*Mind-Bending Math*is called "Cantor's Infinity of Infinities." In this lecture, Dave Kung seeks to answer whether there are any infinities besides the two we know about -- the cardinality of the set of all natural numbers and the cardinality of the set of all real numbers.

But first, Kung describes how

*not*to find new infinities. First, he points out that a small interval of the reals, such as the set of all reals between 0 and 1, has the same cardinality as the set of all reals. A 1-1 correspondence exists between the unit interval and the entire real line, and he demonstrates this by using Geometry, in fact. Here's how Kung does it -- the unit interval is the diameter of a semicircle and the real line is parallel to the interval and tangent to the semicircle. Starting at any point on the real line, we draw a line from that point to the center of the semicircle. At the point where the line intersects the semicircle, we draw a line passing through the intersection point and perpendicular to the unit diameter. Then this is the image of the original point from the real line. This mapping is clearly not an isometry, but it's a 1-1 correspondence, which is all that matters for cardinality.

So any interval, no matter how small, has the same uncountable cardinality as all the reals. Yet the set of all rationals, which span the entire number line, has a smaller countable cardinality. This may seem strange, yet Kung proves it. If the interval has a small length, which Kung labels by the Greek letter

*epsilon*(the traditional name for a small positive number), then we can show that the rationals can be covered by intervals whose lengths add up to epsilon (or less). Here's how he does it -- since the rationals are countable, they fit into Hilbert's Hotel. So take the rational in room 1 and cover it by an interval of length epsilon/2, then take the rational in room 2 and cover it by an interval of length epsilon/4, then take the rational in room 3 and cover it by an interval of length epsilon/8, and so on. Then we can use the trick given back on the day Kung covered Zeno's Paradox -- the lengths all add up to epsilon (actually less than epsilon, since they overlap), which is small.

Kung then ties this back to probability -- given a unit interval from 0 to 1, the probability of choosing a number between 0 and epsilon is epsilon. Yet we covered all the rationals with intervals of length totaling less than epsilon. So the probability of choosing a rational between 0 and 1 is smaller than any small positive number -- in other words, the probability is zero. If you think about it, you can see why this actually makes sense -- let's form a random real by taking a spinner that goes from 0 to 9, spin it (countably) infinitely many times, and then letting those be the digits of our real number. If the spinner is

*random*, what is the probability of it showing a

*pattern*, as in the repeating decimal pattern that's necessary for the number to be rational? Of course it's zero.

After all of this, we still have only two different infinities. So Kung finally shows us how to get some more infinities -- by using the power set. If S is a set, then the power set P(S) is simply the set of all subsets of S. This power set always includes both the empty set and the improper subset, S itself. So here is an example of a power set:

S = {A, B, C}

P(S) = {{}, {A}, {B}, {C}, {A, B}, {A, C}, {B, C}, {A, B, C}}

Now here's the thing -- for any set S, N(P(S)) > N(S). That is, the power set of any set always contains more elements than the set itself. For finite sets this is obvious -- in our example above, S has three elements and P(S) contains 2^3, or eight, elements. The famous set theorist Georg Cantor proved that this holds for infinite sets as well. Cantor's proof is simply another version of his Diagonal Argument -- or Dodgeball, as Kung calls it. For every infinite cardinality, there exists a larger cardinality -- that of its power set. And so we see that far from being only one, or even two infinities, there are in fact

*infinitely*many infinities!

The Quick Conundrum contained an arrow stuck in a bottle -- how did it get there, considering that the arrow is too big to fit in the hole? We see it, yet we can't believe it -- just as Cantor saw all of his many infinities, but couldn't believe it.

We now move on to the next of our concurrency theorems -- the Incenter Theorem. This theorem states that the angle bisectors of a triangle are concurrent as they meet at a point called the incenter -- the center of the inscribed circle.

Last year, I created worksheets for the Circumcenter and Orthocenter Theorems, but not for the Incenter or Centroid Theorems. Instead, I squeezed in proofs of those theorems here on the blog right at the end of the first semester. This year, I allowed a few extra days to make sure that I can do the two missing theorems justice.

It doesn't help, of course, that except for a brief mention of the Circumcenter Theorem, none of the concurrency theorems appear in the U of Chicago text. So instead, I decided to look at another text -- the Houghton-Mifflin Harcourt Integrated Math I text that I've seen at the district where I work.

The concurrency theorems appear in Module 23 (out of 24). So notice that this post is

*not*tagged with the label "subbing," since this isn't a lesson that I taught in any classroom. Since this module appears so late in the HMH text, classrooms will either cover it very late in the year, or -- and I suspect more likely -- not at all.

Lesson 23.1 of the HMH text is on the circumcenter, Lesson 23.2 is on the incenter, and Lesson 23.3 is on the centroid. Notice that I'm covering these three triangle centers in the same order in this text -- the circumcenter yesterday, the incenter today, and the centroid tomorrow. Now you may ask, what happened to the orthocenter? Well, HMH combines the orthocenter with the centroid in Lesson 23.3, whereas I combined it with the circumcenter on yesterday's worksheet.

