Today I’d like to share with you three fake proofs, in increasing order of subtlety, and then discuss what each one of them has to tell us about math.
The first proof is for a formula for the surface area of a sphere, and the way that it starts is to subdivide that sphere into vertical slices, the way you might chop up an orange or paint a beach ball. We then unravel all of those wedge slices from the northern hemisphere so that they poke up like this, and then symmetrically unravel all of those from the southern hemisphere below, and now interlace those pieces to get a shape whose area we want to figure out. The base of this shape came from the circumference of the sphere; it’s an unraveled equator. So its length is 2π times the radius of the sphere. And then the other side of this shape came from the height of one of these wedges, which is a quarter of a walk around the sphere, and so it has a length of π/2 times R. The idea is that this is only an approximation, the edges might not be perfectly straight, but if we think of the limit as we do finer and finer slices of the sphere, this shape whose area we want to know gets closer to being a perfect rectangle, one whose area will be π/2 R times 2π R. Or in other words, π squared times R squared. The proof is elegant. It translates a hard problem into a situation that’s easier to understand. It has that element of surprise while still being intuitive. It’s only fault, really, is that it’s completely wrong. The true surface area of a sphere is 4π R squared. I originally saw this example thanks to Henry Reich, and to be fair, it’s not necessarily inconsistent with the 4π R squared formula, just so long as π is equal to 4.
For the next proof I’d like to show you a simple argument for the fact that π is equal to four. We start off with a circle, say with radius 1, and we ask how can we figure out its circumference. After all π is by definition the ratio of this circumference to the diameter of the circle. We start off by drawing the square whose side lengths are all tangent to that circle. It’s not too hard to see that the perimeter of the square is 8. Then, and some of you may have seen this before it’s a kind of classic argument, the argument proceeds by producing a sequence of curves, all of whom also have this perimeter of 8, but which more and more closely approximate the circle. But the full nuance of this example is not always emphasized. First of all, just to make things crystal clear, the way each of these iterations works is to fold in each of the corners of the previous shape so that they just barely kiss the circle. You can take a moment to convince yourself that in each region where a fold happened, the perimeter doesn’t change. For example in the upper right here, instead of walking up and then left, the new curve goes left and then up. And something similar is true at all of the folds of all of the different iterations, wherever the previous iteration went direction A then direction B, the new iteration goes direction B then direction A. But no length is lost or gained.
Some of you might say, well obviously this isn’t going to give the true perimeter of the circle, because no matter how many iterations you do, when you zoom in it remains jagged. It’s not a smooth curve, you’re taking these very inefficient steps along the circle. While that is true and ultimately the reason things are wrong, if you want to appreciate the lesson this example is teaching us, the claim of the example is not that any one of these approximations equals the curve. It’s that the limit of all of the approximations equals our circle, and to appreciate the lesson that this example teaches us, it’s worth taking a moment to be a little more mathematically precise about what I mean by the limit of a sequence of curves. And then my second claim is that this triangle AEP is the same, or at least congruent, to this triangle down here, ADB.Essentially this follows from the fact that this angle is the same as this angle, and then this length is the same as this length.And so if these two triangles are the same, then their side lengths must also be the same.So this means that the side length AB must be equal to the side length AC.
Let’s say we describe the very first shape, a square, as a parametric function. Something that has an input t, and it outputs a point in 2D space so that as t ranges from 0 to 1, it traces that square. I’ll call that function c_0. And likewise we can parametrize the next iteration with a function I’ll call c_1. As the parameter t ranges from zero up to one, the output of this function traces along that curve. This is just so that we can think of these shapes as instead being functions.
Now I want you to consider a particular value of t, maybe 0.2, and then consider the sequence of points that you get by evaluating the sequence of functions we have at this particular point. Next I want you to consider the limit as n approaches infinity of c_n of 0.2. This limit is a well-defined point in 2D space, in fact that point sits on the circle. And there’s nothing specific about 0.2, we could do this limiting process for any input t, and so I can define a new function that I’ll call c_infinity, which by definition at any input t is whatever this limiting value for all the curves is. So here’s the point: That limiting function c_infinity is the circle. It’s not an approximation of the circle, it’s not some jagged version of the circle, it is the genuine smooth circular curve whose perimeter we want to know. And what’s also true is that the limit of the lengths of all of our curves really is eight, because each individual curve really does have a perimeter of eight. And there are all sorts of examples throughout calculus when we talk about approximating one thing we want to know as a limit of a bunch of other things that are easier to understand, so the question at the heart here is why exactly is it not okay to do that in this example?
Maybe at this point you step back and say, you know it’s just not enough for things to look the same. This is why we need rigor, it’s why we need proofs, it’s why, since the days of Euclid, mathematicians have followed in his footsteps and deduced truths step-by-step from axioms forward. But for this last example I would like to do something that doesn’t lean as hard on visual intuition, and instead give a Euclid-style proof for the claim that all triangles are isosceles.
