Fourier's proof by contradiction

The statement

Theorem. The number $e$ is irrational.

The elementary proof below is attributed to Fourier, who presented it in his lectures around 1815; Euler had already obtained the result in 1737 by an entirely different route, the continued fraction expansion of $e$.

Fourier's proof

Assumption for contradiction. Suppose $e$ is rational, and write $e = \dfrac{p}{q}$ with $p \in \mathbb{N}^*$ and $q \in \mathbb{N}^*$.

A preliminary remark about $q$. Since $2 < e < 3$, the number $e$ is not an integer, so $q \geq 2$.

The key quantity. Set

$$N = q!\,\left(e - S_q\right) = q!\,R_q \qquad \text{where} \qquad S_q = \sum_{k=0}^{q} \frac{1}{k!}$$

Step 1: $N$ is an integer. Write $N = q!\,e - q!\,S_q$ and deal with the two pieces.

On the one hand, the assumption gives

$$q!\,e = q!\,\frac{p}{q} = p \cdot (q-1)! \in \mathbb{Z}$$

On the other hand, for every $k \leq q$, the quotient $\dfrac{q!}{k!} = (k+1)(k+2)\cdots q$ is an integer, so

$$q!\,S_q = \sum_{k=0}^{q} \frac{q!}{k!} \in \mathbb{Z}$$

Since the difference of two integers is an integer, $N \in \mathbb{Z}$.

Step 2: $N$ lies strictly between $0$ and $1$. This is the bound on the remainder established in the previous chapter, multiplied by $q!$:

$$0 < N = q!\,R_q < q! \cdot \frac{1}{q \cdot q!} = \frac{1}{q} \leq \frac{1}{2} < 1$$

Conclusion. We have constructed an integer $N$ satisfying $0 < N < 1$. But there is no integer strictly between $0$ and $1$: contradiction. The initial assumption is therefore false, and $e$ is irrational.

Why the factorial and not something else

The mainspring of the proof is that $q!$ "clears" all the denominators of the partial sum $S_q$ at once, while leaving the remainder $R_q$ tiny. It is the combination of these two effects — an integer on one side, a quantity less than $1$ on the other — that produces the contradiction.

Ingredient	Role in the proof
$e = \sum \dfrac{1}{k!}$	provides an exact expression that can be truncated
$\dfrac{q!}{k!} \in \mathbb{Z}$ for $k \leq q$	makes $q!\,S_q$ an integer
$e = \dfrac{p}{q}$	makes $q!\,e$ an integer
$0 < R_q < \dfrac{1}{q \cdot q!}$	makes $N$ strictly less than $1$

Classic pitfall

Forgetting the strict positivity of $N$. Without it, $N = 0$ would remain possible, and there would be no contradiction: $0$ is a perfectly legitimate integer. But $R_q$ is a sum of terms that are all strictly positive, so $N > 0$ — this line, often rushed in written work, is indispensable.