6
$\begingroup$

Possible Duplicate:
How to show $e^{e^{e^{79}}}$ is not an integer

Is ${^5\pi}$ an integer? It is "obviously" not, right? But can we prove it?

Here ${^5\pi}$ means the result of tetration $\underbrace{\pi^{\pi^{\pi^{\pi^\pi}}}}_{5 \text{ times}}$.

$\endgroup$

marked as duplicate by Andrés E. Caicedo, Gerry Myerson, Asaf Karagila, J. M. is a poor mathematician, t.b. Dec 14 '11 at 5:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 4
    $\begingroup$ If you need to ask whether something can be proved, then it is not obviously true. (Except if your question is "do we need to take this as an axiom or does it follow from simpler principles?"). $\endgroup$ – Henning Makholm Dec 13 '11 at 23:08
  • 4
    $\begingroup$ Excuse my obvious irony, but I find it ironical that the obviously is obviously ironical. $\endgroup$ – Myself Dec 13 '11 at 23:27
  • 2
    $\begingroup$ @HenningMakholm: Of course, that is why I put "obviously" in quotes. I just mean that a possible proof that ${^5\pi}$ is not an integer would hardly be a surprising result for anyone, but the converse would really surprise many people. $\endgroup$ – Vladimir Reshetnikov Dec 13 '11 at 23:27
  • 3
    $\begingroup$ @JonasMeyer: This is the smallest number of the form ${^n\pi}$ for which I do not know the answer. Tetration is the first hyperoperator (after addition, multiplication and exponentiation) for which the question is not trivial. And $\pi$ is just a natural example of a transcendental number. $\endgroup$ – Vladimir Reshetnikov Dec 14 '11 at 1:15
  • 6
    $\begingroup$ Wolfram Alpha says that $(\phi^5-\tau^5)/\sqrt{5}$, where $\phi = (1+\sqrt{5})/2$ and $\tau = (1-\sqrt{5})/2$, is not an integer: wolframalpha.com/input/?i=Is+%28%28%281%2Bsqrt%285%29%29%2F2%29^5+-+%28%281-sqrt%285%29%29%2F2%29^5%29%2Fsqrt%285%29+an+integer%3F -- but it clearly is, because it's the fifth Fibonacci number, namely 5. (WA then gives a "decimal approximation" which is 5 followed by a couple thousand zeroes.) $\endgroup$ – Michael Lugo Dec 14 '11 at 1:38