# How to show $e^{e^{e^{79}}}$ is not an integer

In this question, I needed to assume in my answer that $e^{e^{e^{79}}}$ is not an integer. Is there some standard result in number theory that applies to situations like this?

Much later addendum: when I asked this, I didn't expect it to be an open question. If it is open, if someone could give an answer to that effect, I would accept it simply to close off the question. If some author has stated something similar as an open question, that would be helpful for me to know; I'm not a number theorist and my knowledge of the field is not very deep.

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This is likely to be open, I think. – Qiaochu Yuan Dec 5 '10 at 8:21

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If we don't even know if $\exp(e)$ is algebraic or not... – J. M. Dec 5 '10 at 9:41

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@Qiaochu Yuan and J.M.: I wasn't aware of that. I knew (vaguely) that there are a lot of open problems in transcendence theory, but I was hoping that just the problem of being an integer was easier. – Carl Mummert Dec 5 '10 at 12:56

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It's not particularly feasible; the number of decimal digits is approximately $\log_{10}$ of the number, which is much larger than $10^{80}$, which is supposed to be an estimate for the number of atoms in the observable universe. – Carl Mummert Dec 6 '10 at 23:03

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@Carl This is a really nice question. I don't think it needs to be "closed". There are some really innocuous questions running around the site just because the person who asked them never accepted an answer. But I believe that this is a question that deserves to show up from time to time. Who knows, maybe at some point someone will have something important to say about it, if not to answer it for good =) – Adrián Barquero Mar 20 '11 at 16:29
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## 3 Answers

if $e^{e^{e^{79}}}$ is an integer then $e^{e^{e^{e^{79}}}}$ is not an integer (otherwise $e$ would be algebraic). Perhaps your arguments make sense with this number too.

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This is a good point. Unfortunately, for my example I need to know for sure whether it's an integer or not; knowing that at least one of two numbers is not an integer is not as useful. – Carl Mummert Jun 16 '11 at 0:14

The paper Chuangxun Cheng, Brian Dietel, Mathilde Herblot, Jingjing Huang, Holly Krieger, Diego Marques, Jonathan Mason, Martin Mereb, and S. Robert Wilson, Some consequences of Schanuel’s conjecture, Journal of Number Theory 129 (2009) 1464–1467, shows that $e,e^e,e^{e^e},\dots$ is an algebraically independent set, on the assumption of Schanuel's Conjecture. Maybe a close reading of that paper will suggest a way of applying the result to the 79-question.

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Maybe writing to the authors and telling them about this question directly could prove useful. – Adrián Barquero Jun 16 '11 at 4:38

Perhaps you could do something with the taylor series? Would be pretty nasty but youd only have to worry about the first few terms.

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@remus: Because it is too big perhaps. If you get something useful from Maple or Mathematica, please share. – Jonas Meyer Dec 19 '10 at 23:04

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@Jonas, @remus: Already $e^{e^{79}}$ has approximately as many digits as one cubic kilometre of air contains molecules. So it's already virtually impossible to say how many digits the number in question has. – Hendrik Vogt Dec 20 '10 at 8:57

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@remus: "Quite clearly not an integer"? $10^{0.30102999566398119521373889472449...}$ is 2. – Aryabhata Dec 22 '10 at 1:30

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The probability that the number is an integer is either 100% or 0% depending on whether it is or isn't. There's no random process here; the property is fixed ahead of time and you can't compare it to rolling a dice. – Michael Burge Mar 20 '11 at 20:09

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@Michael Burge: In fact; everything is fixed and everything is unfixed; et c'est la méme. This is why probabilistic is confusing; you cannot assert that an integer; before you see it; is a real number, which is why it could be a likelihood. – awllower Apr 2 '11 at 9:44
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