needs more information

What beta? 07:13, 9 Aug 2004 (UTC)

Any nonzero algebraic number α gives us a set {α} which is trivially a linearly independent set over the rationals, and hence eα is immediately seen to be transcendental.

Is this argument valid? The problem I have with it is that simply because {e^α} is a linearly independent set over the algebraic numbers doesn't mean that the number e^α is transcendental. For instance, {log 2} is linearly independent over the rationals (if a*(log 2) = 0 for rational a, then a = 0.), and also {e^{log 2} = 2} is linearly independent set over the algebraic numbers (if a*2 = 0 for algebraic a, then a = 0.), yet 2 is not transcendental. I also understand the urge to condense everything to be elegant, but I think expressing things out explicitly in terms of linear combinations is still helpful for people who might not be able to immediately mentally untangle "linearly independent" or "algebraically independent". Revolver 04:01, 2 Nov 2004 (UTC)