HCSSiM Workshop day 5
We examined finite sets with addition laws and asked whether they were the “same”, which for now meant their addition table looked the same except for relabeling. Of course we’d need the two sets to have the same size, so we compared and We decided they weren’t the same, but then when we did it for and and decided those were. We eventually decided it worked the second time because the moduli are relatively prime.
We essentially finished by proving the base case of the Chinese Remainder Theorem for two moduli, which for some ridiculous reason we are calling the Llama Remainder Theorem. Actually the reason is that one of the junior staff (Josh Vekhter) declared my lecture to be insufficiently silly (he designated himself the “Chief Silliness Officer”) and he came up with a story about a llama herder named Lou who kept track of his llamas by putting them first in groups of n and then in groups of m and in both cases only keeping track of the remaining left-over llamas. And then he died and his sons were in a fight over whether someone stole some llamas and someone had to be called in to arbitrate. Plus the answer is only well-defined up to a multiple of mn, but we decided that someone in town would have noticed if an extra mn llamas had been stolen.
After briefly discussing finite sets and their sizes, we defined two sets to have the same cardinality if there’s a bijection from one to the other. We showed this condition is reflexive, symmetric, and transitive.
Then we stopped over at the Hilbert Hotel, where a rather silly and grumpy hotel manager at first refused to let us into his hotel even though he had infinitely many rooms, because he said all his rooms were full. At first when we wanted to just add us, so a finite number of people, we simply told people to move down a few times and all was well. Then it got more complicated, when we had an infinite bus of people wanting to get into the hotel, but we solved that as well by forcing everyone to move to the hotel room number which was double their first. Then finally we figured out how to accommodate an infinite number of infinite buses.
We decided we’d proved that has the same cardinality as itself. We generalized to having the same cardinality as and we decided to call sets like that “lodgeable,” since they were reminiscent of Hilbert’s Hotel.
We ended by asking whether the real numbers is lodgeable.
Here’s a motivating problem: you have an irregular hexagon inside a circle, where the alternate sides are the length of the radius. Prove the midpoints of those sides forms an equilateral triangle.
It will turn out that the circle is irrelevant, and that it’s easier to prove this if you actually Circle is entirely prove something harder.
We then explored the idea of size in the complex plane, and the operation of conjugation as reflection through the real line. Using this incredibly simple idea, plus the triangle inequality, we can prove that the polynomial
has no roots inside the unit circle.
Going back to the motivating problem. Take three arbitrary points A, B, C on the complex plane (i.e. three complex numbers), and a fourth point we will assume is at the origin. Now rotate those three points 60 degrees counterclockwise with respect to the origin – this is equivalent to multiplying the original complex numbers by Call these new points A’, B’, C’. Show that the midpoints of the three lines AB’, BC’, and CA’ form an equilateral triangle, and that this result also is sufficient to prove the motivating problem.