LindemannWeierstrass theorem
If ${\alpha}_{1},\mathrm{\dots},{\alpha}_{n}$ are linearly independent^{} algebraic numbers^{} over $\mathbb{Q}$, then ${e}^{{\alpha}_{1}},\mathrm{\dots},{e}^{{\alpha}_{n}}$ are algebraically independent^{} over $\mathbb{Q}$.
An equivalent^{} version of the theorem^{} that if ${\alpha}_{1},\mathrm{\dots},{\alpha}_{n}$ are distinct algebraic numbers over $\mathbb{Q}$, then ${e}^{{\alpha}_{1}},\mathrm{\dots},{e}^{{\alpha}_{n}}$ are linearly independent over $\mathbb{Q}$.
Some immediate consequences of this theorem:

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If $\alpha $ is a nonzero algebraic number over $\mathbb{Q}$, then ${e}^{\alpha}$ is transcendental over $\mathbb{Q}$.

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$e$ is transcendental over $\mathbb{Q}$.

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$\pi $ is transcendental over $\mathbb{Q}$. As a result, it is impossible to “square the circle”!
It is easy to see that $\pi $ is transcendental over $\mathbb{Q}(e)$ iff $e$ is transcendental over $\mathbb{Q}(\pi )$ iff $\pi $ and $e$ are algebraically independent. However, whether $\pi $ and $e$ are algebraically independent is still an open question today.
Schanuel’s conjecture is a generalization^{} of the LindemannWeierstrass theorem^{}. If Schanuel’s conjecture were proven to be true, then the algebraic independence of $e$ and $\pi $ over $\mathbb{Q}$ can be shown.
Title  LindemannWeierstrass theorem 
Canonical name  LindemannWeierstrassTheorem 
Date of creation  20130322 14:19:22 
Last modified on  20130322 14:19:22 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  11 
Author  CWoo (3771) 
Entry type  Theorem 
Classification  msc 12D99 
Classification  msc 11J85 
Synonym  Lindemann’s theorem 
Related topic  SchanuelsConjecutre 
Related topic  GelfondsTheorem 
Related topic  Irrational 
Related topic  EIsTranscendental 