## Proof of the Irrationality of e

In a previous post I proved that $\log_{10} 5$ is irrational.  In this post I prove the irrationality of e.

A proof of the irrationality of e must start by defining e.  There are some different ways to do that.  We’ll take e to be the unique positive number b such that $\dfrac{d}{dx} b^x = b^x$.  (This property itself can be proved from other ways to define e.  See, for example, Fitzpatrick’s Advanced Calculus [1], Section 5.2.)  We’ll also make the assumptions that (1) $e < 3$ and (2) $e^x$ is an increasing function.

Under this definition of e, the Lagrange Remainder Theorem says that, given $x > 0$ and $n \in \mathbb{N}$, there exists $c \in (0,x)$ such that

$\displaystyle e^x = 1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!} + \frac{e^c x^{n+1}}{(n+1)!}.$

Taking $x = 1$, we obtain, for some $c \in (0,1)$,

$\displaystyle e = 1 + 1 + \frac{1}{2} + \cdots + \frac{1}{n!} + \frac{e^c}{(n+1)!}$.

$\displaystyle \implies 0 < e - \left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!}\right] = \frac{e^c }{(n+1)!} \leq \frac{e}{(n+1)!} \leq \frac{3}{(n+1)!}.$

Now, suppose e is rational.  Then there exist $m_0, n_0 \in \mathbb{N}$ with $e = m_0/n_0$.  Therefore,

$\displaystyle 0 < \frac{m_0}{n_0} - \left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!} \right] \leq \frac{3}{(n+1)!}$.

Multiplying by $n!$, this becomes

$\displaystyle 0 < n! \frac{m_0}{n_0} - n! \left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!} \right] \leq \frac{3}{n+1}$.

Since the above inequality is true for all $n \in \mathbb{N}$, it is true for $n = \max \{n_0, 3\}$.  For this value of n,  $\displaystyle n!\frac{m_0}{n_0} - n!\left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!}\right]$ is an integer, and $3/(n+1) \leq 3/4$.

This means that there is an integer in the interval $(0,3/4)$.  However, we know that there is no integer in the interval $(0,1)$.  Contradiction; thus e is irrational.

[1] Patrick M. Fitzpatrick, Advanced Calculus, American Mathematical Society, 2006.