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.