# Pi is irrational

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

# Proof

For any fixed real number \(q\), and any natural number \(n\), let

## Preparatory work

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

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Now,

Also

The first integral term is

The second integral term is

Therefore

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

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 \(0\).

Indeed, we have

## Conclusion

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

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.

Parents:

- Pi
- Irrational number
Real numbers that are not rational numbers