# Integers in a logarithmic relation

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• Jul 31st 2010, 10:39 AM
Demandeur
Quote:

Originally Posted by Also sprach Zarathustra
You misunderstood me, but I don't blame you, my English is very bad, sorry for that...

Not at all. On the contrary, your English is very good. It was my fault.
Quote:

By the way, who it is in your picture avatar...?
The one and only one Carl Friedrich Gauss.
• Jul 31st 2010, 11:00 AM
Also sprach Zarathustra
Quote:

Not at all. On the contrary, your English is very good. It was my fault.
Thank you for that...

Quote:

The one and only one Carl Friedrich Gauss.
I'm ashamed...
• Jul 31st 2010, 01:03 PM
Vlasev
Quote:

Originally Posted by Also sprach Zarathustra

I'm ashamed...

It's ok, he's young in that picture!
• Aug 1st 2010, 01:22 PM
awkward
I don't know whether or not $\displaystyle x = \sqrt{2}^\sqrt{2}$ is rational or not, but there is an interesting proof in which this quantity figures.

Problem: Can $\displaystyle a^b$ be rational if a and b are irrational?

Consider $\displaystyle x = \sqrt{2}^\sqrt{2}$ . If x is rational then we have our example. If not, and x is irrational, then consider $\displaystyle x^{\sqrt{2}} = \sqrt{2}^2 = 2$, and we again have an example of rational $\displaystyle a^b$ with irrational a, b. So whether x is rational or not, we have the required example.

This is an example of (to me) an absolutely infuriating proof. We know either x or $\displaystyle x^\sqrt{2}$ is an example of rational $\displaystyle a^b$ with irrational a and b, but we don't know which, and the proof gives us no clue!

This proof is discussed in "The Princeton Companion to Mathematics".
• Aug 1st 2010, 03:09 PM
Vlasev
The Gelfond-Schneider theorem says that $\displaystyle \sqrt{2}^{\sqrt{2}}$ is transcendental and hence irrational. But I hear you. it's quite a nasty little existence proof!
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