Friday, April 28, 2017

Activity: Constraints and Trig (Day 143)

This is what Theoni Pappas writes on page 118 of her Magic of Mathematics:


That's because page 117 was the last page of Chapter 4. So Chapter 5 begins on page 119, and it's called "Mathematical Magic in Nature." So in this chapter, Pappas ties math to the natural world. In fact, the picture on page 118 is of a spider web.

Taking a glance at the table of contents for this chapter, two of the sections have something to do with, um, the birds and the bees. And that will be all I'll say about this chapter until we begin it in earnest next week.

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

Today's idea comes from an anonymous Washington State teacher who only goes by the username "Alternative Math" -- named for the alternative high school to which this teacher is assigned.

Here is the original post:

I have been busy planning out a unit on trigonometric ratios for my Geometry B course. I have been trying to balance the open ended exploration and project based learning that I prefer with the more typical questions that students will eventually see on state tests or future math classes.
Here is the [Common Core] standard I’m addressing with this lesson: G-MG.3 Apply geometric methods to solve design problems (with a focus on constraints).
I introduce trig with the slope ratio, proportions, and physically measuring before I ever tell them the word tangent. I’m leaning toward using [High Priestess] Kate Nowak’s Introduction to Trig and then running a few Labs where calculate heights and distances of physical things outside before offering this [worksheet].
Afterwards I might show a few ramp fails before giving them a more open ended design problem. I’m still working on the actual formatting piece, but it will be a blueprint showing a door/stoop 5 ft high, but due to size of parking lot also has a restriction on length. Students will figure out it is not possible to use one ramp in that space and will have to figure out how to use two or more ramps to fit the constraints.
Nothing too mind blowing or exciting here, but I figure it gets at what I’m hoping they understand.
Notice that this teacher attributes this activity to yet another teacher -- New Yorker Kate Nowak. Even though I myself found this activity on the Alternative Math page, in today's post I will credit Kate Nowak as the originator of the idea. This is due to the anonymity of the Alternative Math website -- it's far easier for me to write "Nowak" than "the author of Alt Math," and it's easier to write "she" (referring to Nowak) than to write "he or she" over and over. We already know who Nowak is -- I mentioned her blog that same week and explained why she's known as the "High Priestess."

This is what Nowak writes about this activity on her own website:

The children understand that sin, cos, and tan are side ratios. The children! They understand! They are not making ridiculous mistakes, and they can answer deeper understanding questions like, "Explain why sin(11) = cos(79)." I think right triangle trig is a frequent victim of the "First ya do this, then ya do this" treatment -- where kids can solve problems but have no idea what is going on. There's often not a ton of time for it, and it responds well to memorized procedures (in the short term). So, if your Day One of right triangle trig involves defining sine, cosine, and tangent, read on! I have a better way, and it doesn't take any longer.

We see how both Nowak and the author of Alt Math agree that today's activity should be given before the students learn the definitions of sine, cosine, and tangent. And so this is why I switched today's lesson with tomorrow's -- the original plan was for me to cover Lessons 14-3 and 14-4 of the U of Chicago text (where sine, cosine, and tangent are defined) before giving an activity. But I wish to honor Nowak's wishes to give this activity before defining the ratios. If I'm going to post her lesson on this blog, then I should present it the way she suggests it to be taught.

Of course, we observe that Nowak devotes a full week to this activity. She has the students work on only the opposite/adjacent ratio on the first day -- which, interestingly enough, is exactly how the U of Chicago text teaches it (in Lesson 14-3, before 14-4). Not until the fourth day does Nowak reveal the names of the three ratios.

Then again, this is one thing I don't like about the timing of the PARCC and SBAC exams. These tests are given a full month before the last day of school -- thereby forcing us to jump through the second semester material rapidly. The test on Chapter 14 must be next week in order to keep pace. If there were more time, perhaps I really could devote a full week to this activity.

I didn't communicate with Nowak herself, but I was able to speak to the author of Alt Math. Here is my comment followed by the author's response -- I cut out the part where we were commenting on a typo that has since been corrected:

This sounds like an interesting lesson. I see what you’re doing here — just telling the students, “We are going to learn trigonometry now,” leaves the students wondering why they must learn it and being resistant. This approach, on the other hand, introduces a question first, and then they discover that trig is the way to solve it.
I look forward to finding out whether this lesson was successful or not — even more so because I’m especially interested in Geometry lessons.

