Proof that e is Irrational

The proof that $e$ is irrational relies on the following Lemma:

Lemma 1: $\displaystyle{e = \sum_{n=0}^{\infty} \frac{1}{n!}}$.

Theorem 2: The number $e$ is irrational.

• Proof: Suppose instead that $e$ is a rational number. Then there exists positive integers $a$ and $b$ with $b \neq 0$ such that $\displaystyle{e = \frac{a}{b}}$. Consider the number $b!e$:

(1)

\begin{align} \quad b!e &= b! \sum_{n=1}^{\infty} \frac{1}{n!} \\ &= b! \left ( \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + ... + \frac{1}{b!} + \frac{1}{(b + 1)!} + ... \right ) \\ &= \left [\frac{b!}{0!} + \frac{b!}{1} + \frac{b!}{2!} + ... + \frac{b!}{b!} \right ] + \left [ \frac{b!}{(b + 1)!} + ... \right ] \end{align}

• Observe that since $e = \frac{a}{b}$ that then $b!e = a(b - 1)!$. So $b!e$ is an integer. Also observe that the sum $\displaystyle{\frac{b!}{0!} + \frac{b!}{1} + \frac{b!}{2!} + ... + \frac{b!}{b!}}$ is an integer. So the difference, $\displaystyle{b!e - \left [ \frac{b!}{0!} + \frac{b!}{1} + \frac{b!}{2!} + ... + \frac{b!}{b!} \right ]}$ is an integer, call it $M$. From the equation above and using geometric series we have that:

(2)

\begin{align} \quad M &= \frac{b!}{(b + 1)!} + \frac{b!}{(b +2)!} + ... \\ & = \frac{1}{(b + 1)} + \frac{1}{(b+1)(b+2)} + ... \\ & < \frac{1}{(b+1)} + \frac{1}{(b + 1)^2} + ... \\ & < \frac{1}{(b + 1)} \left [ 1 + \frac{1}{(b + 1)} + ... \right ] \\ & < \frac{1}{(b + 1)} \sum_{n=0}^{\infty} \frac{1}{(b+1)^n} \\ & < \frac{1}{(b + 1)} \frac{1}{1 - \frac{1}{b + 1}} \\ & < \frac{1}{(b + 1)} \frac{\left (\frac{b+1}{b+1} \right )}{\left (\frac{b}{b+1} \right )} \\ & < \frac{1}{b} \end{align}

• So $M < \frac{1}{b}$. But $M$ is a positive integer and so this implies that $0 < b < 1$. But $b$ is also a positive integer and there exists no positive integers that are strictly less than $1$. So we have arrived at a contradiction. Hence the assumption that $e$ was rational was false. Hence $e$ is irrational. $\blacksquare$