User:Karlhahn/User e-irrational

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

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

PROOF:

If $\scriptstyle e$ were rational, then

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

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

$qe=p$

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

$q!e=p(q-1)!$

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

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

Multiplying both sides by $\scriptstyle q!$:

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

The first $\scriptstyle q+1$ 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

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

making $\scriptstyle r$ an integer. Observe that

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

So

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

But

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

This means that $\scriptstyle 0<r<{\frac {1}{q-1}}\leq 1$, which requires that $\scriptstyle r$ be an integer between zero and one. That is clearly impossible, hence $\scriptstyle e$ is irrational