Is there a sufficiently reachable plausibility argument that $\pi$ is irrational?

Question

I was teaching someone earlier today (precisely, a twelve-year-old) and we came upon a problem on circles. Little did I know in what direction it would lead. I was able to give a quick plausibility argument that all circles are similar and that they are related by some constant (most conventionally $π$). We came to the actual computation of $π$ and there was no way I could escape mentioning its irrationality. Indeed, being a somewhat bright student, this person asked, after I had written out the partial expansion $3.141592...$ whether she could not continue computing the digits, and how she could do this. Although I mentioned approximation by polygons as one possible approximation algorithm, she assured me that she would continue to find more digits, to which I asked, To what end? Indeed, I went on, it is impossible to finish computing the digits. But it must start to repeat after some digit, no matter how large, she retorted. No, I said. It is an irrational number (we had earlier talked about irrationality and it was easy enough showing a variant of the classic proof of the irrationality of $\sqrt 2,$ being a simple algebraic number). She then asked for reasons why this is the case. Suddenly, I found myself short of explanation as I had myself never bothered to study any of the known proofs in detail. I tried telling her that the known proofs were impossible for her to understand now, but she persisted nevertheless. Then I promised that I would come with it when next we met.

However, I am sure that this would not benefit her in any way. Therefore, I sought for a simple explanation (not necessarily a proof in the usual sense) that was sufficiently convincing, but so far I have found none. All I have found are variants on Lambert's or Hermite's proofs, and those are far from what I'm looking for.

In consequence, I thought to ask here. Perhaps someone has come across a similar situation and had found an argument sufficiently enlightening at that level (that is, a sketch of ideas or plausibility argument that can lead to a proof -- ideas can always be grasped by anyone, after all). In short, do you know any argument that I could present to this person that could at least slake their curiosity for now until they are ready for the classic proofs (if they continue to be interested)? If so please present them.

Thank you.

Answer 1

But it must start to repeat after some digit, no matter how large, she retorted.

This sounds like a misunderstanding that you could do something about by showing her a concretely defined different irrational where it is clear to see that the digits cannot repeat, such as $$ \sum_{n=1}^\infty 10^{-n^2} $$ If you manage to convince her that is is possible for a number to have a decimal expansion that will never start (and continue!) repeating, it might become easier to accept that $\pi$ could be one of those numbers too.

Answer 2

There are a number of things going on, and this will be too long for a comment.

The first is that so far as we know, the first $n$ digits of $\pi$ will be found in order somewhere later in the decimal expansion, whatever $n$ we choose, but it won't be from the $n^{th}$ place (* - see after this para) (or perhaps displaced by $1$ depending on precisely how you are counting), again so far as we know, and it certainly won't be in the form of a recurring decimal. The first part of this is down to $\pi$ conjectured as being a "normal" number, which is a concept worth exploring - this means it is expected that any string of $n$ digits should be found in the decimal expansion with the average frequency which you would expect by chance. The second part is because $\pi$ is known not to be rational.

(*) For small $n$ or for specially constructed bases, or even by blind chance, it is possible that there would be short recurrences by chance or design (see comments). It is certain that these would not persist. There remains the possibility that there is some special feature of $\pi$ (which, after all, is not a random number, but a deliberately chosen one with special properties already known) that could be exploited to do this, but I don't think anyone has yet found one.

It is quite easy to prove that any recurring decimal is rational. If the number $r$ has $n$ digits which eventually recur then $10^nr-r=(10^n-1)r$ will be a terminating decimal and therefore of the form $\frac {p}{10^q}$ for integers $p$ and $q$ and we have $$r=\frac {p}{10^q(10^n-1)}$$

It is possible to prove that almost every real number is normal (Hardy and Wright - Introduction to the Theory of Numbers - do it in their chapter on decimals, which is possibly accessible - certainly for ideas - they also have a proof of the irrationality of $\pi$ which would not be accessible). But the Cantor Diagonal trick to show that most real numbers are not rational is worth introducing. And since all recurrent numbers are rational, most must be not recurrent.

This doesn't exactly answer your whole question, but there is some interesting mathematics lurking here just under the surface.

Answer 3

A rigorous proof may not be possible at this level of mathematics. However, you may be able to get away with a more informal approach.

First, convince her that pi equals some generalized continued fraction, such as: $$ \pi = 3 + \frac{1^2}{6 + \frac{3^2}{6 + \frac{5^2}{6 + \frac{7^2}{\ddots}}}} $$ (or pick another example which converges faster?)

You can do this by computing partial sums and showing how they converge. Now, take some partial sums and try to simplify each one in turn. Write all integers as their prime factorizations, to make it obvious that they are getting progressively more complex and not "nicely cancelling." Now, you can credibly argue that you're not going to reach a regular fraction by following this process, so pi must be irrational.

