Pi is Irrational

Theorem
Pi ($\pi$) is irrational.

Proof
Suppose that $$\pi$$ is rational.

Then $$\pi = \frac a b$$ 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:


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

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


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

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


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

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


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

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:
 * $$(2) \qquad F \left({0}\right)$$ is an integer.

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

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

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

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

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

So is the binomial coefficient $$\binom n {j-n}$$ 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:
 * $$(3) \qquad \frac 1 2 \int_0^\pi f \left({x}\right) \sin x dx = F \left({0}\right)$$

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:
 * $$ \frac 1 2 \int_0^\pi f \left({x}\right) \sin x dx = \frac1 2 \left[{\left({F' \left({x}\right) \sin x - F \left({x}\right) \cos x}\right)}\right]_{x=0}^{x=\pi}$$

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 $$\frac 1 2 \int_0^\pi f \left({x}\right) \sin x dx = F \left({0}\right)$$.


 * 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 $$\left({x - \frac \pi 2}\right)^2 = x^2 - \pi x + \left({\frac \pi 2}\right)^2$$ and so $$x \left({\pi - x}\right) = \left({\frac \pi 2}\right)^2 - \left({x - \frac \pi 2}\right)^2$$

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

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 $$f$$:
 * $$\frac 1 2 \int_0^\pi f \left({x}\right) \sin x dx \le \frac {b^n} {n!} \left({\frac \pi 2}\right)^{2n+1}$$

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.

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