Is there an elementary proof that $2^{\sqrt{2}}$ is irrational?

Asked Jan 22 '24 at 03:54

Active Jan 22 '24 at 15:23

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The Gelfond-Schneider theorem states that if $a$ and $b$ are complex algebraic numbers such that $a \not\in \{0, 1\}$ and $b$ is irrational then $a^b$ is transcendental.

This answered the question of the nature of $2^{\sqrt{2}}$.

So this got me wondering:
Is there an more elementary proof (i.e., not using the G-S theorem) of the seemingly easier question of whether $2^{\sqrt{2}}$ is merely irrational , without going to the transcendental level.

I have played around with this and gotten absolutely nowhere.

That's my question.

Any ideas?

edited Jan 22 '24 at 15:23

Jose Avilez

12,710

asked Jan 22 '24 at 03:54

marty cohen

107,799

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Have you walked through the proof of the GS theorem for this particular special number? – Lee Mosher Jan 22 '24 at 04:10
Many years ago, I looked at a proof in (IIRC) LeVeque volume 2) and understood almost nothing. So, no. Feel free to try. As a great philosopher almost said, "A person's got to know their limitations." – marty cohen Jan 22 '24 at 04:23
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Why did you tag this [tag:complex-numbers]? – J. W. Tanner Jan 22 '24 at 04:36
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No, there is no easier way. The Gelfond-Schneider theorem (I avoid abbreviations like GS whenever possible) is a very deep result. – Peter Jan 22 '24 at 06:44
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The number $2^{\sqrt2}$ is called the Gelfond-Schneider constant. Perhaps this will help when searching through the literature. – Joe Jan 22 '24 at 15:40
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I did some searching, and Mark Sapir apparently has a half page proof of exactly what you want here. This was taken from the MathOverflow post that discusses both $2^{\sqrt2}$ and ${\sqrt2}^{\sqrt2}$. – Joe Jan 22 '24 at 15:44
Of similar context, although almost certainly not helpful for this particular question: Direct proof that $\pi$ is not constructible – Dave L. Renfro Jan 22 '24 at 19:01

Is there an elementary proof that $2^{\sqrt{2}}$ is irrational?

0 Answers0