  1. Basically, I give them lines with an angle measured from the horizontal and have them draw different sized slope triangles. They see or remember that the slope ratio doesn’t change for that line, but any other non parallel line will have a different slope ratio. They find missing pieces given coordinate points or find delta y or a given delta x all for the same line.
    Then I have them draw slope triangles for various other combinations and have them measure the angles and do the process again. Afterwards, I introduce the word tangent and we discuss why it might have a different name, as opposed to always referring to it as slope. This leads to triangles oriented in different directions and then eventually sine and cosine when the hypotenuse is know.
But let's think about what both Nowak and the author of Alt Math are saying here. If I, as a teacher, were to go to a Geometry class and announce, "We are going to learn about sine, cosine, and tangent," imagine what the students' responses might be. We would expect questions like "Why do we have to learn this?" or "When will we ever have to use this?" to be common whenever strange sounding words like "sine," "cosine," "tangent" (or "logarithm") appear in math classes.

And now we can see how Nowak fights this. She provides an activity where students can see why these ratios are useful, and then defines the words "sine," "cosine," and "tangent." Now students are less likely to ask "When will we ever have to use this?" because they'd have already seen how the ratios are useful.

Notice that Nowak's link above itself contains another link -- this link leads to a page titled "Church of the Right Answer." This author criticizes teachers who elevate getting the right answer over understanding the process of getting the right answer or why the answer is right, by comparing them to blind adherents of a church.

I have a special name for adherents of the Church of the Right Answer -- traditionalists. And so this goes right back to the traditionalist debate. Traditionalists, like the ones I mentioned earlier in this post, oppose activities like Nowak's -- especially if they are group tasks, or any activities that span more than one day (as Nowak suggests.) They would prefer just telling the students the definition of "sine," "cosine," and "tangent," and assigning them an individual problem set whether they compute as many trig ratios as possible -- this is the best way to ensure that students get right answers when asked to solve a trig problem.

Of course Nowak is not a traditionalist -- if she were, she wouldn't have posted this activity. Most math teacher bloggers -- especially those who post activities -- are not traditionalists. I myself am sympathetic to traditionalism in the lower grades, but not the higher grades.

Here is the worksheet, which I post intact from its source at Alternative Math.

Thursday, April 27, 2017

Lesson 14-2: Lengths in Right Triangles (Day 142)

This is what Theoni Pappas writes on page 117 of her Magic of Mathematics:

"Every math enthusiast at one time or another has discovered or been delighted by tricks or oddities involving numbers. Here are two for you to explore and hopefully enjoy."

Pappas is "Playing with Numbers" in this section. She begins with summing and squaring:

1 + 2 + 1 = 2^2
1 + 2 + 3 + 2 + 1 = 3^2
1 + 2 + 3 + 4 + 3 + 2 + 1 = 4^2
1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5^2

This reminds us of the sum of odd numbers pattern from two weeks ago. Notice that this and the odd number sums are related -- take a look at that last sum again:

1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5^2

The first term is 1, which is also the first odd number. If we cross it out, the new first and last terms are 2 + 1 = 3, the second odd number. If we cross them out, the new first and last terms are 3 + 2 = 5, the third odd number. If we cross them out, the new first and last terms are 4 + 3 = 7 -- by now, we get the pattern. We obtain:

1 + 3 + 5 + 7 + 9 = 5^2

which is the pattern from page 108.

Another way to think about this pattern is to rotate a 5 x 5 square of dots 45 degrees to form a diamond -- hey, wasn't I writing something about how a diamond is just a square yesterday? We can view this diamond as containing nine rows -- one in the first row, two dots in the second row increasing up to five dots in the fifth row, then four dots in the sixth row decreasing down to one dot in the ninth row. This is a visual justification for:

1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 5^2

Here is the second pattern Pappas mentions today -- the 1's pyramid:

1^2 = 1
11^2 = 121
111^2 = 12321
1111^2 = 1234321
11111^2 = 123454321
111111^2 = 12345654321
1111111^2 = 1234567654321
11111111^2 = 123456787654321

Pappas asks, "When does it stop?" Well, let's try a few more:

111111111^2 = 12345678987654321
1111111111^2 = 1234567900987654321

So that's when the pattern ends. But at this point, we notice that most of the digits of this number still follow the pattern. Indeed, what seems to prevent the pattern is that there's no digit for ten -- and in fact, look at what happens in base 11, where we can use X to represent ten:

1111111111^2 = 123456789X987654321

And in dozenal, we can use E to represent eleven. (I mentioned "dozenal," or base 12, several times on this blog, most recently in my September 22nd post.)

