Wednesday, February 3, 2016

The Pythagorean Theorem and the Distance Formula (Day 96)

Lecture 22 of David Kung's Mind-Bending Math is called "When Measurement Is Impossible." In this lecture, Dave Kung explains that some sets simply aren't measurable.

Kung begins by describing two different types of integration. One is called Riemann integration, named for 19th century German mathematician Bernhard Riemann -- this is what students in Calculus AB and BC are asked to calculate. Students in these classes begin by learning about Riemann sums, and then the shortcuts (power rule, u-substitution, etc.) that allow them to calculate integrals without performing the Riemann sums.

Another type of integration is called Lebesgue integration, named for the French mathematician Henri Lebesgue. Kung thinks of Riemann integration as dividing a region into vertical bars, but of Lebesgue integration as dividing the region into horizontal bars instead. Closely related to the Lebesgue integral is the Lebesgue measure. The Lebesgue measure of a set S equals the integral of a certain function -- f (x) = 1 if x is an element of set S and 0 otherwise.

If a function is Riemann-integrable, then it's also Lebesgue-integrable and the two values of the integral are equal. Yet there exists functions that are Lebesgue-integrable but not Riemann-integrable, such as the function f (x) = 1 if x is rational, 0 otherwise. This function has no Riemann integral, but it does have a Lebesgue integral equal to 0. The set of rationals has Lebesgue measure 0.

The Quick Conundrum involves stacking a wheel above two other wheels, placing a ruler on top, and then sliding the ruler forward. This causes the wheels to spin in the same direction as the ruler, but much faster. Kung says that this is similar to the paradox involving the spool of thread, seen in an earlier lecture.

Now we get to the main part of the lecture -- how can there be a non-measurable set? Kung states that we desire a measure that satisfies four properties. Let's think back to the Area Postulate, mentioned in Lesson 8-5 of the U of Chicago text:

Area Postulate:
a. Uniqueness Property: Given a unit region, every polygonal region has a unique area.
b. Rectangle Formula: The area of a rectangle with dimensions l and w is lw.
c. Congruence Property: Congruent figures have the same area.
d. Additive Property: The area of the union of two nonoverlapping regions is the sum of the areas of the regions.

As it turns out, the four desirable properties of measure that Kung mentions correspond almost exactly to these four parts of the Area Postulate! We want every region to have a measure -- the only difference is that Kung mentions that the measure should be nonnegative. Second, Kung's example is with 1D measure (linear, not square units), so he gives the 1D analog of the Rectangle Formula -- the Interval Formula, which is part of the Ruler Postulate (Point-Line-Plane in the U of Chicago).

Third, we want congruent figures to have the same area, but what does it mean to be congruent? That's easy -- there should exist an isometry mapping one to the other. So we see that Lebesgue used the idea of rigid transformations about a century before there was a Common Core!

We put these all together and come up with four reasonable properties of measure. But, as it turns out, the assumption that such a measure exists leads to a contradiction. As Kung points out, we've seen this several times before -- we come up with a set of reasonable properties for a system to have, and we prove that only some of the properties can be satisfied, not all. We think back to Godel's Incompleteness Theorem, Arrow's Impossibility Theorem, and mapping the globe onto a plane.

The counterexample involves what we call a Vitali set (Italian mathematician Giuseppe Vitali). Kung considers the interval [0, 1), but to make it easier, he places these numbers not on a straight line, but on a circle of circumference 1. He then colors the points such that any two points that differ by a rational have the same color -- this requires uncountably many colors. The Vitali set is formed by choosing (Axiom of Choice) one point of each color.

The Vitali set is uncountable, but the circle consists of countably many such sets -- indeed, Kung shows that simply rotating the Vitali set by a rational number gives another Vitali set, and the circle is the disjoint union of countably many such sets, one for each rational. Each of these infinitely many sets must have the same length since an isometry (a rotation) maps one to the other, yet they must all add up to 1, the circumference of the circle, a contradiction -- the sum of infinitely many zeros is 0, and the sum of infinitely many equal nonzero numbers is infinity. So the Vitali set is nonmeasurable.

