Lindemann–Weierstrass theorem

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In mathematics, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers. It states that if α₁,...,α_n 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,...,α_n 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 e^α 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.

1 Transcendence of e and π
2 p-adic conjecture
3 See also
4 References

[edit] 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 {e^α} is an algebraically independent set; or in other words e^α 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, e^α} = {1, e^α} is linearly independent over the algebraic numbers and in particular e^α 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 e^2πi = 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.

[edit] 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,...,α_n are p-adic numbers which are algebraic over Q and linearly independent over Q, such that $| α i | p < 1 / p$ 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.

[edit] See also

Proof that e is irrational

[edit] References

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

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Categories: Exponentials | Mathematical theorems | Number theory | Pi