Friday, September 29, 2017

Lesson 3-2: Types of Angles (Day 32)

This is what Theoni Pappas writes on page 272 of her Magic of Mathematics:

"Our eyes begin to jump with this puzzle. Determine how many 2 * 1 * 1 blocks make up this structure."

This is the first page of the section "Putting Your Logic to Work." This section contains many puzzles for the reader to solve, the first being a "Rectangular Volume Illusion."

Unfortunately, this puzzle is completely dependent on its picture, which you readers of this blog are unable to see. Indeed, many of these puzzles require pictures. And so I'll only mention some of these questions briefly on the blog.

I do notice that the entire structure fits in a box with dimensions 6 * 6 * 4. So the entire box has a volume of 144 cubic units, and since each block has a volume of two cubic units, exactly 72 blocks fit inside the box. The problem is that some of the blocks are supposed to be missing. The figure is drawn in isometric form, so each block is drawn as three "parallelograms." We distinguish between, say, a 5 * 5 * 5 cube and the same cube with a small 1 * 1 * 1 cube missing by drawing the latter with the parallelograms going "the other way" where the missing unit cube is. This is why Pappas tells us that "our eyes begin to jump" as we try to figure out how many blocks are missing.

Since you can't see the blocks, I might as well skip to the answer. Pappas tells us that there are 61 blocks in the structure. This means that eleven of them are "missing."

Hey, as I'm looking at the answer page anyway, let's give the solutions to yesterday's riddles:

(1) Tom and Jerry are professional baseball players, but play on different teams. This particular day their teams were playing each other. Jerry is a catcher who prevented Tom from reaching home plate.

(2) While walking down the street, Eric had a terrible attack of hiccups. He went into the bar to get a glass of water to see if it would help him. The bartender, realizing Eric's problem, thought he would try to scare the hiccups out of Eric. It worked!

(3) Mary's husband died a number of years ago. She kept his ashes in an urn on her kitchen table. Also on her table was a bowl with goldfish. Mary had left her kitchen window open this particular day. Her neighbor's cat had come in, and in trying to get the goldfish, the cat knocked over the goldfish bowl and the urn. Her husband's ashes were spilled onto the floor.

Lesson 3-2 of the U of Chicago text is called "Types of Angles." In this chapter, students learn about zero, acute, right, obtuse, straight, complementary, supplementary, adjacent, and vertical angles.

In the new Third Edition of the text, this actually corresponds to Lesson 3-3. But the definitions of acute, right, and obtuse are actually combined with yesterday's Lesson 3-1. Only the last four definitions (mainly adjacent and vertical angles) remain in the new Lesson 3-3.

In between these, in the new Lesson 3-2, are rotations. I've mentioned before how strange is this that both the old and new editions define a rotation as the composite of two reflections in intersecting lines, yet the new edition has a section on rotations before defining reflections! The U of Chicago most likely placed this section here so that in introducing rotations, students become more familiar with angles. (Again, I point out that Hung-Hsi Wu of Berkeley, in his recommendations for Common Core Geometry, teaches rotations before reflections, but he defines rotations differently. His lessons have nothing to do with the new Lesson 3-2.)

In fact, Jackie Stone -- the main Blaugust participant last month -- also introduces rotations when teaching her students about angles, just like the U of Chicago text:

What was intended to be a five minute “review” of these skills to launch into the real lesson activity of the day turned into a much more in depth “teaching” of how to use this tool.  Although they might NEVER use a protractor outside of my class again I do find the task of measuring something using a tool useful. The task also spoke to the CCSS Math Practice Standards of attending to precision and using tools strategically. It is so challenging (especially at the beginning of the year) to determine what are appropriate scaffolds to help students work on a task. Moving forward, I plan to assume less which is actually a good thing because then we can talk about refined meanings of things.  For instance, because of their lack of background we were able to really talk about that the measurement in degrees was actually a measurement of a rotation. I think next year my approach might be different.

Again, because I jumbled up the first two lessons of Chapter 3 two years ago, this is what I wrote three years ago on this lesson:

Section 3-2 of the U of Chicago text discusses the various types of angles. It covers both the classification of angles by their measures -- acute, right, and obtuse -- as well as related angles such as vertical angles and those that form a linear pair. Complementary and supplementary angles also appear in this lesson.

As usual, I like to look for other online resources by other geometry teachers. And I found the following blog post from Lisa Bejarano, a high school geometry teacher from Colorado who calls herself the "crazy math teacher lady":

Bejarano writes that in her geometry class, she "starts with students defining many key terms so that we can use this vocabulary as we work through the content." And this current lesson, Section 3-2, contains many vocabulary terms. So this lesson is the perfect time to follow Bejarano's suggestion.

Now two textbooks are mentioned at the above link. One is published by Kagan, and unfortunately I'm not familiar with this book. But I definitely know about the other one -- indeed, I actually mentioned it in my very last post -- Michael Serra's Discovering Geometry.

The cornerstone to Bejarano's lesson is the concept of a Frayer model. Named after the late 20th-century Wisconsin educator Dorothy Frayer, the model directs students to distinguish between the examples and the non-examples of a vocabulary word.

Strictly speaking, I included a Frayer-like model in last week's Section 2-7. This is because the U of Chicago text places a strong emphasis on what is and isn't a polygon. But Serra's text uses examples and non-examples for many terms in its Section 2-3, which corresponds roughly to Section 3-2 in the U of Chicago text.

Here are steps used in Bejarano's implementation of this lesson:

1. I used the widget example from Discovering Geometry (chapter 1). It shows strange blobs and says “these are widgets”, then there is another group of strange blobs and it says “these are not widgets”. I have students define widgets in their groups. Then they read their definition and we try to draw a counterexample. Then we discussed what makes a good definition and we were ready to go!

(Note: Bejarano writes that she found the "widget" example in Chapter 1 of Serra's text, but I found it in Section 2-3 of my copy. Of course, I don't know how old my text is compared to Bejarano's -- mine is the Second Edition of the text, dated 1997. Hers could be the Third Edition or later.)

Notice that "widget" here is a non-mathematical (and indeed, a hypothetical) example that Bejarano uses as an Anticipatory Set. As we've seen throughout Chapter 2, many texts use non-mathematical examples to motivate the students.

3. Students worked in small groups with their 3 terms copying the examples & non-examples, then writing good definitions for each term. I set a timer for 10 minutes. [2017 update: yes, I'm skipping her #2 -- dw]

Now I don't necessarily want to include this as a group activity the way Bejarano does here. After all, I included the Daffynition Game in my last post as a group project [...]

And let's stop right here, because today's an activity day, and I haven't posted that Daffynition game yet this year -- so let me post it today! This is what I wrote three years ago about the activity. (Oh, and if you thought we were done with Serra's text after finishing Chapter 0 last month, think again!):

It's tough trying to find activities that fit this chapter. One source that I like to use for activities is Michael Serra's textbook, Discovering Geometry. Just like most other geometry texts, in Chapter 2 he discusses the concept of definition (Section 2-3, "What is a Widget?") Then the text introduces a project, "the Daffynition Game," where the students take turns making up definitions to real words.

A few comments I'd like to make about the game as introduced by Serra. Step 3 reads:

3. To begin a round, the selector finds a strange new word in the dictionary. It must be a word that nobody in the group knows. (If you know the word, you should say so. The selector should then pick a new word.)

The problem is that this depends on the honor system -- how do we know that a student who knows the word will actually admit it? Rather than depend on the students' honesty, why not make knowing the word an actual strategic move? That student will then earn a point for knowing the word -- and the student can still make up a fake definition in order to earn even more points? This means that the selector must be very careful to choose a word that isn't in the dictionary.

