What is wrong with this proof?

Is $\pi=4?$

Simon Fraser
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Pratik Deoghare
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    The problem is that it doesn't approach circle in a "smooth" way. – J. J. Dec 03 '10 at 13:49
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    Only for very large values of $\pi$ – Ross Millikan Dec 03 '10 at 13:51
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    You may do the same "trick" with a triangle, thereby "countering" Pythagoras theorem. – Fredrik Meyer Dec 03 '10 at 13:55
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    The lengths of the curves certainly form a series which coverges to an upper bound for $\pi$. :P The comments at the source are really funny. – Raskolnikov Dec 03 '10 at 13:57
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    It was almost made law that $\pi=4$ (and $3.2$) in Indiana in 1897: http://en.wikipedia.org/wiki/Indiana_Pi_Bill – Jonas Meyer Dec 03 '10 at 16:05
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    A related question (one dimension higher) is this one: http://math.stackexchange.com/questions/6979/. – Mike Spivey Dec 03 '10 at 16:37
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    I suppose a related question would be: "how long is a slinky?" The answer of course depends on if the slinky is in its "resting/boxed" state, or if it is extended. In the example above, you're measuring the length of the circle with a "compressed" slinky, but you're calling the length of it the length of the "stretched" slinky. It's like a magic trick but where the magic is in the lack of precision in what it *is* you're attempting to measure. – Ryan Budney Dec 03 '10 at 18:13
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    I cannot believe that reddit has found it's way here... – Bey Dec 03 '10 at 21:09
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    The proof fool way to see that doesn't converge to a circle: Its perimeter is 4. – Thomas Ahle Apr 07 '11 at 09:32
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    The funny part is that the picture actually DOES prove that $\pi <4$. Indeed, the area of the circle is $\frac{\pi}{4}$, and after the first iteration, the area of the dodecagon is strictly less than 1. And as areas, the limitation process would be right.... – N. S. Dec 06 '11 at 19:43
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    I guess [this proof](http://demonstrations.wolfram.com/ArchimedesApproximationOfPi/) is more viable. – Red Banana Aug 26 '12 at 20:33
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    Has anybody tried to square the area inside to see what would be the lower bound ? – BlueTrin Nov 08 '12 at 11:38
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    $\pi=arg(i)=3\pi=4$. so $4$ is multivalued just like $\pi$. –  Feb 22 '13 at 04:13
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    How can the answers be out of date?! Someone changed math, and I wasn't informed? – Asaf Karagila Oct 06 '13 at 17:14
  • What if the diameter was not 1, but like 8? Idk just a suggestion – TrueDefault Feb 28 '14 at 02:05
  • @xport I guess the ! in the last picture means " surprise !!!". It was an unfortunate coincidence. – Felix Marin Aug 11 '14 at 18:56
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    My first programming experience, using turtle graphics, involved drawing a circle: "repeat 360: forward 1, right 1". This messed up my the idea of a circle in my 9-year old mind for years. – gerrit Apr 05 '15 at 18:55
  • You can also read about the "Indiana Pi Bill of 1897 in the book "A Brief History of Pi". – DanielWainfleet Oct 17 '15 at 08:46
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    In calculus, the perimeter of the circle wouldn't be approximated by the sums of the sides of the tiny rectangles. Instead, the perimeter of the circle would be approximated by the sums of the hypotenuses of those tiny rectangles. By intuition alone, you can tell that the sums of the hypotenuses are a better approximation for the perimeter of a circle. – Inquisitive Jul 30 '16 at 18:16
  • @RossMillikan I am sorry I don't understand what you mean by " Only for very large values of $\pi$ " – N.S.JOHN Oct 13 '16 at 16:03
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    @N.S.JOHN: it is an allusion to an old joke where you treat a constant, here $\pi$, as a variable. You might say $\sqrt 2 \approx 1.5$ for large values of $2$. – Ross Millikan Oct 13 '16 at 23:48
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    Koch curve, Snowflake, Fractal Weierstrass P.... – Narasimham Oct 30 '16 at 12:03
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    This is NOT a problem for the method of Arhchimedes. The method of Archimedes uses inscribed polygons AND circumscribed polygons such that number of sides of both types of polygons iteratively increase. This puts **pi** as a quantity between two rational numbers *a priori*. – Doug Spoonwood Nov 05 '16 at 18:07
  • The so called "source" has nothing to do with the question and is highly inappropriate, so I removed it. Or it is a broken link. I left it so that it is not directly viewable now. – Simply Beautiful Art Nov 25 '16 at 19:12
  • Are you saying that $\pi = 1\times 2\times 3\times 4$ ?? :) – Mr Pie Feb 25 '18 at 11:48
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    If you take something concrete that's circular (with a radius of 1), and take a string of length 4, the process that you're doing in the picture is essentially bunching the string closer and closer up to the boundary of the circular thing. But you're not "pulling the string tighter" around it. The pulled string is what gives circumference, not the bunched-up one. –  Mar 05 '18 at 09:03
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    ... poof. Moral of the story: you can never make a circle by cutting corners. – Marcos Jun 20 '18 at 20:42

23 Answers23


This question is usually posed as the length of the diagonal of a unit square. You start going from one corner to the opposite one following the perimeter and observe the length is $2$, then take shorter and shorter stair-steps and the length is $2$ but your path approaches the diagonal. So $\sqrt{2}=2$.

In both cases, you are approaching the area but not the path length. You can make this more rigorous by breaking into increments and following the proof of the Riemann sum. The difference in area between the two curves goes nicely to zero, but the difference in arc length stays constant.

Edit: making the square more explicit. Imagine dividing the diagonal into $n$ segments and a stairstep approximation. Each triangle is $(\frac{1}{n},\frac{1}{n},\frac{\sqrt{2}}{n})$. So the area between the stairsteps and the diagonal is $n \frac{1}{2n^2}$ which converges to $0$. The path length is $n \frac{2}{n}$, which converges even more nicely to $2$.

Parcly Taxel
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Ross Millikan
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    The downvote came from me. I added it soon after you posted your answer when there weren't any other answers to complement yours. My rationale was that, based on how I've seen other people (not here) attempt to answer this question, anything less than a completely rigorous demonstration would not suffice simply because almost any intuitive explanation seems plausible in this instance. I later tried to remove the downvote when I realized there were a variety of answers and that yours complemented the others nicely, but because an hour had elapsed I was not (and am not) able to do so. – Zach Conn Dec 03 '10 at 18:33
  • @Zach OK thanks. No problem. – Ross Millikan Dec 03 '10 at 18:53
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    @Zach Conn: Now that Ross has edited his answer, I believe you can remove your downvote if you still wish to do so. – Mike Spivey Dec 04 '10 at 00:32
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    I wasn't so worried about the points as about not satisfying a customer. Zach's comment about there being nice intuitive explanations that lead to the wrong conclusion is well taken. – Ross Millikan Dec 04 '10 at 01:08
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    So in other words custom perimeter in OP's image converges to 4 areas and perimeter of stairsteps built on top of diagonal converges to 2 areas, i.e. areas of rectangle 1 x a (a = square side length). Nitpicking: The area between the stairsteps and the diagonal is actually n*1/(2n^2). – przemoc Jul 12 '11 at 20:23
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    When I saw the question, I was immediately reminded of the "diagonal" paradox is one of Dudney's puzzle books. I don't think I saw what I'd consider the "real" answer there nor here, however, which is that each time one subdivides the problem in half, the amount of error in each piece is cut in half, but *the number of pieces doubles*. In order for a limit computed via problem reduction to be valid, the per-piece error must decrease faster than the number of pieces increases. – supercat Jul 18 '14 at 20:26
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    BTW, on similar note, speaker whom I saw do the trick of cutting a huge hole in a small piece of paper (fold it and make alternating cuts on the inside and outside) gave what I thought was a brilliant explanation: when cutting a "normal" hole, attempts to reduce the surface are of paper required for each lineal inch of parameter also increases the amount of paper wasted in the hole itself and reduces the amount of paper available to fabricate the perimeter. The "trick" cutting technique allows the perimeter to be made thinner without increasing the loss of material. – supercat Jul 18 '14 at 20:29
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    I don't think this answers the question. The OP has correctly computed the length of the staircase and, apparently, he knows that $\pi \neq 4$. So he knows the reasoning is wrong (or at least contradictory) but he is asking which step fails. – Emanuele Paolini Oct 14 '14 at 12:55
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    I am sorry to contradict, and this comment may be stupid, but almost all the answers are propagating how this is wrong, not why it is wrong. –  Mar 16 '16 at 03:30
  • you might be interested in looking at [taxi cab distance](https://en.wikipedia.org/wiki/Taxicab_geometry) – them Apr 10 '16 at 15:23
  • @RossMillikan if each triangle has base 1/n and height 1/n, wouldn't the area of each individual triangle be (1/2)(1/n)(1/n) and not 1/(n^2) as you have indicated? – drzaius7 May 04 '16 at 00:37
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    @drzaius7: yes, there is a factor $2$. That does not change the argument that the sum converges. I will fix. Thanks. – Ross Millikan May 04 '16 at 01:13
  • may be, it depends on how precise we are. – pooja somani May 19 '17 at 12:39
  • But why we cannot say that in limit we get a non-real number, which is neither $2$, nor $\sqrt{2}$, but has modulus $2$ and regularized value $\sqrt{2}$, just as we can say that $(-1)^\infty$ has modulus $1$ but (Cesaro-) regularized value $0$? – Anixx Oct 02 '21 at 17:29
  • @Anixx: We are taking the limit of reals which are bounded. By completeness, the limit exists and is real. – Ross Millikan Oct 02 '21 at 19:45
  • The question here is about what limit we are talking. The limit of the sequence of lengths, or the length of the limit of the curve. The both exist but different. But we can postulate that the length of the resulting curve is neither of these limits, but a limit in more extended sense. Like postulating that $(-1)^\infty$ is neither $0$, nor $1$. – Anixx Oct 02 '21 at 19:53
  • Also, when talking about the limit of the curve we usually understand it in a sense of coordinates, but not in the sense of tangent lines, for instance, or second derivatives. This is a topic of variational calculus, but one can talk about limit of a curve in different senses. – Anixx Oct 02 '21 at 19:57

