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Usuario:Jgomez53/Esbozos2 - Wikipedia, la enciclopedia libre

Usuario:Jgomez53/Esbozos2

De Wikipedia, la enciclopedia libre

In mathematics, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers. It states that if α1,...,αn are algebraic numbers which are linearly independent over the rational numbers \Bbb{Q}, then e^{\alpha_1},\ldots,e^{\alpha_n} are algebraically independent over \Bbb{Q}; in other words the extension field \mathbb{Q}(e^{\alpha_1}, \ldots,e^{\alpha_n}) has transcendence degree n over \Bbb{Q}.

An equivalent formulation, found in reference [1] below, 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 theorem is named for Ferdinand von Lindemann and Karl Weierstrass. Lindemann proved in 1882 that eα is transcendental for every non-zero algebraic number α, thereby establishing that π is transcendental (see below). Weierstrass proved the above more general statement in 1885.

The theorem, along with the Gelfond-Schneider theorem, is generalized by Schanuel's conjecture.

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[editar] Transcendence of e and π

The transcendence of e and π are direct corollaries of this theorem.

Suppose α is a nonzero algebraic number; then {α} is a linearly independent set over the rationals, and therefore by the first formulation of the theorem {eα} is an algebraically independent set; or in other words eα is transcendental. In particular, e1 = e is transcendental. (A more elementary proof that e is transcendental is outlined in the article on transcendental numbers.)

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.

Now, we prove that π is transcendental. If π were algebraic, 2πi would be algebraic too (since 2i is algebraic), and then by the Lindemann-Weierstrass theorem ei = 1 (see Euler's formula) would be transcendental, which is absurd.

A slight variant on the same proof will show that if α is a nonzero algebraic number then sin(α), cos(α), tan(α) and their hyperbolic counterparts are also transcendental.

[editar] p-adic conjecture

The p-adic Lindemann–Weierstrass conjecture is that a p-adic analog of this statement is also true: suppose p is some prime number and α1,...,αn are p-adic numbers which are algebraic over Q and linearly independent over Q, such that | αi | p < 1 / p for all i; then the p-adic exponentials e^{\alpha_1}, \ldots, e^{\alpha_n} are p-adic numbers that are algebraically independent over Q.

[editar] See also

  • Proof that e is irrational

[editar] References

  1. Alan Baker, Transcendental Number Theory, Cambridge University Press, 1975, ISBN 052139791X. Chapter 1, Theorem 1.4.
  2. F. Lindemann, Über die Zahl π, Mathematische Annalen, vol. 20 (1882), pp. 213-225.

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