What does this prove? Some of the most gorgeous visual "shrink" proofs ever invented

the star fails for any regular n-gon where n is a power of two. But not all is lost. again there is a third reduction argument that simply uses the shift property which takes care of all regular n-gons beyond hexagons. This argument actually dates back to 1946 and an article by Swiss mathematician Willy Scherrer. I only found

this

one after doing

some

research. Now here's an animation of Willy Scherrer's

shrink

ing plot. Enjoy. Isn't it absolutely beautiful? As a fairly easy challenge for you, figure out why Scherrer's reduction argument fails for the other regular polygons for hexagons and below. cool stuff, right?

Let's summarize

what

we have so far. so the 2d square grid only contains one type of regular polygons: squares. Alright. The 3D cubic grid contains three types of regular polygons: equilateral triangles, squares, and regular hexagons. I'm not sure if you like

this

stuff, but

what

about higher dimensions? No, nothing new happens there, since these higher dimensional hypercube grids also have the property of displacement. Everything we said for the 3D cubic grid is still true. so they contain equilateral triangles and squares and regular hexagons, but that's all. and, finally, end of the story? No, there's one more chapter left.

We'll answer a question that's been staring at us all school but probably none of us

ever

asked. Ready to dig deeper? Well, last chapter. Ready for

some

tricks? No, wait, come back. There are no masts or ladders against the wall. You'll really like this one, I promise. This is Willy Scherrer's 1946 article on polygon reduction. It's very short and, believe it or not, it

does

n't contain a single diagram. amazing, right? Just by looking at this, you would n

ever

suspect that there is some good

visual

evidence hidden in there. Mathematicians can be really strange sometimes. and it gets stranger Scherrer says that the only reason he published his proof is because another Swiss mathematician, his student Hugo Hadwiger, found a very good application for the contraction argument.

And what was Hadwiger's request? Let's keep moving. There, in the same magazine, just after Scherrer's article, we find Hadwiger's article titled Über die rationalen Hauptwinkel der Goniometrie. Everything clear? what's that? You do not speak German? okay, I forgive you :) Irgendwann mach ich auch mal ein ganzes Mathologer Video auf deutsch. versprochen. Anyway, Germans are not, there are many here who should look very familiar to you. Above are fractions of 2 pi and below I'm sure everyone will recognize the angles in that list 0 30 45 60 90 120 etc. These are the pretty angles we all know from school, the angles that appear in equilateral triangles and squares and regular hexagons, the kind of regular polygons we've been talking about.

The goodness of these angles is also shown in other ways. The

most

obvious thing is that all these angles are rational, they are simple fractions of a complete circle. but, of course, just as there are infinitely many rational numbers, there are also infinitely many rational angles. but what makes those particular rational angles especially nice is trigonometry. Those are the angles for which obtaining the exact values ​​of the trigonometric ratios sine, cosine and tan is really simple. and there is something very special about those trigonometric values ​​that you would have always known but perhaps never focused on. for each of those rational angles above at least one of the trigonometric ratios is also rational. all rational numbers are on the right.

It seems very familiar,

does

n't it? And now comes Hadwiger's idea. If we look at Scherrer's

shrink

ing polygons the right way, Scherrer's argument shows that the angles up there are the only rational angles in which at least one of the trigonometric ratios is also a rational number. So it's really just school stuff and something you probably never knew about. It was definitely new to me. Obviously this is an absolute gem, but unfortunately Hadwiger's article is a bit of a mess. and so, for you and for posterity, let me present to you a proof of Hadwiger's matologerization. I'll focus on the cosine part of the proof, the proof that the only rational angles with rational cosines are the ones that are there along with these obvious companions.

On the right, the cosines of all these simple angles are 0 1 minus 1 1/2 or - 1/2. and the claim is that the cosines of all other rational angles are irrational numbers. That's what we want to demonstrate. Well, just as a reminder, a rational anger is an angle like this. here p over q is a non-negative fraction less than 1 in lower terms, so the integers p and q have no common factor and p is less than q. so the fractions in our short list are just rational fractions with q equal to 2 3 4 and 6. So what we want to show is that all cosines of rational fractions with q equal to 5 7 8 9, etc., are irrational. 5 7 8 9 etc.

Note that these are exactly the numbers corresponding to the polygons for which the displacement property provides us with reduction arguments. not a coincidence :) okay, let me show you how the test works using a specific example. let me show you that the cosine of angle 3/8 of 360 is irrational. As you can probably guess, since the proof is based on our reduction argument, this will also be a proof by contradiction. So let's assume that the cosine of our angle is a rational number. now let's look at the multiples of our angle 2 times alpha, 3 times alpha, 4 times alpha, 5 times alpha and so on.

It turns out, and this is actually a high school thing, that all the cosines of the multiples of our angle alpha can be written as integer polynomials of cos alpha. Above I have listed the first of these polynomials. Let's take a look at the first one. there that one. cos of 2 alpha is equal to negative 1 plus 2 times cos squared alpha. Well, since we assume that cos alpha is a rational number r, that means that cos of 2 alpha is also a rational number, right? If r is a rational number, then negative 1 plus 2 r squared is also a rational number. in fact, in exactly the same way the other cosine polynomials show that all the cosines of these multiples of alpha are rational.

We will use this fact in a moment. Again, the specific angle we want to focus on is 3/8 of 360. Okay, what's next? Well, if we want to apply our reduction argument, we better introduce a regular polygon to reduce it. the denominator of our angle is 8 and so we start with this regular octagon centered at the origin of the XY plane and with outer radius 1. Here is our angle. here there are 2 times this angle and 3 times right and one more 4 times and so on. Because the numerator 3 and the denominator 8 are relatively prime, we reach all the corners of the octagon.

Okay, remember that the cosines of all the multiples of our angle are rational. Because we chose radius 1, this is the same as saying that the x coordinates of all the vertices of our octagon are rational. Now let's make everything whole. good? Rational numbers are good but integers are better. Suppose that all these fractions have a common denominator s. Okay, so scaling this image in s gives an octagon whose vertices have integer x coordinates. let's just do this. Okay, so we scale up and convert all those rational x coordinates to integer x coordinates. everything's fine? and now everything is ready to use shifts to perform the shrink build as before. so let's do this again quickly. do you get it?

The x coordinates of the original octagon are all integers, but because we are using offsets, that means that all the x coordinates of the second smaller octagon must also be integers. All of those types are also integers. and you can see where we're going with this, right? Exactly the same will happen with all subsequent smaller octagons. then all the infinite colored dots that we find in this way represent integers. and of course that is impossible since the space between the points reduces to zero, it is not possible for them all to be integers. And that is our contradiction. tada!! so we have used our contraction regular octagons to show that the cosine of angle 3/8 of 360 is irrational and similarly we can use contraction regular q-gons to show that the cosines of all these general rational angles are irrational. and the sine and tan part of the test are also easy.

Isn't it so beautiful? Okay, leave a note in the comments and let me know what you think of all this. Well, that's all from me for today. until next time.