# 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

$$\frac{A_1}{q} = \pi^2 - (\pi^2 - 4) = 4$$
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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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