Is there a proof that $\pi$ is an irrational number? Most math texts claim that $\pi$ is an irrational number. However, I'm having a little bit of trouble understanding that.
Since nobody has calculated all of the digits of $\pi$, how can we know that either:


*

*one of the digits repeats (as in $\frac{10}{3}$)

*the number eventually terminates



Note: Please be very descriptive in your answers... I don't have anything beyond high school math.

If you know a bit of calculus and have come across induction, then here's an outline of a standard exercise (see Burkill - A first Course in Analysis) to prove $\pi$ irrational.
Let
$$I_n(\alpha)=\int_{-1}^1 (1-x^2)^n \cos \alpha x \textrm{ d}x$$
then integrate by parts to show that for $n \ge 2$
$$\alpha^2 I_n = 2n(2n-1)I_{n-1}-4n(n-1)I_{n-2}.$$
Use induction to show that for positive integer $n$ we have
$$\alpha^{2n+1}I_n(\alpha)=n!(P(\alpha) \sin \alpha + Q(\alpha) \cos \alpha),$$
where $P(\alpha)$ and $Q(\alpha)$ are polynomials of degree less than $2n+1$ in $\alpha$ with integral coefficients.
Show that if $\pi/2 = b/a,$ where $a$ and $b$ are integers, then
$$\frac{b^{2n+1}I_n(\pi/2)}{n!} \quad (1)$$
would be an integer.
Note that
$$I_n(\pi/2) < \int_{-1}^1 (1-x^2)^n \textrm{ d}x < 2 \textrm{ and } 
\frac{b^{2n+1}}{n!} \rightarrow 0 \textrm{ as } n \rightarrow \infty$$
which results in contradiction since $(1)$ is supposed to be an integer but we can show that it is as small as one desires.
This was the first proof of the irrationality of $\pi$ that I came across, and think it is very accessible for those willing to give it a go. 
