# Proof that Pi is constant (the same for all circles), without using limits

Is there a proof that the ratio of a circle's diameter and the circumference is the same for all circles, that doesn't involve some kind of limiting process, e.g. a direct geometrical proof?

• Sounds hard; its being transcendental seems to preclude the existence of a proof that won't appeal to the concept of limits. Commented Aug 24, 2010 at 13:51
• @Chris, the problem is with defining the length of a circle without appealing to a limit! Commented Aug 24, 2010 at 13:54
• If you are going to work 'intuitively', then it is pretty obvious that zooming in or out does not change proportions of lengths, so in particular it does change the proportion between the circumference and the diameter! Now, if you want to actually prove something, you need to define things precisely, and you are more or less stuck with limits. Commented Aug 24, 2010 at 14:20
• Actually it started "Is there a proof ... ?" :-) I'm happy to accept "no" as an answer, if backed up by a convincing argument, e.g. "any such proof would involve defining the length of the circumference and that requires using limits." Commented Aug 24, 2010 at 15:11
• @Chris, what I am asking is: since the length of the circumference is defined in terms of limits, there is no possible way to prove anything about it without invoking limits. If what you want to know is if one can define the length of the circumference without using limits, then your question should ask that :) Commented Aug 24, 2010 at 15:19

Limits are not involved in the problem of proving that $\pi(C)$ is independent of the circle $C$.

In geometrical definitions of $\pi$, to a circle $C$ is associated a sequence of finite polygonal objects and thus a sequence of numbers (or lengths, or areas, or ratios of those) $\pi_k(C)$. This sequence is thought of as a set of approximations converging to $\pi$, but that doesn't concern us here; what is important is that the sequence is independent of the circle C. Any further aspects of the sequence such as its limit or the rate of convergence will also be the same for any two circles.

(edit: an example of a "geometrical definition" of a sequence of approximants $\pi_k(C)$ is: perimeter of a regular $k$-sided polygon inscribed in circle C, divided by the diameter of C. Also, the use of words like limit and approximation above does not reflect any assumption that the sequences have limits or that an environment involving limits has been set up. We are demonstrating that if $\pi(C)$ is defined using some construction on the sequence, then whether that construction involves limits or not, it must produce the same answer for any two circles.)

The proof that $\pi_k(C_1) = \pi_k(C_2)$ of course would just apply the similarity of polygons and the behavior of length and area with respect to changes of scale. This argument does not assume a limit-based theory of length and area, because the theory of length and area for polygons in Euclidean geometry only requires dissections and rigid motions ("cut-and-paste equivalence" or equidecomposability). Any polygonal arc or region can be standardized to an interval or square by a finite number of (area and length preserving) cut-and-paste dissections. Numerical calculations involving the $\pi_k$, such as ratios of particular lengths or areas, can be understood either as applying to equidecomposability classes of polygons, or to the standardizations. In both interpretations, due to the similitude, the results will be the same for $C_1$ and $C_2$.

(You might think that this is proving a different conclusion, that the equidecomposability version of $\pi$ for the two circles is equal, and not the numerical equality of $\pi$ within a theory that has real numbers as lengths and areas for arbitrary curved figures. However, any real number-based theory, including elementary calculus, Jordan measure, and Lebesgue measure, is set up with a minimum requirement of compatibility with the geometric operations of dissection and rigid motion, so once equidecomposability is known, numerical equality will also follow.)

