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What is wrong with this?

Is \pi=4?


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The problem is that it doesn't approach circle in a "smooth" way. –  J. J. Dec 3 '10 at 13:49
Only for very large values of $\pi$ –  Ross Millikan Dec 3 '10 at 13:51
You may do the same "trick" with a triangle, thereby "countering" Pythagoras theorem. –  Fredrik Meyer Dec 3 '10 at 13:55
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 3 '10 at 13:57
The last figure said $\pi=4!$ or $\pi=24$ that is absolutely wrong. :-) –  xport Dec 4 '10 at 3:53

17 Answers 17

up vote 196 down vote accepted

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 $n \frac{1}{n^2}$ which converges to 0. The path length is $n \frac{2}{n}$, which converges even more nicely to 2.

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I would be curious to know why the downvote. –  Ross Millikan Dec 3 '10 at 17:35
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 3 '10 at 18:33
My up vote should compensate nicely :) –  basarat Dec 3 '10 at 20:11
@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 4 '10 at 0:32
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 4 '10 at 1:08

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 3 '10 at 15:50
Curiosity: why did you use the notation ||(x,y|| instead of ||x,y|| or ||(x,y)||? Not an actual problem, unclosed brackets just stick out to me =). –  Malabarba Dec 3 '10 at 17:47
Corrected. It was the result of cut and paste. –  TCL Dec 3 '10 at 19:19

R.I.P. Archimedes

enter image description here

A photogenic answer to such a question!

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This picture is more accurate. –  Jus12 Oct 10 '14 at 5:28
@Jus12: That picture is not accurate either! –  user21820 Jan 4 at 11:50
Makes sense to me! –  itcouldevenbeaboat Jan 22 at 18:12

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 driving 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. you could drive very small, very fast circles around me.

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I prefer your answer, because you've identified the specific false belief that leads immediately to the paradox. –  goblin Jan 19 at 16:44

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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What is wrong with this?

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

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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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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
@phv3773 Leibniz. –  Paul Slevin May 2 '12 at 10:55
Wikipedia links explaining some concepts mentioned here: calculus and (numerical) analysis (classes). Riemann sum. limsup (limit superior). –  Rory O'Kane Nov 13 '12 at 8:14
N.S. isn't talking about numerical analysis; he's referring to en.wikipedia.org/wiki/Mathematical_analysis, which provides the foundations of calculus. –  Ray Mar 20 '13 at 12:44


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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Is the circle a limit curve of the jagged square? –  Muhammad Alkarouri Jan 8 '12 at 17:05
@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

(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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Thus 0.999... != 1 –  elijaheac Jun 1 '13 at 23:26
@Integrator Done :) –  Sawarnik Dec 14 '14 at 12:49

This is an old chestnut, and is usually given in the form mentioned by Ross Millikan in his answer, just as he says.

The concise answer is that topological convergence does not imply convergence in measure. Once you have this correct big picture in your head, the problem disappears. The detailed analysis, such as supplied by Ross Millikan, is just the closing argument for a case already won. Thus, the real purpose of this problem is to bring to the attention of the student the distinction between topological convergence and convergence in measure.

As a bonus, this problem (in the “diagonal of a unit square” version) shows why wiggly coastlines are preferred, in regard to sea-faring vessels: Such coastlines provide much more opportunities for docking/harboring.

edit (18.Dec.2011 Taiyuan China): I suppose that this conundrum can appear in quite a number of forms. One such other form is what we might call the “helicopter paradox”, namely, that if you double the length of the helicopter rotor, but halve its speed, you obtain the same lift, and, if you’re not careful, you also obtain the false conclusion that a rotor of infinite length not rotating would exert the same lift on the helicopter as a normal rotor. The solution to this enigma (“What is the lift of an infinite helicopter rotor at rest?”) is that the convergence alluded to in the statement of the problem is merely topological convergence, but the conclusion of convergence-in-measure is, fallaciously, being drawn from this. A presentation of this conundrum, as a purported “thought experiment” is given at the following link, where there seems to be no hint that the proposer is aware of its real solution:


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Could you please say precisely what you mean by "topological convergence" and "convergence in measure"? –  Jonas Meyer Jun 25 '11 at 22:44
Convergence in measure has a specific technical meaning: see en.wikipedia.org/wiki/Convergence_in_measure. It is in fact a rather weak notion, famously weaker than almost everywhere convergence (at least on a finite measure space). You seem to use "convergence in measure" to mean something else...and in fact your meaning is not very clear. Please note that Jonas Meyer asked you almost six months ago to clarify what you mean. Your recent edit in terms of helicopters and rotors does not do this. –  Pete L. Clark Dec 17 '11 at 22:23

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. –  Heather May 23 '11 at 6:28
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 7:31

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 8:33

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 7:29
@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
@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
@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: "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. –  BlueRaja - Danny Pflughoeft Jul 26 '11 at 22:36

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.

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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 2 '14 at 20:25

This is an interesting question. Although all of the above answers explain it properly, I would like to add a visual to this question. This will provide more insight into this question, hopefully. Thanks! Basically, the converged circle won't be a smooth circle and the perimeter is more!

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My answer is that the value of 4 constantly stays the same, by appearance, however this doesn't truly solve the arc-length of a circle.

I remember I used floor equations for the fun of it, just to solve for the distance line, no matter how close it gets you will always get the same answer, and yet we ignore the approach of truly solving the arclength, your taking a bunch of steps that evenly distributes over a region to keep the same perimeter.

But one should not simply distribute these step-like function because we're not literally solving the arc-length.

As all of the people said above it's by appearance, but appearance can deceive.

Believe or not you can do this same exact process and use it on any function, a line, parabola, circle, you name it, and it will usually have the same arclength, no matter the amount of steps, and always be different from other ways of measuring, on certain domain. This means all the approaches of solving lengths are wrong, or this approach of solving lengths is wrong.;)

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It is impossible to remove an infinite number of corners.

When you remove a very large number of corners, the figure looks like a circle, but it isn't a circle. No matter how many corners you remove, the perimeter is still 4. No matter how many corners you remove, it will not turn into an actual circle.

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protected by Qiaochu Yuan May 30 '11 at 9:01

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