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Set $x_0:=1$ and for every $n\in\mathbb{N}$ set $x_n:=2^{\frac{1}{2}x_{n-1}}$, so $$x_1=\sqrt{2},\; x_2=\sqrt{2}^{\sqrt{2}}, x_3=\sqrt{2}^{\sqrt{2}^{\sqrt{2}}},\ldots$$ Undoubtedly the $x_n$ are irrational for all $n\in\mathbb{N}=\{1, 2,\ldots\}$. (How) can we prove that?

Damian Reding
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  • Definitely x_{0} is rational but I feel that all following terms are irrational. Try proving that by induction. –  Oct 29 '16 at 13:02
  • @Deepak I have edited my answer. – SchrodingersCat Oct 29 '16 at 13:11
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    @Deepak It's not clear to me that Gelfond-Schneider can get us past $n = 2$; $x_3$ is an algebraic raised to a transcendental power, which could possibly be rational (see, for example, $2^{\log_2(3)}$). – Reese Johnston Oct 29 '16 at 13:20
  • @Reese You are correct. The theorem fails quickly here. I overlooked that. – Deepak Oct 30 '16 at 01:01
  • @Rohan: Have you tried proving it by induction? Do you have any reason to think it might work? – TonyK Oct 30 '16 at 13:56

1 Answers1

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Undoubtedly the $x_n$ are irrational for all $n∈N={1,2,…}$

The proof of that statement would earn you a fair amount of fame, since the irrationality of $x_n$ for $n>2$ is still an open problem.

Indeed, one can use the Gelfond-Schneider theorem to prove that $\sqrt 2^{\sqrt 2}$ is transcendental:

Theorem (Gelfond-Schneider). If $a$ and $b$ are algebraic numbers with $a ≠ 0, a ≠ 1$, and $b$ irrational, then any value of $a^b$ is a transcendental number.

However, currently we have no tools to assess the irrationality of numbers such as $\sqrt 2^{{\sqrt 2}^{\sqrt 2}}$ where the exponent is transcendental.

There exist however other related numbers whose rationality is known, for instance the infinite tower

$$ \sqrt 2^{{{\sqrt 2}^{{\sqrt 2}^{\ldots}}}} = \sqrt[\leftroot{-2}\uproot{2}4]{4}^{{\sqrt[\leftroot{-2}\uproot{2}4]{4}}^\ldots} = 2, $$

which is the value of $\displaystyle \lim_{n\to\infty} x_n$. In particular, we have the following lemma.

Lemma. Let $n$ be a positive integer different than $1$, $2$ and $4$. Then the infinite power tower $$\sqrt[\leftroot{-2}\uproot{2}n]{n}^{{\sqrt[\leftroot{-2}\uproot{2}n]{n}}^\ldots}$$ is transcendental.

In particular, $$\sqrt[\leftroot{-2}\uproot{2}3]{3}^{{\sqrt[\leftroot{-2}\uproot{2}3]{3}}^\ldots}$$ is a transcendental number.

Finally, I give the following interesting reference on infinite power towers:

Mladen V-M, Some Results On Infinite Power Towers, Notes on Number Theory and Discrete Mathematics (2010), 3, 18-24.

It is freely accessible at:

http://www.nntdm.net/papers/nntdm-16/NNTDM-16-3-18-24.pdf

Klangen
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