Talk:Lindemann–Weierstrass theorem

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[edit] 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 {eα} is a linearly independent set over the algebraic numbers doesn't mean that the number eα 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 {elog 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.

[edit] The part about Baker's characterization needs a fix.

An equivalent formulation (Baxter 1975, Chapter 1, Theorem 1.4), is the following: If α1,...,αn are distinct algebraic numbers, then the exponentials e^{\alpha_1},\ldots,e^{\alpha_n} are linearly independent over the algebraic numbers.

The part about Baker's theorem needs a fix, extra hypothesis are required (a non real fifth root of unity provides a counter example) —Preceding unsigned comment added by 153.90.244.6 (talk) 18:57, 5 December 2007 (UTC)

A better counterexample: ei + e − i = 0. Unless Baxter's theorem concerns only real numbers this is definitely false.  Grue  22:42, 5 December 2007 (UTC)
But ei + e − i = 2cos(1), not zero. JRSP (talk) 23:47, 23 December 2007 (UTC)
OMG, did I say something stupid on Wikipedia again. Indeed, you are right, and if ex + ey = 0 then x = y + πi + 2πik, so they cannot be both algebraic. Now we need to figure out if the original counterexample holds (probably not, since roots of 1 have transcendental logarithms), and whether the formulation is actually true.  Grue  11:41, 24 December 2007 (UTC)
According to planetmath.org "An equivalent version of the theorem states that if α1,...,αn are distinct algebraic numbers over , then e^{\alpha_1},\ldots,e^{\alpha_n} are linearly independent over \mathbb{Q}." (not over the algebraic numbers). I also found the proof there. (we could add the external link). In the proof they say : "Theorem 3: If α1,...,αn are algebraic and distinct, and if β1,...,βn are algebraic and non-zero, then \beta _1 e^{\alpha _1 } + ... + \beta _n e^{\alpha _n } \ne 0 " which basically says the exponentials are LI over the algebraic numbers. JRSP (talk) 15:01, 24 December 2007 (UTC)
Let's see anon ip's argument: "Take α to be a non-real fifth root of unity. Then α,α2,α3,α4 are distint, but α satisfies 1 + x + x2 + x3 + x4 = 0." This only says that 1,α,α2,α3,α4 are linearly dependent but does not show if e,e^{\alpha},\ldots,e^{\alpha^4} are LD. JRSP (talk) 15:24, 24 December 2007 (UTC)



This example depends on the "second formulation" so I moved it here:

Alternatively, using the second formulation of the theorem, we can argue that if α is a nonzero algebraic number, then {0, α} is a set of distinct algebraic numbers, and so the set {e0,eα} = {1,eα} is linearly independent over the algebraic numbers and in particular eα can't be algebraic and so is transcendental.

JRSP (talk) 01:22, 23 December 2007 (UTC)

The reference in the second paragraph of the article should be to Baker not Baxter.Fathead99 (talk) 12:07, 30 January 2008 (UTC)