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.

Asaf Karagila
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    I don't think there is any simple, intuitive argument for irrationality. As you may be aware the Classical Problem [Squaring the Circle](https://en.wikipedia.org/wiki/Squaring_the_circle) gets at the heart of the matter...but, of course, the Greek Geometers recognized it as a serious open problem. And, I think they had enough confidence to suspect that their inability to solve it meant that it was not solvable. But that sort of argument is not terribly persuasive. – lulu Aug 27 '18 at 14:45
  • Just to clarify, do you need an argument that only rational numbers will have repeating digits, or that pi is irrational? The former is likely to be accessible. If the student is interested in the latter and in calculating digits, I'd suggest showing a simulation of some method for calculating digits to say "at the very least, the first 1000 don't look repeating yet". – Mark S. Aug 27 '18 at 14:49
  • @MarkS. It is the latter question. That is, why is $π$ irrational? Your suggestion in the latter part of your comment is interesting. I tried that approach but she was able to understand that that proves nothing. In particular she thought that even if it would take a practically impossible number of approximations, it should still repeat. I could only assert that it could not. – Allawonder Aug 27 '18 at 14:50
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    I think you can describe that the situation for $\pi$ is not similar to that for $\sqrt{2}$ or $\sqrt[3]{2}$ or for that matter any algebraic number. And convince the student that any techniques based on algebra alone can't help figure out whether $\pi$ is rational or not. Encourage the student to study calculus in coming years and appreciate the proof of irrationality of $\pi$. – Paramanand Singh Aug 27 '18 at 16:12
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    Showing her that $\pi$ is irrational (what to speak of transcendental) is going to be hard work without calculus. I suggest focusing on the easy quadratic irrationals first (and introduce her to the wonderful world of continued fractions), and tell her you'll get to $\pi$ later, once she's had some calculus. – PM 2Ring Aug 27 '18 at 18:26
  • @theDoctor I don't understand your comment as it stands. Could you expand it a bit? – Allawonder Aug 28 '18 at 02:10
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    Wikipedia has a somewhat longer list of proofs, but they all come out to about the same level of difficulty. You could put it this way: the irrationality of $\sqrt2$ was known thousands of years ago, but $\pi$ had to wait until people had been using calculus for a hundred years. So the bad news is she really needs to learn calculus; the good news is it will then take her much less than a hundred years to learn the proof. – David K Aug 28 '18 at 02:58

6 Answers6


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.

hmakholm left over Monica
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    Appreciated. I indeed mentioned an example of this (in particular I gave the counterexample $0.101001000100001...$), which she understood not to be repeating (if continued in the obvious way). There was no way to however even hint at the fact that $π$ could be such a number, its digits having no discernible pattern, so that one couldn't discount its repeating after a large number of digits. – Allawonder Aug 27 '18 at 14:58
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    After the part you quoted, the text in the question says that she asked why it's irrational. So apparently she knows that decimal expansions of irrational numbers don't have to repeat, so this is not really an answer. – JiK Aug 27 '18 at 21:37
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    @Allawonder: Prove to her that $\sqrt{0.101001000100001...}$ is also irrational. She will find that its digits have no discernible pattern. At least this provides a concrete example of a decimal expansion of an irrational that isn't 'irrational for obvious reasons'. – user21820 Aug 28 '18 at 03:29

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 $\pi$.

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.

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.

