65

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?

T..
  • 11,386
  • 2
  • 38
  • 49
Chris Card
  • 1,948
  • 1
  • 15
  • 16
  • 5
    Sounds hard; its being transcendental seems to preclude the existence of a proof that won't appeal to the concept of limits. – J. M. ain't a mathematician Aug 24 '10 at 13:51
  • 22
    @Chris, the problem is with defining the length of a circle without appealing to a limit! – Mariano Suárez-Álvarez Aug 24 '10 at 13:54
  • 1
    well, intuitively you can define the length of the circumference by rolling the circle along a line, but that probably doesn't help much – Chris Card Aug 24 '10 at 14:05
  • Chris: that can be shown to be equivalent to "slicing up" the circle to form a "parallelogram" of appropriate dimensions; unfortunately for you this too involves limits. – J. M. ain't a mathematician Aug 24 '10 at 14:07
  • 21
    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. – Mariano Suárez-Álvarez Aug 24 '10 at 14:20
  • 1
    that was my question and it seems the answer is no. – Chris Card Aug 24 '10 at 14:25
  • 1
    @Chris, Your question was «how can we define the length the circumference without using limits?»? – Mariano Suárez-Álvarez Aug 24 '10 at 14:44
  • 3
    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." – Chris Card Aug 24 '10 at 15:11
  • 6
    @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 :) – Mariano Suárez-Álvarez Aug 24 '10 at 15:19
  • 1
    it's not quite obvious to me, since we're actually interested in a ratio of lengths rather than the length of the circumference itself. Plus, is it obvious that the only way to define the length of the circumference is by using limits? – Chris Card Aug 24 '10 at 15:30
  • @Chris: how else would you define the length of something that isn't straight? You should think very hard about what you think a length is. (Is it something you can measure with a ruler? How do you measure the length of something curved with a straight ruler?) – Qiaochu Yuan Aug 24 '10 at 16:58
  • 1
    you could use a bendy ruler! (not totally serious comment BTW) – Chris Card Aug 24 '10 at 18:07
  • 1
    @Mariano and Qiaochu: underlying the circle-independent limit is a circle-independent sequence of approximations. It is enough to show independence for the sequence and this does not require limits. – T.. Aug 24 '10 at 18:46
  • 1
    @T: but the very definition of "length of the circle" uses a limit. The only reason why your finite polygonal objects can be thought of a "set of aproximations", to use the language in your answer, is because the length of the circle is a limit; and you have to know, for example, that the limit exist for it to even make sense to aproximate it &c. – Mariano Suárez-Álvarez Aug 24 '10 at 19:39
  • 2
    My answer does not depend on any notion of approximation (except to help connect it to other better-known explanations that do involve approximation, that is, convergence of the sequences). Pi as conventionally defined using limits is a limit of *something* and my answer was that the *something*, which you can call a "sequence of approximants" or by any other name, is the same for any two circles. Accordingly, whatever "limits" are and whenever they exist, they would be the same for the two circles. Equality of two limits is easier to prove than existence or evaluation of either one alone. – T.. Aug 24 '10 at 19:51
  • 2
    This is an instance of the common pattern that proving a mathematical object is well-defined (e.g., a sequence associated to a circle does not depend on the choice of circle), is easier than proving more specific properties of the object (e.g., the sequence has a limit, that limit is calculated by a particular integral, the integral has some invariance properties with respect to rotation, the numerical value of the integral is a transcendental number). – T.. Aug 24 '10 at 20:38
  • I don't usually add a new answer when five answers are already there, but I posted one, and up-voted several. – Michael Hardy Aug 18 '13 at 00:41
  • You can avoid introducing any new notion of limit altogether if you just define reals using Dedekind cuts. Define the space of cuts D as all nonempty collections of rationals bounded below. Then define R as the image of D under upward closure (of sets) in Q. Then T.'s proof that \pi_k(C)=\pi_k(C') (say by circumscribed polygons) for all k and all pairs of circles C, C' yields that the map \pi:\mbox{Circles}\rightarrow D defined by \pi(C)=\{\pi_k(C) : k\in \mathbb{N}\} is constant, so its projection onto R consists of a single point. – Marcel Besixdouze Nov 13 '13 at 08:03

7 Answers7

25

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.)