11111111111^2 = 123456789XEX987654321

So it appears that the pattern is limited only by the digits available in our base. To see why it works, we let b be the base of our numeration system. Then we obtain:

11 = b + 1
111 = b^2 + b + 1
1111 = b^3 + b^2 + b + 1
11111 = b^4 + b^3 + b^2 + b + 1

So squaring 11, 111, and so on amounts to squaring polynomials in b:

(b + 1)^2 = b^2 + 2b + 1
(b^2 + b + 1)^2 = b^4 + 2b^3 + 3b^2 + 2b + 1
(b^3 + b^2 + b + 1)^2 = b^6 + 2b^5 + 3b^4 + 4b^3 + 3b^2 + 2b + 1

So now we can see where the coefficients come from. The b^3 coefficient of this last squared polynomial is 4 because there are four b^3 terms (b^3)(1), (b^2)(b), (b)(b^2), (1)(b^3), while the b coefficient is 2 because there are only two b terms (b)(1), (1)(b).

An interesting case occurs when b = 1. Then we obtain:

(1 + 1)^2 = 1 + 2 + 1
(1 + 1 + 1)^2 = 1 + 2 + 3 + 2 + 1
(1 + 1 + 1 + 1)^2 = 1 + 2 + 3 + 4 + 3 + 2 + 1
(1 + 1 + 1 + 1 + 1)^2 = 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1

which should look familiar. That's right -- when b = 1, the second pattern that Pappas gives today reduces to the first pattern!

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

Lesson 14-2 of the U of Chicago text is on lengths in right triangles -- specifically, those lengths that are related to the altitude and involve the geometric mean.

Geometric Mean Theorem:
The geometric mean of the positive numbers a and b is sqrt(ab).
(Note: This may sound like a definition, but actually the U of Chicago defines geometric mean to be the number x such that a/x = x/b, so we need a theorem to get the geometric mean as sqrt(ab).)

Right Triangle Altitude Theorem:
In a right triangle:
a. The altitude of the hypotenuse is the geometric mean of the segments dividing the hypotenuse.
b. Each leg is the geometric mean of the hypotenuse and the segment adjacent to the leg.

In this lesson, I give the proof of the Pythagorean Theorem based on similarity, but this time I gave the proof in the book, which mentions the geometric mean. Let's look at the proof -- as usual, with an extra step for the Given:

Given: Right triangle
Prove: a^2 + b^2 = c^2

Statements                                Reasons
1. Right triangle                       1. Given
2. a geometric mean of c & x,  2. Right Triangle Altitude Theorem
    b geometric mean of c & y
3. a = sqrt(cx), b = sqrt(cy)       3. Geometric Mean Theorem
4. a^2 = cxb^2 = cy                 4. Multiplication Property of Equality
5. a^2 + b^2 = cx + cy               5. Addition Property of Equality
6. a^2 + b^2 = c(x + y)              6. Distributive Property
7. x + y = c                                7. Betweenness Theorem (Segment Addition Postulate)
8. a^2 + b^2 = c^2                    8. Substitution (step 6 into step 7)

It is uncertain whether this is the proof that Common Core intends the students to learn, or whether my earlier proof that avoids geometric means suffices.

Actually, since posting this last year, I've decided to check both the PARCC and SBAC released test questions for those related to the proof of the Pythagorean Theorem. There were a few questions that required use of Pythagoras, but none directly related to the proof. Of course, some people lament that there aren't very many proofs on the Common Core tests.

Speaking of the Pythagoran Theorem, I've been thinking last week about how I presented it to my eighth graders earlier this year:

Monday: Coding
Tuesday: Illinois State project
Wednesday: Traditional lesson on the theorem
Thursday: "Learning Centers" (actually the traditional lesson continued)
Friday: Test

which too many students failed. But as I think about it more, we can see how the students were set up for failure with only two days between introduction and assessment. (As it turned out, even the original project failed because it was on the day that the Hidden Figures movie extra credit was due, and many students focused on finishing that rather than the project.)

It might have been better to teach the theorem on Tuesday, and then the students can apply that knowledge to the project on Wednesday. (And it would have been even better to have the traditional lesson on Monday, but I had no control over the coding teacher's schedule.) The problem is that Illinois State, on its pacing guide, insists that the project be given before the traditional lesson. This in itself leads back to the traditionalist debate, which I'm not getting into in today's post. Another problem is that this might work for sixth and eighth grades, but not for seventh grade, as I had less time with them.

(Speaking of seventh grade, today I taught the volumes of rectangular and triangular prisms to seventh graders.)

Wednesday, April 26, 2017

Lesson 14-1: Special Right Triangles (Day 141)

This is what Theoni Pappas writes on page 116 of her Magic of Mathematics:

"Take the year you were born. To this add the year of an important event in your life. To this sum add the age you will be at the end of 1994. Finally, add to this sum the number of years ago that the important event took place."