Another question is, what affect would this project have on the English learners? I'd say that this would be a great project for them, since they can learn both English and math in this lesson. But English learners might be at a disadvantage in this game, since if they choose a word that a native English speaker knows, the English speaker would earn a point (since I'm not counting on the honor system here). One debate that always comes up in a group activity is whether to group homogeneously or heterogeneously. For this project, it may be best to group homogeneously, but by English, rather than mathematical, ability.

Finally, this project requires students to look up words in a dictionary -- but what dictionary? I threw out the problems in earlier sections that depended on the availability of a dictionary. Perhaps the night before this activity, part of the homework assignment could be to look up a word in the dictionary and write down its definition -- but that assumes that the students will actually do the homework, and besides, there's no guarantee that the students have access to a dictionary at home (or online) either.

My solution is for the teacher to have enough index cards with words and definitions on them. Therefore the selector chooses an index card, not a word from the dictionary. Indeed, the teacher can give an index card to each student even before dividing the class into groups! But the selector should still follow the other steps as originally written in the Serra text.

OK, so let me post the worksheets. I decided to post only the first page of Lesson 3-2 (Lisa Bejarano's lesson) and then go directly to the Daffynition Game.

In cutting Bejarano's second page, I'm dropping some terms that don't appear until later in Chapter 3, but I also dropped "vertical angles" and "angle bisector," which do appear in Lesson 3-2. Teachers can either make sure to write those two dropped terms on index cards in the Daffynition Game, or else go full Bejarano and use the Frayer models as a full group project, just as the Colorado teacher originally intended.

Thursday, September 28, 2017

Lesson 3-1: Angles and Their Measures (Day 31)

This is what Theoni Pappas writes on page 271 of her Magic of Mathematics:

-- "Madre was a midget," [Terri summarized]. "His wife and the trapeze artist sawed off some of his cane, one day when he wasn't using it. When he went to use his cane, it was too short for him. Madre thought that he was growing taller, which aggravated his precarious heart problem. He couldn't get to his medicine in time. How does that sound Mr. Mason?"
-- "That's it!"

OK, and so the girl Terri solves the mystery after all. Since there is still room left on this page, Pappas gives us three logic stumpers for us to try out:

(1) It was Tuesday and Tom and Jerry were at the same job. When it came time for Tom to go home, Jerry wouldn't let him go home. WHY?

(2) Eric walked into a bar, and asked the bartender for a glass of water. The bartender looked at Eric for a moment, then pulled out a gun and pointed it at Eric. Eric was startled for a moment, and then said thank you and walked out of the bar without having drunk the water. WHAT IS THE SCENARIO?

(3) When Mary came home she walked into the kitchen. She suddenly let out a scream when she discovered her dead husband on the floor. Along side was water from a bowl that had been on the table and was now tipped over on the floor. The window over the kitchen table was ajar. WHAT HAD HAPPENED?

Pappas provides the answers in the back of the book. Once again, I think I'll just post the answers right here on the blog -- tomorrow, of course!

By the way, I can keep going on with riddles like these forever. Here is a link to a riddles website:

As for the riddle dated yesterday, maybe Madre was the rider of this elevator:

And here's the riddle dated today:

Two boys wish to cross a river. The only way to get to the other side is by boat, but that boat can only take one boy at a time. The boat cannot return on its own, there are no ropes or similar tricks, yet both boys manage to cross using the boat.

But now here's my own riddle -- why is that these riddles are posted yesterday and today, yet they link to a URL containing /2011/07/ (as in July 2011)? I'll never know the answer to that one.

Anyway, it may be a good idea for math teachers to give these riddles and logic puzzles in class -- particularly Geometry teachers.

Lesson 3-1 of the U of Chicago text is called "Angles and Their Measures." I didn't write much about this lesson two years ago due to a subbing day, so instead we'll go back three years.

This is what I wrote three years ago about today's lesson:

I don't have much to say about the book's treatment of angles. This text is unusual in that it includes a zero angle -- an angle measured zero degrees. Then again, in Common Core we may need to discuss the rotation of zero degrees -- the identity function.

The key to angle measure is what this text calls the Angle Measure Postulate -- so this is the second major postulate included on this blog. Many texts call this the Protractor Postulate -- since protractors measure angles the same way that rulers measure length for the Ruler Postulate. The last part of the postulate, the Angle Addition Property, is often called the Angle Addition Postulate. Notice that unlike the Segment Addition Postulate -- which this text calls the Betweenness Theorem -- the text makes no attempt to prove the Angle Addition Postulate the same way. Notice that Dr. Franklin Mason's Protractor Postulate -- in his Section 1-6, has angle measures going up to 360 rather than 180.

So let me include a few more things in this post. First, here's another relevant video from Square One TV -- the "Angle Dance":

By the way, last year I taught angles to my seventh grade class, and I showed them the "Angle Dance," so let me also reblog what I wrote last year about angles. From my January 10th post:

9:45 -- My sixth graders leave and my seventh graders arrive. In this class, the students are learning about angles, as well as how to draw a triangle given its three angles. I start out by telling the class about a movie I watched over the weekend, Hidden Figures, whose main theme is that black girls can do math, too. I'm offering extra credit points to anyone who watches the movie, brings me the ticket stub, and answers five questions about the movie. I'm hoping that my students -- especially the black female students -- will watch the movie.

Halfway during class, I give the students a "music break" and I sing a song from Square One TV that's appropriate for this lesson, "Angle Dance."

I offer the students a participation point for dancing along with the song. Several students take me up on my offer, including two black girls -- the target demographic of Hidden Figures.

Wednesday, September 27, 2017

Chapter 2 Test (Day 30)

This is what Theoni Pappas writes on page 270 of her Magic of Mathematics:

-- "Perhaps [Madre] didn't know his cane was shortened?" Bob asked.
-- "That's right," Mr. Mason replied.

We're still in the middle of our mystery. Recall that Bob is our shy student who doesn't participate in traditional math lessons, yet is engaged by this logic story. The students continue to ask questions:

-- "Then, when he stood with it he was very upset, but why?" Bob wondered.
-- "Because it didn't fit him. He was too tall for it!" Carol shouted.
-- "So far, so good," the teacher said.
-- "But why would thinking he had grown taller upset him so?" Gary asked the other students.
-- "This time I really got it," Tom yelled. "Madre was a midget."
-- The teacher shouted, "YES! What else?"

At this point the student named Terri believes that she has the complete answer. She begins to tell her story in order for the teacher to confirm it -- but her story doesn't end until page 271.

And so in order to give you, the readers of this blog, one last day to figure out the entire mystery, I'll save all of Terri's explanation until tomorrow. Stay tuned for our thrilling conclusion....


Today is Day 30, and so it marks the end of the first hexter and midpoint of the first trimester. This was relevant last year at my middle school, but this year I'll be subbing mostly at high schools, which don't observe hexters or trimesters.

Before we return to Ogilvy's book, there's something I want to mention here. Today Google is celebrating its 19th birthday by bring back some of its classic games. And guess what that means....

The Fischinger player is back!

When Google first posted the Fischinger player, it was during a time when I wasn't blogging, due to my disappointment when I learned I wasn't hired for the new school year. By the time I felt like blogging again, the Fischinger player had been taken down -- yet I continued to write about the player for weeks afterward, during the Pappas music pages.

