User:Karlhahn/User e-irrational

Usage: 

PROOF:

If $$\scriptstyle e$$ were rational, then


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

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


 * $$q e = p$$

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