This problem illustrates the fact that two functions can be very close: $|f(x)-g(x)|<\epsilon$ for all $x\in [0,1]$, but their derivatives can still be far apart, $|f'(x)-g'(x)|>c$ for some constant $c>0$. In our case, let $x=a(t),y=b(t),0\le t\le 1$ and $x=c(t),y=d(t), 0\le t\le 1$ be the parametrizations of the two curves. By smoothing the corners, we may assume that both are smooth. $$ \|(a(t),b(t))\|\approx \|(c(t),d(t))\|$$ does not imply $$ \|(a'(t),b'(t))\|\approx \|(c'(t),d'(t))\|$$ Therefore $\int_0^1 \|(a'(t),b'(t))\| dt$ need not be close to $\int_0^1 \|(c'(t),d'(t))\| dt.$ Here $\|(x,y)\|$ denotes $\sqrt{x^2+y^2}$.

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    As I say below, I like this answer (+1). However, the "therefore" at the end makes me nervous. Uniform convergence of a sequence of functions is certainly *sufficient* to imply convergence of their integrals, but it is by no means necessary. Indeed, compared to the Lebesgue-style convergence theorems, the "uniform convergence theorem" is almost trivial. – Pete L. Clark Dec 03 '10 at 15:50

The pithy expression for this "paradox" is as follows: let $x_n(t)$ be a sequence of parameterized curves which converges uniformly to a limit curve $x(t)$. Then it need not be the case that the arclengths of $x_n(t)$ approach the arclength of $x(t)$.

[Added after seeing TCL's answer: it is also true that uniform convergence of a sequence of functions does not imply convergence of their derivatives. See Section 3 here for some discussion of this. As TCL points out, since arclength elements are computed using derivatives, the observation about derivatives may be in some sense more fundamental. In other words, I think I like TCL's answer better than mine.]

As Ross Millikan points out, this is more familiarly shown by approximating the hypotenuse of a right triangle by a staircase pattern of horizontal and vertical line segments. I still remember being a senior in high school and having a friend (whom I had had no prior mathematical interactions with) show this to me. I definitely remember thinking that it was not paradoxical but certainly surprising. (And I have mathematically respected this person ever since, even though I haven't seen her since I was a teenager.)

Added much later: if you think about the phenomenon physically rather than geometrically, it seems to me that the surprise disappears. For instance, suppose I'm running and you're riding a motorcycle. It is possible for your speed at every instant to be 25 times (say) faster than mine while maintaining a very small distance from me, e.g. by making very small, very fast circles around me.

Pete L. Clark
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    I prefer your answer, because you've identified the specific false belief that leads immediately to the paradox. – goblin GONE Jan 19 '15 at 16:44
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    "you could drive very small, very fast circles around me" -- this can also happen when walking a dog, either because the dog is on a lead or because you refuse to throw whatever you're holding. No matter how short the lead, the dog will find a way to get significantly more exercise than you. – Steve Jessop Apr 03 '15 at 15:28
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    @PeteL.Clark, the link is offline – Iuri Mielniczuk Sep 01 '18 at 14:19

I am probably going a little off-topic with these comments, so feel free to downvote :)

In my opinion this type of proof emphasizes why it is wrong to teach/take “Calculus” instead of Analysis.

For most of the nice applications of integration, we always use the following approach: take some quantity/expression, break it in many pieces, identify the sum of many pieces as a Riemann sum, and thus our quantity is the limit of the Riemann sums, thus the corresponding integral…

Unfortunately, except in serious Analysis courses, not even once do we go into the subtle details: why is the Riemann sum a good approximation for our quantity, namely why does the error in our approximation go to zero…

Most students who take Calculus end up “understanding” lots of false results, which we don’t have the time to disprove in general: any derivative is continuous, any approximation that looks good is good, …

To come back to this problem, not all approximations that look good are good. We always MUST prove that the errors in our approximations go to zero. And for all the formulas we “prove” in calculus, there is an actual mathematical proof, which is pretty technical (and most non-mathematicians would say boring and stupid, but then without such proofs one cannot really understand why the “proof” from the above picture is wrong). But without going through the formal proofs, one cannot truly understand why that particular approximation works in that case, and more importantly why a different approximation won’t work.

Coming back to the above picture, one way to understand it is the following: we approximate the circle by a sequence of polygons. Let $c_n$ be the length of the $n$th polygon and $c$ be the length of the circle. At each step the error in our approximation is $4-\pi$, which doesn’t go to zero. This means that the arclength of the circle might not be the limit of arclengths of the polygons. The only thing we can conclude is that, if all the quantities and limits that appear in the picture exist, then the limit approximates the arclength of the circle with an error of at most the limsup of the errors. In other words, $4 \approx \pi$ with an error less than or equal to $4-\pi$. Hmm, what is wrong with this?

