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May 14[edit]

e^pi has been proven irrational, but e^e hasn't[edit]

It's kind of surprising that e^pi has been proven to be irrational but e^e hasn't. Any Wikipedia articles that talk about e^e's irrationality in any way?? Georgia guy (talk) 18:23, 14 May 2018 (UTC)[reply]

Not that I know of. We do have an article on the Gelfond–Schneider theorem, which explains the e^π case. I don't think it needs to be all that surprising. There are all sorts of such expressions that are almost certainly transcendental (because there's no reason they should be algebraic), but there simply aren't any known techniques for proving it. --Trovatore (talk) 18:55, 14 May 2018 (UTC)[reply]

What do they include besides e^e?? Georgia guy (talk) 18:56, 14 May 2018 (UTC)[reply]

Oh, for example e+π I'm pretty sure that's still open. Not that I was ever hooked in to that area of research. But it's the sort of thing that would probably show up in the popular press if anyone did manage to settle it. --Trovatore (talk) 18:58, 14 May 2018 (UTC)[reply]

There are several examples in Irrational_number#Open_questions. -- Meni Rosenfeld (talk) 23:08, 14 May 2018 (UTC)[reply]

I'll bet $100US vs. $1 that none of those will be proven to be rational. !!!! — Preceding unsigned comment added by Bubba73 (talk • contribs) 23:39, 14 May 2018 (UTC)[reply]

Well, that isn't hard. As with a lot of "open" questions in this general space, the right answer is not really in doubt. The problem is to find a proof of the answer we already have. --Trovatore (talk) 23:45, 14 May 2018 (UTC)[reply]

There is also Lindemann–Weierstrass_theorem I should mention. Ruslik_Zero 20:42, 14 May 2018 (UTC)[reply]

Also some examples at Transcendental_number#Possibly_transcendental_numbers. --JBL (talk) 00:19, 15 May 2018 (UTC)[reply]

The irrationality of e^e and much more would follow from Schanuel's conjecture.John Z (talk) 22:12, 15 May 2018 (UTC)[reply]

Very cool! Thanks for the link! --Trovatore (talk) 22:19, 15 May 2018 (UTC)[reply]