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?

Solution 1:

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.)

Solution 2:

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.

Solution 3:

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".

Solution 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.

Solution 5:

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.