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?

**Answer**

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

In geometrical definitions of π, 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) πk(C). This sequence is thought of as a set of approximations converging to π, 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 πk(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 π(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 πk(C1)=πk(C2) 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 πk, 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 C1 and C2.

(You might think that this is proving a different conclusion, that the equidecomposability version of π for the two circles is equal, and not the numerical equality of π 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.)

**Attribution***Source : Link , Question Author : Chris Card , Answer Author : T..*