40
$\begingroup$

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?

$\endgroup$
  • 9
    $\begingroup$ See this. $\endgroup$ – David Mitra Jul 18 '13 at 15:32
  • 1
    $\begingroup$ @DavidMitra: many thanks, I was not aware of that theorem. $\endgroup$ – Ron Gordon Jul 18 '13 at 15:43
  • 1
    $\begingroup$ One can find this result, and its proof, in Ivan Niven's book Irrational numbers. $\endgroup$ – David Mitra Jul 18 '13 at 15:46
  • 1
    $\begingroup$ 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.) $\endgroup$ – Andrés E. Caicedo Jul 18 '13 at 15:48
  • 2
    $\begingroup$ @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 cs.nyu.edu/pipermail/fom/2005-March/008846.html $\endgroup$ – Carl Mummert Jul 19 '13 at 11:44
11
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Not a constructive argument (in the sense of mathematical logic), though. $\endgroup$ – Andrés E. Caicedo Jul 19 '13 at 5:00
  • $\begingroup$ @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. :) $\endgroup$ – Shuhao Cao Jul 19 '13 at 5:07
  • 1
    $\begingroup$ I don't know if there is one. It seems a difficult question. $\endgroup$ – Andrés E. Caicedo Jul 19 '13 at 5:11
  • 3
    $\begingroup$ 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. $\endgroup$ – Alex M. Jun 5 '17 at 20:50
  • $\begingroup$ @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 $\sqrt{2}^\sqrt{2}$. $\endgroup$ – GPhys Jun 19 '17 at 6:35

protected by Alex M. Jun 5 '17 at 16:47

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?