# e is irrational

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 Leibniz’ estimate for alternating series (http://planetmath.org/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 $1/(n+1)$.

Title e is irrational EIsIrrational1 2013-03-22 17:02:32 2013-03-22 17:02:32 rspuzio (6075) rspuzio (6075) 8 rspuzio (6075) Proof msc 11J82 msc 11J72 LeibnizEstimateForAlternatingSeries