Euler's Number is Irrational

Theorem

Euler's number $e$ is irrational.

Proof

Aiming for a contradiction, suppose that $e$ is rational.

Then there exist coprime integers $m$ and $n$ (and we can choose $n$ to be positive) such that:

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

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

$\dfrac m n n! = n! \ds \sum_{i \mathop = 0}^\infty \frac 1 {i!} = \paren {\dfrac {n!} {0!} + \dfrac {n!} {1!} + \dfrac {n!} {2!} + \cdots + \dfrac {n!} {n!} } + \paren {\frac {n!} {\paren {n + 1}!} + \dfrac {n!} {\paren {n + 2}!} + \dfrac {n!} {\paren {n + 3}!} + \cdots}$

Hence:

 $\ds m \paren {n - 1}! - \paren {\frac {n!} {0!} + \frac {n!} {1!} + \frac {n!} {2!} + \cdots + \frac {n!} {n!} }$ $=$ $\ds \frac 1 {\paren {n + 1} } + \frac 1 {\paren {n + 1} \paren {n + 2} } + \frac 1 {\paren {n + 1} \paren {n + 2} \paren {n + 3} } + \cdots$ $\ds$ $<$ $\ds \frac 1 {\paren {n + 1} } + \frac 1 {\paren {n + 1} \paren {n + 1} } + \frac 1 {\paren {n + 1} \paren {n + 1} \paren {n + 1} } + \cdots$ $\ds$ $=$ $\ds \sum_{i \mathop = 0}^\infty \paren {\frac 1 {n + 1} }^{\paren {i + 1} }$ $\ds$ $=$ $\ds \frac {\frac 1 {n + 1} } {1 - \frac 1 {n + 1} } = \frac 1 n \le 1$ from Sum of Infinite Geometric Sequence 

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

It must be strictly positive, as it is equal to:

$\dfrac 1 {\paren {n + 1} } + \dfrac 1 {\paren {n + 1} \paren {n + 2} } + \dfrac 1 {\paren {n + 1} \paren {n + 2} \paren {n + 3} } + \cdots$

which is strictly positive.

Thus:

$m \paren {n - 1}! - \paren {\dfrac {n!} {0!} + \dfrac {n!} {1!} + \dfrac {n!} {2!} + \cdots + \dfrac {n!} {n!} }$

must be a strictly positive integer less than $1$.

From this contradiction it follows that $e$ must be irrational.

$\blacksquare$

Historical Note

The proof that Euler's number $e$ is irrational was demonstrated by Leonhard Paul Euler in $1737$.