This is not a "real" proof, and you should stress to her that there are additional pieces which you've left out. In particular, we are omitting a proof that the continued fraction really equals pi, and a proof that the continued fraction is irrational. You may also want to mention the fact that infinite sums need a formal definition.

For added rigor, use an alternating series, so that you can bracket pi from both above and below. This is arguably a "better" proof because it progressively removes from consideration fractions with larger and larger denominators, which means that you can't have "something funny" happen in the limit (cf. $0.999\ldots = 1$ and related limits). Another example suggested by PM 2Ring in the comments is the Wallis product: $$ \prod_{n=1}^\infty \bigg (\frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \bigg ) = \frac{\pi}{2} $$ This brackets $\frac{\pi}{2}$ from above and below because the terms are alternately greater and less than one. It's also "perfectly obvious" that each partial product will have a progressively larger and larger power of two in the numerator (while the denominator is always odd).

Answer 4

I would try to convince her that most numbers are irrational and not even address $\pi$ - showing her that the question of "Do the digits of $\pi$ repeat?" is nontrivial (and, indeed, deep) is probably more valuable than convincing her that they don't, especially since convincing her of $\pi$'s irrationality could make it seem like $\pi$ is special when this is very much not a special property of $\pi$.

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You can argue this fairly simply; the main thing to note is that an irrational plus a rational is irrational - this is easy to show via contradiction using integer ratios and is not too bad to show if you take "rational" as meaning "the digits repeat" either, so you can use whichever is likely to be more comfortable.

Right off the bat, once you get one irrational number, you have, in some sense, that at least $\frac{1}2$ the numbers are irrational, because if $c$ is any irrational number, then $x+c$ for $x$ rational is always irrational. A little more work can convince someone that $x+\alpha c$ for non-zero integer $\alpha$ will always be irrational and that the numbers of the form $x+\alpha c$ are all distinct based on the value of $(x,\alpha,c)$ - which seems to suggest that almost every number is irrational, since given any rational number, I can produce infinitely many irrational numbers!

Now, there's some elision in this argument - namely that it measures the size of sets in a vague way. The usual way to complete this argument would be to consider it mod $1$ (i.e. look at only fractional parts), then consider picking a random number in $[0,1)$ and think of the probability of hitting a rational - then, the formal thing that happens is that you can find infinitely many pairwise disjoint sets that are equally likely to hit as a rational, so the probability of hitting a rational must be zero.

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As pointed out by @R.. in comments, one could rigorously define choosing a random real number in $[0,1]$ by rolling a fair ten sided die repeatedly to get its digits - this both can give an actual model to the argument I suggest and also could be used to give another proof that the probability of getting a rational is $0$ - it's not so hard to convince oneself that, for any fixed $n$ and $m$, the event that you roll $n$ digits and then a pattern with period $m$ occurs with probability $0$ - but then you'd need countable additivity to say that the probability of getting this for any $n$ and $m$ is also zero - and a student might rightfully balk at that reasoning since countable additivity is hard to motivate when uncountable additivity is clearly wrong.

Answer 5

Consider regular polygons with $2^n$ sides inscribed in a circle. To determine the ratio of the area of a $2^n$-gon to that of a square, one truncates Viète’s formula at an appropriate number of factors:

$$ \frac{A_4}{A_{2^n}} =\underbrace{{\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdots}_\text{$n-2$ factors} $$

The limit of the ratio is of course $\dfrac2\pi$. To the student, the factors should look obviously irrational. And while it takes a rigorous proof (Mathologer’s take on Lambert’s proof is probably the most accessible) to exclude the possibility that the product might somehow end up rational, it should be apparent that it would be exceptional for $\pi$ to be rational if its natural approximations are all irrational. At least the student should see that other quantities, clearly related to $\pi$, are indisputably irrational and it can’t be claimed that digits will eventually repeat, and the same can very likely be said for $\pi$ as well.

Answer 6

Maybe student does not understand that irrational numbers do not have any repeats in their decimal representations, while rationals either terminate or end in a repeating cyle. Show her fractions like 12345/99999. So if pi had a repeating decimal pattern, it would be rational. Teach her how to convert between numbers that end with repeating cycles and their equivalent fraction. Teach her APPROXIMATIONS to pi, like 22/7 and 355/113. She could use a calculator to verify that these are not exact.

Explain to the student about algebraic numbers and transcendental numbers. Then show pi/4 as a solution to tan(x)=1. Then show her the power series for tan. Since pi is the solution to an INFINITE series, it is not a solution to a finite polynomial... so it cannot be algebraic.

Not exactly sound math, but it might satisfy the student for now.

Age 12? She must be very bright. What does she have for music lessons? (Math and music seem to go together.)