Euler's Number is Irrational

Theorem
Euler's number $e$ is irrational.

Proof 1
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:

Observe that the quantity on the 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.

Proof 2
Suppose Euler's number $e$ is rational. Then,

$e=\dfrac{m}{n}=\displaystyle\sum_{k=0}^n\dfrac{1}{k!}+\sum_{k=n+1}^\infty\dfrac{1}{k!};\ (m,n)\in\mathbb{Z}_+^2$

Multiplying both sides by $n!$, we get

$m(n-1)!-\displaystyle\sum_{k=0}^n\dfrac{n!}{k!}=\sum_{j=1}^\infty\dfrac{n!}{(n+j)!}<\sum_{j=1}^\infty\dfrac{1}{(n+1)^j}=\dfrac{1}{n}$

The left side is an integer but $0<\tfrac{1}{n}<1$, which contradicts the assumption. Hence, $e$ is irrational. $\blacksquare$