# 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
If we don't even know if $\exp(e)$ is algebraic or not... –  Guess who it is. Dec 5 '10 at 9:41
@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
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
@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

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
Unfortunately Schanuel's Conjecture (that if $n$ complex numbers are linearly independent over the rationals, there are at most $n$ algebraic dependencies between those $n$ numbers and their exponentials) is completely intractable. –  HTFB Jun 3 '13 at 9:20

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 mathematical symbol $e$ is irrational, so raising it to the power of about any other number (except for $0$, of course) would cause it to be totally irrational, plus in a power-of-$10$ representation, it would look like this: $$\Huge{10^{{10^{10}}^{33.94704838165742}}}$$which would probably represent it as an irrational number.

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Before you freak out, I made the exponent part huge by using latex with the word "huge" with a backslash to the left (don't forget the starting and ending dollar signs). –  ReliableMathBoy Feb 8 at 18:49
Before freaking out even more, @MathNoob made it bigger by putting the same word in, except the H has been capitalized there. –  ReliableMathBoy Feb 8 at 18:53
Consider the graph of the function $e^x$. It intersects the line $y = c$ for every positive integer $c$. –  hunter Feb 8 at 18:55
Okay, I'll stop. –  ReliableMathBoy Feb 8 at 18:55
I'm not freaking out about the typesetting; the problem is that, weasel words aside, this answer is just wrong. As Hunter implies in his comment, there are plenty of real numbers $r$ with $e^r$ rational or even integral, and aside from probabilistic arguments we have no reason whatsoever to believe that $e^{e^{79}}$ isn't one of them. –  Steven Stadnicki Feb 8 at 19:14