Now that it's back, let's recall how it works. We can play four different notes -- C, D, E, and G -- in three different octaves. In theory, we should be able to play C, D, E, G in the lowest octave; c, d, e, g in the middle octave, and c', d', e' in the highest octave. But actually, the middle e and g actually play one octave higher as e' and g'. This means that we can play the bass notes C, D, E, G, c, d, and then the treble notes c', d', e', g'. There's a way to change the key from C to any other key, but this tends to mess up the octaves.

Here are some songs that I've been wanting to play on the Fischinger player:

The Fibonacci Song

Here is the Fibonacci sequence:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...

Now it's easy to play the first six numbers as notes in a C major scale:


Notice that these are all Fischinger playable notes. But what should we do with the 13? Others who have converted Fibonacci into music often continue the pattern, so 9 is D (in some octave), 10 is E, 11 is F, 12 is G, and 13 is A. But the note A isn't Fischinger playable in any octave.

Well, the number 13 consists of the digits 1 and 3. These are the notes C and E -- and hey, notice that both of these are Fischinger playable! In fact, we can convert the first eleven Fibonacci numbers into sixteen notes -- which is the number of notes that fit on the player:


Of these 16 notes, 15 of them are Fischinger playable. The lone exception is that F there. Earlier, I suggested that we can just play the note G there instead of F:


This works. But there are still a few changes we can make here. First of all, I've always liked to avoid 89, which takes us past the octave. Instead, we add rests (that is, we play no note at all) between the two G's of 55. This way, it's easier to tell when the player is about to repeat the song:


Second, earlier we changed 34 into 35 since F isn't playable. But after I hear the music, I think it actually sounds better if we change 34 into 23. Then the sequence C-E-F-G (34 and its neighbors), which contains four different notes, becomes C-D-E-G. Of course, if this were a real instrument rather than the Fischinger player, we'd keep it as C-E-F-G in accord with Fibonacci:


Remember, the #1 rule of music is that it should sound good. And to me, these notes above are the best-sounding version of the Fibonacci song.

Notice that these notes are to be played in the bass octave (which is on top). That is, the top line is C, the second line is D, all the way to c on the fifth line. Normally, we like to play the melody on the top octave and the harmony in the bass, but here the melody is in the bass.

So we must use the high octave for the harmony. I've decided to play the harmony on every fourth note, as if this were 4/4 time and the high note is played at the start of each bar. And hey, why don't we use Fibonacci again and use the first four numbers, 1, 1, 2, 3, to give us the treble line?


Don't forget that the bottom line on the player is e', then above that are d' and c'. And voila -- that's our Fibonacci song:

Bass line (melody): C-C-D-E-G-c-C-E-D-C-D-E-G-r-G-r
Treble (harmony):   c' -r -r -r -c' -r -r -r -d' -r -r -r -e' -r -r -r

Body Music: The DNA Song

For the DNA song, let's move (using Modify near the bottom of the screen) to the key of F. The playable notes are now F, G, A, C, and three of these correspond to DNA bases (guanine, adenine, cytosine). The note F corresponds to the last base, thymine, since there is no musical note called T.

To create the song, I decided to choose random numbers 1-5 for the notes of the F major scale. Now the F major scale begins F, G, A, Bb (B-flat), C, but of course Bb isn't playable. So instead, the number 4 corresponds to a rest. Here are the notes I chose:


This time, we put the melody in the highest octave. Notice that Fischinger now plays the c' as the lowest note in the octave and a' as the highest note (as opposed to C major, where the highest note was g' despite the bottom line playing e'). This should at least make the melody lines easier to find, as now a' is the bottom line, and counting up from it gives g', f', and c'.

To form the harmony, we use the DNA rules for base pairs -- G pairs with c', C pairs with g', A pairs with f' (since there's no t'), and F pairs with a':

Bass line (harmony): G-C-G-C-r-C-G-G-r-C-C-A-C-r-A-F
Treble line (melody): c'-g'-c'-g'-r-g'-c'-c'  -r-g'-g'-f'-g'-r-f'-a'

Again C plays as the lowest note in the octave, even though the top line is F. To make it easier, we start with G as the second line from the top, and then always keep five blank notes between the melody and harmony (so the harmony is the melody translated up six steps).

This will end up making the last harmony note sound as middle f instead of bass F. We can live with this, or just remember to go to the top line for the last bass note. Alternately, we can just leave the last note off. Then there will be two rests near the end of the song, just as with 55 in the Fibonacci song -- and moreover the last note played in the melody is the tonic F. But then the melody and harmony contain only three of the four DNA bases each. It seems so much better to have each part contain all four DNA bases.

The Unit Rates Song

I wish to play one of my songs from last year on the Fischinger player. Of course, most of the songs contain all seven notes of the scale and so aren't Fischinger playable. A few of the songs I wrote had only six notes -- for example, the Benchmark Tests song had only C, D, E, F, G, A. In practice, I ended up dropping the F when I sang it, to leave the pentatonic scale C, D, E, G, A. But neither G nor A appear in the first two lines, which are thus Fischinger playable (in the key of C):


For the bass line, let's just play the note C on every fourth beat (beginning with the first note).

But I decided that I want a full song that is Fishinger playable. Today is Day 30, and so I choose the song from Day 29 last year, the Unit Rates song. This isn't the same as the UCLA fight song parody, which contains the entire major scale and so is Fischinger unplayable. (Also, I didn't have a song on Day 30 last year, as Day 30 both last year and this year are Wednesdays, when I had no music break.)

My original version of the song was written in the key of G major. I often used G major since this key was convenient for the guitar, where the three main chords G, C, D are easy to play. So we will use Modify to move to the key of G. Like the Benchmark Tests song, the Unit Rates song was originally written with six notes (here G, A, B, C, D, E). But the last two lines contain only Fischinger playable notes, which I reproduce here:


Again the note d' is played lower than the b' here. My original version actually had the D note as the highest note. But I like keeping the lower d' here, since the descending major sixth here makes the song sound different from the ordinary.

For our bass line, this time I decided to play G on every second beat, except we change the sixth downbeat to D (when the melody is playing a', as these two notes sound more harmonious).

Bass line (Harmony): G-r-G-r  -G-r-G-r  -G-r-D-r  -G-r-G-r
Treble line (Melody): b'-b'-b'-b'-d'-d'-d'-r-g'-g'-a'-a'-g'-g'-g'-r

Here are the lyrics to the Unit Rate song from last year, in case you want to sing along. Just sing the whole song to the two lines that keep repeating:


If you want to find unit rates,
There's one thing you must know.
To find a unit rate,
All you do is divide!
To see if it's proportional,
All you do is divide!
Write it as a fraction,
Reduce it then you're fine.
Graph it at (0, 0),
Then just draw a line.

If you want to find square roots,
There's one thing you must know.
To find an estimate,
4 and below, round down!
To find an estimate,
5 and above, round up!
1 place for tenths, 2 for hundredths,
3 for thousandths, you're fine.
Graph it between two values,
Right on the number line.

Recall that the second verse was for the eighth graders, who were learning about square roots rather than unit rates at the time.

If I had never left my old classroom and were writing a new song this week, the subject of the song might not be about unit rates, but about the playground installed at the new campus. Playworks (which I mentioned last year) helped support the new playground, and there was even a visit from our local pro soccer team, the LA Galaxy. (Is there an LA Galaxy fight song? Of course if there is, it's probably not Fischinger playable anyway.)

There are other musical games available on Google's birthday besides the Fischinger player. One of them allows users to recreate Beethoven's symphonies. Another one is a "theremin," an electronic instrument played by Clara Rockmore. The theremin is set up to play a full G major scale, with no notes left out. (So the original version of the Unit Rates song, as well as many of my other songs, would be theremin playable, if not Fischinger playable.)

