> [!NOTE] Theorem > [[Euler's Number]] is irrational. >*Proof*. Suppose $e = \frac{p}{q} \in \mathbb{N}$ such that $\gcd(p,q)=1$. Consider the expression $T= q! \left( e -\sum_{n=0}^{q} \frac{1}{n!} \right) = \frac{1}{q+1} + \frac{1}{(q+1)(q+2)} +\dots $ >The first term of $q! \left( e -\sum_{n=0}^{q} \frac{1}{n!} \right)$ is $p(q-1)!$ which is an integer. The terms of the sum are also integers because $q!$ is divisible by $n!$ for $n\leq q$. Deduce that $T$ is an integer. >On the other hand $\frac{1}{q+1} + \frac{1}{(q+1)(q+2)} +\dots < \frac{1}{q+1}+ \frac{1}{(q+1)^{2}}+\dots = \frac{1}{q+1} \left( \frac{1}{1- \frac{1}{q+1} } \right) = \frac{1}{q} $ so $0<T< \frac{1}{q} \leq 1$ which is contradiction. >