• If we say (for example) that $\pi(C)$ is the Cauchy sequence ($\pi_k(C)$), that does not use limits, and the proof of the equation $\pi(C_1) = \pi(C_2)$ also does not use limits. A proof is limit-free if it has no epsilon-delta arguments, $O()$ notation, or other arguments about asymptotic equality-in-the-limit (do you agree?). This is avoided for the question of $\pi$ being circle-independent, because there one has exact, term by term, non-asymptotic equality of the sequences.
– T..
Commented Aug 25, 2010 at 18:25
• @Tsuyoshi: in the Cauchy (convergent sequence) model of real numbers, the numbers are sequences, and some proofs of equality can be performed directly on the sequences (even for transcendental numbers, such as the statement "$e - e = 0$") without using limits. This is the case for Pi(C_1) = Pi(C_2). Limits would be involved in proving that those Pi sequences are convergent, but here we are only claiming that whatever "limits" and "convergence" are, any statements involving those concepts that hold for one circle are true for any other, i.e., Pi (under any one definition) is well-defined.
– T..
Commented Aug 26, 2010 at 16:48
• It should be emphasized that not using limits is a well-defined technical concept, not a matter of personal feeling or whether words like "convergence" are used. If the proof does not contain statements with quantifier structure All-Exists-All (such as "for all (Epsilon>0), there exists (N), such that for all (m > N)") then it unequivocally does not use limits. This is the case for proving the equality of the Pi-related sequences associated to two circles, or other purely formal statements like e=e, or equalities between power series that work coefficient by coefficient.
– T..
Commented Aug 26, 2010 at 16:56
• I do not interpret “not using limits” that way, but you can do so if you want to. Commented Aug 27, 2010 at 12:07
• The infinite sum has a standard definition of what it "means", i.e., what mathematical object one can associate to the notation Sum(a_i), irrespective of whether the sum converges. Convergence is an additional property of an already meaningful object -- the sequence of partial sums -- and not a requirement for the sum to have meaning. Infinite sums of real numbers (or matrices, or other objects of some type, call it X) are not themselves of type X; extracting an X from the sum is a different question and here limits do appear, just as asymptotics appear in the case of divergent series.
– T..
Commented Sep 1, 2010 at 1:23

Intuitively, all circles are similar and therefore doubling the diameter also doubles the circumference. The same applies to ratios other than 2.

To make this rigorous, we have to consider what we mean by “the length of the circumference.” The usual rigorous definition uses integration and therefore relies on the notion of limits. I guess that any rigorous definition of the length of a curve ultimately requires the notion of limits.

Edit: Rephrased a little to make the connection between the two paragraphs clearer.

• Of course, this does not explain why (in Euclidean space) doubling the radius also doubles the circumference. But as you note, any answer will have to describe how the circumference is defined in the first place, and probably as a limit, in which case the standard limit arguments will do. Commented Aug 24, 2010 at 13:58
• @Tsuyhoshi: I agree that the core concept is "similarity": but the very fact that similarity is a pertinent concept could be said to be a property of Euclidean space. Or is "similarity" a similarly-useful concept in spherical or hyperbolic geometry? Commented Aug 24, 2010 at 14:17
• @Tsuyoshi: (1) The usual rigorous definition of length of a curve is not as an integral, but as the supremum (least upper bound) of the lengths of finite inscribed polygonal arcs. This does not assume differentiability or a curve in Euclidean space. For smooth enough curves in Euclidean space (or in a Riemannian manifold) it is true that this equals the integral. (2) From the definition it follows, rigorously and without limits (using only the scaling of lengths of the finite polygons) that doubling the size of a circle doubles its length, and the same for any other curve.
– T..
Commented Sep 1, 2010 at 6:30
• @T..: As for (1), I did not know that. Thanks. Taking supremum is arguably a more basic operation than taking limit, but I would argue that considering an infinite set whose supremum defines π to deduce something about π counts as a “kind of limiting process.” As for (2), as you can imagine, I would argue that the proof uses the notion of supremum (which is a limiting process). Commented Sep 1, 2010 at 11:17
• @T..: As I indicated in a comment to you answer (which I posted in response to your question), both of us already clarified the difference of opinion about what counts as a “proof without limits,” and I do not find it fruitful for either of us to try to convert the other. Commented Sep 1, 2010 at 11:18

It seems -as far as I can understand what he was doing- that even Euclid used some sort of limiting process (the principle of exhaustion): http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII2.html . What Euclid is proving here is the following: let $d_1$, $d_2$ be the diameters of two circles and $A_1$, $A_2$ their areas. Then

$$\frac{A_1} {A_2} = \frac{d_1^2}{d_2^2} \ .$$

Which is the same as saying that the proportion between the area of a circle and the square of its radius is constant: if $r_1$ and $r_2$ are the radii of our circles

$$\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2} \quad \Longleftrightarrow \quad \frac{A_1}{r_1^2} = \frac{A_2}{r_2^2}$$

That is to say, "$\pi$ is constant".

