Pi is irrational

The number \(\pi\) is not rational.


For any fixed real number \(q\), and any natural number \(n\), let \($A_n = \frac{q^n}{n!} \int_0^{\pi} [x (\pi - x)]^n \sin(x) dx\)$ where \(n!\) is the factorial of \(n\), \(\int\) is the definite integral, and \(\sin\) is the sin function.

Preparatory work

Exercise: \(A_n = (4n-2) q A_{n-1} - (q \pi)^2 A_{n-2}\).

We use integration by parts.

show this <div><div>

Now, \($A_0 = \int_0^{\pi} \sin(x) dx = 2\)$ so \(A_0\) is an integer.

Also \($A_1 = q \int_0^{\pi} x (\pi-x) \sin(x) dx\)$ which by a simple calculation is \(4q\).

Expand the integrand and then integrate by parts repeatedly:
$$\frac{A_1}{q} = \int_0^{\pi} x (\pi-x) \sin(x) dx = \pi \int_0^{\pi} x \sin(x) dx - \int_0^{\pi} x^2 \sin(x) dx$$

The first integral term is \($[-x \cos(x)]_0^{\pi} + \int_0^{\pi} \cos(x) dx = \pi\)$

The second integral term is \($[-x^2 \cos(x)]_{0}^{\pi} + \int_0^{\pi} 2x \cos(x) dx\)$ which is \($\pi^2 + 2 \left( [x \sin(x)]_0^{\pi} - \int_0^{\pi} \sin(x) dx \right)\)$ which is \($\pi^2 -4\)$

Therefore \($\frac{A_1}{q} = \pi^2 - (\pi^2 - 4) = 4\)$ <div><div>

Therefore, if \(q\) and \(q \pi\) are integers, then so is \(A_n\) inductively, because \((4n-2) q A_{n-1}\) is an integer and \((q \pi)^2 A_{n-2}\) is an integer.

But also \(A_n \to 0\) as \(n \to \infty\), because \(\int_0^{\pi} [x (\pi-x)]^n \sin(x) dx\) is in modulus at most \($\pi \times \max_{0 \leq x \leq \pi} [x (\pi-x)]^n \sin(x) \leq \pi \times \max_{0 \leq x \leq \pi} [x (\pi-x)]^n = \pi \times \left[\frac{\pi^2}{4}\right]^n\)$ and hence \($|A_n| \leq \frac{1}{n!} \left[\frac{\pi^2 q}{4}\right]^n\)$

For \(n\) larger than \(\frac{\pi^2 q}{4}\), this expression is getting smaller with \(n\), and moreover it gets smaller faster and faster as \(n\) increases; so its limit is \(0\).

We claim that \(\frac{r^n}{n!} \to 0\) as \(n \to \infty\), for any \(r > 0\).

Indeed, we have \($\frac{r^{n+1}/(n+1)!}{r^n/n!} = \frac{r}{n+1}\)$ which, for \(n > 2r-1\), is less than \(\frac{1}{2}\). Therefore the ratio between successive terms is less than \(\frac{1}{2}\) for sufficiently large \(n\), and so the sequence must shrink at least geometrically to \(0\). <div><div>


Suppose (for contradiction) that \(\pi\) is rational; then it is \(\frac{p}{q}\) for some integers \(p, q\).

Now \(q \pi\) is an integer (indeed, it is \(p\)), and \(q\) is certainly an integer, so by what we showed above, \(A_n\) is an integer for all \(n\).

But \(A_n \to 0\) as \(n \to \infty\), so there is some \(N\) for which \(|A_n| < \frac{1}{2}\) for all \(n > N\); hence for all sufficiently large \(n\), \(A_n\) is \(0\). We already know that \(A_0 = 2\) and \(A_1 = 4q\), neither of which is \(0\); so let \(N\) be the first integer such that \(A_n = 0\) for all \(n \geq N\), and we can already note that \(N > 1\).

Then \($0 = A_{N+1} = (4N-2) q A_N - (q \pi)^2 A_{N-1} = - (q \pi)^2 A_{N-1}\)$ whence \(q=0\) or \(\pi = 0\) or \(A_{N-1} = 0\).

Certainly \(q \not = 0\) because \(q\) is the denominator of a fraction; and \(\pi \not = 0\) by whatever definition of \(\pi\) we care to use. But also \(A_{N-1}\) is not \(0\) because then \(N-1\) would be an integer \(m\) such that \(A_n = 0\) for all \(n \geq m\), and that contradicts the definition of \(N\) as the least such integer.

We have obtained the required contradiction; so it must be the case that \(\pi\) is irrational.