*Mind-Bending Math*is called "Impossible Sets." But Dave Kung tells us that he is discussing one impossible set in particular.

I've mentioned the British mathematician Bertrand Russell a few times on the blog before. Kung begins by describing a few details of Russell's life. In addition to a mathematician, Russell was also a peace activist and a Nobel Prize laureate. But in mathematics, he is especially known for a paradox, Russell's Paradox:

-- Does the set of all sets that don't contain themselves contain itself?

Kung explains that most sets don't contain themselves as elements -- for example, the set of all prime numbers is not itself a number, much less a prime number. But the set of abstract ideas is an abstract idea, so it does contain itself as an element.

The professor then points out that Russell's Paradox is the same as the Barber's Paradox, except replacing "shaves" with "contains (as an element)." The barber shaves himself if and only if it doesn't shave himself, and the Russell set contains itself if and only if it doesn't contain itself.

So Kung devotes the rest of the lecture to the mathematicians' quest to come up with a list of axioms for set theory -- one which avoids Russell's Paradox. He compares the axiomatization of set theory to another famous axiomatization -- that of Euclidean geometry. Recall that an "axiom" in set theory just means the same thing as a "postulate" in Geometry -- it's a statement that is assumed, not proved.

According to Kung, a list of axioms should first be consistent -- that is, it doesn't lead to any contradictions such as Russell's Paradox. Second, it should be complete -- that is, everything that is true should be provable. Third, it should not be redundant -- that is, an axiom that is provable from the other axioms should be proved (as a theorem), not assumed (as a postulate).

For centuries, some mathematicians suspected that Euclid's Fifth Postulate -- best known as Uniqueness of Parallels, or Playfair's Parallel Postulate -- was in fact redundant. But all attempts to prove the theorem failed. Instead, non-Euclidean geometry was discovered. Kung mentions spherical geometry -- or should I say,

*elliptic*geometry (and we've discussed the subtle differences between elliptic and spherical geometry here on the blog) -- as well as hyperbolic geometry. Of course he uses a sphere as a model of elliptic geometry, and he actually had a physical (embroidered) model of hyperbolic geometry. Kung demonstrated on his hyperbolic model that the sum of the angles of a triangle were clearly less than 180 -- and that as the triangles grew in area, the sum of their angles actually decreased.

After Russell discovered his paradox, he and his teacher, Alfred Whitehead came up with something called type theory, in which each set was assigned a type number. Each set was assigned a higher type than the types of its elements, meaning that no set could contain itself as an element, thereby avoiding Russell's Paradox. Russell and Whitehead proved many results using type theory in their famous work

*Principia Mathematica*, named after Isaac Newton's

*Principia*. I've actually mentioned Russell and Whitehead earlier on the blog -- they were the ones who took hundreds of pages to prove something as simple as 1 + 1 = 2. Now we can begin to appreciate why the proof was so long -- numbers had to be represented as

*sets*, and the authors had to make sure that the sets were of the correct type before they could use them. (That didn't stop one traditionalist from invoking the name of Whitehead to argue that things should be as

*simple*as possible!)

Nowadays, we don't use Russell and Whitehead's type theory. Instead, as Kung explains, we use something called Zermelo-Fraenkel set theory, named for two German mathematicians. And just as one of Euclid's axioms (the Parallel Postulate) was more controversial than the rest, so one of the axioms used by Zermelo and Fraenkel was controversial -- the Axiom of Choice.

The Axiom of Choice sounds as if it should be obvious. If we have a collection of sets, it should be possible to choose one element from each set and combine them into a larger set. Yes, this does work for finite sets, but not necessarily for

*infinite*sets unless we use the axiom. Kung mentions an example of this -- it's easy to choose one shoe from each of infinitely many pairs of shoes without the Axiom of Choice (always choose the left shoe), but it's impossible to choose one

*sock*from each of infinitely many pairs of socks (where we can't distinguish left from right) without AC. It's impossible to prove the Axiom of Choice from the other ZF axioms, just as we can't prove Euclid's Fifth Postulate from the first four.

