Is $e^e$ irrational?

Question

I was surprised to find out that the following question is open: Is $e^e$ transcedental?

According to Wikipedia, a positive answer to Schanuel's conjecture implies "yes" to the above question.

My questions:

1) Can we at least prove that $e^e$ is irrational? Or is this also open?

2) Given that $e^e$ is irrational, does it follow that $e^e$ is transcedental?

Added comment: For (2) I mean "Does the knowledge that $e^e$ is irrational help with the proof that $e^e$ is transcedental?"

I doubt that this can be proved in an easy way.What do you mean in (2)? — Konstantinos Gaitanas, Feb 29 '16 at 19:15
@Paul: What do you mean? It is false? But then the only options for $e^e$ are either rational, or algebraic. — Ioannis Souldatos, Feb 29 '16 at 19:16
Good question. Trying for 1, and yes, for 2, not necessarily. — ABcDexter, Feb 29 '16 at 19:16
@KonstantinosGaitanas: I mean "Prove that $e^e$ is transcedental, given the assumption that it is irrational" — Ioannis Souldatos, Feb 29 '16 at 19:16
@Paul: My understanding of (2) was that he was wondering if $e^e$ might have some special property that would allow us to conclude that $e^e$ is transcendental if we knew that $e^e$ was irrational. I don't know the answer, but I very strongly suspect that the answer is NO (i.e. we don't at present know that $e^e$ has any such special property). — Dave L. Renfro, Feb 29 '16 at 19:18
Sidenote: According to this mathoverflow answer by Matt Papanikolas, at least one of $e^e$ and $e^{e^2}$ is transcendental. — Peter Woolfitt, Feb 29 '16 at 19:23
Wow Stephan Schanuel was a professor at my university before he passed away — Triatticus, Feb 29 '16 at 19:27

NoetherianCheese · Accepted Answer · 2016-02-29T19:45:50.497

8

About (1), it is still unknown whether $e^e$ is irrational or not, according to Wikipedia.

https://en.wikipedia.org/wiki/Irrational_number#Open_questions

Even more interesting, according to Gelfond's Theorem, $a^b$ is transcendental (therefore irrational) if $a$ is algebraic (and $\not\in\{0,1\}$) and if $b$ is irrational and algebraic.

http://mathworld.wolfram.com/GelfondsTheorem.html

This theorem can be used to prove that $e^\pi$ is transcendental and therefore irrational.

edited Feb 29 '16 at 19:45

answered Feb 29 '16 at 19:25

NoetherianCheese

506

This answers (1). Let me wait to see if anyone has any ideas about (2) and I will mark the answer correct. – Ioannis Souldatos Feb 29 '16 at 20:51
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Lindemann–Weierstrass theorem: If $a\neq 0$ is algebraic, then $e^a$ is transcendental. It is written in the list of transcendental numbers. – user236182 Feb 29 '16 at 21:32
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How does Gelfond's theorem show $e^\pi$ is irrational? $\pi$ is certainly not rational, so it doesn't apply in an obvious way. – Richard Rast Apr 06 '16 at 17:17
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@Richard $e^\pi = (-1)^(-i)$ – NoetherianCheese Apr 06 '16 at 22:00

Is $e^e$ irrational?

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