I can prove using the Gelfond–Schneider theorem that the positive root of the equation $x^{x^x}=2$, $x=1.47668433...$ is an irrational number. Is it possible to prove it is transcendental?
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Vladimir Reshetnikov
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2I couldn't find a solution but from all the things I tried, the only one that made me feel like I might be on the right way was noticing that $\left(x^x\right)^{x^x}=x^{xx^x}=\left(x^{x^x}\right)^x=2^x$ which, if you assume $x$ is irrational algebraic, is transcendental. – xavierm02 Apr 26 '13 at 22:52

1I can't help at all with your question, but I'm curious how you used GelfondSchneider to prove $x$ is irrational. If $x$ and $x^x$ both happen to be rational, what's the problem? – Jason DeVito Apr 26 '13 at 23:13

5For detailed proof see [Marshall, Ash J., and Tan, Yiren, "A rational number of the form $a^a$ with $a$ irrational", Mathematical Gazette 96, March 2012, pp. 106109.](http://condor.depaul.edu/mash/atotheamg.pdf) – Vladimir Reshetnikov Apr 27 '13 at 18:12
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I believe, it is a known open problem. Ditto for ${^3 x}=3$, ${^3 x}=4$, ${^3 x}=5$ (left superscript denotes tetration).
Oksana Gimmel
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In addition to the other answer and comments, the following might prove to be useful. Define the sequence $s_n$ as follows: $s_0:=1$ and for all positive integers $n\geq 1$ let $s_n:=2^{\frac{1}{2}s_{n1}}$. We have
$$ s_1=\sqrt{2},\; s_2=\sqrt{2}^{\sqrt{2}}, s_3=\sqrt{2}^{\sqrt{2}^{\sqrt{2}}},\ldots $$
The limit of this sequence is
$$ \lim_{n\to\infty}s_n=\sqrt 2^{{{\sqrt 2}^{{\sqrt 2}^{\ldots}}}}=2. $$
The solution of the power tower $x^{x^{{x}^{\ldots}}}=2$ is therefore $x=\sqrt 2=1.41421356237309\ldots$ which is irrational but not transcendental.
Klangen
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