Chapter 10 of Stanley Ogilvy's Excursions in Number Theory is called "Continued Fractions." But I've decided to skip to Chapter 11, the final chapter, "Fibonacci Numbers," in honor of the return of the Fischinger player and my Fibonacci song.

By the way, continued fractions are related to two of my favorite topics. One of them is music theory, but only indirectly. Indeed, continued fractions are about rational approximations of irrational numbers, while music theory uses irrational approximations (such as 2^(7/12)) of rational numbers (such as 3/2).

The other is calendar reform. In fact, today is Google's 19th birthday -- or first Metonic birthday -- and continued fractions explain why a Metonic lunisolar cycle has 19 years. In fact, there exists a pure lunar calendar -- the Yerm Calendar -- that was constructed wholly from continued fractions. I have decided, therefore, to postpone our reading Chapter 10 until the end of the year, when I have my usual Calendar Reform posts.

Ogilvy actually opens Chapter 11 with a little more on continued fractions, and so let's just skip that part and go directly to the sequence itself:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...

named for the 13th century Italian mathematician Leonardo (da Pisa) Fibonacci. We can see that every number in the sequence equals the sum of the two previous numbers:

F_n = F_(n - 1) + F_(n - 2)

Notice that Ogilvy actually uses the same notation for Fermat numbers and Fibonacci numbers. I've decided to write a roman F for Fibonacci and same the italicized F for Fermat.

Ogilvy begins by finding the sum of the first n Fibonacci numbers. He writes:

F_1 = F_3 - F_2
F_2 = F_4 - F_3
F_(n - 1) = F_(n + 1) - F_n
F_n = F_(n + 2) - F_(n + 1)

We add up these equations, and many terms are telescoping, or cancelling out. We are left with the sum as F_(n + 2) - F_2. But F_2 = 1, so the final sum is F_(n + 2) - 1.

The next property to prove is:

(F_(n + 1))^2 = F_n F_(n + 2) + (-1)^n

For example:

n = 6: (F_7)^2 = F_6 F_8 + 1; 13^2 = 8 * 21 + 1
n = 7: (F_8)^2 = F_7 F_9 - 1; 21^2 = 13 * 34 - 1

Ogilvy proves this by induction. The initial case is trivial, for n = 1, we have 1^2 = 1 * 2 - 1. So now we assume it for n = k:

(F_(k + 1))^2 = F_k F_(k + 2) + (-1)^k

Add F_(k + 1)F_(k + 2) to both sides:

(F_(k + 1))^2 + F_(k + 1)F_(k + 2) = F_k F_(k + 2) + F_(k + 1)F_(k + 2) + (-1)^k

F_(k + 1)(F_(k + 1) + F(k + 2)) = F_(k + 2)(F_k + F_(k + 1)) + (-1)^k.

Using the recursive definition of Fibonacci as F_k + F_(k + 1) = F_(k + 2):

F_(k + 1)F_(k + 3) = (F_(k + 2))^2 + (-1)^k
(F_(k + 2))^2 = F_(k + 1)F_(k + 3) - (-1)^k
(F_(k + 2))^2 = F_(k + 1)F_(k + 3) + (-1)^(k + 1)

which proves the n = k + 1 case. QED

By the way, this demonstrates why Fibonacci bases like 21 and 34 tend to have divisibility rules for many different primes. In base F_n, we can show that any prime divisor of F_(n - 2), F_(n - 1), F_n, F_(n + 1), and F_(n + 2) has a divisibility rule of type divisor, alpha, omega, or SPD (based on the square alpha or omega). Of course, the SPD rules in this range are usually impractical.

After telling us that the sum of rising diagonals (22.5 degrees) in Pascal's triangle add up to the Fibonacci numbers, Ogilvy now connects Fibonacci to Geometry. If we have a rectangle of sides consecutive Fibonacci numbers, then we can cut out a square and be left with another rectangle with Fibonacci sides. For example, an 8 * 13 rectangle gives us:

1^2 + 1^2 + 2^2 + 3^2 + 5^2 + 8^2 = 8 * 13.


(F_1)^2 + (F_2)^2 + (F_3)^2 + ... + (F_n)^2 = F_n F_(n + 1)

Ogilvy continues with Geometry. We wish to find the point x on a segment of unit length such that the longer part is the geometric mean of the unit length and the shorter part:

1/x = x/(1 - x)
x^2 + x - 1 = 0
x = (-1 + sqrt(5))/2 (discarding the negative solution)

Now the ratio of the unit segment to the longer part is:

1/x = 2/(-1 + sqrt(5))
      = 2/(-1 + sqrt(5)) * (1 + sqrt(5))/(1 + sqrt(5))
      = 2(1 + sqrt(5))/(-1 + 5)
      = (1 + sqrt(5))/2

which is Phi, the golden section. Earlier, Ogilvy uses continued fractions to show us that the ratio of consecutive Fibonacci numbers converges to Phi. (as in 1/1, 2/1, 3/2, 5/3, 8/5, and so on).

Indeed, if we have a Golden rectangle of dimensions Phi * 1 and take away the unit square, the remaining rectangle has sides in ratio 1/(Phi - 1). But then we start with the known equation for Phi:

Phi^2 - Phi - 1 = 0
Phi^2 - Phi = 1
Phi(Phi - 1) = 1
Phi = 1/(Phi - 1)

and so the remaining rectangle is also a Golden rectangle.

Ogilvy's final trick is to do the same with a 36-72-72 triangle ABC (vertex angle at A). We bisect angle C to create a new 36-72-72 triangle CBD, and then we see that:

AB/BC = BC/BD (corresponding sides of similar triangles)
AB/AD = AD/BD (as AD = DC = BC)

and so D divides AB in a golden section. The author tells us that this "Golden triangle," as it's often called, appears as part of a regular pentagon. Indeed, the ratio of the diagonal to the side of a regular pentagon is Phi.

Ogilvy ends the book as follows:

"The path is endless, but many rewards are offered along the way. One could do worse than follow the gleam of numbers."

Before I end this post, let me link to Fawn Nguyen's most recent post:

Apparently, there's a "Global Math Week" coming up. Notice that the date is October 10th, which is also Metric Week (as in 10/10). Nguyen writes about something called "Exploding Dots," which somehow connects whole number operations to polynomial operations. Just as Ogilvy tells us that integer arithmetic is just polynomial arithmetic with x = 10, Nguyen informs us that we can even have fractional bases if we let x equal a fraction (but she doesn't specify which fraction she tried out).

Here is the Chapter 2 Test:

Tuesday, September 26, 2017

Chapter 2 Review, Continued (Day 29)

This is what Theoni Pappas writes on page 269 of her Magic of Mathematics:

-- "Wrong! Gary," Mr. Mason replied.

That is the teacher's response to the question asked at the bottom of the last page in Pappas. We are in the middle of a murder mystery. Gary is a student who asks Mr. Mason whether Madre's failure to take his medicine was the cause of his heart attack, and he replies that Gary is wrong.

The other students now ask each other whether they've considered all the initial facts of the case. And Pappas continues:

-- "No," Bob, who rarely talked in class, replied. "We haven't considered why the sawdust information was mentioned, what his job was at the circus, and what about the cane, and Madre's wife and the trapeze artist."

The other students agree with Bob. One of them, Tom, thinks that the sawdust was left when someone shortened the cane with a saw. And so Pappas ends the page with his question:

-- "Perhaps Madre got upset because someone wrecked his cane."
-- "NO"

The lone image on this page is a picture of the circus. I assume that there are no clues hidden in this picture -- but who knows? Maybe there are...


