# Theorem of Lindemann-Weierstrass

In Mathematical, the theorem of Lindemann-Weierstrass establishes that if $\ alpha_1, \ cdots, \ alpha_n \,$ are algebraic numbers which are Linéairement independent S on the rational numbers, then $e^ \left\{\ alpha_1\right\} \ cdots e^ \left\{\ alpha_n\right\} \,$ are algebraically independent on the algebraic numbers; in other words, the unit $\left\{e^ \left\{\ alpha_1\right\} \ cdots e^ \left\{\ alpha_n\right\}\right\} \,$ has the Degré of transcendence N on $\ Bbb \left\{Q\right\}$. An equivalent formulation of the theorem is the following one: if $\ alpha_1, \ cdots, \ alpha_n \,$ is distinct algebraic numbers then $e^ \left\{\ alpha_1\right\} \ cdots e^ \left\{\ alpha_n\right\}$ are linearly independent on the algebraic numbers.

The theorem was named thus in the honor of Ferdinand von Lindemann, which proved the particular case of the transcendence of $\ pi \,$, and Karl Weierstrass.

## Transcendence of E and π

The transcendence of E and that of $\ pi \,$ is immediate corollaries of this theorem. Let us suppose that $\ alpha \,$ is an algebraic number different from zero; then {$\ alpha \,$} is a linearly independent unit on the rational numbers, and consequently {$e^ \left\{\ alpha\right\} \,$} has a degree of transcendence one on the rational numbers; in other words $e^ \left\{\ alpha\right\} \,$ is transcendent. By using the other formulation, we can argue that if {$0, \ alpha \,$} is a whole of distinct algebraic numbers, then the unit {$e^0, e^ \left\{\ alpha\right\} \,$} = {$1, e^ \left\{\ alpha\right\} \,$} is linearly independent on the algebraic numbers, and thus $e^ \left\{\ alpha\right\} \,$ is immediately seen as being transcendent. In particular, $e^1 = E \,$ is transcendent. Therefore, if $\ beta = e^ \left\{I \ alpha\right\} \,$ is transcendent, then its real part and its imaginary part:

$\ cos \left(\ alpha\right) = Re \left(\ beta\right) = \ frac \left\{\ beta + \ beta^ \left\{- 1\right\}\right\} \left\{2\right\} \,$ and
$\ sin \left(\ alpha\right) = Im \left(\ beta\right) = \ frac \left\{\ beta - \ beta^ \left\{- 1\right\}\right\} \left\{2i\right\}$ is to it aussi.
Consequently, if $\ pi \,$ was algebraic, $\ cos \left(\ pi\right) = - 1 \,$ and $\ sin \left(\ pi\right) = 0 \,$ would be transcendent, which proves by the absurdity that $\ pi \,$ is not algebraic, in other words that it is transcendent.

The p-adic conjecture of Lindemann-Weierstrass affirms that this result is true for the numbers p-adic: if $\ alpha_1, \ cdots, \ alpha_n \,$ is a whole of linearly independent algebraic numbers on the rational numbers such as $|\ alpha_i|_p < 1/p$ for a certain prime number p , then exponential p-adic the $e^ \left\{\ alpha_1\right\} \ cdots e^ \left\{\ alpha_n\right\}$ is transcendent algebraically independent.