User:Karlhahn/User e-irrational

<math>e \ne \frac {p} {q}</math>This user can prove that the number, e, is irrational.

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

PROOF:

If <math>\scriptstyle e</math> were rational, then

<math>e = \frac {p} {q}</math>

where <math>\scriptstyle p</math> and <math>\scriptstyle q</math> are both positive integers. Hence

<math>q e = p</math>

making <math>\scriptstyle qe</math> an integer. Multiplying both sides by <math>\scriptstyle (q-1)!</math>,

<math>q! e = p(q-1)!</math>

so clearly <math>\scriptstyle q! e</math> is also an integer. By Maclaurin series

<math>e = \sum_{k=0}^\infty \frac {1} {k!}</math>

Multiplying both sides by <math>\scriptstyle q!</math>:

<math>q! e = \sum_{k=0}^\infty \frac {q!} {k!}</math>

The first <math>\scriptstyle q+1</math> 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

<math>r = \sum_{k=1}^\infty \frac {q!} {(q+k)!}</math>

making <math>\scriptstyle r</math> an integer. Observe that

<math>\frac {q!} {(q+k)!} < \frac {q!} {q! q^k} = \frac {1} {q^k}</math>

So

<math>r = \sum_{k=1}^\infty \frac {q!} {(q+k)!} < \sum_{k=1}^\infty \frac {1} {q^k}</math>

But

<math>\sum_{k=1}^\infty \frac {1} {q^k} = \frac {1} {q-1}</math>

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