual basis, maximum modulus principle, .... Aug 1 '13 at 3:07
• @GerryMyerson: All that is not needed if you deal with the function $2^x$, and only that function is needed for the question. I think the whole proof would just be 2 pages long. The idea is very straightforward (I had that theorem on my oral qual exam for graduate school 35 years ago).
– user6976
Aug 1 '13 at 3:16
• @Mark Sapir. A brilliant proof and a clear flow of ideas. The only problem is that almost every particular formula has a misprint in it, starting with $k<r$, which should be $k<n$, and ending with the outlandish choice of $R$ at the culmination moment. If you could correct all these stupid typos that severely restrict the list of potential readers of your nice opus, I'll make all graduate students in my class read it :-). Sep 5 '13 at 1:08
• @fedja: Thanks, but the proof is Lang's (possibly with a co-author), and your students should read Lang. In fact the main ideas go back to Hermite and it would be useful for students to compare Hermite's proofs with Lang's. The misprints are the results of me trying to use LaTeX in wordpress (comparing to MO, this turned out to be difficult).
– user6976
Sep 5 '13 at 1:43
• @Mark Sapir Archimedes, Mark, Archimedes! The guy knew everything, just said half of it, wrote quarter of it, and the monks took care of destroying most of his writings ;). Yes, they should read Lang, Hermite, etc., but after they see that the stuff is neat and not over their heads, and your proof is a perfect preparatory text for that (minus misprints, of course). :-). Sep 5 '13 at 2:08