The way this will work is we’ll take any particular triangle and make no assumptions about it. I’ll label its vertices A, B and C. And what I would like to prove for you is that the side length AB is necessarily equal to the side length AC. Now to be clear, the result is obviously false, just in the diagram I’ve drawn you can visually see that these lengths are not equal to each other. But I challenge you to see if you can identify what’s wrong about the proof I’m about to show you. Honestly, it’s very subtle, and three gold stars for anyone who can identify it.
The first thing I’ll do is draw the perpendicular bisector for the line BC. So that means this angle here is 90 degrees, and this length is by definition the same as this length. And we’ll label that intersection point D. And then next I will draw the angle bisector at A, which means by definition this little angle here is the same as this little angle here. I’ll label both of them α. And we’ll say that the point where these two intersect is P.
And now like a lot of Euclid-style proofs, we’re just going to draw some new lines, figure out what things must be equal, and get some conclusions. For instance, let’s draw the line from P which is perpendicular to the side length AC, and we’ll label that intersection point E. And likewise we’ll draw the line from P down to the other side length AB, again it’s perpendicular, and we’ll label that intersection point F.
My first claim is that this triangle here, which is AFP, is the same, or at least congruent, to this More specifically, we can say they share a side length, and then they both have an angle $\alpha$, and both have an angle of 90 degrees, so it follows by the side-angle-angle congruent relation. Maybe my drawing is a little bit sloppy, but the logic helps us see that they do have to be the same.
Next I’ll draw a line from $P$ down to $B$, and then from $P$ down to $C$, and I claim that this triangle here is congruent to its reflection across that perpendicular bisector, which will be $CPD$. Again the symmetry maybe helps make this clear, but more rigorously they both have the same base, they both have a 90 degree angle, and they both have the same height, so it follows by the side-angle-side relation. So based on that first pair of triangles, I’m going to mark this side length here as being the same as this side length here, marking them with double tick marks. And based on the second triangle relation I’ll mark this side length here as the same as this line over here, marking them with triple tick marks.
And so from that, we have two more triangles that need to be the same, namely this one over here, and the one with corresponding two side lengths over here. And the reasoning here is they both have that triple ticked side, a double ticked side, and they’re both 90-degree triangles, so this follows by the 90-degree side-side-angle congruence relation. And all of those are valid congruence relations, I’m not pulling the wool over your eyes with one of those. All of this will basically be enough to show us why $AB$ has to be the same as $BC$. That first pair of triangles implies that the length $AF$ is the same as the length $AE$, those are corresponding sides to each other. I’ll just color them in red here. And then that last triangle relation guarantees for us that the side $FB$ is going to be the same as the side $EC$. I’ll kind of color both of those in blue.
And finally the result we want basically comes from adding up these two equations. The length $AF$ plus $FB$ is clearly the same as the total length $AB$. And likewise the length $AE$ plus $EC$ is the same as the total length $AC$. So all-in-all, the side length $AB$ has to be the same as the side length $AC$. And because we made no assumptions about the triangle, this implies that any triangle is isosceles. Actually, for that matter, since we made no assumptions about the specific two sides we chose, it implies that any triangle is equilateral. So this leaves us, somewhat disturbingly, with three different possibilities: All triangles really are equilateral, that’s just the truth of the universe, or you can use Euclid-style reasoning to derive false results, or there’s something wrong in the proof. But if there is, where exactly is it? The main problem with the sphere argument is that when we flatten out all of those orange wedges, if we were to do it accurately in a way that preserves their area, they don’t look like triangles, they should bulge outward. This is due to the fact that the width of the wedge varies according to a sine curve, and when we try to interlace all of the wedges from the northern hemisphere with those from the southern there’s a meaningful amount of overlap between those non-linear edges. This overlap persists as you take finer and finer subdivisions and it ultimately accounts for the difference between the false answer of π squared and the true answer of four π. It is similar to a rearrangement puzzle where you have a number of pieces and just by moving them around you can seemingly create area out of nowhere. The answer reveals itself if we carefully draw the edges of the triangle and zoom in close enough to see that the pieces don’t actually fit inside the triangle, they bulge out ever so slightly. However, when rearranged, they dent inward ever so slightly and that very subtle difference between the bulge out and the dent inward accounts for all of the difference in area. You have to be wary of lines that appear to be straight, but have not been explicitly confirmed as such. The geometry of a curved surface is fundamentally different from the geometry of a flat surface, and flattening out a sphere without losing geometric information is not possible. When applying limiting arguments that relate to pieces on a sphere, they must get smaller in both directions in order to be valid.
The subtlety of limiting arguments is demonstrated by a limit of jagged curves that approaches a smooth circular curve. The limiting curve is a circle and the limiting value for the length of the approximations is 8, but the limit of the lengths of the curves is not necessarily the same as the length of the limits of the curves.
The example of proving that all triangles are isosceles also demonstrates the need for care when applying limiting arguments. When constructing the angle bisector more carefully, the relevant intersection point sits outside of the triangle and the full side length AC is not equal to AE plus EC for many triangles. It is quite subtle, isn’t it? Visual intuition and proofs can be helpful in explaining complex concepts, however, they do not replace the need for critical thinking. In mathematics, it is important to be mindful of any hidden assumptions and potential edge cases.