## Appendix

### 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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