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N. S.
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    To some extent, this puzzle illustrates the arc of mathematics from Archimedes to Newton. Archimedes (who would not have made this error) knew about approximation by tiny increments, but he did not have the formal theorems that are supposed to keep us out of trouble. That was the program that Newton and Leibnetz (or Leibnez?) finished. – phv3773 Jun 29 '11 at 19:51
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    @phv3773 Leibniz. – Paul Slevin May 02 '12 at 10:55
  • Wikipedia links explaining some concepts mentioned here: [calculus](http://en.wikipedia.org/wiki/Calculus) and [(numerical) analysis](http://en.wikipedia.org/wiki/Numerical_analysis) (classes). [Riemann sum](http://en.wikipedia.org/wiki/Riemann_sum). [limsup (limit superior)](http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior). – Rory O'Kane Nov 13 '12 at 08:14
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    N.S. isn't talking about numerical analysis; he's referring to http://en.wikipedia.org/wiki/Mathematical_analysis, which provides the foundations of calculus. – Ray Mar 20 '13 at 12:44
  • There may be a tendency among students of calculus (and perhaps also among teachers) to consider only the lower Riemann sum; that is, to look only at rectangles inscribed under a curve. If you look at both the upper _and_ lower Riemann sums, however, and can prove they converge to the same value, what's wrong with that? – David K Sep 25 '18 at 11:31
  • @DavidK Nothing. But in this picture that is NOT a Riemann sum :) – N. S. Sep 25 '18 at 15:36
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    Of course it is not. But this answer does mention Riemann sums. Using that topic as an example, I wonder if there is a useful level of rigor that may often be missing from some "calculus" courses that can be accomplished without turning the "calculus" course into a full-blown real analysis course. – David K Sep 25 '18 at 17:27
  • @DavidK Yes and no... I think that the main problem is that most students and even some instructors don't really get all subtilities, and everything which could go wrong. Most of the times they only look as you said at left hand points vs right hand points, instead of looking at upper/lower (Darboux sums). Also, and this is part of the problem, I am not aware any simple way of proving that if the upper minus lower Darboux sum converges to 0 then the function is Riemann integrable. Last but not least, and this is another subtle problem, – N. S. Sep 25 '18 at 17:47
  • showing that regular partitions are sufficient is highly non-trivial :)To put it simple, yea we could show them how to calculate using properly Riemann sums in Calculus, but we could not explain them why it works.....From my experience, most students which take Calculus struggle to calculate $\int_0^3 x^2+1 dx$ using left end points, imagine asking them to do something where Darboux sums are really necessary. – N. S. Sep 25 '18 at 17:48
  • The problem is that it's a problem we assume that π < 4 at the first place or not. If we don't, we also can't say that the error is a non-zero constant toward infinite iteration. But if we assume that π < 4, so the proof is still a contradiction itself... – tia Dec 02 '19 at 05:45
  • @N.S. I agree with the spirit of your answer, but I don't think it supports the conclusion that it is wrong to teach "calculus" rather than "analysis." There is a perfectly good one-line response to this question that any calculus student *should* be able to give, which I will post now as a separate answer. The problem isn't calculus versus analysis, but thinking (in this case, defining the quantity you are trying to calculate) versus not thinking. – sasquires Jun 12 '20 at 01:43
  • @sasquires I really dissagree that any calculus student should be able to figure this out, unless your calculus students are really smart. – N. S. Jun 12 '20 at 04:29
  • @N.S. Perhaps you're right and the first-year calculus students wouldn't answer this correctly. But my point is basically that it's really easy to refute the paradox using the language of the calculus of infinitesimals. You just need to know how to define the arc length of a differentiable curve (i.e., $\int_C \sqrt{ dx^2 + dy^2}$), which is something they teach in first-year calculus. But I want to re-iterate that I also appreciate your point, which is that learning how to write a rigorous error bound is essential for avoiding stupid mistakes. – sasquires Jun 13 '20 at 01:00

Hilarious! Of course, the circumference is not approximated by the sum of lengths of the lines constructed as shown, but by the sum of the hypotenuses of each of the right-angle triangles formed around the edge of the circle (forming a polygon with vertices on the circle).

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Dan Christensen
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What is wrong with this?

Fundamentally, that you have jumped in without a definition of the length of an arc.

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    Implicitly, this __is__ someone's definition on defining the length of a circle, and we can use existing knowledge to counter (eg, they implicitly accept the meaning of length of straight lines). Further, a definition in math need not coincide with any reality. – Jose_X Mar 05 '15 at 16:24

This is simply another example of why the "limit of the sum" is not the "sum of the limit."

(Length of curves are a subset of Sums/Integrals which are really the same thing in my mind. If you like, in this case "the limit of the lengths of the curves " is not the "length of the limit curve")

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Eric Naslund
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    Is the circle a _limit curve_ of the jagged square? – Muhammad Alkarouri Jan 08 '12 at 17:05
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    @MuhammadAlkarouri: Yes it is. (Formally: parametrize the $n$th jagged square such that as $t$ varies over $[0, 1]$, we have $(x_n(t), y_n(t))$ exactly traversing the jagged square. This is what a "curve" means. Then in the limit as $n\to\infty$, if we let $x(t)=\lim_{n\to\infty}x_n(t)$ and $y(t)=\lim_{n\to\infty}y_n(t)$ we get a curve $(x(t),y(t))$ which traverses the circle as $t$ goes from $0$ to $1$. In other words: the points on the jagged square _do_ approach the points on the circle, but the lengths of the curves remain $4$ and don't approach the length $\pi$ of the limit curve.) – ShreevatsaR Mar 30 '14 at 10:18
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    This should be the accepted answer. It gives the mathematical reason why this cannot be done. Other long or 'intuitive' explanations with pictures are not sufficient. They just add to the noise. – AnandJ May 16 '19 at 00:03

Correct answer: Nothing is wrong with this, as long as your space is defined using a Manhattan metric. Normal Euclidean space is defined using a Euclidean metric.

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    using Manhattan metric, l=2piR is Wrong. – cnd May 23 '11 at 06:28
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    This is a cop-out answer: you change the definition of $\pi$ if you do this. That is not in the spirit of the question and does not explain why the limiting process described in the question does not converge to the expected answer. – Qiaochu Yuan May 23 '11 at 07:31
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    +1 This seems to explain it fine. Two different metrics give two different numbers. Everything else assumes arc lengths are defined by infinitesimal hypotenuses, which was what the question is clearly asking about. So the above are tautological. – Zach466920 Jun 13 '15 at 16:13
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    The underpinning of this question is why are arc lengths defined the way they are. The above discuss this assuming Euclidean is the space. Of course we know that $\pi$, Euclidean, rather than $4$, Manhattan, is more useful in hindsight, but if we lived in Manhattan, perimeters of 4 could actually match up with reality! ;) – Zach466920 Jun 13 '15 at 16:22
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    I don't think it's a cop-out. I think it nicely complements the other answers by suggesting that whether a limiting process "works" is relative to the metric you use. This leads to both an answer to the original question (the limiting process does *not* work in the case of a Euclidean metric) and an interesting side note (it doesn't mean that such a limiting process is invalid for all metrics). – 6005 Jul 23 '16 at 20:47

Intuitive Response (for those who don't understand the more analytical responses)

The answer is easy. We just have to zoom in. Π &ne 4 solution image

We can see at low zoom how the (purple) staircase hugs circle, but higher zoom shows it always remains a crude approximation to the circle's shrinking matching segments except near 0, π/2, π, and 3π/2. [In contrast, the (green) inscribed polygon is an increasingly good approximation and equally good at all angles.]
-- see "Simple Geometrical Explanation" below for longer but still simple explanation. The Updates at bottom add more insight once the simple geometrical explanation is not good enough for you. [Need to add more pics to clarify some aspects better.. ultimately potentially leading into something approaching a formal proof.]

The javascript code used to make the picture frames of the gif follows at the bottom. The code can be used as a starting point to make your own improved gif/animation or just single png frame. [may try to clean js code up later on as well as make more running time efficient]. I then clicked through to each pic, carefully screen captured the same bordered region for each pic, and saved to file. I integrated them into a gif using http://gifcreator.me/ (most frames got 250ms delay, but the first and the last of each of the 6 sequences got 750ms). I took that final gif and uploaded to stackexchange https://meta.stackexchange.com/questions/75491/how-to-upload-an-image-to-a-post

In case the above very short explanation + pic is not enough, here is a longer re-explanation (leveraging pic):

Simple Geometrical Explanation:

[To get a simple explanation, we have to have a simple approach. A circle is a simple, easy-to-make shape, and this problem was studied ages ago with simplified reasoning.]

The question posed is why can't we approximate the length of a circle [PI = the length of a circle of diameter 1] by measuring the length of a "staircase" path that hugs the circle tightly?

The answer is simple:

If we aim to find the length of some near straight object from point A to point B, we want to measure as closely as possible to a straight path from A to B (see green/red quasi-overlap). We won't get the correct answer if instead, like the staircase approach above (purple), we measure from A to a point far off to the side and then from that point to B. This is very intuitive.

Now, to approximate the length of a circle, we replace the whole circle with many little straight paths following closely the shape of the circle (green). We do use a single direct connecting (green) piece between every two adjacent points A and B (A and B, not pictured, would be where adjacent gray lines intersect red circle) instead of using the inaccurate 2-piece (purple) step. Do observe a key point that makes this work out: any little arc of a circle, as with any small section of any simple curve, becomes nearly indistinguishable from a similarly sized line segment when these are short enough.

[Recap:] So, at any angle around the circle, for large N, a small green line segment ≈ small red arc. Meanwhile around most of circle 2 right angled purple line segments are clearly > matching red arc, no matter N. This is why the green approximation gets very close to π while the purple approximation is way off at 4. [Note: green π = N sin (pi/N) and is easily derivable from basic geometry by summing 2*N pieces that are opposite radial triangles with hypotnuse .5 and central angles 2π/(2N).]

[Finally, I apologize if you can't discern green from red. I may change colors later but found these convenient and generally easy to differentiate.]

