What is wrong with this proof?

Is $\pi=4?$

  • 169
    $\begingroup$ The problem is that it doesn't approach circle in a "smooth" way. $\endgroup$
    – J. J.
    Commented Dec 3, 2010 at 13:49
  • 832
    $\begingroup$ Only for very large values of $\pi$ $\endgroup$ Commented Dec 3, 2010 at 13:51
  • 91
    $\begingroup$ You may do the same "trick" with a triangle, thereby "countering" Pythagoras theorem. $\endgroup$ Commented Dec 3, 2010 at 13:55
  • 37
    $\begingroup$ 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. $\endgroup$ Commented Dec 3, 2010 at 13:57
  • 57
    $\begingroup$ I cannot believe that reddit has found it's way here... $\endgroup$
    – Bey
    Commented Dec 3, 2010 at 21:09

24 Answers 24


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 $2$.

  • 105
    $\begingroup$ 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. $\endgroup$
    – Zach Conn
    Commented Dec 3, 2010 at 18:33
  • 9
    $\begingroup$ @Zach Conn: Now that Ross has edited his answer, I believe you can remove your downvote if you still wish to do so. $\endgroup$ Commented Dec 4, 2010 at 0:32
  • 45
    $\begingroup$ 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. $\endgroup$ Commented Dec 4, 2010 at 1:08
  • 25
    $\begingroup$ 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. $\endgroup$ Commented Oct 14, 2014 at 12:55
  • 9
    $\begingroup$ 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. $\endgroup$
    – user185498
    Commented Mar 16, 2016 at 3:30

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 $\sqrt{x^2+y^2}$.

  • 13
    $\begingroup$ 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. $\endgroup$ Commented Dec 3, 2010 at 15:50
  • $\begingroup$ The trouble is, once you smooth the corners you can no longer assume that the expression |f'(x) - g'(x)| remains > 𝜖 for all x. $\endgroup$
    – Dan Asimov
    Commented Oct 14, 2022 at 20:10

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.

  • 6
    $\begingroup$ I prefer your answer, because you've identified the specific false belief that leads immediately to the paradox. $\endgroup$ Commented Jan 19, 2015 at 16:44
  • 46
    $\begingroup$ "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. $\endgroup$ Commented Apr 3, 2015 at 15:28
  • 2
    $\begingroup$ @PeteL.Clark, the link is offline $\endgroup$
    – user561334
    Commented Sep 1, 2018 at 14:19
  • $\begingroup$ AFAICT the notes linked in the answer are now part of Pete L. Clark's text Honors Calculus (Wayback Machine). $\endgroup$ Commented Jul 5, 2022 at 10:53

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?

  • 6
    $\begingroup$ 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. $\endgroup$
    – phv3773
    Commented Jun 29, 2011 at 19:51
  • 13
    $\begingroup$ @phv3773 Leibniz. $\endgroup$ Commented May 2, 2012 at 10:55
  • 1
    $\begingroup$ Wikipedia links explaining some concepts mentioned here: calculus and (numerical) analysis (classes). Riemann sum. limsup (limit superior). $\endgroup$ Commented Nov 13, 2012 at 8:14
  • 13
    $\begingroup$ N.S. isn't talking about numerical analysis; he's referring to en.wikipedia.org/wiki/Mathematical_analysis, which provides the foundations of calculus. $\endgroup$
    – Ray
    Commented Mar 20, 2013 at 12:44
  • 2
    $\begingroup$ 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. $\endgroup$
    – David K
    Commented Sep 25, 2018 at 17:27

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).

  • $\begingroup$ This is the best answer! The image in the problem only approximates the area of the circle, not its circumference. If the side lengths of the square and its corners are infinitely small, then their combined lengths equal the sum of the hypotenuses of each right-triangle formed by the corners, which is okay. But if the side lengths have a finite value, then their sum will always be greater than the sum of the hypotenuses, so it becomes wrong to say this approximates the model with infinitely small side lengths. $\endgroup$
    – VV_721
    Commented May 1 at 17:55

What is wrong with this?

