# User:Karlhahn/User e-irrational

 ${\displaystyle e\neq {\frac {p}{q}}}$ This user can prove that the number, e, is irrational.

Usage: {{User:Karlhahn/User e-irrational}}

PROOF:

If ${\displaystyle \scriptstyle e}$ were rational, then

${\displaystyle e={\frac {p}{q}}}$

where ${\displaystyle \scriptstyle p}$ and ${\displaystyle \scriptstyle q}$ are both positive integers. Hence

${\displaystyle qe=p}$

making ${\displaystyle \scriptstyle qe}$ an integer. Multiplying both sides by ${\displaystyle \scriptstyle (q-1)!}$,

${\displaystyle q!e=p(q-1)!}$

so clearly ${\displaystyle \scriptstyle q!e}$ is also an integer. By Maclaurin series

${\displaystyle e=\sum _{k=0}^{\infty }{\frac {1}{k!}}}$

Multiplying both sides by ${\displaystyle \scriptstyle q!}$:

${\displaystyle q!e=\sum _{k=0}^{\infty }{\frac {q!}{k!}}}$

The first ${\displaystyle \scriptstyle q+1}$ terms of this sum are integers. It follows that the sum of the remaining terms must also be an integer. The sum of those remaining terms is

${\displaystyle r=\sum _{k=1}^{\infty }{\frac {q!}{(q+k)!}}}$

making ${\displaystyle \scriptstyle r}$ an integer. Observe that

${\displaystyle {\frac {q!}{(q+k)!}}<{\frac {q!}{q!q^{k}}}={\frac {1}{q^{k}}}}$

So

${\displaystyle r=\sum _{k=1}^{\infty }{\frac {q!}{(q+k)!}}<\sum _{k=1}^{\infty }{\frac {1}{q^{k}}}}$

But

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{q^{k}}}={\frac {1}{q-1}}}$

This means that ${\displaystyle \scriptstyle 0, which requires that ${\displaystyle \scriptstyle r}$ be an integer between zero and one. That is clearly impossible, hence ${\displaystyle \scriptstyle e}$ is irrational