The Quick Conundrum is all about falling leaves. Kung discovers that in the fall, leaves don't just change colors and fall off the tree. In reality, the tree stops giving the leaves nutrients -- that is, the tree

*pushes*its leaves off.

Yes, I know that I already discussed probability last week, after having just watched Kung's lectures on probability. But I'm discussing it again this week because probability has been in the news lately -- specifically, the probability of winning Powerball tonight. This will be the first billion-dollar jackpot in world history.

Sometimes I wonder whether it's a good idea to discuss gambling in classrooms where most of the students are underage. Indeed, many probability texts use a strange-sounding phrase, "number cubes," just to avoid mentioning the word

*dice*to students who are too young to gamble. I've decided that I will mention the lottery here on the blog anyway. If we want to convince our students that what we are teaching them has relevance in the real world, we must seize on such opportunities as the largest ever Powerball jackpot to make the connection.

It is well known that the lottery has a negative expected payout. It's easy to calculate this -- for every $2 Powerball ticket purchased, only about $1 is paid back to winners. So the expected payout per ticket is negative one dollar. The other dollar goes to system maintenance as well as the education fund in many states. It's because of this that the lottery is derided as a "tax on stupidity" -- if people were intelligent, they wouldn't ever play a game with a negative expectation such as the lottery.

I saw someone calculate that with the jackpot so high, perhaps for the first time ever, the lottery has a slight positive expectation. This person factored in the fact that the cash value payout is always less than the announced jackpot, as well as the federal and state taxes that must be paid. There are two more things the poster didn't account for -- the smaller prizes that players can win if they match fewer than all the numbers (which would make the expectation even more positive), and the fact that if multiple players match all the numbers, they only split the jackpot.

This is harder to calculate. In general, as the jackpot rises, many more people participate, thereby raising the probability of a split jackpot. But, as Kung mentioned in his probability lectures last week, humans are terrible at making random choices. Many people choose birthdays for their winning numbers -- meaning that all the numbers chosen are 31 or less. Now that the Powerball has raised the total number of balls to 75, birthday players limit themselves to less than half the range. It could be that for tonight's record-breaking jackpot, birthday players have a negative expectation as they're likely to split the jackpot, while those who choose numbers from the full range have a slightly positive expectation, as they're more likely to scoop the entire jackpot.

Still, the Powerball has its detractors. I've seen some people say that the probability of winning is so small that it should be rounded off to exactly zero. I disagree with this -- if the probability were zero, then no one in the history of the lottery would have ever won. I only round off to zero things that have never happened in the history of the world (if I round them off at all). But it is correct to say that lottery winners are a statistically

*insignificant*subset of the population.

Oh, and the Powerball jackpot is only enough to give about four or five dollars to each American...

I also saw someone say that you're more likely to be elected President of the United States than you are to win the lottery. This is also false -- there have been only 44 presidents (counting Grover Cleveland twice), yet more than 44 people have won the Powerball jackpot. But it is correct to say that you're more likely to be struck by lightning

*twice*than you are to win the lottery. It's definitely a small probability -- one mathematicians would describe as an

*epsilon*. Of course, you're still more likely to win the lottery than you are to choose a rational number randomly from the unit interval (which is exactly zero).

Here's what the mathematician R.D. Silverman has to say about the lottery:

http://www.mersenneforum.org/showthread.php?t=10795

On the other hand, if you are a

**competent**mathematician, you will

realize that different people have different economic utility functions

and that the utility of winning a large amount of money more than

compensates for the small

**expected value loss**.

[emphasis Dr. Silverman's]

And if the lottery is good enough for Dr. Silverman, it's good enough for me. To say that the lottery is a "tax on stupidity" is to imply that a person shouldn't play

*any*game unless there is a positive expected value of winning money. Very few games exist -- since after all, the game owner has to make a living somehow! But there actually are a few gambling games with positive expected value -- such as certain versions of poker. Professional poker players exist because they are extremely skilled at making the correct choices when playing cards.

