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In mathematics, the Lindemann-Weierstrass theorem states that if α1,...,αn are algebraic numbers which are linearly independent over the rational numbers, then <math>e^{\alpha_1} \cdots e^{\alpha_n}<math> are algebraically independent over the algebraic numbers; in other words the set <math>{e^{\alpha_1} \cdots e^{\alpha_n}}<math> has transcendency degree n over <math>\Bbb{Q}<math>. An equivalent formulation of the theorem is that if α1,...,αn are distinct algebraic numbers then <math>e^{\alpha_1} \cdots e^{\alpha_n}<math> are linearly independent over the algebraic numbers.
The theorem is named for Ferdinand von Lindemann, who proved the particular result that π is transcendental, and Karl Weierstraß.
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 {eα} has transcendence degree one over the rationals; or in other words eα is transcendental. Using the other formulation we can argue that if {0, α} is a set of distinct algebraic numbers, then the set {e0, eα} = {1, eα} is linearly independent over the algebraic numbers, and so eα is immediately seen to be transcendental. In particular, e1</sub> = e is transcendental. Also, if β = eiα is transcendental, so are the real and imaginary parts of β, Re(β) = (β + β−1)/2 and Im(β) = (β − β−1)/2i, and hence cos(α) = Re(β) and sin(α) = Im(β) are also. Therefore, if π were algebraic, cos(π) = −1 and sin(π) = 0 would be transcendental, which proves by contradiction π is not algebraic, and hence is transcendental.
p-adic conjecture
The p-adic Lindemann-Weierstrass conjecture is that this conjecture is also true for a p-adic analog: if α1,...,αn are a set of algebraic numbers linearly independent over the rationals such that <math>|\alpha_i|_p < 1/p<math> for some prime p, then the p-adic exponentials <math>e^{\alpha_1} \cdots e^{\alpha_n}<math> are algebraically independent transcendentals.
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