You may notice that today's post has been labeled "traditionalists." Technically, the traditionalists post shouldn't be until tomorrow, the day of the Chapter 2 Test. But I'm doing traditionalists a day early -- and no, it's not an excuse to bring up politics, race, and sports that have dominate the news lately. No, it's because there are things on today's Pappas and Ogilvy pages that address the traditionalist cause.

Let's start with Pappas. We back up a few pages to the beginning of this mystery, when she writes:

"These logic stories always seem to spark even the shyest student."

And of course, that "shyest student" is Bob. According to Pappas, Bob "rarely talked in class," yet he is definitely engaged by the logic story.

But traditionalists, of course, would consider telling logic stories in class a waste of time. So I ask, what's the likelihood that Bob would be equally engaged by any lesson that the traditionalists would prefer to teach? Bob should consider himself lucky that his teacher, Mr. Mason, isn't a traditionalist.

Last night was the premiere of the CBS sitcom Young Sheldon. It is a spin-off of the long-running series Big Bang Theory. I'm not a BBT fan, and so I won't be watching Young Sheldon either. And so all I know about the show is what I gleaned from watching the commercials and promos all summer.

Apparently, the young Sheldon is a nine-year-old prodigy. He lives with his parents, his older brother, and his twin sister. As the series begins, it's the first day of school, and Sheldon has been allowed to skip several grades and start the year as a high school freshman. One promoted scene asserts that he is now in the same grade as his older brother, and thus no longer in the same grade as his twin sister.

The young Sheldon definitely enjoys learning. One scene shows him watching some program, most likely on PBS. His sister wants to change the channel to Duck Tales, but Sheldon doesn't let her, since her favorite show isn't educational. Her response is, "It's TV. You're not supposed to learn!"

But the one scene I want to focus on is just before school starts. The young Sheldon looks forward to learning at his new school. He imagines how the other high school students will treat him once they find out how smart he is. Indeed, he fantasizes, "Maybe they'll make me their leader."

Now I never watched the actual episode, but it's safe to say that nothing of the sort happens. Here's what probably happens when Sheldon arrives at the school -- the other students are not too impressed by Sheldon's smarts and act hostile towards him. At the least, they make fun of him as a "nerd," and at the worst, they might even beat him up.

The problem, of course, is Sheldon's naivete. He wrongly assumes that the other high school students value book smarts in their friends. The only way for a nine-year-old to be "made the leader" of a group of high school guys is for the youngster to be large for his age and good enough to be made the "leader" -- or captain -- of a sport, especially if it's football or basketball. (Oops, so I did mention sports in this traditionalist post anyway.)

But Sheldon is only nine, so he doesn't know any better. My problem is that many traditionalists seem to think that every student is like Sheldon. They promote the idea that every student is interested in learning, and that it's the "progressive pedagogues" who separate students from what they "want."

Also, we know that traditionalists are enamored with the idea of tracking. Students should be divided not by age but by knowledge, and so students like Sheldon ought to be promoted to high school if they are intelligent enough. But "progressive pedagogues" wish to hold Sheldon back and force him to be in the same class as his age-mates, like his twin sister, in order to spare the latter's "self-esteem."

But there are other reasons to oppose tracking besides "self-esteem." If we promote Sheldon to high school, then we expose him, as I mentioned above, to being beaten up by the others. So it's not just about the broken feelings of the students on the low track but the broken bones of the students on the high track. (Let's hope that Sheldon wasn't beaten up too badly last night!)

Speaking of nine-year-olds, I point out that I actually taught a nine-year-old last year -- of course, he was only in sixth grade, not ninth like Sheldon. His tenth birthday was in late September (so in fact he turns 11 very close to today's date). Of course, this was back when California had a later cut-off date (as the first cohort of transitional kindergartners hasn't quite made it to middle school yet). He told me that when he was four, his parents tried to enroll him in kindergarten due to his fifth birthday being before the old cut-off date -- except he was so bright that they placed him in first grade instead.

I'm sure that a student like Sheldon would be engaged by a traditionalist lesson. But many students would be more engaged by lessons other than the direct instruction of math -- like Bob. And so traditionalist lessons work well for the Sheldons of the world, but we also need nontraditional lessons like logic problems for the Bobs of the world. Some students look forward to working on the problem sets, while others just want to watch Duck Tales when they get home.

Logic problems teach students logic, which is important to math and especially Geometry. Sure, maybe Bob learns very little by solving the logic problem, but he learns nothing at all from a traditional lecture that he doesn't pay attention to. And Sheldon's sister may learn a little from a project in her math class, but not from a problem set -- if she ignores and just watches Duck Tales.

Does skipping all of middle school help the young Sheldon in the long run? I don't know, since his "future" is of course his character on the main series, which I don't watch. I do know that Sheldon is supposed to be the ultimate "nerd" on a program that's all about "nerdity" -- after all, the show is named after a theory in physics. Then again, Sheldon isn't too "nerdy" to have a girlfriend -- a promo for the main series actually shows Sheldon trying to propose marriage to her.

Of course, I must also look at myself when criticizing the young Sheldon's naivete. Last year, I told students that they must do the work in order to maintain a high grade. This assumes that the students are motivated by high grades -- Sheldon probably is, but many others aren't. My mistake, of course, was when I didn't answer "Because I said so!" when answering why they had to do the work.

Chapter 9 of Stanley Ogilvy's Excursions in Number Theory is called "Calculating Prodigies and Prodigious Calculations." It begins:

"From time to time there have come to public notice youngsters with extraordinary powers of performing mental calculations."

Such students are, of course, the exact opposite of "drens." And now you can see why I'm tying today's Ogilvy chapter to traditionalism as well. Traditionalists are worried about the youngsters who lack even ordinary powers of performing mental calculations -- the "drens."

Ogilvy begins by writing about Zerah Colburn, a nineteenth-century whiz-kid from Vermont. He tells us that at the age of nine -- yes, the same age as the young Sheldon -- he was taken by his father to England, where his math skills impressed an audience. (Speaking of Sheldon, I don't watch either BBT or YS enough to know whether he could do hard math problems in his head or not.)

According to Ogilvy, the young Zerah could do all of the following in his head:

-- raised one-digit numbers to the tenth power
-- raised two-digit numbers to the sixth, seventh, and eighth power
-- extracted a square root and a cube root, with the answer in each case a three-digit integer
-- factored the fifth Fermat number, 2^32 + 1, with the factors having three and seven digits

Each time the author tells us an impressive feat of one of these human calculators, let's compare it to something that a "dren" can't do. For example, Zerah could raise one-digit numbers to the tenth power, while some drens can't even raise one-digit numbers to the second power (though generally, even drens often find squares easier than other multiplications). In my own class, I taught my eighth graders the first ten squares and first six cubes, and most of them did well. In fact, in the future I may be bolder and teach them the first sixteen squares and first ten cubes. (The reason I keep stopping with numbers that end in 6 is that all of their powers end in 6, so these should be slightly easier than the powers of numbers ending in 7.)

Ogilvy now moves on to the German Zacharias Dase. While in his twenties, Dase mentally:

-- multiplied two eight-digit numbers in under a minute
-- multiplied two 20-digit numbers in six minutes
-- multiplied two 40-digit numbers in 40 minutes
-- multiplied two 100-digit numbers in under nine hours
-- extracted the square root of a 100-digit number in under one hour

On the other hand, a dren can't even multiply two two-digit numbers. Well, we don't expect students to multiply two-digit numbers mentally anyway, but let's say that our two digit numbers both end in zero, which is indeed something students should be able to do mentally, but some drens can't.

