The proof of the day is: $e$ is irrational
Theorem: $e \not\in \mathbb{Q}$
Proof:
By definition, $e = \sum\limits_{n=0}^{\infty}\frac{1}{n!}$. Define the partial sum
$S_N = \sum\limits_{n=0}^{N}\frac{1}{n!}$ and let $L = N!e - N!S_N$. Since both sums
have only positive terms, $L>0$. So we have $L=\sum\limits_{n=N+1}^{\infty}\frac{N!}{n!}$.
Then we can reorder the indexes, having $L=\sum\limits_{k=0}^{\infty}\frac{N!}{(N+k+1)!}$.
Now, note that $\frac{(N+k+1)!}{N!} = (N+k+1)\cdot(N+k)\cdots(N+1) > N$, since $N+k+1 > N$.
Furthermore, $(N+k)\cdots(N+1)$ is the product of $k$ consecutive numbers, i.e., it is
divisible by $k!$. In particular, it is greater than or equal to $k!$. Now, we have
$\frac{(N+k+1)!}{N!} > Nk!$. So, $0 < N!e - \sum\limits_{n=0}^{N}\frac{N!}{n!} < \sum\limits_{k=0}^{\infty}
\frac{1}{Nk!} = \frac{e}{N}$.
Now, suppose that $e=\frac{p}{q}$ with $p,q \in \mathbb{Z}^{+}$, let $\mathcal{L}=max\{3,q\}$
and consider the relation we just obtained, letting $N=\mathcal{L}$. We end up with
$0 < \mathcal{L}!\frac{p}{q}-\sum\limits_{n=0}^{\mathcal{L}}\frac{\mathcal{L}}{n!} < \frac{e}{\mathcal{L}}$.
And we have $q \mid \mathcal{L}!$, so $\mathcal{L}!\frac{p}{q} \in \mathbb{Z}$, so
$\mathcal{L}!\frac{p}{q}-\sum\limits_{n=0}^{\mathcal{L}}\frac{\mathcal{L}}{n!} \in \mathbb{Z}$.
But for $\mathcal{L} \geq 3$, $\frac{e}{\mathcal{L}} < 1$. Since there is no integer between
$0$ and $1$, contradiction in supposing that $e \in \mathbb{Q}$.