From Taylor's theorem, \(\exists\xi\in(0,1)\) such that \(e=\sum_{n=1}^{N}\frac{1}{n!} + R_{N}\) where \(0<R_{N}=\frac{e^{\xi}}{(N+1)!}<\frac{3}{(N+1)!}\). Assume \(e\) is rational: \(e=\frac{a}{b}, a,b\in\mathbb{Z}, b\neq 0\) & let \(N>\max(b,3)\): \(\frac{N!a}{b}-\sum_{n=1}^{N}\frac{N!}{n!} = N!R_{N}\). Since \(N>b\), \(N!\) must contain \(b\) and \(N!R_{N}\in\mathbb{Z}\) . Since \(N>3\): \(0<N!R_{N}<\frac{3}{N+1}<\frac{3}{4}\) which is a contradiction.

#proofinatoot that there are infinitely-many Pythagorean triples (that is, integers \(x,y,z\) such that \(x^{2}+y^{2}=z^{2}\)):

Let \(a,b\) be any integers, then:

\((a^{2}+b^{2})^{2}=(a^{2}-b^{2})^{2} + (2ab)^{2}\) \(\blacksquare\)

Sometimes a mathematician, sometimes a physicist, depending on who asks.

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Joined May 2017