T..
  • 11,386
  • 2
  • 38
  • 49
  • Could you provide a reference for this finite sequence $\pi_k (C)$? Thank you. – Agustí Roig Aug 24 '10 at 18:45
  • Take any reference that defines Pi as a limit of geometrical calculations on regular polygons with K vertices. For each such definition of $\pi$ there is a unique sequence $\pi_k$, but different definitions will have different sequences. For example, we could define the k'th approximant using an inscribed or circumscribed polygons, calculating a particular quantity for each polygon, such as (circumference/diameter) or (circumference^2)/(4*area) or area/(diameter^2). – T.. Aug 24 '10 at 19:09
  • @T.. Sorry, I misunderstood your "finite" in the second paragraph. But, anyway, you are computing limits there – Agustí Roig Aug 25 '10 at 02:31
  • @Agusti: where is there a limit computation in what I wrote? I am asserting that any ordinary geometrical definition of pi(C) using polygonal objects will be as a function of a sequence $\pi_k(C)$ and equality of the functions follows from equality of the sequences ($\pi_k(C_1) = \pi_k(C_2)$, for all $k$). The latter does not require limits either to state or to prove within Euclidean geometry, because it has a cut-and-paste proof, which is stronger than an equation of real numbers (the latter of course would use limits). "Limits of identical sequences are equal" is not a use of limits. – T.. Aug 25 '10 at 02:53
  • @T.. I agree that you can proof the equality of the sequences $\pi_k (C_1) = \pi_k (C_2)$ whithout refering to limits at any time. But, also: *where* is $\pi$ in all these equalities? At some point you must say that it's the *limit* of those $\pi_k(C)$, don't you? So, I don't see how a proof that starts defining $\pi$ as a limit of some quantities $\pi_k (C)$ can be called "limit free". – Agustí Roig Aug 25 '10 at 08:22
  • 2
    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.. Aug 25 '10 at 18:25
  • 1
    To me, this proof is a typical proof which uses limits and avoids the mere use of the words “limits” or “converge.” I would say that you prove some property about Cauchy sequences converging to π(C_1) and π(C_2), but even if you avoid the word “converging” by saying that the Cauchy sequence is the very definition of the real number, a proof working on the defining Cauchy sequences sounds like a proof using limits to me. – Tsuyoshi Ito Aug 26 '10 at 13:55
  • 2
    @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.. Aug 26 '10 at 16:48
  • 3
    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.. Aug 26 '10 at 16:56
  • 3
    I do not interpret “not using limits” that way, but you can do so if you want to. – Tsuyoshi Ito Aug 27 '10 at 12:07
  • @Tsuyoshi: does proving the statement "e - e = 0" require limits in your interpretation, where e = 2.718.... = sum( 1/n!)? What about 2X = 1+X where X = 1/2 + 1/4 + 1/8 + ... is the sum of reciprocals of powers of 2? Another good example is proving Y=0 where Y=sum (1/(2^k)) with the sum taken over an enumeration of counterexamples to Fermat's Last Theorem (so that the sum is empty due to Wiles proof of FLT, but it is easy to prove without FLT that it is bounded). – T.. Aug 27 '10 at 15:52
  • “does proving the statement "e - e = 0" require limits in your interpretation, where e = 2.718.... = sum( 1/n!)?” I do not know why you ask me this question (I do not think that we can come to agreement on what “not using limits” means), but my answer is “No, if you assume that the sum is well-defined.” To establish the fact that the sum is well-defined, we need the notion of limits. – Tsuyoshi Ito Aug 28 '10 at 03:45
  • That's not the usual meaning of "well-defined". There is no possible ambiguity involving the sum, just as there is no ambiguity about the partial sums of divergent series such as 1 + 2 + 4 + (etc). Whether the sum defines a real number (e.g., does it satisfy the definition of convergence or that of being a Cauchy sequence) is a separate question that of course involves limits, asymptotics, O( ) notation or the equivalent. However, e - e, like Pi(C_1) - Pi(C_2) is already equal to (0,0,0,0,...) as a sequence and no asymptotics are involved in proving that this difference is equal to zero. – T.. Aug 28 '10 at 04:06
  • Ok, it was a mistake to use the word “well-defined.” Instead my answer is “No, if you assume that the sum means anything at all.” To argue that the infinite sum means anything at all, we need the notion of limits. Are you satisfied? – Tsuyoshi Ito Aug 31 '10 at 22:54
  • 1
    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.. Sep 01 '10 at 01:23
7

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.

Tsuyoshi Ito
  • 1,928
  • 11
  • 18
  • 1
    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. – Niel de Beaudrap Aug 24 '10 at 13:58
  • Niel's comment was made in response to my mangled one above (reproduced here with tex corrected): See [arc length](http://en.wikipedia.org/wiki/Arc_length) in cartesian coordinates. So $S=rθ$. When the circumference C is drawn out by an angle $\theta = 2\pi$, then $C=2\pi r$. (But, of course arc length is defined through integration.) – Tom Stephens Aug 24 '10 at 14:06
  • @Niel de Beaudrap: I would say that it is part of the “intuitive” understanding of the notion of similarity. This can be explained by noting that mathematical facts are independent of the choice of unit length. But I am no expert of mathematical education, and I am not too sure what is intuitive and what is not. – Tsuyoshi Ito Aug 24 '10 at 14:12
  • @Tom: actually, my comment was made with reference to Tsuyoshi's answer. I should have made that clearer: sorry about that. – Niel de Beaudrap Aug 24 '10 at 14:14
  • 1
    @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? – Niel de Beaudrap Aug 24 '10 at 14:17
  • @Niel: I do not know whether similarity is useful in non-Euclidean geometry. Probably I should have restricted my claim to mathematical facts about Euclidean geometry (or to mathematical facts about the length of circumference!). – Tsuyoshi Ito Aug 24 '10 at 14:24
  • My minimal knowledge of non-euclidean (spherical and hyperbolic) geometry tells me that similarity of triangles (AAA) is the same as congruence. But similarity in Euclidean geometry is really just dilations in disguise, and as far as I know hyperbolic and spherical geometry don't have dilations in the same vector space-y spirit. – Vladimir Sotirov Aug 25 '10 at 19:19
  • 1
    @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.. Sep 01 '10 at 06:30
  • 1
    @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). – Tsuyoshi Ito Sep 01 '10 at 11:17
  • 1
    @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. – Tsuyoshi Ito Sep 01 '10 at 11:18
6

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".

Agustí Roig
  • 16,909
  • 2
  • 40
  • 82
4

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.

user126154
  • 7,237
  • 12
  • 21
4

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.

Ambesh
  • 3,152
  • 2
  • 22
  • 45
2

I came across this proof once http://mathforum.org/kb/message.jspa?messageID=68371&tstart=0, though I'm not 100% sure that writing pi in terms of the tan of a given angle isn't circular.

wheresrhys
  • 129
  • 3
  • 1
    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 – jimjim Feb 09 '13 at 02:57
2

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.

Michael Hardy
  • 1
  • 30
  • 276
  • 565