Lindemann-Weierstrass theorem LindemannWeierstrassTheorem 2004-04-21 18:33:11 2006-03-28 12:05:10 Theorem % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here If $\alpha_1,\ldots,\alpha_n$ are linearly independent algebraic numbers over $\mathbb{Q}$, then $e^{\alpha_1},\ldots,e^{\alpha_n}$ are algebraically independent over $\mathbb{Q}$. An equivalent version of the theorem \PMlinkescapetext{states} that if $\alpha_1,\ldots,\alpha_n$ are distinct algebraic numbers over $\mathbb{Q}$, then $e^{\alpha_1},\ldots,e^{\alpha_n}$ are linearly independent over $\mathbb{Q}$. Some immediate consequences of this theorem: \begin{itemize} \item If $\alpha$ is a non-zero algebraic number over $\mathbb{Q}$, then $e^{\alpha}$ is transcendental over $\mathbb{Q}$. \item $e$ is transcendental over $\mathbb{Q}$. \item $\pi$ is transcendental over $\mathbb{Q}$. As a result, it is impossible to ``square the circle''! \end{itemize} 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 Lindemann-Weierstrass theorem. If Schanuel's conjecture were proven to be true, then the algebraic independence of $e$ and $\pi$ over $\mathbb{Q}$ can be shown.