    <table style="border:3px solid black;"><tbody>
    <tr><td colspan="2"><center><b><font size="4"><span style="color:red;">&#960; = 3.141592...</span></font></b></center></td></tr>
    <tr><td><center><b><span id="sp1" style="color:purple;">N = 4</span></b></center></td><td style="margin-left:50px;"><center><b><span id="sp2" style="color:purple;">&#960; = 4</span></b></center></td></tr>
    <tr><td><center><b><span id="sp3" style="color:green;">N = 4</span></b></center></td><td style="margin-left:50px;"><center><b><span id="sp4" style="color:green;">&#960; = 3.1111...</span></b></center></td></tr>
    <center style="position:relative;"><span id="ssp1" style="background-color:white; position:absolute; top:6px; border:2px solid red; padding:3px; z-index:1;">1x</span></center>
    <center style="position:relative;"><span id="ssp1b" style="background-color:; position:absolute; top:36px; left:145; border:0px; padding:0px; z-index:1;">hugs</span></center>
    <center style="position:relative;"><span id="ssp1c" style="background-color:; position:absolute; top:56px; left:145; border:0px; padding:0px; z-index:1;">circle</span></center>
    <svg id="svg1" height="200px" width="200px" style="margin-left:0px; border:2px solid black;">
    <!--<svg viewBox="0 0 200 200">-->
    <circle cx="51" cy="51"  r="50" stroke-width="1" stroke="green" fill="transparent"/>
    <center style="position:relative;"><span id="ssp2" style="background-color:white; position:absolute; top:6px; border:2px solid red; padding:3px; z-index:1;">5x</span></center>
    <svg id="svg2" height="200px" width="200px" style="margin-left:0px; border:2px solid black;">
    <!--<svg viewBox="0 0 200 200">-->
    <circle cx="51" cy="51"  r="50" stroke-width="1" stroke="green" fill="transparent"/>
    <tr style="margin:20px; border:20px solid blue;"><td>
    <center style="position:relative;"><span id="ssp3" style="background-color:white; position:absolute; top:6px; border:2px solid red; padding:3px; z-index:1;">20x</span></center>
    <center style="position:relative;"><span id="ssp3b" style="background-color:white; position:absolute; top:36px; left:145; border:0px; padding:0px; z-index:1;">bad</span></center>
    <center style="position:relative;"><span id="ssp3c" style="background-color:white; position:absolute; top:56px; left:145; border:0px; padding:0px; z-index:1;">approx</span></center>
    <center style="position:relative;"><span id="ssp3dd" style="background-color:white; position:absolute; top:36px; left:145; border:0px; padding:0px; z-index:1;">ok</span></center>
    <center style="position:relative;"><span id="ssp3d" style="background-color:white; position:absolute; top:36px; left:145; border:0px; padding:0px; z-index:1;">good</span></center>
    <center style="position:relative;"><span id="ssp3e" style="background-color:white; position:absolute; top:56px; left:145; border:0px; padding:0px; z-index:1;">approx</span></center>
    <svg id="svg3" height="200px" width="200px" style="margin-left:0px; border:2px solid black;">
    <!--<svg viewBox="0 0 200 200">-->
    <circle cx="51" cy="51"  r="50" stroke-width="1" stroke="green" fill="transparent"/>
    <center style="position:relative;"><span id="ssp4" style="background-color:white; position:absolute; top:6px; border:2px solid red; padding:3px; z-index:1;">100x</span></center>
    <center style="position:relative;"><span id="ssp4e" style="background-color:white; position:absolute; top:36px; left:25; border:0px; padding:0px; z-index:1;">a<sup>2</sup> + b<sup>2</sup></span></center>
    <center style="position:relative;"><span id="ssp4f" style="background-color:white; position:absolute; top:56px; left:25; border:0px; padding:0px; z-index:1;">&#8775; c<sup>2</sup></span></center>
    <center style="position:relative;"><span id="ssp4b" style="background-color:white; position:absolute; top:36px; left:145; border:0px; padding:0px; z-index:1;">circle</span></center>
    <center style="position:relative;"><span id="ssp4c" style="background-color:white; position:absolute; top:56px; left:145; border:0px; padding:0px; z-index:1;">looks</span></center>
    <center style="position:relative;"><span id="ssp4d" style="background-color:white; position:absolute; top:76px; left:145; border:0px; padding:0px; z-index:1;">straight</span></center>
    <svg id="svg4" height="200px" width="200px" style="margin-left:0px; border:2px solid black;">
    <!--<svg viewBox="0 0 200 200">-->
    <circle cx="51" cy="51"  r="50" stroke-width="1" stroke="green" fill="transparent"/>
    <input type="button" onclick="doprev();">Prev</input>
    <input type="button" onclick="donext();" style="margin-left:30;">Next</input>

    alert('js syntax ok');
    function xxx(iter,first,second,third,fourth) {
        xxxcore(iter,"svg1",first,second,third,fourth,1,document.getElementById('ssp1') );
        xxxcore(iter,"svg2",first,second,third,fourth,5,document.getElementById('ssp2') );
        xxxcore(iter,"svg3",first,second,third,fourth,20,document.getElementById('ssp3') );
        xxxcore(iter,"svg4",first,second,third,fourth,80,document.getElementById('ssp4') );

    function xxxcore(iter,svgid,first,second,third,fourth,mult,ssp) {

    var i,j,ktf;
    //var iter=3;
    var alpha1=Math.PI*2/40;
    var alpha2=Math.PI*2*3/16+0.000;
    var steps=Math.pow(2,(iter+2));  //(iter+1)*4;
    var delta=Math.PI*2/steps;
    //var first=true;
    //var second=true;
    //var third=true;
    var cx0=100;
    var cy0=100;
    var r0=50;
    var cx=cx0+(mult-1)*r0*Math.cos(alpha2)-30 ; //351;
    var cy=cy0-(mult-1)*r0*Math.sin(alpha2)-10 ;  //-401;
    var r=r0*mult;
    var geostr1="";
    if (first!=0)
        geostr1+="<circle cx='"+cx+"' cy='"+cy+"'  r='"+r+"' stroke-width='1' stroke='red' fill='transparent'/>";
    for (i=0,j=(Math.PI*2/steps); i<steps; i++) {
      if (second!=0) {
        if (second!=0&&i*j==alpha2) {  //floating variation? 
            geostr1+="<path d='M "+cx+" "+cy+" L "+(cx-r*Math.cos(i*j))+" "+(cy+r*Math.sin(i*j))+"' stroke-width='2' stroke='blue' fill='transparent'/>"
        } else if (0&&     (i-1)*j==alpha2) {
            geostr1+="<path d='M "+cx+" "+cy+" L "+(cx-r*Math.cos(i*j))+" "+(cy+r*Math.sin(i*j))+"' stroke-width='1' stroke='blue' fill='transparent'/>"
        } else if (1||0) {
            geostr1+="<path d='M "+cx+" "+cy+" L "+(cx-r*Math.cos(i*j))+" "+(cy+r*Math.sin(i*j))+"' stroke-width='1' stroke='gray' fill='transparent'/>"
      if (third!=0) {
        geostr1+="<path d='M "+(cx-r*Math.cos(i*j))+" "+(cy+r*Math.sin(i*j))+" L "+(cx-r*Math.cos((i+1)*j))+" "+(cy+r*Math.sin((i+1)*j))+"' stroke-width='1' stroke='green' fill='transparent'/>";
      if (fourth!=0) {
        geostr1+="<path d='M "+(cx-r*Math.cos(i*j))+" "+(cy+r*Math.sin(i*j))+" L "+(ktf?cx-r*Math.cos(i*j):cx-r*Math.cos((i+1)*j))+" "+(ktf?cy+r*Math.sin((i+1)*j):cy+r*Math.sin(i*j))+" L "+(cx-r*Math.cos((i+1)*j))+" "+(cy+r*Math.sin((i+1)*j))+"' stroke-width='1' stroke='purple' fill='transparent'/>";
    }   //also stroke-opacity  fill-opacity
    var pistr=new Number(steps*Math.sin(Math.PI/steps)).toPrecision(7);
    if (third!=0) {
        document.getElementById("sp3").innerHTML="N = "+steps;
        document.getElementById("sp4").innerHTML="&#960; = "+pistr+"...";
       if (fourth==0) {
        if (steps==8) {
        if (steps>=16) {
    if (fourth!=0) {
        document.getElementById("sp1").innerHTML="N = "+steps;
        document.getElementById("sp2").innerHTML="&#960; = 4";  //&pi;
       if (third==0) {
        if (steps>=32) {
    //    if (steps>=256) {  //don't bother adding a,b,c labels and just keep invisible.. else fix "circle looks straight" to "circle (hypot) looks straight" but 
                             // keep in mind that c is not c but approx straight. etc. so avoid imprecision and just use visual pic.
    //        document.getElementById("ssp4e").style.visibility="visible";
    //        document.getElementById("ssp4f").style.visibility="visible";
    //    }

