## March 14, 2017

### How to Not Prove that Pi = 4

Today is March 14, which, due to 3.14 being the most common approximation for the most well-known mathematical constant, is celebrated as Pi Day. Some mathematicians frown upon that, since they know that Pi isn't actually equal to 3.14 and think that impresses someone, but any excuse to write about math is fine by me.

So before we get into anything else, we need to ask - what is Pi? A lot of people know that it's the ratio between the circumference and the diameter of a circle, but who's to say that ratio is the same for all circles? You can measure some circles and see that the ratio is pretty much the same, but who says that it's exactly the same? Who says we didn't just get lucky and pick a few circles that happen to have that ratio?

In other words, we want to prove that Pi is well-defined - that the definition we gave of Pi as said ratio actually means something.

Well, there's good news and bad news on that front. The good news is that there's a basic, intuitive proof that doesn't require a whole lot of mathematical background to understand. The idea is as follows:

Image shamelessly pilfered from ProofWiki

1. Position two circle so that they're centered at the same point, and in particular, the one with the smaller radius lies inside the one with the larger radius.
2. Choose some integer, n, which is equal to at least 3, and divide both circles to n equal parts, like slices of pizza, i.e. by drawing straight lines from the center to the perimeter of the outer circle.
3. Now, for each circle, connect each points where the straight lines from step 2 intersect it with its neighbor.
4. We now have two families of triangles, one for each circle. The sides of each triangle are the length of the corresponding circle's radius, and the angle between the sides is the same for all triangles - namely, it is equal to 360 degrees divided by n, the number of triangles we divided our circles into.
5. In particular, each small triangle is contained in a large triangle such that the ratio of the sides is the ratio of the smaller radius to the larger radius, and the angle between these sides is equal. Laws of triangle similarity dictate that the ratio of the "bases" - the third edge formed by connecting two points on the perimeter of the circle - is also the ratio of the smaller radius to the larger radius.
6. Now the magic comes in: as you can clearly see from the image up there, the larger n is, the more the sum of the bases of the triangles approach the circumference of the circle they're contained in. But we've seen that the bases of the triangles have the same length ratio as the radii, and since this is true for every n, in particular, the ratio of circumferences thus approximated must also be the ratio of the radii.
7. Step 6 shows us that, if we mark the circumference of the smaller circle by $P_1$ and its radius by $R_1$, and do the same with index 2 for the larger circle, we have $$\frac{P_1}{P_2}=\frac{R_1}{R_2}$$ which is the same as saying $$\frac{P_1}{2R_1}=\frac{P_2}{2R_2}$$ thus showing that, indeed, the ratio between circumference and diameter is constant.
Those were the good news. The bad news is that, alas, our approach can be used to render all of mathematics meaningless, for as the follow image shows, it can be used to prove that Pi is not roughly 3.14 but is, in fact, 4:

Which is a shame. I liked math. It's too bad that it has to Go Away Forever.

Well, as you can guess, the problem isn't so much with math as with our reasoning. Like I said, our approach can be used to show that Pi equals 4, but that just shows that our approach is wrong. Specifically, the issue lies in step 6, where we said that the lines formed by the bases of our triangles approach the perimeters of the circles and that therefore their lengths must also be the same. This reasoning is dead wrong, even if our conclusion was correct. Actually, math is full of examples of correct results one can reach with bad reasoning - my favorite is $$\require{cancel} \frac{64}{16}=\frac{\bcancel{6}4}{1\bcancel{6}}=4$$ Don't get me wrong - the approach of calculating a curve's length by approximating its length with a sum of lengths of straight lines that becomes ever-closer can be used correctly. In fact, it's pretty much what we're going to do. But as we can see, sometimes it works, and sometimes it doesn't, and to understand when it can actually work, we need to be a bit more rigorous. In fact, this is probably the simplest example I know of showing just how important mathematical rigor is.

So how do we show that Pi is well-defined? For this, just like with all good things in life, we're going to use Calculus - specifically, line integrals. The theory of integration is more than I can cover in a single blog post, but the idea is the same - calculate areas, lengths of curves or anything else by making these calculations for more basic shapes, and show that these basic shapes can be used to approximate the complex ones. The key is in the last step, though - understanding which approximations work and which don't, and most importantly, why. If I've made that point in this post, then my work here is done.

Just to be cool, though, let's show our calculus proof in full. We will calculate the length of half the perimeter of a circle with radius $R$ using the following parameterization: $$x(t) = t, y(t) = \sqrt{R^2-t^2}, -R \le t \le R$$ Why not use the more standard parametrization using cosine and sine? I'll leave that for you to answer, but think about how you would go about justifying each step of the following calculation using that parametrization.

Onwards, then. Let's use $P$ to denote the circumference of the circle. The integral we must now calculate is $$\int _{-R} ^{R} \sqrt{x'(t)^2+y'(t)^2}dt$$ which, after some calculations, is shown to be $$R \cdot \int _{-R} ^{R} \frac{1}{\sqrt{R^2-t^2}}dt$$ Now here comes the real magic. In this integral, substitute $v = \frac{t}{R}$. This gives us the integral $$R \cdot \int _{-1} ^{1} \frac{1}{\sqrt{1-v^2}}dv$$ But with $P$ as above, we have $$\frac{P}{2}=R \cdot \int _{-1} ^{1} \frac{1}{\sqrt{1-v^2}}dv$$ or equivalently $$\frac{P}{2R}= \int _{-1} ^{1} \frac{1}{\sqrt{1-v^2}}dv$$ Now check out this beauty! On the left hand side we have our coveted ratio, but on the right we have a definite integral - a number - which more importantly than anything else, does not depend on $R$, It's also kinda important that the function we're integrating is continuous and bounded over the closed interval of integration, which means that our integral exists.

Some of you may be tempted to say the primitive function here is $arcsin(v)$ and conclude that the left-hand side is indeed equal to $\pi$. But that would be naughty, as we have only now shown that Pi is well-defined, and that makes any reasoning in that direction fishy. But the important part is that the right-hand side is a constant, and from here, all the wonders of geometry, trigonometry and much of calculus are ours for the taking.