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Irrationality Of E

Problem

Prove that $e \approx$ 2.7182818284... is irrational.


Solution

We begin by writing $e$ using the standard infinite series expansion:

   $e = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{n!} + R$

In other words, we can write $e$ as a finite sum plus some remainder term, $R \gt 0$.

$\therefore R = e - \left(1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{n!}\right)$

Let us assume that $e$ is rational and can be written as $\dfrac{a}{b}$, where $a$, $b$ are positive integers. Furthermore we shall write the finite series up to the $\dfrac{1}{b!}$ term.

$\therefore R = \dfrac{a}{b} - \left(1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{b!}\right)$

Multiplying both sides by $b!$ we get:

   $b!R = \dfrac{b!a}{b} - \left(b! + \dfrac{b!}{1!} + \dfrac{b!}{2!} + \dfrac{b!}{3!} + \cdots. + \dfrac{b!}{b!}\right)$

It can be seen that all the terms on the right hand side are integer, so $b!R$ must be integer.

$\begin{eqnarray}R &=& \dfrac{1}{(b+1)!} + \dfrac{1}{(b+2)!} + \dfrac{1}{(b+3)!} + \cdots\\\therefore b!R &=& \dfrac{b!}{(b+1)!} + \dfrac{b!}{(b+2)!} + \dfrac{b!}{(b+3)!} + \cdots\\&=& \dfrac{1}{b+1} + \dfrac{1}{(b+1)(b+2)} + \dfrac{1}{(b+1)(b+2)(b+3)} + \cdots\\&\lt& \dfrac{1}{b+1} + \dfrac{1}{(b+1)^2} + \dfrac{1}{(b+1)^3} + \cdots\\\end{eqnarray}$

Using the sum to infinity for a geometric progression with both the first term, $a$, and common difference, $r$, as $\dfrac{1}{b+1}$:

$S_{\infty} = \dfrac{a}{1-r} = \dfrac{\dfrac{1}{b+1}}{1 - \dfrac{1}{b+1}} = \dfrac{1}{b + 1 - 1} = \dfrac{1}{b}$

$\therefore 0 \lt b!R \lt \dfrac{1}{b} \lt 1$

Clearly no integer lies between 0 and 1. Thus by reductio ad absurdum we prove that $e$ cannot be written as the ratio of two integers and is irrational. Q.E.D.

Problem ID: 377 (17 Oct 2010)     Difficulty: 4 Star

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