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 $e^ {\ alpha_1} \ cdots e^ {\ alpha_n} \,$ are algebraically independent on the algebraic numbers; in other words, the unit ${e^ {\ alpha_1} \ cdots e^ {\ alpha_n}} \,$ has the Degré of transcendence N on $\ Bbb {Q}$ . An equivalent formulation of the theorem is the following one: if $\ alpha_1, \ cdots, \ alpha_n \,$ is distinct algebraic numbers then $e^ {\ alpha_1} \ cdots e^ {\ alpha_n}$ 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 { $e^ {\ alpha} \,$ } has a degree of transcendence one on the rational numbers; in other words $e^ {\ alpha} \,$ is transcendent. By using the other formulation, we can argue that if { $0, \ alpha \,$ } is a whole of distinct algebraic numbers, then the unit { $e^0, e^ {\ alpha} \,$ } = { $1, e^ {\ alpha} \,$ } is linearly independent on the algebraic numbers, and thus $e^ {\ alpha} \,$ is immediately seen as being transcendent. In particular, $e^1 = E \,$ is transcendent. Therefore, if $\ beta = e^ {I \ alpha} \,$ is transcendent, then its real part and its imaginary part:

\ cos (\ alpha) = Re (\ beta) = \ frac {\ beta + \ beta^ {- 1}} {2} \,

and

\ sin (\ alpha) = Im (\ beta) = \ frac {\ beta - \ beta^ {- 1}} {2i}

is to it aussi.

Consequently, if

\ pi \,

was algebraic,

\ cos (\ pi) = - 1 \,

and

\ sin (\ pi) = 0 \,

would be transcendent, which proves by the absurdity that

\ pi \,

is not algebraic, in other words that it is transcendent.

Conjecture p - adic

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 $e^ {\ alpha_1} \ cdots e^ {\ alpha_n}$ is transcendent algebraically independent.

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