    } //end func

    var ii=0;
    var jj=0;

    sz=5; //of each line below
    var xxxarr=[
    0, 1,0,0,0,  //hold a bit
    0, 1,1,0,1,
    1, 1,1,0,1,
    2, 1,1,0,1,
    3, 1,1,0,1,
    4, 1,1,0,1,
    5, 1,1,0,1,
    6, 1,1,0,1,
    7, 1,1,0,1,
    //0, 1,0,0,1,
    //1, 1,0,0,1,
    //2, 1,0,0,1,
    //3, 1,0,0,1,
    //4, 1,0,0,1,
    //5, 1,0,0,1,
    //6, 1,0,0,1,
    //7, 1,0,0,1,
    0, 1,1,1,0,
    1, 1,1,1,0,
    2, 1,1,1,0,
    3, 1,1,1,0,
    4, 1,1,1,0,
    5, 1,1,1,0,
    6, 1,1,1,0,
    7, 1,1,1,0,
    0, 1,1,1,1,
    1, 1,1,1,1,
    2, 1,1,1,1,
    3, 1,1,1,1,
    4, 1,1,1,1,
    5, 1,1,1,1,
    6, 1,1,1,1,
    7, 1,1,1,1,
    0, 1,0,0,1,
    1, 1,0,0,1,
    2, 1,0,0,1,
    3, 1,0,0,1,
    4, 1,0,0,1,
    5, 1,0,0,1,
    6, 1,0,0,1,
    7, 1,0,0,1,
    //0, 1,0,0,1,
    //1, 1,0,0,1,
    //2, 1,0,0,1,
    //3, 1,0,0,1,
    //4, 1,0,0,1,
    //5, 1,0,0,1,
    //6, 1,0,0,1,
    //7, 1,0,0,1,
    0, 1,0,1,0,
    1, 1,0,1,0,
    2, 1,0,1,0,
    3, 1,0,1,0,
    4, 1,0,1,0,
    5, 1,0,1,0,
    6, 1,0,1,0,
    7, 1,0,1,0,
    0, 1,0,1,1,
    1, 1,0,1,1,
    2, 1,0,1,1,
    3, 1,0,1,1,
    4, 1,0,1,1,
    5, 1,0,1,1,
    6, 1,0,1,1,
    7, 1,0,1,1,
    var xxxstr="";
    //for (i=0; i<1; i++) {
        //keep in sync with below
        for (j=0; j<sz-1; j++) {
    //        xxx(xxxarr[1*ii+jj]);
        eval (xxxstr);

    function donext () {
        if (++ii==xxxarr.length/sz)
        //common with below
        for (j=0; j<sz-1; j++) {
        eval (xxxstr);

    function doprev() {
        if (--ii<0)
        //same as above;
        for (j=0; j<sz-1; j++) {
        eval (xxxstr); //alert(xxxstr)

    alert('initialization done');


Update 1:

After seeing the picture (eg, green/red overlap zoom 80x for large N), we still might wonder (and not outright accept) why the two sides of right triangle don't equal the hypotenuse, why doesn't a+b=c?

Well, with basic Euclidean geometry, we can prove Pythagoras Theorem; thus, we are asking, given a2+b2=c2, why doesn't a+b=c? Well, a simple counterexample of 3,4,5 shows that Pythagoras holds where the other simpler equation doesn't (3+4≠5), so a+b=c is not true generally. That being the case, we can't conclude PI=4.

Update 2:

The main issue with the right triangles is this, no matter how small they get and how many, those within a given region (in neighborhood of a given angle), even as you go to infinitely many of them, adding the lengths of the (purple) legs of each one will be a significant fraction extra than by taking the straight (green) hypotenuse path. This fraction goes to a given number (say 30%-50% extra near the +-45 degree region.. as a lower bound) that is clearly not zero. This is for every single triangle in that region no matter how many you make, so it factors out of all of them (distributive property). 4 is an upper bound all right. Any shape you use (a saw/staircase), inside or outside the circle, will converge to a higher number as long as it isn't a straight path distance as you get closer and closer to the circle. Distance is defined as the smallest path from point A to B. Every other path shape that does not approach it in value (below epsilon for all n>N0) but instead stays above some lower bound difference against that straight line path (within some "wide" angle region of circle) cannot come arbitrarily close to specifying π. ..this response could use another pic that shows calculations of length and how clearly all "right triangles" within a "wide" region of circle (except tightly at N, E, S, W) will add extra length. A given region of circle, and given path definition, can have a higher-than-zero lower bound plucked out (distributed out) of that region. Being a clear value higher than zero higher than pi in a "wide" region of circle is a sure way to not approach π.

Analysis goes further than Euclidean geo ever could.. but you are still getting at the same basic thing

To get π, you use a straight line path connecting the many points on circle. Tiny circular arcs straighten out and approach that path ("difference" bounded below only by 0) any other poly path that clearly does not approach straight line ("difference" bounded below by number higher than zero) will thus not approach circle. Problem is that there is no real definition of length of a curve we are going on. Euclidean geom defines length more loosely. It gives definite values for some shapes, including lines, circles, etc. These match the physical world notion of length. Analysis (and there are different variations, some of which go even further) goes further than Euclidean Geo and define more generally a distance definition for arbitrary curves. To prove using those tools, you have to first know precisely how length is defined there and then build the formal argument upon it. With Euclidean (intuitive) view and not diving more formally than that, you are limited to a certain amount of hand waving. You really must define length for a curve precisely if you want a precise argument.

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  • I said the code is draft (but works). That is an alert I keep around to catch other mistakes I might make as I code up the javascript. Some types of mistakes will result in that alert not showing up (I have another at the beginning of js, which catches syntax issues right away). – Jose_X Oct 30 '16 at 13:09
  • alert('initialization done'); ? Come on, you really posted 3 answers in the same question ? – reuns Oct 30 '16 at 13:09
  • Each one is different. I tried to provide insight not easily seen in other responses, especially if you don't have a background in analysis. This last one also adds picture/animation and code to the pictures. This helps clarify that circle arcs become indistinguishable from straight lines, making it much easier to see that the arc is roughly a hypotenuse, so cannot be approx by the sum of the lengths of the other 2 legs. This is why inscribed poly perimeter approaches the limit (Nsin(pi/N)) while the staircase is always a signif fraction off virtually everywhere no matter how small the pieces. – Jose_X Oct 30 '16 at 14:35
  • I misunderstood you. No, I have one answer here. I separated the Simple Geometric Explanation to keep a very short answer at top, since that might be sufficient for reader. [Yea, I got hooked on this question a bit.. which is why this is actually my third distinct answer.] – Jose_X Oct 30 '16 at 14:44

The fundamental concept here is discontinuity. The arclength of a curve is a discontinuous function of its path, in the sense that two paths can be arbitrarily close (in the visual or point-by-point sense) but have dramatically different arclengths.

You can take any discontinuous function and build a dumb apparant-paradox in the same style.

  • The sign of a number is discontinuous. Here's a dumb apparent paradox:

1 is positive. 0.1 is positive 0.01 is positive. Repeat to infinity and you conclude that 0 is positive! OMG!

  • The "is rational" mapping is discontinuous. Here's a dumb apparent paradox:

3.14 is rational. 3.141 is rational. 3.1415 is rational. Repeat to infinity and you conclude that pi is rational! OMG!

  • The "is equal" mapping is discontinuous. Here's a dumb apparent paradox:

When I'm 50% of the way to my destination, I'm not there yet. When I'm 75% of the way there, I'm not there yet. When I'm 87.5% of the way there, I'm not there yet. Repeat to infinity and you conclude that I'll never get there! OMG!

(Isn't that Zeno's paradox or something?)

