e is irrational EIsIrrational2 2007-05-03 21:21:54 2007-05-04 06:08:39 Proof e is irrational 0 % 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 We have the series \[ e^{-1} = \sum_{k=0}^\infty {(-1)^k \over k!} \] Note that this is an alternating series and that the magnitudes of the terms decrease. Hence, for every integer $n > 0$, we have the bound \[ 0 < \left| \sum_{k=0}^{n} {(-1)^k \over k!} - e^{-1} \right| < {1 \over (n+1)!}, \] by the \PMlinkname{Leibniz' estimate for alternating series}{LeibnizEstimateForAlternatingSeries}.\, Assume that $e = n/m$, where $m$ and $n$ are integers and $n > 0$.\, Then we would have \[ 0 < \left| \sum_{k=0}^{n} {(-1)^k \over k!} - {m \over n} \right| < {1 \over (n+1)!} . \] Multiplying both sides by $n!$, this would imply \[ 0 < \left| \sum_{k=0}^{n} {(-1)^k n! \over k!} - m (n-1)! \right| < {1 \over n+1} , \] which is a contradiction because every term in the sum is an integer, but there are no integers between $0$ and $1/(n+1)$.