Pi is irrational

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

Proof

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>

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

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

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