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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)

I see...you're secretly still using the fact that the number is algebraic. I was seeing that ({a} LI over Q) does not imply ({a} AI over A, which is true, but if you add a being alg. to hypothesis, it does go through. Revolver 02:51, 9 September 2005 (UTC)

[edit] needs date

If Lindemann proved pi is trancendental, we should cite the published proof, and at least give a date.