Is there a proof that $\pi$ is an irrational number?

Question

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

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

Answer 1

There are quite a number of proofs on Wikipedia. Not sure how accessible you will find them, though -- feel free to ask about one of them that you might be able to understand.

Answer 2

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.