"The most promising young mathematician known to the present writers has some of this ability."

And this is what this mathematician wrote on the blackboard:

e^(pi sqrt(163)) = 262,537,412,640,768,743.999999999999250...

I'm tempted to say that this student must be Srinivasa Ramanujan, whom I mentioned back in my May 1st post. But Ramanujan died in 1920, so he couldn't be the young student known to the authors Ogilvy and John Anderson, who wrote this book in the 1960's. Nonetheless, the number mentioned here is known as Ramanujan's constant, since the famous Indian mathematician first noticed that it is within one-trillionth of an integer.

The next thing mentioned in this chapter is the calculation of the digits of pi. In 1962, pi was calculated to 100,000 digits. Strangely enough, there's no mention of the digit memorization record, which would seemingly belong in a chapter on math prodigies. I don't know what the record was at the time the book was written, but now the record belongs to Akira Haraguchi. He first recited 100,000 digits of pi in 2006 -- that's right, all the digits known in 1962 were recited by one man merely 44 years later!

Ogilvy now moves on to large numbers. He begins with the number described by Edward Kasner, who asked his nine-year-old nephew (young Sheldon's age again!) to name it:

One googol = 10^100.

So the numbers that Dase multiplied and found the square root of were on the order of a googol. Now Kasner went a step further and proposed yet another name:

Googolplex = 10^googol

Ogilvy now compares googol and googolplex to the Fermat numbers. If F_n is the n'th Fermat number, that is, F_n = 2^2^n + 1, then the author obtains the following results, which he includes in his table. Here I include only part of his table, but add on prime-related discoveries made after the book was published:

-- F_8 has 78 digits.
-- Googol has 100 digits. (Well, I guess it's the first 101-digit number -- 1 one, 100 zeros.)
-- 1000! has 2568 digits.
-- 2^11213-1 has 3376 digits. (This is the largest known Mersenne prime at publishing.)
-- F_17 has approx. 40,000 digits. (Smallest Fermat of unknown primality at publishing.)
-- F_36 has approx. 20 billion digits. (Smallest Fermat of unknown primality now.)
-- Googolplex has 10^100 digits.
-- F_1945 has approx. 10^600 digits. (Largest Fermat with known factor at publishing.)
-- F_9448 has approx. 10^3000 digits. (Largest Fermat with known factor now.)
-- Skewes' number has approx. 10^10,000,000,000,000,000,000,000,000,000,000,000 digits.

Again, Ogilvy reminds us that Skewes' number is:

S = e^e^e^79 = approx. 10^10^10^34.

To answer this, Ogilvy provides us with a formula that tells us how many factors of a prime p appear in the factorization of n!:

[n/p] + [n/p^2] + [n/p^3] + ...

Here [x] is the greatest integer function. It is the same function as INT, which we encountered last week in our BASIC programming. Technically, this sum is infinite, but all except finitely many terms are zero, so in practice we stop after the last nonzero term.

Ogilvy finds the number of factors of two in 9! as an example:

[9/2] + [9/2^2] + [9/2^3] + [9/2^4] STOP
= 4 + 2 + 1 + 0 = 7

He checks this by multiplying it out:

9! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9
    = 1 * 2 * 3 * (2 * 2) * 5 * (2 * 3) * 7 * (2 * 2 * 2) * (3 * 3)
    = 1 * 2^7 * 3^4 * 5 * 7.

So there are indeed seven factors of two. This means that 9! ends in seven zeros in binary. He explains why this works informally as follows -- every even factor provides a factor of two. Every multiple of four provides an additional factor of two (beyond the factor already counted by the number being even), every 8th multiple provides an extra factor of two, and so on. If we replace 2 by p we have the general proof. QED

To find the number of zeros in 1000! (in decimal), we can't use the formula since it only works for prime factors, and 10 isn't prime. Ogilvy explains that the factors of two dominate the factors of five (as there are always more of the former), and so we use 5 in the formula above:

[1000/5] + [1000/5^2] + [1000/5^3] + [1000/5^4] + [1000/5^5]
= 200 + 40 + 8 + 1 + 0 = 249.

Hence 1000! has 249 terminal zeros.

By the way, this question becomes trickier in dozenal. Suppose we wanted to know how many zeros the factorial of 1000 (great gross) ends in dozenal. The problem is that the base is 2^2 * 3, and while there are always more factors of two than three, it's not obvious whether there are twice as many factors of two as three. So we must use the formula for both 2 and 3.

[c] (default dozenal -- that is, all calculations below are in dozenal):

[1000/2] + [1000/2^2] + [1000/2^3] + ... + [1000/2^9] + [1000/2^a] + [1000/2^b]
= 600 + 300 + 160 + 90 + 46 + 23 + 11 + 6 + 3 + 1 + 0 = bb8

Notice that half of this is [bb8/2] = 5ba, which is how many factors of four (that is, pairs of factors of 2, since we're in base 2^2 * 3) we have.

[1000/3] + [1000/3^2] + [1000/3^3]+ [1000/3^4] + [1000/3^5] + [1000/3^6] + [1000/3^7]
= 400 + 140 + 54 + 19 + 7 + 2 + 0 = 5ba

We see that there are exactly twice as many factors of two as three! This isn't a coincidence -- this phenomenon occurs only in certain bases. Dozenal is the first -- the next is 39 (written in dozenal -- the base is 45 in decimal). So the answer is that there are 5ba zeros. (This is 862 in decimal.)

Ogilvy ends the chapter as follows:

"Having stretched our minds and our credulity in the chapter by strolling down a few of the lesser lanes and byways, some of which do not properly belong to number theory, we end our digression and return now to the main highway."

Well for today, the main highway is our second day of the Chapter 2 review. I've decided that when we have these two-day reviews, I'll use the second day to post interesting activities that I find from other sources.

My usual go-to sites, Fawn Nguyen and Sarah Carter, are out for Geometry, since the former is a middle school teacher and the latter is an Algebra I teacher. But hey -- the most well-known Geometry site is probably now Shaun Carter, ever since he married Sarah. (That's right -- just as physics nerd Sheldon is about to marry a fellow nerd, Shaun Carter married a fellow math teacher.)

The following worksheet is all about the three undefined terms (point, line, plane). Recall that the first question on the test asks students to identify these undefined terms. The page didn't print well for me, so you might want to find this worksheet at the original source:

Oh, and while I'm posting links, let's get back to the traditionalist Barry Garelick. Here's a link to his most recent post:

In this post, Garelick writes about progressive classes that either lack texts altogether, or use texts that are anti-traditionalist. He mentions CPM and the U of Chicago elementary texts, and also EngageNY, an online curriculum that I mentioned years ago on the blog. The EngageNY texts are based on Common Core, and I've seen some suggestions from Dr. Hung-Hsi Wu of Berkeley incorporated into the EngageNY Geometry curriculum.

The "convoluted logic" mentioned in Garelick's title is this -- progressives want to abolish textbooks because they aren't helpful in learning. But from Garelick's perspective, of course their progressive texts aren't helpful. To him, the only helpful texts are the traditionalist texts -- those which contain short explanations and lots of problem sets. Instead of abandoning textbooks for online curricula, teachers should return to traditionalist texts. Then maybe there would be a lot more Zerah Colburns in our generation today -- or at least more young Sheldons.

Monday, September 25, 2017

Chapter 2 Review (Day 28)

This is what Theoni Pappas writes on page 268 of her Magic of Mathematics:

-- A student asked, "What did Madre do at the circus?"
-- "Remember, only yes or no questions," the teacher cautioned.