Here Pappas is showing us some "Number Magic." Notice that 1994 is the year in which she first wrote this book. The following is the example that she gives:

1953 -- Year born
1980 -- Year Pappas (claimed to have) traveled into outer space. (Really, Pappas?)
    41 -- Age as of 1994
    14 -- Years elapsed since 1994 of important event
3988 -- Total

According to Pappas, the answer is always 3988. Of course, this isn't really "magic" at all -- notice that there are two pairs that add up to 1994. The year of her birth and her age add up to 1994, as do the important event and the number of years since that event. So it's not surprising that the numbers always add up to 1994 * 2 = 3988.

Let me try another example. This time, I'll change it to the current year, 2017, so my four numbers should add up to 2017 * 2 = 4034. The year of my birth is 1980 -- hey, that's the same year that Pappas (supposedly) traveled into space. And as for the year of my important event -- I just think I'll choose the year 1994. After all, it's the year I took my favorite math class, Geometry. Oh, and a lady with the initials TP wrote one of my favorite books in that year.

1980 -- Year born
1994 -- Year of important event
    37 -- Age at the end of this year
    23 -- Years elapsed since important event
4034 -- Total

Unlike Pappas, here's something that really did travel into space -- Cassini. There was a special Google Doodle marking the space probe's arrival at the rings of Saturn. Again, I'm a science teacher, so I like to point out important scientific events. In this case, researchers are excited over the possibility that there is life on at least one of the ringed planet's moons. Cassini will orbit Saturn 22 times to gather data from the Cronian moons.

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

Chapter 14 of the U of Chicago text is on Trigonometry and Vectors. Here's the plan:

Today, April 26th -- Lesson 14-1: Special Right Triangles
Tomorrow, April 27th -- Lesson 14-2: Lengths in Right Triangles
Friday, April 28th -- Activity (includes Lesson 14-3: The Tangent Ratio)
Monday, May 1st -- Lesson 14-4: The Sine and Cosine Ratios
Tuesday, May 2nd -- Lesson 14-5: Vectors
Wednesday, May 3rd -- Lesson 14-6: Properties of Vectors
Thursday, May 4th -- Lesson 14-7: Adding Vectors Using Trigonometry
Friday, May 5th -- ActivityMonday, May 8th -- Review for Chapter 14 Test
Tuesday, May 9th -- Chapter 14 Test

So the plan for this chapter is straightforward. The one thing to note is how the day that Lesson 14-4 would have occurred, there is a planned activity day. I've noticed how many texts, including the U of Chicago, discuss the tangent ratio in a separate lesson from sine and cosine. I suppose that in many ways, sine and cosine are alike in a way that tangent isn't. The sine or cosine of any real number is between -1 and 1, while the tangent can be any real number. Therefore the graphs of sine and cosine resemble each other. The tangent ratio involves two legs, while the sine and cosine ratios involve one leg and the hypotenuse. Even the name "cosine" includes the word "sine," while the name "tangent" doesn't include "sine."

Yet I will end up covering sine, cosine, and tangent all on the same day. In the past, I've seen many teachers simply teach SOH-CAH-TOA all in the same lesson, and then when they come to me for tutoring, they look at each triangle in the homework to determine whether sine, cosine, or tangent is needed to solve the problem. But as it turns out, all of the questions require tangent because the student is actually reading the tangent lesson in the text! If the student is going through all of that, then we might as well have all three trig ratios in the same lesson.

And so this is exactly what I'll do. This will then free a day for an activity. My planned activity will actually be one that I found off of another teacher's website. (Actually, I'm still debating whether to do the activity on Friday or on Thursday, since this teacher presents this activity before teaching the students about sine, cosine, and tangent.)

[2017 Update: The activity day is now on Friday and the trig ratios are on Monday. Thus the other teacher's original intent has been restored.]
But that's for later this week -- how about today's lesson? Lesson 14-1 of the U of Chicago text is on Special Right Triangles -- that is, the 45-45-90 and 30-60-90 triangles. The text emphasizes how these triangles are related to the regular polygons. In particular, the 45-45-90 and 30-60-90 triangles are half of the square and the equilateral triangle, respectively. We can obtain these regular polygons, in true Common Core fashion, by reflecting each right triangle over one of its legs. The regular hexagon is also closely related to the 30-60-90 triangle.

The questions that I selected from the text refers to these regular polygons and using the triangles to measure lengths related to the regular polygons. I mentioned today how I like to watch baseball over summer break -- well, a baseball "diamond" (really a square) appears on the worksheet. Also, a honeycomb, with its hexagonal bee cells, also appears.

The review questions that I selected are also preview questions. Two of the questions involve similar right triangles in preparation for geometric means in Lesson 14-2, and the other one is about how to simplify radicals, so we can explain in Lesson 14-4 why the sine and cosine of 45 degrees are usually written as sqrt(2)/2.

