Fourier's Proof that e Is Irrational

Janot de Stainville ascribes this proof to Joseph Fourier.^[1] Start with:

The reason for the inequality is that each factorial is greater than a product of as many twos as there are factors in the factorial, discounting the one; for example, $4! = 1 \cdot 2 \cdot 3 \cdot 4 \gt 1 \cdot 2 \cdot 2 \cdot 2 = 2^3.$ The final step results from summing the geometric series. It follows that $2 \lt e \lt 3$ and in particular, that $e$ is not a whole number.

We want to prove that $e$ is irrational. Assume contrarily that $e = p /q = \sum_{n=0}^\infty{1 / n!}$, where $p$ and $q \geq 2$ are positive whole numbers. Multiply through by $q!$:

$$\begin{align*} p \cdot (q-1)! &= q! \left(1 + {1 \over 1!} + {1 \over 2!} + \cdots + {1 \over q!} + {1 \over (q+1)!} + {1 \over (q+2)!} + {1 \over (q+3)!} + \cdots\right)\\ &= \underbrace{q! \left(1 + {1 \over 1!} + {1 \over 2!} + \cdots + {1 \over q!}\right)}_{\LARGE A} + \underbrace{q!\left({1 \over (q+1)!} + {1 \over (q+2)!} + {1 \over (q+3)!} + \cdots\right)}_{\LARGE B}. \end{align*}$$

Each of the summands in $A$ is an integer because $q!$ cancels all the denominators; it follows that $A$ itself is an integer. The left side of this equation is also an integer; consequently $B$ must be an integer. But:

$$\begin{align*} B &= q!\left({1 \over (q+1)!} + {1 \over (q+2)!} + {1 \over (q+3)!} + \cdots\right)\\ &= {1 \over {q+1}} + {1 \over{(q+1)(q+2)}} + {1 \over{(q+1)(q+2)(q+3)}} + \cdots\\ &\leq {1 \over {q+1}} + {1 \over {(q+1)}^2} + {1 \over {(q+1)}^3} + \cdots\\ &\leq {1 \over 3} + {1 \over 3^2} + {1 \over 3^3} + \cdots\\ &= {1 \over 2}. \end{align*}$$

So $B$ is an integer with $0 \lt B \leq \frac{1}{2}$, which is impossible. Consequently the original assumption, that $e$ is rational, is not tenable. QED.

Mike Bertrand

Madison, WI

Jan 28, 2018

^ 1. Mélanges d'analyse algébrique et de géométrie, by Janot de Stainville (Veuve Courcier, Paris, 1815).