• Doesn't the premise $A_1/A_2=d_1^2/d_2^2$ already imply the result?
– Ovi
Commented Apr 27, 2016 at 23:38
• How does that prove that $\frac{C}{d}$ is constant?
Commented Mar 9, 2019 at 16:33

The idea is this:

Similar polygons inscribed in circles have areas that are proportional to the squares of the diameters of the circles. By approximating circles closely by similar polygons of more and more sides, the proportion is carried over to the circle as a limit.

This is the way that it was proved by Euclid.

Let me start by claiming that this is just simply "not true", (while see below if you want a proof.)

In Hyperbolic spaces, the ratio between the circunference and the radius is exponential

In a round Sphere, the ratio between the circunference and the radius is sinusoidal.

So, this means that the ratio between the circunference and the radius is not something that can be easily done by means of simple geometric tools. For instance, the above examples shows that you cannot prove it without the use of the fith Euclid postulate.

Of course, the proportionality between the circunference and the radius is trivially true if you accept that the procedure of scaling a geometric figure by $\lambda$ scales all the one-dimensional length by $\lambda$.

What you can easily do with all the standard geometric tools is prove it for polygons: For triangles that's just Tale's Intercept Theorem, and a polygon can be easily subdivided in triangles.

Now, if you want to use Thale's theorem for building a propotionality principle for circunferences, than you are forced to introduce limits.

Note that if you don't want to use limits, then your big problem is to define the lenght of a curve, rather that prove that in the Euclidean space this is scale-multiplicative.

Finally, if you are more interested in a proof that "hides" limits (for instance for didattic purposes) here is a paper-and-cisor proof of the doubling properties for circunferences:

Consider a disk of paper of radius R. Its circumference has some length L. Now, cut the disk in two halfs: No one has problems in accepting that the two half circunferences have equal lenghts L/2.

If you glue the two radii of the half-disks, you get two identical cones. Now you have to convince you audience that if you put one of this cones on the table and look it "from the same level of the table" (i.e. you do a projection) then you see an equilateral triangle!!!

For doing that put your two cones on the table in two different ways: one with the basis on the table, the other with a radius on the table: they will look the same.

This means that the circumference at the base of the cone has radius R/2.

Since we know from the beginning that the base-circumference has length L/2, We have "proved" that if the circumference of radius R has length L, than the circumference of radius R/2 has lenght L/2.

By changing the cone angle you get multiplicative constants different from 2 or 1/2, but now convincing your audience will be more tricky.

One should learn that the ratios of distances in any geometric figure remain constant if the shape remains constant, despite changes in the size. Similarly the ratios of areas are as the squares of the ratios of distances, so that, for example, $$\text{area enclosed by a circle} = \text{constant}\times\text{radius }^2,$$ and "constant" means not depending on the size. And $$\text{area enclosed by a sphere} = \text{constant}\times\text{radius }^3,$$ and again "constant" means not depending on the size (this time the "constant" is $4\pi/3$, as Archimedes showed).

Likewise $$\text{area enclosed by a pentagon} = \text{constant}\times\text{length of diagonal }^2$$ and with a bit of work one could evaluate the constant. In every case, finding the constant may (or may not) take a lot of work.

How does one demonstrate this general proposition? Maybe I'll say someting about that on another occasion.

I came across this proof once Link, though I'm not 100% sure that writing pi in terms of the tan of a given angle isn't circular.

• It can be made not to be circular at all, in the book "Intermediate Real Analysis" by Fischer, it is shown that the value $\pi_1$ from circle and geometry and the value that makes the definition $sin x$ with infinite series ( without any reference to circle ) , coincide ( of course only under Euclidean Geometry). Questioning your own answer is a good sign of willingness to progress +1 Commented Feb 9, 2013 at 2:57