Tuesday, April 25, 2017

Chapter 13 Test (Day 140)

This is what Theoni Pappas writes on page 115 of her Magic of Mathematics:

"It seems the further out one ventures in the field of whole numbers, the primes become rarer and rarer. One might think that because they appear less and less frequently, that perhaps they end somewhere. As early as about 300 B.C. Euclid provided the first proof that the prime numbers are infinite."

Euclid -- hey, that name sounds familiar to readers of this Geometry blog! Yes, the same Euclid after whom Euclidean geometry is named is also the one who proved the infinitude of primes.

Pappas continues, "He used indirect reasoning..." -- that's right, as in indirect proof. It's interesting that just as we cover the U of Chicago's Chapter 13 on indirect reasoning, we read about so many indirect proofs in Pappas.

Notice the similarity between Cantor's Diagonal Argument and Euclid's infinitude of primes. Cantor assumes that the reals are countable and derives a contradiction, so the reals are uncountable. And so Euclid assumes that the primes are finite and derives a contradiction, so the primes are infinite.

Pappas assumes that the finite list 2, 3, 5, 7, and 11 contains all the primes. She then multiplies these primes and adds 1 to obtain 2311. But this number can't be divisible by 2, 3, 5, 7, or 11 since each leaves a remainder of 1. Therefore 2311 must be a new prime! Pappas then generalizes this to the case where n is the largest prime. In each case we generate a new prime not on the list, and so n is not the largest prime -- a contradiction, therefore there is no largest prime. The primes are infinite. QED

By the way, tonight's the first episode of Genius, about Albert Einstein, on National Geographic. I'm a science teacher, so I'll watch the miniseries, but I may or may not write about it on the blog.

Notice that Euclid's proof generates a much larger prime that's not on the list, in this case 2311. It doesn't necessarily give the actual next prime, which is 13. Oh, and speaking of 13...:

This is what I wrote two years ago about today's test. As it turned out, I originally posted this test in 2015 on Friday the 13th, and I mentioned this fact throughout the test. Unfortunately, this year today is neither Friday nor the 13th. But if it's any consolation, this is my 13th post blog post in April:

Here's an answer key for the test:

1. a. 90 degrees. I could have made this one more difficult by choosing a heptagon, or even a triskaidecagon, but I just stuck with the easy square.

b. Here is the Logo program:

Notice that the side length is 13. I'll still find a way to sneak 13, if possible, into each problem.

2. a. If a person is not a Rhode Islander, then that person doesn't live in the U.S.
b. If a person doesn't live in the U.S., then that person isn't a Rhode Islander.
c. The inverse is false, while the contrapositive is true.

Notice that Rhode Island is the thirteenth state.

3. y = 10.

4. There is a line MN. (M is the thirteenth letter of the alphabet.)

5. Every name in this list is melodious.

6. The equation has no solution. (This question references 13, as 13x appears in the expansion.)

7. a. 13, 11, 9 (descending odds).
b. 13, 17, 19 (increasing primes -- of course, Euclid proved that this sequence is infinite).

8. a = 2, b = 1, c = 3.

9. kite.

10. I discussed this problem earlier this week. It is the same as the problem from the Glencoe text, except that I only drew half of the figure -- the part where a contradiction appears.

Assume that the figure is possible. Then ABC is isosceles, therefore angles A and C are each 40 degrees (as the third angle of the triangle is 100). Then ABO is isosceles (as it has two 40 degree angles), so AO = BO = 3. Then by the Triangle Inequality, 3 + 3 > 8, a contradiction.

11. Through any two points, there is exactly one line. (This is part of the Point-Line-Plane Postulate.)

12. a. KML measures 13 degrees.
b. K measures less than 167 degrees.
c. L measures less than 167 degrees. (This is the TEAI, Exterior Angle Inequality.)

13. a. Law of Ruling Out Possibilities.
b. You forgot to rule out another possibility -- that nothing bad will happen to you today. Hopefully, this will be true for you.

Monday, April 24, 2017

Review for Chapter 13 Test (Day 139)

Note: This spring break post has been edited so that I can write about Eugenia Cheng's new book, Beyond Infinity. The "Eugenia Cheng" label has been added to this post. Today I selected a particular chapter to discuss -- the one that lines up best with the Pappas page.

This post is dated April 23rd, one week after Easter and the last day of spring break. This is the 113th day of the year, and so we read page 113 of Pappas.

This is what Theoni Pappas writes on page 113 of her Magic of Mathematics:

"Georg Cantor's set theory and transfinite numbers were brilliant accomplishments. His proof of the countability of the rational numbers was elegant."

