Does some real non-integral $x$ exist such that $x^x$ equals a natural number?
Thanks, Tom
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1Try $x = 2.3161454958785612301325503$. With greater precision than I can give here, you can get $x^x$ very close to exactly $7.0$. – Mr. Brooks Jul 23 '14 at 21:35
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The function $x^x$ is continuous, and becomes very large for large $x$. It follows by the Intermediate Value Theorem that every integer $n\ge 1$ is $x^x$ for some real $x$.
André Nicolas
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Thanks, this is so obvious now.
Will x always be transcendental or could it be rational was probably what I was thinking?
– TomCV Jul 23 '14 at 21:31 -
2@TomCV If $x>0$ and $x^x\in\mathbb N$, then $x$ is either rational or transcendental; that is, it cannot be an irrational algebraic number. See the Gelfond–Schneider theorem. – triple_sec Jul 23 '14 at 21:34
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You are welcome. Your question, as mentioned by triple_sec, is settled by Gelfond-Schneider. – André Nicolas Jul 23 '14 at 21:38
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@AndréNicolas I'm still curious about whether such an $x$ can ever be a non-integer rational, or must it always be either whole or transcendental. – triple_sec Jul 23 '14 at 21:40
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1Yeah, it seems like it should be impossible for non-integer rationals, but I may be missing an obvious case. @triple_sec – Thomas Andrews Jul 23 '14 at 21:42
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2The real question is, when is $$e^{W(\ln{n})}$$ rational? With $n \in \mathbb{N}$ – ClassicStyle Jul 23 '14 at 21:45
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@AndréNicolas I see it now. If $x$ were a non-integer rational and $x^x=k$ for some $k\in\mathbb N$, then $x=m/n$ for some positive coprime integers $m,n$ such that $n\neq1$. But then, $m^m=n^m\cdot k^n$, so $m$ and $n$ must have at least one common prime factor—a contradiction. Is this correct? – triple_sec Jul 23 '14 at 21:51
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