Thursday, February 4, 2016

Slope: Based in part on Lesson 3-4 (Day 97)

Lecture 23 of David Kung's Mind-Bending Math is called "Banach-Tarski's 1 + 1 = 1." As Dave Kung points out, this is the strangest paradox of all, one that seems to refute kindergarten-level math.

The main result of this lecture is to demonstrate that it's possible to divide a sphere into finitely many parts and put them together to create two spheres, each as large as the original. That is, for each piece of the sphere, we can use an isometry to map it to a piece of one of two new spheres.

Kung points out that the proof of Banach-Tarski is very long -- so long that he devotes this entire lecture to just this one proof. I will only give parts of his proof -- the parts that are the most relevant to Geometry.

Kung begins his proof by considering groups of "words" consisting of only a and b. There exist countably infinitely many such words. Now he wants these words to correspond to transformations, with a representing a 120-degree rotation and b a 180-degree rotation. Then ab represents the composite of a rotation of 120 degrees following one of 180 degrees -- that is, the 180-degree rotation is done first. Also, aaa, also written a^3, is the composite of three 120-degree rotations, and b^2 is the composite of two 180-degree rotations. Both of these are equivalent to a rotation of 360 degrees, which is the identity, also written as 1.

Here's an aside to discuss the notation. Why do we write ab to perform b first? Notice that the U of Chicago text does the same thing -- a o b means "a following b." After all, (a o b)(x) = a(b(x)) as opposed to b(a(x)) -- it's just how function notation works. So when the U of Chicago writes the notation r_m o r_l (in the Two Reflections Theorems for Translations and Rotations), the text really means to reflect over line l first.

But this leads to another question -- why write ab for a o b? As Kung points out, the set of all words form a group, and group theorists tend to write the group operation as multiplication -- this is why they also write the identity as 1, the multiplicative identity. But this is an endless source of confusion for trig students, where sin^2 implies multiplication as sin^2(x) = (sin x)(sin x), while sin^-1 implies composition as (sin o sin^-1)(x) = x.

Anyway, notice that ababa is not the identity 1, as when we apply these to the sphere, the centers of rotation are different. Notice that in 3D, the center of rotation is an axis. The axis for a might as well be the actual axis of the globe (from North Pole to South Pole), while the axis for b can be any angle that is not a rational multiple of 360 degrees from the axis of a. Notice that 1 radian works. (My house is approximately 1 radian from the North Pole.)

Kung proceeds by assigning every word a color (navy blue, orange, purple) as follows. Let W be a word consisting of a and b as follows:

The empty word (which is 1) is colored navy.
If W is navy, then aW is orange.
If W is orange, then aW is purple.
If W is purple, then aW is navy.
If W is either orange or purple, then bW is navy.
If W is navy, then bW is either orange or purple via a more complicated rule.

The rest of the proof is complicated, so I'll try to summarize it quickly. Not only do the words get colors, but so do the points on a sphere. Just as with the Vitali set formation from yesterday, we color two points the same color if there exists a composite of a and b mapping one to the other (where there are uncountably many colors). Then we choose (Axiom of Choice) one point of each color. This gives us a new unmeasurable set to which we can apply the words -- these map each point to another point of the same color. Finally, another trick allows us to make two copies of each point based on the color of the words -- for example, it's possible to use bW to map both all the orange words to all the navy words, and all the purple words to all the navy words, so that there are two copies of navy. If we do this enough times we generate two copies of the entire sphere.

In the Quick Conundrum Kung shows a trick for tying a knot, one that I've seen before -- start with the arms already twisted, grab each end of a rope, then untwist the arms.

Today will be my traditionalist topic for the week. Notice that the proof of Banach-Tarski is ultimately based on rotations -- which traditionalists dislike. Indeed, it may be nice for Geometry teachers and traditionalists alike to watch Kung's lectures, starting from Lecture 19 (the start of the fourth disc), or possibly even Lecture 17, to see the relationship between translations, rotations, reflections and higher-level mathematics.

I want to mention a local high school student who's been in the news lately -- Cedrick Argueta, who is currently a senior. Last year as a junior, not only did he earn a 5 on the AP Calculus exam, but he earned a perfect score:

I  also want to comment on the continued backlash to the Common Core tests -- in particular, the compromise of using the SAT as the end-of-year test for juniors:

Each time I read an article like this one, or a similar story about Colorado's transition from ACT to SAT as the end-of-year test, I'm starting to believe that ACT is better than SAT. But even ACT doesn't avoid all of the problems mentioned in the article -- for example, the ACT also has an optional writing section (Problem #4 from the article).

This is what I wrote last year about slope:

Section 3-4 of the U of Chicago text is where slope -- that very important concept -- is defined. The definition in the text is simple enough and is typical for most high school math texts. Here it is, rendered in ASCII:

The slope of the line through (x_1, y_1) and (x_2, y_2), with x_1 != x_2, is (y_2 - y_1) / (x_2 - x_1).

But this isn't good enough for Common Core. Let's look at the relevant standard:

Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point).
[emphasis mine]

So we have to prove that slope works. It isn't good enough just to say that slope formula works just because we defined it to work. To see why, consider the following definition:

The favorite food of the class containing the student S is the food that S likes to eat the most.

And now you instantly see the problem with this "definition." Naturally, not every student in the class likes the same food. The food that student S loves to eat may be a food to which student T is indifferent and a food that absolutely disgusts student U. The "favorite food" of a class depends on which student we ask to name a favorite food. Simply writing the word "Definition:" and setting the phrase "favorite food" in bold doesn't magically force everyone to have the same favorite food.

Likewise, writing the word "Definition:" and setting the word "slope" in bold doesn't magically force the slope to be the same no matter which two points we choose. It could be that if we choose points (x_1, y_1) and (x_2, y_2), we get a different slope from if we choose  (x_3, y_3) and (x_4, y_4), just as if we choose student S we get a different favorite food (or favorite song) from if we choose student T instead of S.

In mathematics, we would say that "favorite food of a class" is not well-defined. In order for slope to be well-defined, we must prove that the slope is independent of which two points we choose to plug into the formula. We can't prove that the favorite food (or song) is independent of which student we choose -- indeed, it's trivial to find a counterexample: simply declare one student's favorite to be the class's favorite, and watch all the counterexample students call out how much they don't like the declared favorite!

And so the main theorem for today is to prove that slope is well-defined. We want to prove that slope of a line is independent of which two points we choose to plug into the formula. The trick for doing this will remind us of the proof of the Distance Formula.

Let's end this post with a joke. What's the anagram of Banach-Tarski?

The answer is Banach-Tarski Banach-Tarski! Stay tuned for the last paradox ... of all paradoxes.

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