Now here's the thing I looked for in the HMH text -- do the lessons on the triangle centers require use of the coordinate plane? Well, it depends on which center. The incenter lesson is

*never*presented on a coordinate plane, as it's hard to find angle bisectors on the coordinate plane without trigonometry. For the circumcenter, HMH only uses right triangles with sides parallel to the coordinate axes. Then the perpendicular bisectors of the legs are also parallel to the axes, so it's easy to find the circumcenter.

As for the centroid, HMH sets up the triangles so that two of the

*medians*are parallel to the coordinate axes! This means that none of the

*sides*are parallel to the axes, but this is more convenient for finding the centroid. But the orthocenter is the trickiest of all. The only way for a triangle to have two altitudes parallel to the axes is for it to be a right triangle -- but in that case, each leg is in fact an altitude to the other leg. This implies that the orthocenter of a right triangle is simply the vertex of its right angle. HMH wanted to avoid this special case -- lest the students believe that the orthocenter of

*any*triangle is always a vertex. So the text was forced to use oblique triangles -- although to make it easier, one side can still be parallel to an axis.

Notice that when I said I want to avoid the coordinate plane in the first semester, I really mean to avoid the Slope and Distance Formulas. We can't avoid the Slope Formula for the orthocenters given in the text, but we do avoid it for the circumcenters. As for the centroids, we don't need the Slope Formula, but we do need the Midpoint Formula.

Now the Midpoint Formula isn't a difficult formula at all -- just take the average of the coordinates of the vertices of the segment. The problem is that

*proving*the Midpoint Formula requires both the Slope and Distance Formulas -- at least if we use the proof given in Lesson 11-4 of the U of Chicago text.

I think I'll save a proof of the Midpoint Formula for tomorrow, since we need it for tomorrow's Centroid Theorem, but it has nothing to do with today's Incenter Theorem. Instead, I include some few questions taken directly from the HMH text. There are a mixture of questions where students have to fill in a value and those where they must complete the proofs.

There's one problem that I must change from that text -- the proof of the Incenter Theorem itself. I wrote a version of the proof last year on the blog (but not on any worksheet):

Angle Bisector Concurrency Theorem:

The three angle bisectors of a triangle meet at a point, called the incenter of the triangle. The incenter is the unique point equidistant from the three sides.

Given: Ray

*AE*bisects angle

*BAC*, Ray

*BD*bisects angle

*ABC*, Rays

*AE*and

*BD*intersect at

*I*

Prove: Ray

*CI*bisects angle

*ACB*,

*I*equidistant from all three sides

Proof:

Statements Reasons

1. Ray

*AE*bisects angle

*BAC*, ... 1. Given

2.

*I*equidistant from

*AC*and

*AB*2. Angle Bisector Theorem

3.

*I*equidistant from

*BA*and

*BC*3. Angle Bisector Theorem

4.

*I*equidistant from

*CA*and

*CB*4. Transitive Property of Equality

5. Ray

*CI*bisects angle

*ACB*5. Converse of Angle Bisector Theorem

The HMH text's version of the proof places points on each of the three sides (

*X*,

*Y*, and

*Z*) to make it easier to see what exactly it means to be "equidistant from" the sides. But here's the problem with the paragraph proof that the text provides -- the first line of the proof is:

Let

*P*be the incenter of Triangle

*ABC*...

In other words, to prove that the incenter exists, we begin by assuming -- that the incenter exists! No proof should ever begin by assuming what we want to prove! Otherwise, what's to stop me from doing the following?

I've discovered that every quadrilateral has a point which is equidistant from its four sides. Since I'm the first to discover it, I'll name it after myself -- the

*davidcenter*. This implies that a circle can be inscribed in every quadrilateral. Yes, it may appear that it's impossible to inscribe a circle inside some quadrilaterals, such as a rectangle (that isn't a square -- and no, an ellipse doesn't count). Yet I can prove that there really is a circle inscribed in the rectangle. Here is my proof:

Given: Quadrilateral

*ABCD*

Prove: There exist

*P*(interior) and

*W*,

*X*,

*Y*,

*Z*(one on each side) such that

*PW*=

*PX*=

*PY*=

*PZ*.

Proof:

Let

*P*be the davidcenter of

*ABCD*. Then

*PW*=

*PX*=

*PY*=

*PZ*. QED

Of course this proof is invalid -- but it's no less silly than the incenter proof given in the text. So of course I corrected this on my new worksheet. (Oh, and about that davidcenter -- since it doesn't really exist, let's pretend that I was naming it after David Kung all along...)

Speaking of David Kung, he ends today's lecture with a quote from yet another David -- Hilbert: "No one will drive us from the paradise which Cantor has created."

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