As it turns out, a new section begins on page 113 -- "Cantor & the Uncountable Real Numbers." Here "Cantor" refers to the great 19th century German mathematician, Georg Cantor.

We don't need to read page 113 to figure out what Pappas is talking about here. Eugenia Cheng writes about this in Chapter 6 of her latest book, Beyond Infinity. The name of this chapter is "Some Things Are More Infinite than Others." She begins:

Children sometimes have this kind of argument:
"I'm right."
"I'm more right."
"I'm right times a hundred."

"I'm right times a million."
"I'm right times a billion."
"I'm right times infinity!"

"I'm right times two infinity!"
"I'm right times infinity squared!"

However, as Cheng points out, "two infinity" and "infinity squared" are equal to just infinity. This is implied in Cantor's proof that there exist only countably infinitely many rationals (that is, just as many rationals as whole numbers). In a sense, there are "infinity squared" rationals, since there are infinitely many numerators and infinitely many denominators. But "infinitely squared" is infinity, and so there are only (countably) infinitely many rationals.

Cheng continues:

"There are more irrational people than rational people in the world, probably. In fact, most people are a bit irrational -- that, to me, is an important part of being human and not a computer."

She uses this as an analogy about numbers -- just as more people are irrational than rational, more numbers are irrational than rational. In some ways, this is counterintuitive, because if one were to think of a number, that number most likely be a rational number -- indeed a natural number.

And so Cheng asks the reader to find the area of a circle whose radius is the chosen number. She points out that if the original chosen number was rational, the resulting area must be irrational:

"If you multiply a rational number by an irrational number, the result is irrational. Likewise, if you add a rational and an irrational number, the result is rational."

She points out that in order for the area to be rational, we must choose some contrived irrational number for the radius, something like 1/sqrt(pi):

pi * (1/sqrt(pi))^2 = 1.

Bringing this back to Pappas, notice that on her Mathematical Calendar, she often comes up with such contrived problems in order to make the answer be the date. So on the first of the month, she might indeed ask for the area of a circle with radius 1/sqrt(pi).

Cheng also shows us how to find the radius of any circle whose area is the rational number a/b:

r = sqrt(a/(pi b))
pi r^2 = pi(sqrt(a/(pi b)))
           = (pi a)/(pi b)
           = a/b.

She suggests that the reason that we must come up with such contrived radii just to make the area rational is that there are so many more irrational numbers than rational numbers.

Cheng proceeds:

"We're going to 'count' the irrational numbers, except that we don't have a way of saying what the irrational numbers are. Irrational numbers are defined by what they're not: they're the real numbers that are not rational."

She adds that she'll actually define irrational numbers in a subsequent chapter. (I still remember the college class where I first learned about what irrational numbers are.)

Cheng lists the following things that we already know:

1. The rational numbers are countable.
2. If you put two countable sets together, you get another countable set (as in the two-floor Hilbert Hotel).

The Hilbert Hotel is the dominant analogy used to explain the sizes of infinite sets. Indeed, the "room" I mentioned last year when describing a David Kung DVD lesson is in a Hilbert Hotel:

We begin as Cantor did -- let's imagine what a solution would look like. We would have every decimal in a room -- and we'd be able to prove it by giving a list of all of the decimals. So all we have to do is check the list -- if all the decimals are listed then we have a 1-1 correspondence, and if there's a decimal missing then we don't have a 1-1 correspondence.

And Cheng is getting ready to do the same thing in her book. She concludes that as soon as we know how many reals there are, then we also know how many irrationals there are:

rationals                  if reals           then irrationals
countable (known)  countable      countable
                                uncountable  uncountable

She continues:

"The real numbers are difficult to pin down, but for now let's say they are all the possible decimal numbers, where the decimals are allowed to go on forever, repeating or not repeating. (In fact, all decimals go forever if you put 0's on the end; we just don't usually bother writing all those 0's.)

Our goal is to determine whether it's possible to move every guest from a hotel whose rooms are numbered with all the reals, to one whose rooms are numbered all the rationals. She eliminates simple possibilities, such as placing the smallest real in room 1 (because there is no smallest real), and just reversing the real number's digits to form a natural number (because some real numbers have infinite representations such as 1/6 = 0.16666..., but there is no infinite natural number ...66661).

And so Cheng comes up with a list and applies Cantor's famous diagonal argument:

new room no.  old room no.
1                      0.238795317...
2                      0.984718573...
3                      0.389716438...
4                      0.777362889...
5                      0.444317895...
6                      0.879000001...
7                      0.892225673...
8                      0.191919234...