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

  • 15
    $\begingroup$ 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. $\endgroup$
    – Jose_X
    Commented Mar 5, 2015 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")

  • 9
    $\begingroup$ Is the circle a limit curve of the jagged square? $\endgroup$ Commented Jan 8, 2012 at 17:05
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    $\begingroup$ @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.) $\endgroup$ Commented Mar 30, 2014 at 10:18
  • 6
    $\begingroup$ 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. $\endgroup$
    – AnandJ
    Commented May 16, 2019 at 0:03

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.

  • 10
    $\begingroup$ using Manhattan metric, l=2piR is Wrong. $\endgroup$
    – cnd
    Commented May 23, 2011 at 6:28
  • 56
    $\begingroup$ 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. $\endgroup$ Commented May 23, 2011 at 7:31
  • 6
    $\begingroup$ +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. $\endgroup$
    – Zach466920
    Commented Jun 13, 2015 at 16:13
  • 4
    $\begingroup$ 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! ;) $\endgroup$
    – Zach466920
    Commented Jun 13, 2015 at 16:22
  • 4
    $\begingroup$ 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). $\endgroup$ Commented Jul 23, 2016 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>
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    <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.

  • $\begingroup$ 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). $\endgroup$
    – Jose_X
    Commented Oct 30, 2016 at 13:09
  • $\begingroup$ alert('initialization done'); ? Come on, you really posted 3 answers in the same question ? $\endgroup$
    – reuns
    Commented Oct 30, 2016 at 13:09
  • $\begingroup$ 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. $\endgroup$
    – Jose_X
    Commented Oct 30, 2016 at 14:35
  • $\begingroup$ 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.] $\endgroup$
    – Jose_X
    Commented Oct 30, 2016 at 14:44

The fundamental concept here is discontinuity. The arc length 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 arc lengths.

You can take any discontinuous function and build a dumb apparent-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 apparent-paradoxes as you want. Get creative! Impress your friends! :-)

  • 8
    $\begingroup$ Not sure why this got downvoted. This answer is better than some of the answers with 50+ score. $\endgroup$ Commented Sep 20, 2016 at 5:03
  • $\begingroup$ 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. $\endgroup$ Commented Nov 5, 2016 at 17:18
  • 9
    $\begingroup$ @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). $\endgroup$ Commented Mar 20, 2020 at 1:36
  • 1
    $\begingroup$ 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 😄. $\endgroup$ Commented Nov 1, 2021 at 10:40

(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.


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.

  • 1
    $\begingroup$ ..., 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. $\endgroup$
    – knedlsepp
    Commented Nov 2, 2014 at 20:25

$\newcommand{\Reals}{\mathbf{R}}$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.

  • 1
    $\begingroup$ 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: maa.org/programs/maa-awards/writing-awards/…. $\endgroup$ Commented Sep 15, 2019 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.

  • 4
    $\begingroup$ @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. $\endgroup$ Commented May 23, 2011 at 7:29
  • 7
    $\begingroup$ @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. $\endgroup$ Commented May 23, 2011 at 17:56
  • 5
    $\begingroup$ @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. $\endgroup$ Commented May 23, 2011 at 20:13
  • 4
    $\begingroup$ @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. $\endgroup$ Commented May 23, 2011 at 20:30
  • 35
    $\begingroup$ @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 - 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. $\endgroup$ Commented Jul 26, 2011 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.

  • 2
    $\begingroup$ 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? $\endgroup$ Commented Jul 23, 2017 at 6:09
  • $\begingroup$ @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: imgh.us/Screen_Shot_2017-07-23_at_18.20.26.png $\endgroup$
    – MCCCS
    Commented Jul 23, 2017 at 15:17
  • 1
    $\begingroup$ Hmm. An $n$- sided regular polygon circumscribed around a unit circle has perimeter not $n$, but $2n\tan(\pi/n)$. $\endgroup$ Commented Sep 15, 2019 at 14:20
  • $\begingroup$ @ZachTeitler Thanks for the feedback, updated. $\endgroup$
    – MCCCS
    Commented Apr 17, 2020 at 13:37
  • 1
    $\begingroup$ 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 $x \geq 2n \tan(\pi/n)$. $\endgroup$
    – epimorphic
    Commented Apr 18, 2020 at 18:35

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.