Milo Brandt
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    +1 for the initial paragraph alone. In particular, I think showing a loose and fast pseudo-argument for the irrationality of $\pi$ could do more harm than good here. Better try to turn her curiosity into a motivator for learning enough to understand a real proof someday. (After all, it took the brightest heads on the planet 2000 years to find _any_ proof that $\pi$ is irrational; it's not unreasonable to have to study for 10 years or more to understand what they eventually came up with). – hmakholm left over Monica Aug 27 '18 at 18:57
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    Following the second part of the advice properly (without being another loose-and-fast pseudo-argument) would probably require getting into cardinalities -- but that's still more approachable for a bright 12-year-old than calculus would be, I think. – hmakholm left over Monica Aug 27 '18 at 19:06
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    While most numbers are irrational, most numbers we deal with in daily life are not, so that argument might not be so convincing. – Barmar Aug 27 '18 at 20:15
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    @Milo Brandt "that x+αc for non-zero integer α will always be zero..." do you mean irrational rather than zero? – Tyberius Aug 27 '18 at 22:12
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    To make "most numbers are irrational" somewhat rigorous, play a dice game with a D10. Keep rolling to get the next digit. – R.. GitHub STOP HELPING ICE Aug 27 '18 at 22:40
  • Thank you for mentioning the possibility of this approach. However, I don't see how this could be plausibly done. In particular refer to your third paragraph, near the end. Couldn't one define a similar extension of the positive integers by using unit fractions in place of $c$ and positive integers in place of $x$ and $\alpha$? I don't know that she'll be able to raise this objection -- at least not immediately -- but I do not intend to deceive her. Or is there something I'm missing here? – Allawonder Aug 28 '18 at 02:30
  • @Allawonder If I understand you correctly, then not quite - that has the issue that $2+\frac{3}3$ is again an integer, or that it is non-unique in that $2+\frac{1}2=1+\frac{9}6$, which is not the case with the proposed representation. You could observe that $n+\frac{1}m$ gives, for each integer, infinitely many rationals (by a mere translation!), so there should be many more rationals than integers (and so the probability of picking an integer at random from $[0,1)$ is definitely $0$) - though this runs into the issue that you can't really pick a random rational number in the first place. – Milo Brandt Aug 28 '18 at 03:26
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    @Allawonder: Ask her whether she expects dice throws to eventually repeat themselves. Her answer would guide the subsequent responses. – user21820 Aug 28 '18 at 03:32
  • By diagonalizing the rationals, you could prove to her that it’s possible to count them, so only a countable subset of numbers can be rational. Then you just have to convince her that there are uncountably many real numbers! – Davislor Aug 28 '18 at 04:58

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).

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  • The third paragraph is sufficiently cool and seems to solve this problem. Although tedious, it shouldn't be hard to carry out the program therein advised. Thanks. – Allawonder Aug 27 '18 at 16:00
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    However, I have a question. Does the fact that the rational approximations are not "nicely" simplifying *actually* tell us anything about whether the limit is rational or not? I don't see how it does. – Allawonder Aug 27 '18 at 16:30
  • @Allawonder: As I said, this is not a rigorous proof. However, the trick with the rationals is that, although they are dense, the denominators get bigger as you zoom in more finely. So if you can set both upper and lower bounds (e.g. with an alternating series), then you can exclude progressively larger and larger denominators, and that can be made rigorous. – Kevin Aug 27 '18 at 16:41
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    A nice example of the fraction of $\pi$ not cancelling is the [Wallis product](https://en.wikipedia.org/wiki/Wallis_product), but of course showing that $\pi/2$ equals the Wallis product isn't exactly trivial. – PM 2Ring Aug 27 '18 at 18:21
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    @PM2Ring: I *like* that example! – Kevin Aug 27 '18 at 18:47
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    @PM2Ring 3Blue1Brown has a video on the Wallis product that might be accessible to a bright 12 year old. https://www.youtube.com/watch?v=8GPy_UMV-08 – Roman Odaisky Aug 27 '18 at 22:50
  • @barto: As I've said three or four times now, this is not a rigorous proof. – Kevin Aug 29 '18 at 17:10

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.