She explains, "Then the digits we ask for when we knock on each person's door are the ones in bold here, along the diagonal."

new room no.  old room no.
1                      0.238795317...
2                      0.984718573...
3                      0.389716438...
4                      0.777362889...
5                      0.444317895...
6                      0.879000001...
7                      0.892225673...
8                      0.191919234...

Cheng then adds one to each digit marked in bold, to produce the following number:


She adds, "and we can show that this person is not in any room n by looking at their nth digit, which is the different from the person who is actually in room n. So this person has not been evacuated to any room, and the smarty-pants [who claims to have matched every real with a natural] has failed."

And this is a contradiction, completing the indirect proof that the set of real numbers is uncountable.

Most of the time, the uncountability of the real numbers is the end of the story. But Cheng moves on with another possible example of an uncountable set:

"Another way in which something can be 'more infinite' than the natural numbers is more subtle, and perhaps relates more to the word 'uncountable'"...another way in which we can fail to evacuate a hotel is if we get decision fatigue."

She assumes that the original hotel has double rooms, and the destination hotel has single rooms. We would like to be able to give the following instruction:

"If you're currently in room n, then one of you go to room 2n and the other to room 2n - 1."

but there isn't necessary any way to determine which guest goes to 2n and which goes to 2n - 1.

The following analogy is usually attributed to the British mathematician Bertrand Russell. We first assume that the guests in the rooms are pairs of shoes -- left shoes and right shoes. Then it's easy to empty them into a single room hotel. We could go:

-- left shoe 1, right shoe 1, left shoe 2, right shoe 2, left shoe 3, right shoe 3, ...

and so on. Written mathematically:

-- The left shoe n will go to position 2n - 1.
-- The right shoe n will go to position 2n.

But if these were socks, then there's no such thing as a "left sock" or a "right sock." And indeed, as obvious as it might seem, it's impossible to prove that we can make infinitely arbitrary choices unless we assume an extra postulate, or axiom, in set theory. As Cheng explains:

"The question of whether or not this is possible is not exactly resolved in mathematics, but is a tricky and subtle point that still worries mathematicians. It is called the Axiom of Choice."

She tells us that if we assume the Axiom of Choice, then the socks fit into the hotel -- and so there are only countably many socks. But without AC, then the socks might not fit -- they'd be uncountable!

As it happens, AC is set theory's Parallel Postulate. Trying to prove AC from the other axioms of set theory is as futile as trying to derive Euclid's Fifth Postulate from the first four. There are some versions of set theory in which AC holds, and others in which it fails -- just as there are some versions of geometry in which parallels uniquely exist, and others in which they aren't unique.

But let's return to the main result of this chapter. There exist uncountably many reals, yet all but countably many of them are irrational. So the irrationals must be uncountable as well. Actually, because of AC issues, it might be that the rationals are countable and the irrationals are countable, yet their union, the reals, are uncountable. As it turns out, this isn't the case, because we can prove that the irrationals are uncountable without using AC. Here's how:

One of my favorite questions is, is there a 1-1 correspondence from R to R \ Q? As it turns out, there does exist such a 1-1 correspondence, and we can prove it. Let's think of in terms of a Hilbert's Hotel -- except it's an uncountable hotel with a room for every irrational. Now a bus arrives containing all of the rational numbers. How can we fit them into the hotel?

Well, let's take our favorite irrational, pi, and so we add pi to every rational. We see that the sum of a rational and an irrational is irrational, and so we can ask each rational to go to the room where that irrational is presently sitting. And where do those irrationals already in those rooms go? By now we should know the Hilbert's Hotel pattern -- they add pi to themselves to get new irrationals, and then go to the room where those new irrationals are. In other words:

The number x + n * pi (x, rational, n a whole number) goes to room x + (n + 1)pi.

Of course, we must be careful that x + (n + 1)pi isn't accidentally rational. But there is an indirect proof that it's always irrational: assume that x + (n + 1)pi = y is rational. Then rearranging the equation, we obtain pi = (y - x) / (n + 1), the quotient of two rationals, so is itself rational. So we get that pi is rational -- a contradiction since pi is irrational. Therefore x + (n + 1)pi is irrational. QED

We notice that in this case, there are uncountably many rooms, yet only countably many occupants need to change rooms. Most irrationals don't have to move -- in particular, sqrt(2) doesn't move, neither does e, and neither does -pi.

Here's another thing I've noticed -- sets of numbers that are special, or fit some sort of pattern, tend to be countable, while sets of numbers that don't fit patterns tend to be uncountable. The natural numbers fit the ultimate pattern -- 1, 2, 3, so they are countable. The rational numbers are countable (they fit the pattern a / b), while the irrationals are countable. The algebraic numbers are countable (they fit into polynomials), while the transcendental numbers are uncountable.

