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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.

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