Is \(e\) rational?

It turns out that \(e\) is irrational. We can give a proof of this fact.

Suppose to the contrary that \(e\) can be written as a fraction, so

\[ e = \dfrac{a}{b}, \] where \(a\) and \(b\) are positive integers. Using the series for \(e\), we then have \[ e = \dfrac{a}{b} = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dotsb. \]

The key idea of the proof is to consider this series up to the term \(\dfrac{1}{b!}\). So we think of the above sum as

\[ \dfrac{a}{b} = \Bigl( 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dotsb + \dfrac{1}{b!} \Bigr) + \Bigl( \dfrac{1}{(b+1)!} + \dfrac{1}{(b+2)!} + \dotsb \Bigr). \]

Now multiply through by \(b!\). This gives

\[ b! \, \dfrac{a}{b} = \Bigl( b! + \dfrac{b!}{1!} + \dfrac{b!}{2!} + \dotsb + \dfrac{b!}{b!} \Bigr) + \Bigl( \dfrac{b!}{(b+1)!} + \dfrac{b!}{(b+2)!} + \dotsb \Bigr), \] and so \[ (b-1)! \, a = \Bigl( b! + \dfrac{b!}{1!} + \dotsb + \dfrac{b!}{b!} \Bigr) + \Bigl( \dfrac{1}{b+1} + \dfrac{1}{(b+1)(b+2)} + \dfrac{1}{(b+1)(b+2)(b+3)} + \dotsb \Bigr). \]

All the terms are positive. The left-hand side is an integer, and all the terms in the first bracket on the right-hand side are also integers, as the denominators cancel against \(b!\). Therefore, the sum \(x\) of the terms in the second bracket must be a positive integer.

However, all the terms in the second bracket are rather small. The denominators \((b+1)\), \((b+1)(b+2)\), \((b+1)(b+2)(b+3)\) increase very quickly. So \(x\) is a small positive integer \(\dots\) suspiciously small. Noting that \((b+1)(b+2) \dotsm (b+k) > (b+1)^k\), for all \(k \geq 2\), we can estimate \(x\) as

\begin{align*} x &= \dfrac{1}{b+1} + \dfrac{1}{(b+1)(b+2)} + \dfrac{1}{(b+1)(b+2)(b+3)} + \dotsb\\ &< \dfrac{1}{b+1} + \dfrac{1}{(b+1)^2} + \dfrac{1}{(b+1)^3} + \dotsb. \end{align*}

This is a geometric series, and its sum is

\[ \dfrac{\dfrac{1}{b+1}}{1 - \dfrac{1}{b+1}} = \dfrac{1}{b+1} \cdot \dfrac{b+1}{b} = \dfrac{1}{b}. \]

So \(x\) is a positive integer with \(x < \dfrac{1}{b} \leq 1\). There are not many positive integers less than one! This is a contradiction, and so our initial assumption must have been wrong. We conclude that \(e\) is irrational.

In fact, it can be shown that, like \(\pi\), the number \(e\) is transcendental: it is not the root of any polynomial with rational coefficients.

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