Talk:Skewes Number

Transcendentality
Schanuel's conjecture would imply the transcendentality of both Skewes Numbers:

Let a be equal to either 79 or 7.705.

Firstly, a is a nonzero algebraic number, so the Lindemann–Weierstrass theorem does imply the transcendence of e^a.

Secondly, a and e^a are linearly independent over Q, so Q(a, e^a, e^a, e^(e^a)) has transcendence degree of at least 2 over Q. Therefore, e^a and e^(e^a) are algebraically independent over Q.

Thirdly, a, e^a, and e^(e^a) are linearly independent over Q, so Q(a, e^a, e^(e^a), e^a, e^(e^a), e^(e^(e^a))) has transcendence degree of at least 3 over Q. Therefore, e^a, e^(e^a), and e^(e^(e^a)) are algebraically independent over Q. With a = 79, the transcendentality of the first Skewes Number follows.

And fourthly, a, e^a, e^(e^a), and e^(e^(e^a)) are linearly independent over Q, so Q(a, e^a, e^(e^a), e^(e^(e^a)), e^a, e^(e^a), e^(e^(e^a)), e^(e^(e^(e^a)))) has transcendence degree of at least 4 over Q. Therefore, e^a, e^(e^a), e^(e^(e^a)), and e^(e^(e^(e^a))) are algebraically independent over Q. With a = 7.705, the transcendentality of the second Skewes Number follows. --84.61.186.220 18:45, September 29, 2014 (UTC)