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(somewhat related to my earlier question)

Let ${^n}a$ denote tetration $\underbrace{a^{a^{.^{.^{.^a}}}}}_{n \text{ times}}$ (or, defined recursively, ${^1}a=a$, ${^{n+1}}a=a^{({^n}a)}$).

The equation ${^4}x=2$ has a positive root $x=1.4466014324...$

Is it possible to prove it irrational?

Community
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Vladimir Reshetnikov
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1 Answers1

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It is a known open problem. It is also open for all equations of the form ${^n q}=2$ or ${^n q}=3$ for integer $n>3$. For $n=3$ the roots are known to be irrational, but not known to be algebraic or transcendental.

Liu Jin Tsai
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