# Proof that e is irrational

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This article covers much about the mathematical constant e, Euler's number, concluding with the result that it is irrational.

## IntroductionEdit

The mathematical constant e was first found by Bernoulli with the formula

$e=\lim_{n\to\infty}\left(1+\frac{1}{n} \right)^{n}$

We will use this formula to determine a new formula for e and then we will use it to prove e's irrationality.

## LemmasEdit

Lemma 1. The sequence $\textstyle \left(1+\frac{x}{n} \right)^{n}$ increases.

Proof. We need to show

$\left(1+\frac{x}{n+1} \right)^{n+1}>\left(1+\frac{x}{n} \right)^{n}$

which is equivalent to

$\left(1-\frac{x}{n}+\frac{x}{n}+\frac{x}{n+1} \right)^{n+1}>\left(1+\frac{x}{n} \right)^{n}$

Simplifying, we get

$\left(1+\frac{x}{n}-\frac{x}{n(n+1)}\right)^{n+1}>\left(1+\frac{x}{n} \right)^{n}$

If we change parameters and set $\textstyle a=1+\frac{x}{n}$, we get

$\left(a-\frac{x}{n(n+1)}\right)^{n+1}>a^{n}$

which simplifies to

$1 > \frac{1}{a}$

which, of course, holds.

Lemma 2. The sequence $\textstyle \left(1+\frac{x}{n} \right)^{n}$ has an upper bound of $\textstyle e^{x}$.

Proof. It follows directly:

$\left(1+\frac{x}{n} \right)^{n}\leq \left(1+\frac{x}{xn} \right)^{xn}=\left(1+\frac{1}{n} \right)^{xn}\leq e^{x}$

Corollary 3. The sequence $\textstyle \left(1+\frac{x}{n} \right)^{n}$ converges.

Proof. Follows directly from Lemmas 1 and 2.

Lemma 4. The expression $\textstyle \lim_{n\to\infty} \left( 1+\frac{k}{n} \right)^{n}$ is equal to $\textstyle e^{k}$.

Proof. Since the first expression is equal to $\textstyle \lim_{n\to\infty} \left( 1+\frac{k}{kn} \right)^{kn}=\lim_{n\to\infty} \left( 1+\frac{1}{n} \right)^{kn}$ due to the Bolzano-Weierstrass theorem, it immediately follows.

Lemma 5. $\textstyle\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n-1}$

Proof. It is obvious:

$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n-1}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n-1}$

Lemma 6. The Taylor series expansion for $\textstyle e^{x}$ is

$\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$

Proof. We know an expression for $\textstyle e^{k}$, so we will differentiate it to obtain a result. It is obvious that this expression is 1 when k=0, so if we make the definition $\textstyle \exp (x)=e^{x}$, $\exp(0)=1$. Now we differentiate our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}\lim_{n\to\infty}\left(1+\frac{x}{n} \right)^{n} =\lim_{n\to\infty}\frac{\mathrm{d} }{\mathrm{d} x}\left(1+\frac{x}{n} \right)^{n}=\lim_{n\to\infty}\frac{1}{n}n\left(1+\frac{x}{n} \right)^{n-1}=\lim_{n\to\infty}\left(1+\frac{x}{n} \right)^{n}$

As we can see, $\textstyle \exp'(x) = \exp(x)$, so all of the other derivatives will be 1 when evaluated at 0. This yields the following Taylor series for our function:

$\sum_{k=0}^{\infty}\frac{\exp^{(k)}(0)}{k!}x^{k}=\sum_{k=0}^{\infty}\frac{1}{k!}x^{k}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$

And the proof is complete.

Corollary 7. An infinite sum representation of e is $\textstyle \sum_{k=0}^{\infty}1/k!$

Proof. We take x=1 in the previous lemma to obtain this.

## The proof of the theoremEdit

We will use a proof by contradiction. Let's assume e is rational and for $\textstyle p,q\epsilon\mathbb{N}^{>0}$, it can be written as p/q. Observe the following equalities:

$q!e=q!\frac{p}{q}=(q-1)!p$

Since the expression on the RHS is a positive integer, so must be the expression on the LHS. This yields:

$q!e=\sum_{k=0}^{\infty}\frac{q!}{k!}=\sum_{k=0}^{q}\frac{q!}{k!} + \sum_{k=q+1}^{\infty}\frac{q!}{k!}$

The last term of the last sum must be an integer because the sum equals a positive integer and its first term is a positive integer. This is important, the contradiction follows from here. Let's call this integer R. Now, observe these:

$R=\sum_{k=q+1}^{\infty}\frac{q!}{k!}=\sum_{k=q+1}^{\infty}\frac{1}{(q+1)(q+2)...(k-1)k}<\sum_{k=q+1}^{\infty}\frac{1}{(q+1)^{k}}<\sum_{k=1}^{\infty}\frac{1}{(q+1)^{k}}$

Therefore:

$R<\sum_{k=1}^{\infty}\frac{1}{(q+1)^{k}}=\sum_{k=1}^{\infty}\left(\frac{1}{q+1}\right)^{k}=\frac{1}{q+1}\sum_{k=0}^{\infty}\left(\frac{1}{q+1}\right)^{k}=\frac{1}{q+1}\frac{1}{1-\frac{1}{q+1}}=\frac{1}{q}<1.$

Now also note that q > 0, so all the terms in R are strictly positive, therefore R > 0. So we have 0 < R < 1, but we earlier established that R was a positive integer. As there are no integers between 0 and 1, we have a contradiction. Hence, it is impossible to express e as a ratio of two integers, so it must be irrational. Proof complete.