What about the set of all numbers consisting of the digits 0 and 1 only -- is it countable? Even though this may sound like a pattern, it really isn't. After all, in binary, all reals consist of only two digits, so consisting of two digits isn't special enough to make the set countable. The set of reals with only two digits is uncountable, regardless of which digits or the base. (In the special case of 0 and 1, we can use Cantor's diagonal argument to see why.) If the digits are 0 and 2, and the base is changed to ternary (base 3), we obtain a famous uncountable set -- the Cantor fractal set!

The idea, though, that most real numbers are irrationals is profound. After all, at one time, ancient mathematicians were surprised that irrational numbers exist at all.

Here is a link to a common indirect proof that sqrt(2) is irrational:

Neither the U of Chicago nor Glencoe gives the proof outright. But both hint at it -- I just mentioned the U of Chicago's square root proofs. The Glencoe text asks the students to prove that if the square of a number is even, then it is divisible by four. As we can see at the above link, this fact is directly mentioned in the irrationality proof.

I remember once reading the proof of the irrationality of sqrt(2) in my textbook back when I was an Algebra I student. Until then, I had always heard that sqrt(2) was irrational, but I never realized that it was something that could be proved. So I was fascinated by the proof. Naturally, the text only included this as an extra page between the main sections, so it was something that the teacher skipped and most students probably ignored.

The irrationality of sqrt(2) has an interesting history. It goes back to Pythagoras -- he was one of the first mathematicians to use sqrt(2), since his famous Theorem could be used to show that the diagonal of a square has length sqrt(2). The website Cut the Knot, which has many proofs of the Pythagorean Theorem, also contains many proofs of the irrationality of sqrt(2):

Now there is a famous story regarding sqrt(2) and Pythagoras. At the following link, we see that Pythagoras was the leader of a secret society, or Brotherhood:

Now Pythagoras and his followers believed that only natural numbers were truly numbers. Not even fractions were considered to be numbers, but simply the ratios of numbers -- numberhood itself was reserved only for the natural numbers. In some ways, this attitude resembles that of algebra students today -- when the solution of an equation is a fraction, they often don't consider it to be a real answer, even though modern mathematics considers fractions to be numbers. (The phrases real number and imaginary number reflect a similar attitude about 2000 years after Pythagoras -- that some numbers aren't really numbers.) So of course, the idea that there were "numbers" that weren't the ratio of natural numbers at all was just unthinkable.

Pythagoras and his followers must have spent years searching for the correct fraction whose square is 2, but to no avail. Finally, one of his followers, Hippasus, discovered the reason that they were having such bad luck finding the correct fraction -- because there is no such fraction! And, as the story goes, Pythagoras was so distraught, afraid that the secret that sqrt(2) was irrational would be revealed, that he ordered to have poor Hippasus drowned at sea!

But as I said, nowadays students simply complain when they have a fractional, or worse irrational, answer to a problem. No one has to drown any more just because of irrational numbers.

Question 10 on my test review, therefore, is actually the final step of that proof, since that's the step where the contradiction occurs. They are given a triangle with sides of length 3 and 8, and two angles each 40 degrees (one of which is opposite the side of length 3). The students are to use the Converse of the Isosceles Triangle Theorem to show that the missing side must also be of length 3, and then the Triangle Inequality to show that 3 + 3 must be greater than 8, a contradiction.

When I wrote this problem, I had trouble deciding how difficult I wanted my indirect proof to be. For example, I considered giving 100 as the measure of the angle opposite the side of length 8, and give only one 40-degree angle instead. Then the students would have to use the Triangle Angle-Sum Theorem to find the missing angle as 40 degrees before applying the Isosceles Converse.

Or, to go even further, we can derive a contradiction without making the angle isosceles at all. For example, we could make the angle opposite the 8 side to be, say, 90 degrees instead of 100. Then the missing angle would be 50 instead of 40. If the triangle is drawn so that 50 degrees is opposite the 3 side, then by the Unequal Angles Theorem, the missing side would be less than 3, so the sum of the two legs would still be less than the longest side.

But this might confuse the students even more -- especially if the 90-degree angle is marked with a box (to indicate right angle) rather than "90." A right triangle might lead a student to use the Pythagorean Theorem to find the missing leg. Although this still eventually leads to contradiction -- the missing side would be sqrt(55), which isn't less than 3 -- that irrational side length might still cause some students to drown.

And so I wrote my Question 10 on the review so that it will actually help the students prepare for the corresponding question on the test. I balance out this tough question with some easier questions about logic (converse, inverse, etc.). Hopefully the test won't be too hard for the students.