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.

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    $\begingroup$ Have you looked at en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational ? $\endgroup$ – InterestedGuest Feb 8 '11 at 19:38
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    $\begingroup$ Digits are an unnatural way of representing numbers. We know ways of representing pi that have nothing to do with digits, and those ways tell us, in some sense, "exactly" what pi is (at least well enough to prove that it's irrational). $\endgroup$ – Qiaochu Yuan Feb 8 '11 at 22:09
  • $\begingroup$ I notice that all your concerns about $\pi$ also apply to $\sqrt {2}.$ Since it is MUCH easier to prove $\sqrt 2$ is irrational than to prove $\pi$ is irrational (at least at the present time), I suggest first looking at proofs that $\sqrt 2$ is irrational. $\endgroup$ – Dave L. Renfro Jul 7 '14 at 16:29

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.

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    $\begingroup$ I just looked over the proofs there quickly -- "Cartwright's proof" seems the most accessible to me. $\endgroup$ – joriki Feb 8 '11 at 20:00
  • $\begingroup$ Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. $\endgroup$ – Grant Thomas Nov 23 '11 at 8:40
  • $\begingroup$ Whilst this is theoretically correct, in this case Derek has already written out the answer. $\endgroup$ – joriki Nov 23 '11 at 8:48

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.


$$I_n(\alpha)=\int_{-1}^1 (1-x^2)^n \cos \alpha x \textrm{ d}x$$

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.

  • 1
    $\begingroup$ This is "Cartwright's proof" that I referred to in the comment to my answer. $\endgroup$ – joriki Feb 8 '11 at 20:42

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