The number [49r $\pi$] 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: $A_n = (4n-2) q A_{n-1} - (q \pi)^2 A_{n-2}$. %%hidden(Show solution): We use [-integration_by_parts].

[todo: show this] %%

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 $4q$. %%hidden(Show calculation): 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 $$\frac{A_1}{q} = \pi^2 - (\pi^2 - 4) = 4$$ %%

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 $0$. %%hidden(Formal treatment): 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 $0$. %%

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.