With this template, you can build as many dumb apparant-paradoxes as you want. Get creative! Impress your friends! :-)

Steve Byrnes
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    Not sure why this got downvoted. This answer is better than some of the answers with 50+ score. – Eric Wofsey Sep 20 '16 at 05:03
  • No, you can't conclude that 0 is positive from the first sentence you put in italics. You never actually reach 0 in that sequence. The same for your second sentence. You never reach pi, you only have *lower* rational approximations at each step. Neither of the conclusions follow and thus those are not paradoxes. The last sentence in italics though does imply that you'll never get there. That is a paradox. – Doug Spoonwood Nov 05 '16 at 17:18
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    @DougSpoonwood You can, if you assume the same reasoning that the OP is assuming, which is basically that $\lim f(x_n) = f(\lim x_n)$ (i.e., continuity). – Luke Collins Mar 20 '20 at 01:36
  • To be a bit more specific: The path length as a function from, say, $C^0$ paths equipped with $C^0$-norm to $[0,\infty]$ with the usual topology is discontinuous. But if you change $C^0$ for instance to $C^1$ and equip it with the $C^1$-norm, then the path length becomes a continuous function . – Maximilian Janisch Nov 01 '21 at 10:40

The picture shows a sequence of curves $\gamma_n$ which approach (in what is called "uniform distance") the circumference of a circle $\gamma$. Then the picture says that the length of these curves is always the same: $\ell (\gamma_n) = 4$. If the function $\ell$ were a continuous function you would get the stated result: $$ 4 = \lim_{n\to \infty} \ell(\gamma_n) = \ell(\gamma) = \pi. $$

Unfortunately $\ell$ is not a continuous function, and this example is a proof of this fact.

(added) As suggested by @knedlsepp in the comments: the functional $\ell$ is continuous with respect to $C^1$ convergence (i.e. whenever both $\gamma_k$ and $\gamma'_k$ converge to $\gamma$ and $\gamma'$). In this case it is easy to see that the curves $\gamma_k$ do not converge in $C^1$ because the derivatives $\gamma'_k$ are always either horizontal or vertical vectors, while the limit curve $\gamma$ can have any intermediate slope.

Emanuele Paolini
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  • ..., but $\ell|_{\mathcal{C}^1}$ is. So if the curves $\gamma_n$ were to approach via $\gamma_n \xrightarrow{\mathcal{C}^1} \gamma$, then $\lim_{n \to \infty} \ell(\gamma_n) = \ell(\gamma)$ would be true. – knedlsepp Nov 02 '14 at 20:25

(non rigorous) If you repeat the process a million times it "seems" (visually) that the perimeter approaches in length to the circumference, but if you magnify the picture of a single "tooth" to full screen, you will notice a big difference from the orthogonal segments and the arc of the circumference. No matter how many times you repeat the process that difference will never fade.

ADDED: A visual example of what I meant is folding of a rope. If you imagine the rope not having thickness, you can fold it so many times that you can tend to a point (zero length?). If you unfold it, it will return to its original shape. In the example the perimeter will always be of total length = 4, but it only appears to blend with the circumference.

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Mario Stefanutti
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$\newcommand{\Reals}{\mathbf{R}}$There are a number of good answers to this primordial question, but none mentions the usual definition of arc length: If $\gamma:[a, b] \to \Reals^{n}$ is a continuous path, the arc length of $\gamma$ is the supremum, taken over all partitions $(t_{i})_{i=0}^{n}$ of $[a, b]$, of $$ \sum_{i=1}^{n} \|\gamma(t_{i}) - \gamma(t_{i-1})\|. \tag{1} $$ This supremum always exists as a positive extended real number. If $\gamma$ is piecewise continuously-differentiable, the arc length is finite, and given by $$ \int_{a}^{b} \|\gamma'(t)\|\, dt. $$ (The integral is computed by partitioning $[a, b]$ into finitely many intervals on which $\gamma'$ is continuous and summing the contributions.)

Here, the circle can be (smoothly) parametrized by $$ \gamma(t) = (\tfrac{1}{2}\cos t, \tfrac{1}{2}\sin t),\qquad 0 \leq t \leq 2\pi. $$

The point is, the troll's approximation by "Manhattan" polygons does not give the supremum of (1), nor is there any reason to expect it should since none of the vertices lie on the circle.

On the subject, there is a genuinely vexing issue for surfaces in $\Reals^{3}$: The analog of (1) (form a triangular approximation whose vertices lie on the surface, sum the areas of the triangles, and take the supremum) is infinite even for a bounded portion of a right circular cylinder, a surface as smooth as one could hope. Spivak's Comprehensive Introduction to Differential Geometry has a nice diagram illustrating what goes wrong. Intuitively, crush a paper cylinder so it looks like an accordion bellows, and imagine this is done in such a way that the vertices of the crushed cylinder lie on another cylinder whose area we wish to approximate. The area of the bellows can be made as large as we like.

Andrew D. Hwang
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    Excellent explanation, thank you (although I think that *half* of the staircase’s vertices lie on the circle. But it doesn’t matter. The point is that *not all of them* lie on the circle). An award-winning article about the surface area paradox is here: https://www.maa.org/programs/maa-awards/writing-awards/surface-area-and-the-cylinder-area-paradox. – Zach Teitler Sep 15 '19 at 14:12

Summary: The construction at the top (pi=4) merely shows an upper bound. It's an upper bound but it is very easy to come up with a lower but still upper bound to the circumference. The =?4 construction doesn't prove or disprove very much more about the length of the circle besides providing an upper bound pi<4. It certainly doesn't prove =4. Finally, we always should rely on physical "experimentation" to support the meaningfulness of any construction.

If we use a hexagon circumscribed around the circle, we can find a different upper bound rather easily that would be lower than 4; hence proving that the construction above is merely an upper bound (of which there are an infinite number.. including pi=?10000, by using a path that weaves all over a small area between a square and circle) but certainly not the lowest upper bound.

To get closer to discovering pi, we can then also use a mirrored complementary approach of increasing lower bounds via inscribed polygons of an increasing number of sides. However, arguably we could "weave" as well with the incribing to create such a "lower" bound approaching say 4 from below! [Ie, by weaving, we can inscribe a weaved path that ends up being arbitrarily large in measurement despite nicely fitting inside the tightening space between the circle and polygons.]

Ultimately, one key to getting sanity is to postulate/believe that the shortest distance between 2 points is a line (Euclid did this a long time ago). We observe, for example, that the accepted height of a person involves a procedure of using a taught measuring stick or, generally, measuring curves by tightening a flexible measuring device as far as it can be tightened while still hugging (remaining within a certain area of) the curvy item being measured. This procedure is very easy to do for a circle made of an iron ring, for example, and would give a very close approximation to pi. This exercise also shows that math is not reality. Math is built upon postulates and definitions (something must be accepted as being true), but these postulates need not match our physical world in order to remain logically consistent. To say meaningful things about the physical world, we must judge the reasonableness of mathematical postulates and definitions [phv3773 noted in an answer how definitions were missing, while others have also noted collectively many of these points]; we must determine just what is a reasonable set of postulates to start with in order to reach a reasonable meaning of length of a circle (ie, of the value of pi). We might conclude for example, that the usual inscribed/circumscribed approach relies upon a framework closer to reality because in fact it approaches the tape measure "experimental" results arbitrarily close.

I googled an excellent essay that goes into detail on Archimedes' essay http://www.ams.org/samplings/feature-column/fc-2012-02. It appears to have been written for the American Mathematical Society but maybe its author (Bill Casselman) can be persuaded to make a contribution here.

[Below is my old response]

What if the measurement we use, patterning it after a string wrapped around this circle, weaves back and forth? Essentially, we can find a series of connected line segments with length that total $1000000000$ and yet "hug" the circle very closely. A string analogy follows closely though line segments have width $0$ so we can fit arbitrarily many.

This is why not just any reasoning about infinity will do. Mathematicians have developed well reasoned arguments and axioms that correlate well in many cases with reality (see also this argument).

So the question of why doesn't $\pi = 4$ is best answered by asking, "Why should it?" We can just as well have used the ridiculous construction above to suggest $\pi =$ any number $> 3.15$.

The approach we take to argue convincingly that the sum of the line segments approaches the "length of the curve" is to find sequences (from series partial sums) that match to functions (note the question example and the weaving example do not constitute a function because of its multiple values at a given "$x$") which have certain characteristics. For example we might use a lower and upper bounding pair of sequences that correspond to function values of line segment endpoints for such created polygons where one remains on one side of the curve and the other on the other side at all times and where these two sequences approach the same limiting value. We might use the Mean Value Theorem or related results to help prove our final answer. In any case, mathematicians leverage a convincing set of arguments and assumptions and don't just ad hoc throw a bunch of twisted string at a problem and claim the amount of string used proves the unprovable.

