2 Mar
2014
2 Mar
'14
10:57 p.m.
Well, e^(pi*i) = -1 is algebraic not in {0,1}, and -i*sqrt(163) is an irrational algebraic. So Gelfond-Schneider applies. —Dan On Mar 2, 2014, at 8:52 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
We were discussing raising a transcendental number (e) to a transcendental power (pi*sqrt(163)). So I don't know what theorem applies, if any. Maybe e^(pi*sqrt(n)) really can be an integer for some nonzero positive integer n? It is an integer for n = -1, after all.