So is the Area Postulate of Lesson 8-5 wrong? Notice that the text is careful to write that every polygonal region has a unique area. Without the word "polygonal," the region could be constructed similar to a Vitali set and the claim would be false.

By the way, you may be wondering, where is the MTBoS Week 4 post? Actually, I'm having a problem with Week 4. You see, the Week 4 prompt is supposed to be all about an actual lesson that we're teaching this week. As a sub, I'm not guaranteed to teach any math lessons this week. In fact, let's look at my subbing load this week:

Monday: Subbed in a special ed English/history class for the second day -- first day was last Friday.
Tuesday: Rejected an elementary school assignment. I'm not comfortable teaching the youngest students -- especially kindergartners, and the assignment was at a K-3 school. (I mentioned last year how some districts, including my own, separate K-3 and 4-5 at certain schools.)
Wednesday: Subbed in a high school biology class. This is another teacher who will be on paternity leave -- his lesson plans extend until February 19th. I'm at least assigned to cover this class for the rest of the week -- perhaps a biology specialist may be found to cover the next two weeks.

This bio teacher has sixth period conference, and I was assigned to cover for an Algebra II teacher (grandfathered for older students -- she also teaches Integrated Math II) who had to leave early for a dentist appointment. I ended up covering only half of the period. This class was learning about radicals, but there was a review worksheet on factoring, and I was able to help one student on this factoring worksheet.

Not only that, but today's biology class involved some math. They were working on a predator-prey model where they had to simulate the interaction between two species (sea lions and sardines). I was confused with how the question was worded, and so I was only able to help the students with graphing the data.

So I covered half a period of actual math, plus mathematical modeling in biology where I didn't fully understand the model. The bio teacher also has a period of AVID, and twice a week (Tuesdays and Thursdays) college tutors come in to help the students with different subjects. Sometimes I join in when I see them working on math. Otherwise I'm locked into the bio job, so there's no chance that I'll cover a real math class this week.

Putting it all together, this doesn't sound appropriate for a MTBoS post. I'll have to decide soon what I'm going to do about this week's post.

Now let's get back to Geometry class. It's been some time since I mentioned Theoni Pappas and her Mathematical Calendar, but I just can't help mentioning today's question:

(Figure shows AE and BD intersecting at C with right angles ABC, EDC.)

Triangle ABC can be shown to be similar to Triangle EDC most easily by:

(1) SSS~
(2) SAS~
(3) AA~
(4) not possible

The answer is (3) -- and notice that today's date is the 3rd. Pappas likes to set up the questions so that the correct answer is the date. Today's question fits perfectly as a warm-up, since it follows directly from yesterday's lesson on SSS~ and AA~.

This is what I wrote last year about today's lesson:

Lesson 8-7 of the U of Chicago text is on the Pythagorean Theorem, and Lesson 11-2 of the same text is on the Distance Formula. I explained yesterday that I will cover these two related theorems in this lesson.

The Pythagorean Theorem is, of course, one of the most famous mathematical theorems. It is usually the first theorem that a student learns that is named for a person -- the famous Greek mathematician Pythagoras, who lived about 2500 years ago -- a few centuries before Euclid. I believe that the only other named theorem in the text is the Cavalieri Principle -- named after an Italian mathematician from 400 years ago. Perhaps the best known named theorem is Fermat's Last Theorem -- named after the same mathematician Fermat mentioned in yesterday's lecture. (We discuss some other mathematicians such as Euclid and Descartes, but not their theorems.)

It's known that Pythagoras was not the only person who knew of his named theorem. The ancient Babylonians and Chinese knew of the theorem, and it's possible that the Egyptians at least knew about the 3-4-5 case.

