# Proof for Pi

1. Aug 22, 2008

### mite

can we prove the ratio of circumference to diameter is same for all circles & is equal to pi?

2. Aug 22, 2008

### CompuChip

Well, the circumference of a circle is $L = 2\pi r$ source and by definition the diameter is twice the radius (d = 2r). So $L/d = 2 \pi r / (2 r) = \pi$.

3. Aug 22, 2008

### tiny-tim

Hi mite!

Euclid regarded the similarity of two circles as an axiom, so there was nothing to prove!

And π is defined as the ratio.

4. Aug 22, 2008

### HallsofIvy

Staff Emeritus
Historically, the fact that the ratio of circumference to diameter is a constant was a numerical observation. The Greeks proved it by calculating the ratio of the perimeter of a regular n-gon to its "diameter" and then seeing what happened as n got larger and larger (a limit process). As for the fact that that ratio is equal to pi- that's essentially the definition of pi.

A modern proof would be something like this: Since sin2(t)+ cos2(t)= 1 for all t, x= Rcos(t) and y= Rsin(t) are parametric equations for a circle of radius R. The circumference, then, is given by
$$\int_0^{2\pi}\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}dt$$
$$= \int_0^{2\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= \int_0^{2\pi}Rdt= 2\pi R$$
Since the circumference is $2\pi R$ and the diameter is 2R, the ratio of circumference to diameter is $2\pi R/(2R)= \pi$.

(The fact that sin(t) and cos(t) have period $2\pi$, which is critical to this proof, can be shown by using the fact that the second derivative of sin(t) is -sin(t) and the second derivative of cos(t) is -cos(t).)

5. Aug 22, 2008

### tiny-tim

Hi HallsofIvy!

No, surely that's how they calculated π …

they were already convinced that it was the same for all circles?

6. Aug 22, 2008

### HallsofIvy

Staff Emeritus
I didn't say that was how they calculated it. I said that was how they proved it was the same ratio for all circles. It was Archimedes who did that. I am sure that Greeks before that just assumed it was a constant.

7. Aug 22, 2008

### tiny-tim

And they were right to do so!

From their point of view, because it was axiomatic and/or obvious …

from our point of view, because of the scalar symmetry of Euclidean space.

(Of course, circumference/diameter isn't a constant in non-Euclidean space. )

(If you'd said to them "you've proved that it's a constant", they'd have replied "no we haven't, we've only calculated the constant … we implicitly used a symmetry theorem on polygons in the course of that calculation, and that applies to circles anyway" )

8. Aug 22, 2008

### dynamicsolo

The arclength integration doesn't really constitute a proof. The choice of integration limit for the "angle parameter" of $$2 \pi$$ is based on the definition of angle as the ratio of arclength to radius for a circle: $$\theta = \frac{s}{R}$$. So it is completely unsurprising that the result of the integration for the circumference of a circle is $$2\pi R$$.

I haven't explored the history of the number thoroughly (although there are at least two or three histories of pi out there now), but I believe that pi is simply defined as the ratio of circumference to diameter for a circle (the specific letter was chosen somewhere around the 18th Century), it having already been understood in antiquity that the ratio is a constant. So there is no proof involved for this. (One of the continuing mysteries is why $$\pi$$ is so deeply imbedded in the structure of mathematics and turns up in other relations which has little to do with circles...)

9. Aug 22, 2008

### granpa

real world circles or mathematical circles? the circumference of the latter depends on your metric.