Talk:Gelfond–Schneider theorem

A question .... do there exsist transcendental numbers that are not of the form a^b (with a real, b not both real and rational)? In other words, does tehconverse of the theorum hold? I feel the answer to this should be included in the main article - though I apreciate that it may currently be an open question. Tompw 14:15, 23 February 2006 (UTC)

There must be, since there are only countably many numbers of the a^b with a and b algebraic, but there are uncountably many transcendental numbers. Ben Standeven 21:37, 10 April 2007 (UTC)

A valid nonconstructive proof. Ironic. —Preceding unsigned comment added by 216.163.255.2 (talk) 21:15, 13 December 2010 (UTC)

Cantor's diagonal argument, as well as enumeration of algebraic numbers by integers, are perfectly constructive.—Emil J. 10:59, 14 December 2010 (UTC)

@Tompw : The correct spelling is "theorem". 2607:F140:6000:6:301E:C1F5:1DC0:CFEB (talk) 10:19, 22 April 2016 (UTC)

The case α=1

I really think it's more perspicuous to rule out this case than have it lurking behind the non-zero logarithm criterion: I for one was confused. And it doesn't detract from the power of the theorem, since if you want it for exp(2πin) just apply the theorem to (-1) instead. Doctor Adachi (talk) 22:26, 20 November 2009 (UTC)

External link

The link to a "proof" of the theorem does not contain a proof, it is a set of class notes and the first sentence is "we begin by stating some results without proof". Patronus Potter (talk) 06:02, 9 July 2010 (UTC)

If you read it past the first sentence, you will find that it actually does include a fairly complete proof of the Gelfond–Schneider theorem (it starts in the middle of page 27), albeit under additional assumptions that α and β are real and α > 0. The "results without proof" are some variants and generalizations of the theorem which are only mentioned in the introductory part to give more context.—Emil J. 13:40, 9 July 2010 (UTC)

Proof

The article should contain a synopsis of the proof written to be accessible to a general audience. — Preceding unsigned comment added by 81.28.91.31 (talk) 13:16, 26 January 2015 (UTC)

Is $b$ irrational, or just not rational?

Since this theorem seems to apply for any value of $b\in \mathbb {C} \setminus \mathbb {Q}$, shouldn't the statement of the theorem, state this? It currently says that $b$ is irrational, which means it is a real number, and so wouldn't allow numbers with nonzero imaginary part. --Lukeuser (talk) 03:30, 2 March 2020 (UTC)