We begin with the proof of the Pythagorean Theorem -- but which one? One of my favorite math websites, Cut the Knot (previously mentioned on this blog), gives over a hundred proofs of Pythagoras:

The only other theorem with many known proofs is Gauss's Law of Quadratic Reciprocity. Here is a discussion of some of the first few proofs:

Proof #1 is Euclid's own proof, his Proposition I.47. Proof #2 is simple enough, but rarely seen. Proofs #3 and #4 both appear in the U of Chicago, Lesson 8-7 -- one is given as the main proof and the other appears in the exercises. Proof #5 is the presidential proof -- it was first proposed by James Garfield, the twentieth President of the United States. I've once seen a text where the high school students were expected to reproduce Garfield's proof.

So far, the first five proofs all involve area. My favorite area-based proof is actually Proof #9. I've tutored students where I've shown them this version of the proof. Just as the Cut the Knot page points out, Proof #9 "makes the algebraic part of proof #4 completely redundant" -- and because it doesn't require the students to know any area formulas at all (save that of the square), I could give this proof right now. In fact, I was considering including Proof #9 on today's worksheet. Instead, I will wait until our next activity day on Friday to post it.

But it's the proof by similarity, Proof #6, that's endorsed by Common Core. This proof has its own page:

Here is Proof #6 below. The only difference between my proof and #6 from the Cut the Knot webpage is that I switched points A and C, so that the right angle is at C. This fits the usual notation that c, the side opposite C, is the hypotenuse.

Given: ACB and ADC are right angles.
Prove: BC * BC + AC * AC = AB * AB (that is, a^2 + b^2 = c^2)

Statements                                Reasons
1. ADC, ACBCDB rt. angles   1. Given
2. Angle A = A, Angle B = B     2. Reflexive Property of Congruence
3. ADCACBCDB sim. tri.     3. AA Similarity Theorem
4. AC/AB = AD/AC,                 4. Corresponding sides are in proportion.
    BC/AB = BD/BC
5. AC * AC = AB * AD,           5. Multiplication Property of Equality
    BC BC = AB * BD
6. BC * BC + AC * AC =         6. Addition Property of Equality
    AB * BD + AB * AD
7. BC * BC + AC * AC =         7. Distributive Property
    AB * (BD + AD)
8. BC * BC + AC * AC =         8. Betweenness Theorem (Segment Addition)
    AB * AB

I mentioned before that, like many converses, the Converse of the Pythagorean Theorem is proved using the forward theorem plus a uniqueness theorem -- and the correct uniqueness theorem happens to be the SSS Congruence Theorem (i.e., up to isometry, there is at most one triangle given three side lengths). To prove this, given a triangle with lengths a^2 + b^2 = c^2 we take another triangle with legs a and b, and we're given a right angle between a and b. By the forward Pythagorean Theorem, if the hypotenuse of the new triangle is z, then a^2 + b^2 = z^2. (I chose z following the U of Chicago proof.) Thenz^2 = c^2 by transitivity -- that is, z = c. So all three pairs of both triangles are congruent -- SSS. Then by CPCTC, the original triangle has an angle congruent to the given right angle -- so it's a right triangle. QED

Interestingly enough, there's yet another link at Proof #6 at Cut the Knot, "Lipogrammatic Proof of the Pythagorean Theorem." At that link, not only is Proof #6 remodified so that it's also an area proof (just like Proofs #1-5), but, as its author points out, slope is well-defined without referring to similar triangles!

The Common Core Standards only require that the Pythagorean Theorem be proved using similarity -- not the concept of slope (which we'll cover here on the blog tomorrow). So now I'm wondering whether it might be easier for the students to understand this derivation of slope.

It was David Joyce who pointed out that slope requires similarity to prove. He also criticized the area-based proof of the Pythagorean Theorem given in the Prentice-Hall text -- but this is because he wanted the Parallel Tests and Consequences to be taught first. (According to Cut the Knot, the Pythagorean Theorem is equivalent to the Parallel Postulate.)

For now, I think I'll stick to my plan to use similarity to prove slope -- but my proof may still be based on the one given at the above link.

Thus ends this post. Stay tuned for when Kung solves the mystery of 1 + 1 = 1...

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