This is the second page of the first section in the games chapter, "Mathematical Mystery Tales." As it happens, on page 267, Pappas begins telling us a story -- but we missed that part of the story, since the 267th day was over the weekend.

Actually, this mystery is a type of logic problem. To that end, this fits perfectly with the current Chapter 2 of the U of Chicago text, which is also on logic. We might as well go back to the beginning of the problem on page 267, since there's point in proceeding with the questions on page 268 without reading the setup on the previous page:

It was one of those weeks at the circus, when everything seemed to go wrong. First, one of the horses for the acrobat dancers went lame. Next the head clown threw a fit because the fat lady's child got into his makeup. Then Madre and his wife had an argument over the trapeze artist. The final blow came when Madre was found dead under the big top. Next to his body was his cane, which he occasionally used. An overturned glass of water was on his desk, and a minute pile of sawdust was near his body.

Pappas presents this story from the view of a teacher, Mr. Mason, reading this to his students. The class is to solve the mystery by asking the teacher yes or no questions. This is what the students discover by asking questions on page 268:

-- Madre was not a manager.
-- There was no sign of violence.
-- Something else was near the glass, and it's important to know what it was.
-- It was a pill.
-- There wasn't a pencil sharpener on his desk.
-- He died from a heart attack.

At this point, the students realize that they haven't found out what caused his heart attack yet. The student Gary replies, "Sure we have, he didn't take his medicine. Right Mr. Mason?"

Unfortunately, Mr. Mason's response is on page 269, so you'll have to wait until tomorrow before we can proceed with this mystery. I know, it's just like a TV show that ends, "To Be Continued...."

Chapter 8 of Stanley Ogilvy's Excursions in Mathematical Theory is "The primes as leftover scrap." I begin reading:

"'The prime numbers go forever'; but their frequency decreases. One of the major achievements of analytic number theory has been the discovery and proof of the Prime Number Theorem, which describes the asymptotic density of the primes."

Ogilvy starts this chapter by stating the Prime Number Theorem. According to this theorem, the number of primes less than x is approximately x/log x, and that the probability that a number about the size of x is prime is 1/log x. Actually, these are natural logs, so we should write x/ln x and 1/ln x.

According to Ogilvy, it's easy to find arbitrarily long sequences of composite numbers. For example, here are 99 consecutive composite numbers:

100! + 2, 100! + 3, 100! + 4, ..., 100! + 100.

For the first is certainly divisible by 2, since 100! is; the second by 3; and so on.

The author now asks whether there exists an arithmetic "progression" -- or arithmetic "sequence," as we're likely to call it in a Common Core Algebra I class -- consisting of only composites. There is:

10, 15, 20, 25, 30, ...

all of whose terms are multiples of 5. It's easy to see that if the first term of the sequence is a (or perhaps a_1 with a subscript of 1, as we may see in Algebra I) and the common difference is d, and a and d have a factor in common, then the sequence consists of all composites. Every member of the sequence has that same factor in common.

Just because a conditional is true, it doesn't mean that its converse must be true. According to Ogilvy, it was the 19th century German mathematician P.G.L. Dirichlet who proved the converse -- if every member of an arithmetic sequence is composite, then a and d must have some factor in common. In fact, he proved that not only must a sequence with a and d relatively prime must contain a prime, but it must contain infinitely many primes.

Ogilvy tells us that the number of primes less than x is approximately given by the function Li(x), called the "logarithmic integral" of x (since it's defined by an integral). The author now produces a table, which I don't reproduce here. The table gives, for x the third through seventh powers of 10, the value Li(x) as well as the actual number N of primes less than x and the error. Even though I don't print the table here, I will give its first and last rows:

x = 10^3, N = 168, Li(x) = 178, d = Li(x) - N = 10, d/N = rel. error = .060
x = 10^7, N = 664,579, Li(x) = 664,918, d = 339, d/N = rel. error = .0005

For small values of x, Li(x) is an overestimate and so the error d is positive. The author tells us that the error will become negative at some point before the Skewes' number, which is humongous:

S = e^e^e^79

Note, since Ogilvy's book was written, the upper bound has been reduced significantly from the one found by the South African mathematician Stanley Skewes. The current upper bound is about e^728 (or about 10^316), while the lower bound is 10^19. In my research of the current bounds, many sources use the symbol pi(x) for the what Ogilvy calls N -- the exact count of primes less than N. We notice that "pi" and "prime" both start with the letter p -- which is most likely why it was chosen.

Ogilvy now presents us with the Sieve of Eratosthenes, a classical method of finding primes. We consider a list of numbers from 1 to 100 and cross out all the multiples of 2. Traditionally, we would sieve out multiples of 3 next, but it's easier in decimal to sieve out 5, since this would leave us with numbers ending in 1, 3, 7, and 9. We then cross out multiples of 3 and then 7 -- and since this is the last prime less than sqrt(100), we stop here. All the numbers that remain -- the "leftover scrap" mentioned in the title -- are primes.

The author shows us an alternative way to find the primes. He begins by creating another table, the first row and column of which are the arithmetic sequence with a = 4 and d = 3. The second row and column are the sequence with a = 7 and d = 5, the third row and column are the sequence with a = 10 and d = 7, and so on -- that is, the differences d are successive odd numbers.

Then he claims that if x is any integer greater than 2, x is prime if and only if (x - 1)/2 does not appear in the table. Here is his proof:

n = (x - 1)/2
x = 2n + 1

And so he recreates the table by doubling and adding 1 to each entry. The resulting table resembles a multiplication table -- the upper-left corner is 9 = 3 * 3, and the rows and columns correspond to multiplying by odd numbers (except for 1). Therefore all entries are odd composites, and all the missing odds are prime. (Of course, we don't forget the even prime 2.) QED

Ogilvy wraps up the chapter by defining the lucky numbers. He tells us that they are found using a sieve similar to that of Eratosthenes. I've been omitting the tables for the primes since we know what they are, but let's actually show the sieve for the lucky numbers.

We begin by sieving out all the even numbers. Therefore all lucky numbers are odd:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57

It doesn't make sense to sieve out 1, so we sieve out 3 by crossing out every third number:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57
1, 3, 7, 9, 13, 15, 19, 21, 25, 27, 31, 33, 37, 39, 43, 45, 49, 51, 55, 57

The next number is 7, which we sieve out by crossing out every seventh number:

1, 3, 7, 9, 13, 15, 19, 21, 25, 27, 31, 33, 37, 39, 43, 45, 49, 41, 55, 57
1, 3, 7, 9, 13, 15, 21, 25, 27, 31, 33, 37, 43, 45, 49, 51, 55, 57

The next number is 9, which we sieve out by crossing out every ninth number:

1, 3, 7, 9, 13, 15, 21, 25, 27, 31, 33, 37, 43, 45, 49, 51, 55, 57
1, 3, 7, 9, 13, 15, 21, 25, 31, 33, 37, 43, 45, 49, 51, 55

Let's cross out the thirteenth, and then the fifteenth number that remains:

1, 3, 7, 9, 13, 15, 21, 25, 31, 33, 37, 43, 45, 49, 51, 55
1, 3, 7, 9, 13, 15, 21, 25, 31, 33, 37, 43, 49, 51 (63, 67, 69, 73, 75, 79, 87, 93, 99)

And these are the lucky numbers. The mathematician Stanislaw Ulam, who first named these the "lucky numbers," perhaps named them so since they are "lucky" enough to avoid being sieved out. I notice that 7, the traditional number associated with luck, is a lucky number -- but then again, so is 13, the traditionally unlucky number.

