# $e$ is irrational.
Using the definition that $$
e = \sum_{k=0}^{\infty} \frac{1}{k!}
$$we can show $e$ is irrational as follows. Note for any positive integer $N$, we have $$
\begin{align*}
e &= \sum_{k=0}^{N} \frac{1}{k!} +\sum_{k=N+1}^{\infty} \frac{1}{k!} \\
& = \sum_{k=0}^{N} \frac{1}{k!} + \frac{1}{(N+1)!} \left( 1 + \frac{1}{N+1} + \frac{1}{(N+1)(N+2)} + \cdots \right) \\
& \le \sum_{k=0}^{N} \frac{1}{k!} + \frac{1}{(N+1)!} \sum_{k=0}^{\infty} \frac{1}{(N+1)^{k}} \\
& = \sum_{k=0}^{N} \frac{1}{k!} + \frac{1}{(N+1)!} \frac{1}{1- \frac{1}{N+1}} \\
\end{align*}
$$So this implies $$
\begin{align*}
0 < e-\sum_{k=0}^{N} \frac{1}{k!} &\le \frac{1}{N(N!)} \\
\implies 0 < N! e-\sum_{k=0}^{N} \frac{N!}{k!} &\le \frac{1}{N}
\end{align*}
$$
Note $\sum_{k=0}^{N} \frac{N!}{k!}$ is always an integer for any positive integer $N$.
If to the contrary that $e = \frac{a}{b}$ is a rational number, for some integers $a,b$, then there exists a large enough positive integer $N$ such that $N!e$ is an integer. But $\frac{1}{N} < 1$ for all large $N$. This shows there is an integer strictly between $0$ and $1$, a contradiction!