ProofWiki:Sandbox

Theorem: Euler's number $$e$$ is irrational.

Proof by Contradiction:
Assume that $$e$$ is rational. Then, there exist coprime integers $$m$$ and $$n$$ such that: $$\frac{m}{n} = e = \sum_{i=0}^{\infty}\frac{1}{i!}$$ Multiplying both sides by $$n!$$, observe that: $$\frac{m}{n}n! = n!\sum_{i=0}^{\infty}\frac{1}{i!} = (\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \ldots + \frac{n!}{n!}) + (\frac{n!}{(n+1)!} + \frac{n!}{(n+2)!} + \frac{n!}{(n+3)!} + \ldots)$$ $$m(n-1)! - (\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \ldots + \frac{n!}{n!}) = \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \ldots$$

Observe that the quantity on the left must be a positive integer. However, using the sum of a geometric series: $$\frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} + \ldots < \frac{1}{(n+1)} + \frac{1}{(n+1)(n+1)} + \ldots = \sum_{i=0}^{\infty}(\frac{1}{n+1})^{(i+1)} = \frac{\frac{1}{n+1}}{1-\frac{1}{n+1}} = \frac{1}{n} < 1$$

Thus,$$\frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \ldots$$ must be a positive integer less than 1, a contradiction, so $$e$$ must be irrational.

Q.E.D.