The Irrationality of e

The proof that $\sqrt{2}$ is irrational is part of the standard high-school curriculum. The same cannot be said for the proof that $e$, the base of the natural logarithm, is irrational as well. Yet the proof is short, simple, elegant.

The proof is indirect and proceeds by contradiction: assume that $e$ is rational, so that there are two integers, $u$ and $v$, such that:

\[ e = \frac{u}{v} \qquad \text{with $u, v \in \mathbb{N}$ and $u, v > 0$} \]

Now consider the following rational approximation to $e$:

\[ e \approx \sum_{n=0}^v \frac{1}{n!} \]

Since $e = \sum_{n=0}^\infty \frac{1}{n!} > \sum_{n=0}^v \frac{1}{n!}$, we have:

\[ 0 < e - \sum_{n=0}^v \frac{1}{n!} \]

Now write $e = u/v$ (per our assumption) and multiply through with $v!$:

\[ 0 < v! \left[ \frac{u}{v} - \sum_{n=0}^v \frac{1}{n!} \right] \]

Notice that the quantity on the right is an integer, because the factor of $v!$ cancels all denominators, leaving only an integer quantity (which is strictly greater than zero).

Now use the definition of $e = \sum_{n=0}^\infty \frac{1}{n!}$ and perform the subtraction, then cancel the factor $v!$, leaving:

\begin{align*} 0 & < v! \sum_{n=v+1}^\infty \frac{1}{n!} \\ & = \frac{1}{v+1} + \frac{1}{(v+1)(v+2)} + \frac{1}{(v+1)(v+2)(v+3)} + \dots \end{align*}

Each term in this sum can be bounded from above as follows:

\[ \frac{1}{v+1} + \frac{1}{(v+1)(v+2)} + \frac{1}{(v+1)(v+2)(v+3)} + \dots < \frac{1}{v+1} + \frac{1}{(v+1)^2} + \frac{1}{(v+1)^3} + \dots \]

But the expression on the right is simply a geometric series in $\frac{1}{1+v}$ therefore equal to

\[ \frac{1}{1 - 1/(1+v)} = \frac{1}{v} \]

Hence we have:

\[ 0 < v! \sum_{n=v+1}^\infty \frac{1}{n!} < \frac{1}{v} \]

But we already established that the expression in the middle is an integer (greater than zero), whereas the expression on the right is less than or equal to $1$ for all integers $v > 0$. Since it is impossible to have an integer that is simultaneously strictly greater than zero and strictly less than one, we have a contradiction, and $e$ cannot be written as a ratio of integers.

Notes

This proof of the irrationality of $e$ is apparently due to Fournier; I have taken it, fairly directly, from the wonderful book “Making Transcendence Transparent” by Burger and Tubbs. (See here for a review.)

Of course, $e$ is not merely irrational, but transcendent. The proof of this is longer and more technical; it can be found in the reference mentioned.