# 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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