And besides, if there were no lottery, we'd have to find another example of probability to mention in our math classes today -- one that might not interest students as much. Before leaving the topic of the lottery, let me post the following YouTube video. In honor of the first billion-dollar jackpot, here's a song from

*Square One TV*, performed by the Fat Boys -- "One Billion Is Big":

But of course, I'm not teaching probability today on the blog, but Geometry. It's time for us to get to the Centroid Concurrency Theorem.

There are several theorems to prove beforehand. First, we must prove the Midpoint Formula -- and we must do so

*without*using the Slope or Distance Formulas. We wish to calculate the midpoint of the segment whose endpoints are (

*a*,

*b*) and (

*c*,

*d*).

We begin with a right triangle without coordinates. Let's take the perpendicular bisector of one of its legs -- since this is a right triangle, this leg is perpendicular to the other leg. So we have two lines perpendicular to the first leg -- its perpendicular bisector and the other leg. Thus by the Two Perpendiculars Theorem, the perpendicular bisector of one leg is parallel to the other leg.

But notice that by the Midpoint Connector Theorem (which we already proved on the blog), the line joining the midpoints of two sides of a triangle is parallel to the third side. So the line joining the midpoints of one leg and the hypotenuse of the triangle is parallel to the other leg.

This means that we have

*two*lines, each passing through the midpoint of the first leg, that are parallel to the second leg -- the perpendicular bisector and the midpoint. But by Playfair, there is only

*one*line passing through the midpoint of the first leg and parallel to the second. Thus the perpendicular bisector and the midpoint connector are really the same line -- that is, the perpendicular bisector of the first leg passes through the midpoint of the hypotenuse.

Of course, it didn't matter which leg we started with above. The conclusion is that the perpendicular bisectors of

*both*legs pass through the midpoint of the hypotenuse. And of course, the perpendicular bisector of the

*hypotenuse*trivially passes through its own midpoint. So

*all three*perpendicular bisectors pass through the midpoint of the hypotenuse -- that is, the midpoint of the hypotenuse is the

*circumcenter*of the right triangle! (I know, it's strange for me to start talking about circumcenters when we're supposed to be finding the

*centroid*, but bear with me.)

Once we have this result, we can find the midpoint of the segment from (

*a*,

*b*) to (

*c*,

*d*) as follows. We set up a right triangle whose vertices are (

*a*,

*b*), (

*a*,

*d*), and (

*c*,

*d*). (This is the same trick that we use to derive the Distance Formula.) The perpendicular bisector of the base is halfway between the lines

*x*=

*a*and

*x*=

*c*, so it's

*x*= (

*a*+

*c*) / 2, and the perpendicular bisector of the height is halfway between the lines

*y*=

*b*and

*y*=

*d*, so it's

*y*= (

*b*+

*d*) / 2. Then the midpoint of the hypotenuse (which goes from (

*a*,

*b*) to (

*c*,

*d*)) is where those perpendicular bisectors intersect -- ((

*a*+

*c*) / 2, (

*b*+

*d*) / 2).

Now we can calculate the centroid of any triangle -- but we're going to follow a trick established in the Houghton Mifflin Harcourt Integrated Math I text. We will choose vertices for the triangle so that two of the medians turn out to be parallel to the axes. Here's an example of such a triangle (inspired by one of the triangles I placed on last week's Euler Line worksheet):

*A*(16, 16),

*B*(0, 0),

*C*(8, 32). We see that the midpoint of

*C*is (8, 32), the median from

*C*is

*x*= 8. And the midpoint of

*is (4, 16), and since*~~BC~~

*A*is (16, 16), the median from

*A*is

*y*= 16. So the centroid must have the coordinates (8, 16). Notice that is this case, finding the median from

*B*isn't too terrible -- the midpoint of

*B*is (0, 0), so the equation of the median is

*y*= 2

*x*. But it's much easier to use the medians from

*A*and

*C*.

