I'm tutoring a girl in 8th grade (so she is 14 years old) and she recently had a mathematics chapter about numbers. In the last paragraph they introduced the difference between rational and irrational numbers. After that they gave two examples of an irrational number, namely $\pi$ and $\sqrt 2$. In the book it wasn't proved these numbers really were irrational.
The exercises started with a few easy questions, but then the following was asked:
Is $\pi^2$ rational or irrational?
She immediately thought it had to be irrational because $\pi$ is. I explained to her this argument is false since $\sqrt{2}^2=2\in\mathbb{Q}$. I remembered that $\pi$ is transcedental so $\pi^2$ cannot be rational. However, since she is only in 8th grade and the notion of irrational was just introduced I couldn't talk about fields, minimal polynomials and such.
Does anyone know an elementary proof of the fact that $\pi^2$ is not rational?
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$\begingroup$Maybe the author made a mistake, and meant to ask something like "is $\sqrt{\pi}$ irrational?"? Or maybe the author just intends to spark open-ended curiosity.
If there was an elementary standalone proof that $\pi^2$ was irrational, then it would imply $\pi$ was too. But I don't think there is a straightforward proof for $\pi$.
Making use of the given that $\pi$ is irrational doesn't help either of course.
$\endgroup$ 5 $\begingroup$Hermite's proof that $\pi^2$ is irrational is here (found via a Google search for proof that pi squared is irrational ):
$\endgroup$ 2 $\begingroup$The identity: $$ \pi^2 = 18\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}} \tag{1}$$ comes from the Euler series acceleration method and it can be used to prove the irrationality of $\pi^2$ and even more, for instance providing a (rather crude) upper bound for the irrationality measure of $\pi^2$. In fact, the existence of an identity similar to $(1)$ is the key of Apery's proof about the irrationality of $\zeta(3)$. However, I wouldn't try to prove $(1)$ or to explain a good portion of the technicalities of diophantine approximation to an $8$th-grader, no matter how brilliant he/she is.
$\endgroup$ 2 $\begingroup$Ivan Niven's proof is the simplest one (IMHO) with math level of medium-to-high-graders. Here it is.
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