Pi is Irrational

Theorem
Pi ($\pi$) is irrational.

Proof
Suppose that $\pi$ is rational.

Then $\displaystyle \pi = \frac a b$ for some $a, b \in \Z, b > 0$ from Canonical Form of Rational Number.

Let $n \in \Z: n > 0$.

We define the polynomial function:


 * $\displaystyle \forall x \in \R: f \left({x}\right) = \frac {x^n \left({a - b x}\right)^n} {n!}$

We differentiate this $2n$ times, and then we build:


 * $\displaystyle F \left({x}\right) = \sum_{j=0}^{n} \left({-1}\right)^j f^{\left({2j}\right)} \left({x}\right) = f \left({x}\right) + \cdots + \left({-1}\right)^j f^{\left({2j}\right)} \left({x}\right) + \cdots + \left({-1}\right)^n f^{(2n)}(x)$

... that is, the alternating sum of $f$ and its first $n$ even derivatives.


 * First we show that:
 * $(1): \quad F \left({0}\right) = F \left({\pi}\right)$.

From the definition of $f \left({x}\right)$, and our supposition that $\pi = \dfrac a b$, we have that:


 * $\displaystyle \forall x \in \R: f \left({x}\right) = b^n \frac {x^n \left({\pi - x}\right)^n} {n!} = f \left({\pi - x}\right)$

Using the Chain Rule, we can apply the Principle of Mathematical Induction to show that, for all the above derivatives:


 * $\forall x \in \R: f^{\left({j}\right)} \left({x}\right) = \left({-1}\right)^j f^{\left({j}\right)} \left({\pi - x}\right)$

In particular, we have:


 * $\forall j \in \left\{{1, 2, \ldots, n}\right\}: f^{\left({2j}\right)} \left({0}\right) = f^{\left({2j}\right)} \left({\pi}\right)$

From the definition of $F$, it follows that $F \left({0}\right) = F \left({\pi}\right)$.


 * Next we show that:
 * $(2): \quad F \left({0}\right)$ is an integer.

We use the Binomial Theorem to expand $\left({a - bx}\right)^n$:
 * $\displaystyle \left({a - bx}\right)^n = \sum_{k \mathop = 0}^n \binom n k a^{n-k} (-b)^k x^k$

By substituting $j = k + n$, we obtain the following expression for $f$:
 * $\displaystyle f \left({x}\right) = \frac 1 {n!} \sum_{j \mathop = n}^{2n} \binom n {j-n} a^{2n-j} \left({-b}\right)^{j-n} x^{j}$

Note the following:
 * The coefficients of $x^0, x^1, \ldots, x^{n-1}$ are all zero;
 * The degree of the polynomial $f$ is at most $2n$.

So we have:
 * $\forall j < n: f^{\left({j}\right)} \left({0}\right) = 0$
 * $\forall j > 2n: f^{\left({j}\right)} \left({0}\right) = 0$.

But for $n \le j \le 2n$, we have:
 * $\displaystyle f^{\left({j}\right)} \left({0}\right) = \frac {j!} {n!} \binom n {j-n} a^{2n-j} \left({-b}\right)^{j-n}$

Because $j \ge n$, $\displaystyle \frac {j!} {n!}$ is an integer.

So is the binomial coefficient $\displaystyle \binom n {j-n}$ by its very nature.

As $a$ and $b$ are both integers, then so are $a^{2n-j}$ and $\left({-b}\right)^{j-n}$.

So $f^{\left({j}\right)} \left({0}\right)$ is an integer for all $j$, and hence so is $F \left({0}\right)$.


 * Next we show that:
 * $\displaystyle (3): \quad \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \, dx = F \left({0}\right)$

As $f \left({x}\right)$ is a polynomial function of degree $n$, it follows that $f^{\left({2n + 2}\right)}$ is the null polynomial.

This means:
 * $F'' + F = f$

Using the Product Rule and the derivatives of sine and cosine, we get:
 * $\left({F' \left({x}\right) \sin x - F \left({x}\right) \cos x}\right)' = f \left({x}\right) \sin x$

By the Fundamental Theorem of Calculus, this leads us to:
 * $\displaystyle \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \, dx = \frac1 2 \left[{\left({F' \left({x}\right) \sin x - F \left({x}\right) \cos x}\right)}\right]_{x = 0}^{x = \pi}$

From Sine and Cosine are Periodic on Reals, we have that $\sin 0 = \sin \pi = 0$ and $\cos 0 = - \cos \pi = 1$.

So, from $F \left({0}\right) = F \left({\pi}\right)$ (see $(1)$ above), we have $\displaystyle \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \, dx = F \left({0}\right)$.


 * The final step:

On the interval $\left({0 \,.\,.\, \pi}\right)$, we have from Sine and Cosine are Periodic on Reals that $\sin x > 0$.

So from $(2)$ and $(3)$ above, we have that $F \left({0}\right)$ is a positive integer.

Now, we have that $\displaystyle \left({x - \frac \pi 2}\right)^2 = x^2 - \pi x + \left({\frac \pi 2}\right)^2$ and so $\displaystyle x \left({\pi - x}\right) = \left({\frac \pi 2}\right)^2 - \left({x - \frac \pi 2}\right)^2$

Hence $\displaystyle \forall x \in \R: x \left({\pi - x}\right) \le \left({\frac \pi 2}\right)^2$.

Also, from Boundedness of Sine and Cosine, $0 \le \sin x \le 1$ on the interval $\left({0 \,.\,.\, \pi}\right)$.

So, by the definition of $f$:
 * $\displaystyle \frac 1 2 \int_0^\pi f \left({x}\right) \sin x \, dx \le \frac {b^n} {n!} \left({\frac \pi 2}\right)^{2n+1}$

But this is smaller than $1$ for large $n$, from Power Series over Factorial.

Hence, for these large $n$, we have $F \left({0}\right) < 1$, by $(3)$.

This is impossible for the positive integer $F \left({0}\right)$.

So our assumption that $\pi$ is rational must have been false.

Historical Note
This proof was first published by Ivan M. Niven in 1947, and is considered a classic.