Mark Bennet
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  • The first paragraph here appears to claim that there is no base in which the first two digits of $\pi$ (after the point) are equal, or at least no such base that is a power of ten? I don't think that follows from $\pi$ being irrational. – hmakholm left over Monica Aug 27 '18 at 15:43
  • @HenningMakholm You are right, of course - I appreciated I was being informal. I haven't tested - and I wasn't explicit either about whether the initial $3$ should be included in the digits which recur. – Mark Bennet Aug 27 '18 at 15:52
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    I don't see how going from something difficult to prove (pi is irrational), to something that hasn't been proved at all (pi is normal) helps the situation. – Acccumulation Aug 27 '18 at 15:57
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    @Acccumulation If you are talking about mathematics to an intelligent and interested twelve-year-old, I think normal numbers are interesting things to think about. "What are most numbers like?" and "What are the most familiar numbers like?" turn out to be different questions. I would have put this in a comment had it not been so long - but for me it is not about just answering the presenting question, but about provoking interest and helping the formulation of better questions in future. – Mark Bennet Aug 27 '18 at 16:16
  • I'm still confused by the "for small $n$ and specially constructed bases" qualifier here. Who's to say that the first quadrillion digits of pi do not repeat exactly as the _second_ quadrillion digits (but the _third_ quadrillion is completely different)? I would have thoughts such claims are beyond our current knowledge. – hmakholm left over Monica Aug 27 '18 at 19:01
  • @HenningMakholm I thought I had put enough uncertainty into the way I had expressed it - my intention was that a small $n$ might work by chance and a large $n$ might work by design [if $\pi$ has some special features which can be exploited] ... if you think it is not quite right, I will change it, since it is really important not to overstate the case, and my main point is that these might be ideas worth introducing. – Mark Bennet Aug 27 '18 at 19:08
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    @MarkBennet: Ah, okay then. I misunderstood the sense of your "and", I think. It could still be blind chance, though. For example, I don't think there is any "special feature" of $\pi$ that causes [six consecutive nines](https://en.wikipedia.org/wiki/Six_nines_in_pi) to show up as early as they do, except for the fact that they happen to do! – hmakholm left over Monica Aug 27 '18 at 19:09
  • @HenningMakholm I have changed and expanded the comment. Thanks for the care you take so often on the site - much appreciated, – Mark Bennet Aug 27 '18 at 19:19
  • @HenningMakholm The Feynmann point is cute, OTOH, there are the [Schizophrenic numbers](https://en.wikipedia.org/wiki/Schizophrenic_number) ;) – PM 2Ring Aug 27 '18 at 23:11
  • To the very first comment by @HenningMakholm: It appears that `2, 8, 114, 227, 340, 453, 566, 679, 33103, 66318, 265382, 1360121, 1725034, 25510583, 78256780, ...` is the list of bases for which $\pi$, when expressed in that base, has the first two digits after the point equal. I wonder if such a sequence says anything about the number (here $\pi$) considered. – Jeppe Stig Nielsen May 19 '19 at 21:37
  • @JeppeStigNielsen: To the OEIS-mobile! :-) – hmakholm left over Monica May 19 '19 at 22:06

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.)

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    The rational number $0$ is a "root" of the infinite "polynomial" $\sum_{n\ge 1} x^n.$ – Allawonder Sep 01 '18 at 11:57
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    A solution to an infinite series could also be a solution to a finite polynomial (even if irrational). Say $x=\frac1{\sqrt2}$ which is a solution to $2=1+\frac12+\frac14+\frac18+...=1+x^2+x^4+x^6+...$ , as well as to $2x^2-1=0.$ – Mirko Jul 02 '21 at 15:30
  • As I said, "Not exactly sound math..." – richard1941 Jul 04 '21 at 18:15

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.

Roman Odaisky
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    1/2 + 1/sqrt((n+2)!) is a series of approximations for 1/2 that are all irrational. So this relies on "natural" being convincing; which it is not. – Yakk Aug 28 '18 at 01:03
  • @Yakk the question was how to dispel the student’s misconception. The student seems to think, “There must be some repeating pattern in those digits, even if I can’t see it at a glance”. After a series of related values is provided which can be easily proven to be irrational and thus certainly have no repeating pattern, the frame of mind shifts to “well, many things are irrational even if I don’t like that”. That’s an opportunity for the educator to explain almost all numbers are irrational, give some rigorous proofs and so on. – Roman Odaisky Aug 28 '18 at 01:52