Euler's Number is Irrational

Theorem

Euler's number $e$ is irrational.

Proof by Contradiction

Assume that $e$ is rational. Then, there exist coprime integers $m$ and $n$ such that:

$\displaystyle \frac m n = e = \sum_{i=0}^\infty \frac 1 {i!}$ from the definition of Euler's number.

Multiplying both sides by $n!$, observe that:

$\displaystyle \frac m n n! = n! \sum_{i=0}^\infty \frac 1 {i!} = \left({\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \cdots + \frac{n!}{n!}}\right) + \left({\frac{n!}{(n+1)!} + \frac{n!}{(n+2)!} + \frac{n!}{(n+3)!} + \ldots}\right)$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle m(n-1)! - \left({\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \ldots + \frac{n!}{n!} }\right)\)	\(=\)	\(\displaystyle \frac 1 {(n+1)} + \frac 1 {(n+1)(n+2)} + \frac 1 {(n+1)(n+2)(n+3)} + \ldots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \frac 1 {(n+1)} + \frac 1 {(n+1)(n+1)} + \frac 1 {(n+1)(n+1)(n+1)} + \ldots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=0}^\infty \left ({\frac 1 {n+1} }\right)^{(i+1)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{\frac 1 {n+1} } {1 - \frac 1 {n+1} } = \frac 1 n < 1\) from Sum of Infinite Geometric Progression \(\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Observe that the quantity on the left must be an integer, as it is composed entirely of sums and differences of integral components.

It must be positive, as it is equal to $\displaystyle \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \ldots$, which is strictly positive.

Thus, $\displaystyle m(n-1)! - \left({\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \ldots + \frac{n!}{n!}}\right)$ must be a positive integer less than $1$, a contradiction, so $e$ must be irrational.

$\blacksquare$

Sources

• George F. Simmons: Calculus Gems (1992): Chapter $\text{B}.17$: Appendix