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    @Jose_X: I hate to dampen your enthusiasm, but it was not really necessary to post a new answer to this thread. There are already plenty of good answers and this thread is fairly old already. – Qiaochu Yuan May 23 '11 at 07:29
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    @Qiaochu Yuan, I read the top answer and it talks about the length being some wrong answer via a particular construction but simply states it is wrong. It gives no explanation why pi should be something besides 4. It also talks about area which has nothing to do with the problem. We can't assume A=pi*r*r, where that pi is the same as Circumference/pi. Now, we can resolve these issues, but they simply aren't "obvious" and are not addressed by this question. The second highest answer mentions an important point, about derivatives, but it's in isolation. A derivative is not length either. – Jose_X May 23 '11 at 16:40
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    @Jose_X: the accepted answer is terse, but it answers the question. The basic observation here is that length is not a continuous function of the "obvious" topology on rectifiable curves in the plane. The second highest answer gives a short explanation of why this makes sense for the special case of $C^1$ curves. – Qiaochu Yuan May 23 '11 at 17:56
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    @Qiaochu Yuan, I don't see how the first answer address the issue of curve length. Area has nothing to do with curve length. Especially as a person that might see this question on google, I hardly think that answer is a good one. It's not for me to accept or reject, but I do see how the bit I added to the conversation is -1 while that one is 40. This is also a site for the wider public interested in mathematics, right? The question itself suggests nothing that you need a math undergrad degree to be able to appreciate the contributed answers. Any particular thm the 1st answer was invoking? – Jose_X May 23 '11 at 20:07
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    @Jose_X: the highest-voted answer is trying to address a reasonable intuition (which is false) that if you have a curve and a sequence of curves such that the area between them goes to zero, then their lengths approach each other. The point here is that area (which behaves continuously when a region in the plane is deformed) behaves differently from length (which can behave erratically after a small deformation) and this is an important point to make because one might naively think that they behave similarly. – Qiaochu Yuan May 23 '11 at 20:13
  • Very well, I accept it was drawing upon relationship between curve length and area (if that is what it was doing), but this website is for a wider audience than math majors (or grads). – Jose_X May 23 '11 at 20:22
  • If we are going to talk intuition, by bunching up a measuring string, anyone will tell you that you won't get to the correct answer. Area, as an intuitive concept to explain away a problem measuring length, plays no role here (eg, we can have a very long string "take up 1mm^2" very easily). – Jose_X May 23 '11 at 20:24
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    @Jose_X: I hardly think that math majors are the only people who think about area and length. If you have taken a standard calculus course, you have probably computed the area of a region by approximating it with rectangles, and if you didn't pay too much attention in your standard calculus course, you might think that you can also compute perimeters of regions by approximating them with rectangles. That's a context it's fair to say many people who've been through college or did well in high school are familiar with. – Qiaochu Yuan May 23 '11 at 20:30
  • @Qiaochu Yuan, I see what you mean that area might be taken to correspond to length even if that is not how we measure it and is totally ridiculous when you consider how you can bunch up miles of thin string into a tiny area and that a triangle and square (each "easily" measurable) have different ratios of area to perimeter. In this context, that the top answer addresses one particular possible misunderstanding of how length and area relate, I don't see why addressing other concerns => < 0 score? Anyway, thanks for clarifying the first answer for me. I really wasn't thinking about that confusi – Jose_X May 23 '11 at 20:39
  • @Qiaochu Yuan, thanks for taking the time to address my concerns. My "answer" isn't that clear anyway. – Jose_X May 23 '11 at 21:01
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    @Qiaochu: *"it was not really necessary to post a new answer... [this thread] is fairly old already."* - I disagree with that statement. Posting in old threads (even those with accepted answers) is not only allowed, it's [encouraged](http://math.stackexchange.com/badges/17/necromancer) - we don't want to become another Yahoo Answers. That said, I am not upvoting this answer for a different reason: I don't think it answers the question. – BlueRaja - Danny Pflughoeft Jul 26 '11 at 22:36

Proof by contradiction

What's the perimeter of a regular polygon with an inscribed circle of unit diameter?

It has apothem equal to radius (distance between a side and the center): $\frac{1}{2}$

The central angle in front of a side is $\frac{360^\circ}{n}$.

Next to 180 it says "deg"

So a whole side is of length $2\cdot \left(\frac{1}{2}\tan{\frac{180^\circ}{n}}\right)$

There are $n$ sides, so it has perimeter ${n\tan{\frac{180^\circ}{n}}}$

The folding method (details at the end of the answer) preserves perimeter. So the polygon's perimeter equals the circle's circumference

Dividing by the diameter (the definitions of $\pi$), which is $1$, we get $\pi = n\tan{\frac{180^\circ}n}$

(which is actually only correct when $\lim_{n\to \infty}$! Think of how different choices of $n$ can change the calculated value of $\pi$!)

The contradiction

Replace the square with a triangle, and apply the same steps and you'll see that $\pi = 3\tan{\frac{180^\circ}{3}} \approx 5.196$

Replace the square with a hexagon, and apply the same steps and you'll see that $\pi = 6\tan{\frac{180^\circ}{6}} \approx 3.464$

Replace the square with a enneadecagon (19-gon), and apply the same steps and you'll see that $\pi = 19\tan{\frac{180^\circ}{19}} \approx 3.171$


This way to calculate $\pi$ by assuming $\text{circumference} = \text{perimeter}$ is invalid, as it contradicts with itself.

Further description

The green line bisects the distance between the angle and the arc. It represents folding. If we repeat this multiple times (applying to newly formed vertices in each step), we'll find that the circle's circumference is equal to the polygon's perimeter, because, after every step:

  • all vertices become closer to the circle
  • the number of vertices doubles
  • the perimeter stays constant

The polygon-part in the image is from a pentagon. No matter what the angle is or how many sides the polygon has, we can fold each vertex infinite times and find "circumference = perimeter", but since every polygon would have a different perimeter, we have self-contradiction.

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    I loved how all the other answers were long and complicated a lil bit and then u have an answer amongst them so short and sweet. But according to your contradiction, does this mean that a circle has $π$ sides? – George N. Missailidis Jul 23 '17 at 06:09
  • @GeorgeN.Missailidis > "Apply the same steps" < = After bending the corners inf times, you'll see that $\pi = n$, e.g. 4 for square. The problem is "We can't reach infinity." See this: http://imgh.us/Screen_Shot_2017-07-23_at_18.20.26.png – MCCCS Jul 23 '17 at 15:17
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    Hmm. An $n$- sided regular polygon circumscribed around a unit circle has perimeter not $n$, but $2n\tan(\pi/n)$. – Zach Teitler Sep 15 '19 at 14:20
  • @ZachTeitler Thanks for the feedback, updated. – MCCCS Apr 17 '20 at 13:37
  • I downvoted without thinking through how "apply the same steps" could work (I did realize my error but not until the vote was locked in), so thanks for proactively clarifying for this idiot here. Sorry about that. – epimorphic Apr 18 '20 at 18:35
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    To expand on Zach Teitler's comment though: For $n \geq 5$, the regular $n$-gon with side length $1$ no longer circumscribes the circle with diameter $1$. This has the following implications: (1) Especially for larger $n$, there may be multiple steps of folding at the beginning that doesn't touch the circle. (2) There's nothing special geometrically about having perimeter $n$; we can choose any side length, as long as the resulting regular polygon can contain the circle. So in fact, using a regular $n$-gon, we can actually show "$\pi = x$" for any real number $x \geq 2n \tan(\pi/n)$. – epimorphic Apr 18 '20 at 18:35
  • One assertion that _is_ incorrect in your answer: "Replace the square with a triangle, and apply the same steps and you'll see that $\pi = 3$" is false, because if a path always stays within distance $\epsilon$ of the circle, then the path length must be at least $(1-2\epsilon)\pi$, the circumference of the inner bound. – epimorphic Apr 18 '20 at 18:43
  • @epimorphic You're absolutely right, I misunderstood Zach's comment and didn't see my mistake. I'm editing now. Thanks a lot for explaining it – MCCCS Apr 18 '20 at 18:53

Ah, the old engineer vs mathematician thought process.

Place an engineer and a mathematician at one end of a room. At the other end is a beautiful woman. At each "step", they can each move half of the remaining distance between their current position and the woman. The mathematician will say you'll never reach her. The engineer will say you can get close enough.