But you may notice that the Euler Line worksheet gives a different method to find the centroid -- simply take the average of all of the coordinates. Notice that this method works for the triangle we gave above -- ((16 + 0 + 8) / 3), (16 + 0 + 32) / 3) = (8, 16). So now we might wonder, why doesn't the HMH text simply give that formula rather than force us to find the equations of two medians?

I suspect it's because that formula simply hasn't been

*proved*yet. The proof of the general formula for the centroid requires some rather nasty algebra, and so we don't prove it. So in order to avoid mentioning formulas before proving them, we just skip the formula altogether.

So now you ask, why did I set up the Euler line worksheet the way that I did? Recall that all of those worksheets were set up mostly for the benefit of the student I was tutoring. My student was working out of a different text -- the Glencoe text -- and that text asked him to calculate centroids when none (or at most one) of the medians were parallel to the axes. I don't recall how the Glencoe text directed the students to find centroids, but when I saw him struggle with finding the equations of the medians and solving the system to calculate their intersection, I couldn't resist simply telling him to find the average of the coordinates.

Without providing the Centroid Formula, it's difficult to give a simple example to calculate the full Euler line. To make the median easy to find, we need to choose a triangle with a horizontal median and a vertical median. This will never be a triangle whose circumcenter is easy to find -- a right triangle with a horizontal

*side*and a vertical

*side*. So we can never have a triangle where all three centers on the Euler line are easy to calculate.

Well, at least I can provide the proof of the Centroid Theorem, as I gave last year:

Median Concurrency Theorem:

The three medians of a triangle meet at a point G, called the centroid of the triangle. On each median, the distance of G to the vertex is twice the distance of G to the midpoint of the opposite side.

Given: In triangle ABC, C', B', and A' are midpoints of AB, AC, and BC respectively.

(This way, the medians are AA', BB', and CC' as per Wu's notation.)

Prove: CC' meets BB' at a point G such that BG = 2GB'.

(Without loss of generality, the same proof will work if we change B or C to A.)

Proof:

Statements Reasons

1. bla, bla, bla 1. Given

2. C'B' | | BC, C'B' = 1/2 BC 2. Midpoint Connector Theorem (ABC)

3. Let M, N be midpoints of BG, CG 3. Ruler Postulate (Point-Line)

4. MN | | BC, MN = 1/2 BC 4. Midpoint Connector Theorem (GBC)

5. C'B' | | MN 5. Transitivity of Parallelism Theorem

6. C'B' = MN 6. Transitive Property of Equality

7. MNB'C' is a parallelogram 7. Sufficient Conditions (pgram test), part (d)

8. MB' and NC' bisect each other 8. Properties (pgram consequence), part (c)

9. BM = MG, MG = GB' 9. Definition of midpoint

10. BM = GB' 10. Transitive Property of Equality

11. BM + MG = GB' + GB' 11. Addition Property of Equality

12. BG = 2GB' 12. Segment Addition (Betweenness)

I was considering taking the questions for today's worksheet directly from Lesson 23.3 of the HMH text, just as I took yesterday's from Lesson 23.2. But Lesson 23.3 doesn't prove the Midpoint Formula (since it assumes that the students already know it), plus it squeezes in questions to calculate orthocenters as well. which are never easy.

(Actually, the orthocenter of a right triangle is easy to find -- it's just the vertex of the right angle! I suspect that HMH wanted to avoid this special case. But then you can argue that the circumcenter of a right triangle is also trivial to find, as it's just the midpoint of the hypotenuse.)

Instead, I decided to post proofs of the Midpoint Formula and the Centroid Theorem, and then a few examples of its use.

At the end of the lecture, Kung points out how Hilbert, Russell, and Whitehead all wanted to come up with axioms to prove all of mathematics. But as it turns out, this is doomed to fail -- just as I'm doomed not to win the Powerball tonight...

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