# The Number e, Part 1: e is Irrational

(Or would "The Number 1, Part e" be more interesting?!)

Let's talk about the number $e$, my favourite number. Of course, to talk about it we need a definition, so we define $e$ as

$e = 1/0! + 1/1! + 1/2! + 1/3! + \cdots.$

Exercise: check that this converges! In Part 2, we shall see Hermite's argument that $e$ is transcendental. In order to warm up, let us prove that $e$ is not an integer first, and then let us prove that $e$ irrational.

### $e$ is not an Integer

To prove that $e$ is not an integer, we make the observation that $2^n < n!$ for all $n \geq 4$. Indeed, $2^4 = 16 < 24 = 4!$. Thus it is true for $n = 4$. Now, if the statement is true for some $n$, then $2^{n+1} = 2\cdot 2^n < 2\cdot n! \leq (n+1)\cdot n! = (n+1)!$. Thus by induction the statement is true. Next, by using our definition of $e$, we see that $e > 2$. On the other hand using that $2^n < n!$ for all $n \geq 4$,

$e = 1 + 1 + 1/2! + 1/3! + \cdots < 1 + 1 + 1/2 + 1/6 + 1/16 + 1/32 + \cdots = 67/24 < 3.$

So we have proved that $2 < e < 3$. Thus $e$ is not an integer.

### $e$ is not Rational

Showing that a number is not an integer is much easier than showing that it is not rational. Some numbers can be proven irrational by using unique factorisation in $\mathbb{Z}$, such as the numbers as $\sqrt{2}$ and $\log_2 3$. The irrationality of $e$ is only slightly trickier though, and the proof we will now see uses estimates to reach a contradiction. I learned this proof from Rudin's "Principles of Mathematical Analysis".

We define $e_n = 1/0! + 1/1! + \cdots + 1/n!$; that is, $e_n$ is the $n$th partial sum of the series defining $e$. Next, we observe that

$e – e_n = 1/(n+1)! + 1/(n+2)! + \cdots$
$= 1/(n+1)!(1 + 1/(n+2) + 1/[(n+2)(n+3)] + \cdots)$
$< 1/(n+1)!(1 + 1/(n+1) + 1/(n+1)^2 + \cdots)$ $= 1/(n!n)$

Now we suppose that $e$ is rational so that $e = a/b$ where $a,b$ are positive integers. In this case, we can take $n = b$ in the above estimate to get

$0 < a/b - (1 + \cdots + 1/b!) < 1/(b!b)$.

Multiplying this inequality by $b!$ shows that there is an integer strictly between $0$ and $1$. This is a contradiction, so the assumption that $e$ was rational cannot be true.

### Algebraic and Transcendental Numbers

If a real number is not the solution to any nonzero polynomial with integer coefficients, it is called transcendental. Otherwise, it is called algebraic. Since the real numbers are uncountable but the algebraic numbers are countable, there must be transcendental numbers. In Part 2 we will actually see that $e$ is such a number.