This problem is similar. A unit square's outermost corners are being "bent" inward to touch a 1/2-unit circle until there are so many corners that the square is, at this zoom level, indistinguishable from the circle itself (similar to using rectangular pixels). Repeated "to infinity" the two shapes would have the same area. However, this process will never yield a mathematical circle; only an engineer's approximation ("close enough") This will always produce the same perimeter measurement even as the areas of the two shapes converge. If instead you were to measure around the hypotenuses as you iterated this shape definition, the perimeter WOULD begin to approach that of the circumference of the half-unit circle, $\pi$.

The fallacy of the proof is illustrated if you consider the shape made by any two line segments that intersect at a point other than on the circle. These two lines will inscribe an arc length as they each intersect a different point on the circle. For simplicity, you can think of the resulting shape as a right triangle. The proof is basically claiming that the sum of the length of the two legs of that triangle is equal to the hypotenuse. This is never true, because the Pythagorean Theorem of $a^2+b^2=c^2$ never holds for any $a,b,c > 0$ where $a+b=c$.

The only way it can work is for an $a$ or $b$ that is zero and thus the area of the shape is zero; this never happens in the construction being generated, at any interval, because by the definition of the construction we have two points that lie on the circle and one point lying outside the circle, and from geometry, any three non-colinear points will always inscribe a shape within a plane of non-zero area.

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    Do you perhaps mean 90% of the remaining distance? Or are you saying that 1/9 of the way there is close enough for the engineer? :-) – Jesse Madnick Nov 13 '12 at 08:33
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    The unreachable woman problem is a wrong argument for this problem. This because if the sequence of perimeters converged to the one of the circle, then not only pi could be made as close to 4 as possible but also pi IS 4(contrary to the unreachable woman problem, where the limit is not attained by any element in the sequence). This beacuse pi is a fixed number, and if a,b are reals, $|a-b|<\epsilon$ for every $\epsilon$ implies $a=b$ – Diego Sep 10 '16 at 03:31

We can say that the process depicted actually demonstrates that $$\frac{\sqrt{2}}{2} 4 < \pi < 4$$

G Cab
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Simple Geometrical Explanation

Consider the approximation to the circumference (to Pi) that is suggested by the question. Call that a step-wise approach.

Consider the approximation of using an inscribed or circumscribed regular polygon to approximate the circumference. Call that a polygon approach.

Let's note the following.

1: A step approach relies entirely on 90 degree angles.

2: Meanwhile a regular polygon approach has the angles obtuse and actually approach 180 degrees as the number of sides increases.

3: If we zoom in, we can no longer discern a tiny segment of the circle from a straight line.

In the step construction, when we are at a very high iteration count, each step starts at one end of a tiny segment and ends at the other end. In some cases as we go around the circle, the vertical step component will be very short and then a long horizontal component. In these cases, the step construction will approach the circle's "nearly straight" tiny segment in length. But in many other cases, the up step will be longer. In the extreme case, the up vertical part is exactly as long as the horizontal part. In this extreme case, we are approximating the length of the "flat" hypotenuse of an isosceles right triangle to be equal to the sum of the 2 legs. Clearly that is a bad approximation. These cases with many fairly bad approximation occur regularly.

Contrast with the polygon approach. Here we approximate the length of the longest side of an isosceles triangle to be the sum of the lengths of the 2 equal sides. In all cases, this is a good approximation, since the polygon must necessarily be almost 180 degrees to achieve "continuity" in slope to the next segment. (It's a convex polygon).

Summary: The step approach must use 90 deg pseudo-triangles where the short 2 legs are use to approximate the long "nearly straight" leg. This is clearly insufficient in many cases, particularly when the triangle is near isosceles. In contrast, in the polygon approach, the triangle always has one angle approaching 180 degrees so that sum of the 2 short legs is necessarily about the same (in terms of relative percentages) as the length of the "nearly straight" leg.

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    Can the person who downvoted please explain themselves? (foot tapping and arms crossed.. lol) – Jose_X Sep 19 '16 at 18:56

Many of the answers above have covered this in a rigorous way, so I'll try to put some intuition behind it.

Let $ C_n $ be your parameterization's circumference. You assume that $ \pi = \lim_{n \to \infty} C_n $. The problem lies in that $ \pi $ isn't defined as such limit. In fact, the argument is fundamentally flawed by assuming $ [\forall n \in \mathbb N, \,\,P(S_n)] \implies P(\lim_{n \to \infty} S_n) $, that is, the if every item of a sequence satisfy some property $ P $, the limit of the sequence itself satisfy $ P $.

Applying a similar argument, you can show that $ \mathbb R = \mathbb Q $. Let $ r \in \mathbb R $. Let $ c_n $ be the Cauchy sequence of $ r $. Since $ c_n \in \mathbb Q $, $ r \in \mathbb Q $. Since $ \mathbb Q \subseteq \mathbb R $, we have $ \mathbb R = \mathbb Q $.


Although there are many answers I would like to add the following simple non-rigorous explanation, if this could be called an explanation at all, which uses only the notions of countability and uncountability.

Let us denote the curve at which we arrive after the $n$ "removal steps" as $A_n$ and let us denote the circle which we are approximating as $C$. For every $n \in \mathbb N$ we have that the set $A_n \cap C$ is finite because there is finite number of points which lie both on the curve and the circle.

In the limit $\lim_{n \to \infty} A_n \cap C=A \cap C$ we have that the limiting curve $A$ and the circle $C$ have an infinite number of points that are both on the curve $A$ and on the circle $C$ but the set of all such points is countable and the circle has an uncountable number of points.

So with this simple and elementary analysis we see that the limiting curve $A$ and the circle $C$ do not coincide, in fact, almost all points of the curve $A$ will not be on the circle $C$ so it is not such a big surprise that their lengths will be different.

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  • This seems to me to miss the point. First, the same argument applies to an inscribed or circumscribed regular $n$-gon, but now "the answer comes out as expected". Conversely, if you modify the OP's approximating curve by replacing a portion with an arbitrarily short circular arc, the length of the "approximation" doesn't appreciably change, but the approximating curve now intersects the circle at uncountably many points. – Andrew D. Hwang Apr 09 '16 at 23:07
  • @AndrewD.Hwang You are right, I was just adding simple observation that in the limit the limiting curve will not coincide with the circle so that there is no paradox. In this case the curve at each step has the same length but in the case of inscribed or circumscribed regular $n$-gons the length decreases and converges to $\pi$ for circumscribed ones and increases and converges to $\pi$ for inscribed ones, but in this "paradox" it is constant at every step, that is the difference. – Farewell Apr 09 '16 at 23:15

If I were asked this question by a first-year calculus student, then the first thing I would do is write the following on the chalkboard: $$ \sqrt{dx^2 + dy^2} \ne |dx| + |dy| $$ (This is still a common enough mistake among high school math students, but by the time anyone learns calculus, I hope that they would be aware that it is wrong.)

If you consider any infinitesimal element of the circle, the length is $$d\ell = \sqrt{dx^2 + dy^2}$$ The only relationship between the circle and the "approximating" curve is that they have the same value of $|dx| + |dy|$, which is not a meaningful quantity when talking about length (at least, as has been pointed out, when you are using a Euclidean metric). The value of $d\ell$ is not equal for any corresponding infinitesimal elements of the two curves, so there is no reason to believe it will be the same for the whole polygon.

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I am not satisfied with any of the answers so far, so here is mine:
Let $S(n)$ be the shape we have after $n$ "foldings". (For example $S(0)$ is the square at the beginning. )

The construction of the "paradox" could claim one of the following:

  1. Since the area of $S(n)$ is decreasing, the perimeter of $S(n)$ approaches the area of $C$.


  1. Since the area of $S(n)$ is decreasing, the area of $S(n)$ approaches the area of $C$.

The first one does not even make sense, since the area of a shape is not the same with the perimeter. There is no reason to connect the area with the perimeter.

The second one is correct. After one million repetitions, the area of the polygon we obtain from "folding" $S$ will be approximately equal to $\pi$. But this has nothing to do with the number $4$, since this was the perimeter of $S$.

Konstantinos Gaitanas
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Actually, the answer to this question depends on compactification chosen.

There can be compactification in which the limit of your perimeter is neither $\pi$, nor $4$. It is a non-real number, which has modulus $4$ but regularized (real) part $\pi$.

You can imagine another such number if you consider $(-1)^\infty$. It has limit of the modulus $1$ but (Cesaro- or Abel-) regularized value is $0$.

The regularized values do not necessarily need to be smaller than modulus, there are opposite examples as well.

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