---
tag: 710DZ9P29V
kind: proof
name: Proof of the irrationality of e, after Fourier
msc: [11J72]
proves: 0W4GQSG7B6
uses: [0MVCFMW7S4, 0G4D3ZKTP8, 0WFCGVMRRC, 0RAM4E07NR, 0MX1HZBBXJ, 01X5CVV1Q5, 080T2TXVQV]
sources:
  - url: https://en.wikipedia.org/wiki/Proof_that_e_is_irrational
    locator: Fourier's proof
---

## Proof

We argue by contradiction. We build an integer that lies strictly between $0$ and $1$.

1. By the [[series for $e$|0G4D3ZKTP8]], $e = \sum_{n=0}^{\infty} \frac{1}{n!}$, where [[$e$|0MVCFMW7S4]] is $\exp(1)$.
2. Assume that $e$ is not [[irrational|080T2TXVQV]]. Then $e = a/b$ for an integer $a$ and an integer $b \ge 1$.
3. Let $x = b! \left( e - \sum_{n=0}^{b} \frac{1}{n!} \right)$.
4. $x$ is an integer. First, $b! \, e = b! \, a / b = a \cdot (b-1)!$, an integer. Second, for $0 \le n \le b$, the definition of the [[factorial|0WFCGVMRRC]] gives $\frac{b!}{n!} = (n+1)(n+2) \cdots b$, an integer.
5. For $N > b$, $\sum_{n=0}^{N} \frac{1}{n!} - \sum_{n=0}^{b} \frac{1}{n!} = \sum_{n=b+1}^{N} \frac{1}{n!}$. As $N \to \infty$, the left side tends to $e - \sum_{n=0}^{b} \frac{1}{n!}$, by the [[limit of a sum|0RAM4E07NR]].
6. Multiply by $b!$ and use the algebra of limits. Then step 5 gives $x = \sum_{k=1}^{\infty} t_k$, where $t_k = \frac{b!}{(b+k)!}$.
7. All $t_k$ are positive. By the [[result on series of nonnegative terms|0MX1HZBBXJ]], $x$ is the supremum of the partial sums, so $x \ge t_1 > 0$.
8. Let $q = \frac{1}{b+1}$, so $0 < q \le \frac12$. For $k \ge 1$, $t_k = \frac{1}{(b+1)(b+2) \cdots (b+k)} \le q^k$, since each factor is at least $b + 1$.
9. For $k \ge 2$, the factor $b + 2$ exceeds $b + 1$. So $t_2 < q^2$.
10. By the [[geometric series|01X5CVV1Q5]] with first term $q$ and ratio $q$, $\sum_{k=1}^{\infty} q^k = \frac{q}{1 - q} = \frac{1}{b}$.
11. Let $d_k = q^k - t_k \ge 0$. The series $\sum d_k$ converges, with sum $\frac{1}{b} - x$, by step 10 and the limit of a sum.
12. By the result on series of nonnegative terms, $\sum d_k \ge d_1 + d_2 \ge d_2 > 0$, using step 9. So $x < \frac{1}{b} \le 1$.
13. By steps 7 and 12, $x$ is an integer with $0 < x < 1$. No such integer exists. So $e$ is irrational.

## Idea

If $e = a/b$, multiplying $e$ by $b!$ makes it an integer, and so does multiplying the first $b + 1$ terms of its series. So the scaled tail of the series is an integer. But the factorials make the tail smaller than a geometric series with sum $1/b$, and it is positive, so it cannot be an integer.