  • 4
    $\begingroup$ 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? :-) $\endgroup$ Commented Nov 13, 2012 at 8:33
  • 2
    $\begingroup$ 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 $a=b$ $\endgroup$
    – Diego
    Commented Sep 10, 2016 at 3:31

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


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.

  • 1
    $\begingroup$ Can the person who downvoted please explain themselves? (foot tapping and arms crossed.. lol) $\endgroup$
    – Jose_X
    Commented Sep 19, 2016 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 $.


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.


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.

  • $\begingroup$ 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. $\endgroup$ Commented Apr 9, 2016 at 23:07
  • $\begingroup$ @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. $\endgroup$
    – Farewell
    Commented Apr 9, 2016 at 23:15

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 $S$.


I remember thinking a lot about what $\pi$ actually means, why isn't a staircase a circle? Why can't we reproduce it with a finite combination of lines? What is a circle?

This is a great visual proof, but there a couple problems with the staircase paradox:

Why isn't it a staircase a circle?

$(A)$ A circle is symmetric all around. A staircase isn't.

$(B)$ The equations of a circle is $$y^2+x^2=r^2\implies \not\exists x, y, h \neq 0\ni (x+h)^2+y^2 = x^2+y^2$$ In every new quadrant, the function for every unique $x$, there is unique $y$.

Notice that a staircase has an element with the same $x$, but not the same $y$ which contradicts with the statement above.

Why can't we reproduce the same with a finite combination of lines?

Suppose we could: then to check if every point is on a circle we would get the value of the function of the circle in the given quadrant.
The length of the lines should be uniform, since otherwise if we rotate it, it won't be the same. And any shape with a finite amount of lines could be rotated such it wouldn't be the same of. The closest we could come is using a regular polygon

What is a circle?

If we picture some regular polygon with $n$ sides and compare with the circumference of the circle. $$ P_n = 2\pi r $$ We know what the perimeter of a $n$-sided regular polygon is: $$ P_n = 2 r \sin\left(\frac\pi n\right)n\implies \\\pi = \sin\left(\frac{\pi}{n}\right)n $$ Which is a transcendental equation: $$ \pi = \frac{n}{2i}\left(e^{i\pi\over n}-e^{-i\pi\over n}\right)\implies\left(e^{i\pi\over n}-e^{-i\pi\over n}\right)n=0 $$ So either $n=0$ or: $$ i\pi n^{-1}=-i\pi n^{-1}\implies n = \infty $$ So a circle is a regular polygon with $0$ or an infinite amount of sides


The approximation of circumference or arc length by the sum $(\mathrm dx+\mathrm dy)$ is not a correct way, unlike the area approximation by $(\mathrm dx~\mathrm dy)$. The proper way to approximate it is the length of hypotenuse $\mathrm ds=\sqrt{(\mathrm dx)^2+(\mathrm dy)^2}$. That is because the area approximation by the trapeziums/trapezia is equivalent to that by the rectangles in 2D, analogous to the volume (solid of revolution) approximation by the conical frusta, which is equivalent to the approximation by the cylindrical disks in 3D. However, this is not true for the arc length in 2D and the surface area of revolution in 3D. My related answer, which includes proof, can be read here.

In 2D, $$A=\iint_R\mathrm dA=\iint_R\mathrm dx~\mathrm dy;\ L=\int_0^L\mathrm ds=\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}~\mathrm dx$$

In 3D, $$V=\iiint_R\mathrm dV=\iiint_R\mathrm dx~\mathrm dy~\mathrm dz;\ S=\int2\pi y~\mathrm ds=\int_a^b2\pi y\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}~\mathrm dx$$

Intuitively, unlike the area, the arc length or circumference does not behave appropriately under small deformations of 2D regions in 2D. Similarly, unlike the volume, the surface area does not behave appropriately under small deformations of 3D regions in 3D.


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