current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 18 down vote favorite
12

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?

share|cite|improve this question
5  
See this. – David Mitra Jul 18 '13 at 15:32
1  
@DavidMitra: many thanks, I was not aware of that theorem. – Ron Gordon Jul 18 '13 at 15:43
1  
One can find this result, and its proof, in Ivan Niven's book Irrational numbers. – David Mitra Jul 18 '13 at 15:46
1  
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 Caicedo Jul 18 '13 at 15:48
1  
@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 – Carl Mummert Jul 19 '13 at 11:44

3 Answers 3

up vote 6 down vote

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.

share|cite|improve this answer
2  
Not a constructive argument (in the sense of mathematical logic), though. – Andrés 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
1  
I don't know if there is one. It seems a difficult question. – Andrés Caicedo Jul 19 '13 at 5:11
up vote 0 down vote

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

share|cite|improve this answer
    
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
up vote 0 down vote

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.

share|cite|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.