Ogilvy tells us that lucky numbers are similar to prime numbers. The probability that a number near x is lucky is again 1/ln x. The number of twin luckies less than N is similar to the number of twin primes less than N, and there is a Goldbach-like conjecture that every even number is the sum of two lucky numbers.

Here are two links to Lucky Numbers. The first is Wolfram, and the second is the OEIS:

Ulam, who first came up with lucky numbers, suggests that it's the nature of the sieving process that explains why primes and luckies are so similar. As Ogilvy concludes, primes and luckies are both:

"Leftover scrap, perhaps; but it is interesting scrap indeed that organizes itself into two (and possibly more) different and yet strangely similar heaps."

Today marks the start of the review for the Chapter 2 Test. Notice that Chapter 1 of the U of Chicago text had nine sections, and so there was no true review day for the Chapter 1 Test. But Chapters 2 through 6 each have only seven sections, and so the opposite happens -- there are two review days for each section.

In the past, I kept juggling around how I wanted to assess the first three chapters. The worksheet from two years ago prepares the students for a test with 11 questions -- and some of these questions are from Chapter 1, not just Chapter 2. But there's no harm in retesting Chapter 1 material again.

With two days of review, some teachers may use the extra day differently. Some, for example, might choose to cover Lesson 2-7 from the Third Edition of the text today. This section, "Conjectures," doesn't appear in my old Second Edition of the text. There's even an activity -- the Conjectures Game (or "Who Am I?") that I refer to several times on the blog -- that fits here. (How ironic is that? I have a Conjectures Game even though my book doesn't have a conjectures section.) Then the review worksheet can be given tomorrow instead.

This is what I wrote two years ago about today's worksheet:

Here is the rationale for which questions I decided to include on this review worksheet -- just as I did for the Chapter 1 Quiz, these problems come directly from the "Questions on SPUR Objectives" appearing at the end of each chapter.

For Chapter 1, I begin with Question 21, the three undefined terms (pointline, and plane), and then move on to Questions 26 and 32, two of the properties from arithmetic/algebra (Multiplication Property of Inequality and Substitution Property of Equality). Next are Questions 36-37, order on the number line -- except that I made the distances whole numbers, not decimals, and also I omitted point V from the second question, which serves no purpose other than to confuse and frustrate the students. Question 39 directs students to find the two points R on the number line that are the right distance from Q, and Question 41 is another distance question. Finally, I jumped to Question 61, another absolute value question similar to one that appeared on the Chapter 1 Quiz.

For Chapter 2, I begin with Question 16, which asks why the following definition is not a good definition of triangle: "A triangle is a closed path with three sides." The problem is, what exactly is a "closed path"? We're not allowed to give definitions containing words that also themselves need definitions. Question 20 asks the students to rewrite a statement in if-then form, then Question 30 reminds students that just because a conditional p=>q is true, it doesn't mean that its converse q=>p must be true.

Friday, September 22, 2017

Lesson 2-7: Terms Associated with Polygons (Day 27)

This is what Theoni Pappas writes on page 265 of her Magic of Mathematics -- a table of contents for Chapter 11: "The Spell of Logic, Recreation & Games: The Three Mathematical Musketeers."

Table of Contents:
-- Mathematical Mystery Tales
-- Putting Your Logic to Work
-- The Games Mathematicians Play
-- Some Mathematical Recreations
-- Magic Squares & More Recreations
-- The Konigsberg Bridge Problem Update
-- Checkerboard Mania
-- A Few Oldies

This chapter sounds like fun! But unfortunately, we've reached the weekend, and so we have to wait until next week to play any of these games. Instead, let's celebrate the fall equinox today by continuing in Ogilvy's book.

Chapter 7 of Stanley Ogilvy's Excursions in Number Theory is called "Number Curios." It begins:

"It is assuredly curious that 142857, when multiplied respectively by 2, 3, 4, 5, and 6 yields always a product consisting of a cyclic permutation of the original six digits. But we analyzed this behavior in Chapter 5, answered the question 'why,' and examined the problem in connection with other allied problems."

Ogilvy tells us that a number "curio" is a "curiosity" -- something interesting that happens with numbers as a mere coincidence. He explains that the property of 142857 is not a coincidence, and he explained why in an earlier chapter (i.e., it's the period of 1/7). Instead, his first true curio is:

a^b * c^d = abcd (read as four digits, not multiplication)

The lone solution was found by mathematician and author Joseph de Grazia:

2^5 * 9^2 = 2592

which he finds by trial and error. Ogilvy tells us that some curios are base-dependent -- so the above example depends on the decimal system. It's obvious, for example, that a^b * c^d = abcd has no solution in binary -- the only digits are 0 and 1, and the only powers of 0 and 1 are 0 and 1, so we can never get a four-digit answer.

Ogilvy gives us the following fraction:

16/64 = 1/4

where the 6's "cancel," and asks whether there are any more examples aside from the trivial:

44/44 = 4/4 = 1

where the 4's "cancel." To solve this, Ogilvy lets the two-digit numbers be xy and yz, where y is the digit that "cancels." We write this "cancellation" formally as:

(10x + y)/(10y + z) = x/z,

which says that

9xz = y(10x - z).

Now it's possible that the 9 on the left-hand side divides the (10x - z) on the right-hand side. Then:

10x - z == 0 (mod 9)
10x == z (mod 9)

But then

10 == 1 (mod 9)
10x == x (mod 9)

which implies that z == x (since both are nonzero digits), and this is trivial (like 44/44). So the only interesting case is if 9 doesn't divide (10x - z). Since 9 isn't prime, this doesn't imply that 9 must divide the other factor y on the right-hand side, but it does imply that 9 and y have a factor in common, such as 3. We then use trial-and-error on all cases, since only 3's, 6's, and 9's can cancel:

26/65, 19/95, and 49/98

The author also gives us an example where two digits cancel:

143185/1701856 = 1435/17056.

Even though Ogilvy doesn't give examples in other bases, I want to consider them now. Yes, I know it's amazing that I keep returning to alternate number bases over and over in this book. Over the summer, I spend a day or two on number bases in reading the Pappas book. It's one of my favorite topics, yet I didn't do it justice over the summer, so I'm squeezing it in during Ogilvy's book.

The equation from earlier:

9xz = y(10x - z)

generalizes to other number bases as:

omega * xz = y(base * x - z)

Except for trivial solutions, y and omega (that is, base - 1) have a factor other than 1 in common. In fact, I notice that in the case where y = omega, the decimal examples 19/95 and 49/98 generalize to many bases. For each divisor of the base (other than 1 and the base itself), let x be one less than the divisor and z be the divisor subtracted from the base. This works in all bases other than prime bases, since prime bases have no divisors other than 1 and the base itself:

Quaternary: 13/32 = 1/2
Senary: 15/53 = 1/3, 25/54 = 2/4
Octal: 17/74 = 1/4, 37/76 = 3/6
Dozenal: 1b/b6 = 1/6, 2b/b8 = 2/8, 3b/b9 = 3/9, 5b/ba = 5/a
14-mal: 1d/d7 = 1/7, 6d/dc = 6/c

All of these bases have prime omegas, and so no others are possible in these bases.

For bases with composite omegas, there are other possibilities for y, due to the requirement that the y and omega not be relatively prime. In quinary, y = 2 is possible, but a quick check of all possible numbers shows us that 2 doesn't work. In septimal, y = 2, 3, 4 are all possible, but as it turns out, none of these work either. In nonary, something interesting happens. Since the base itself is composite, it has an example where the omega 8 cancels, 28/86 = 2/6. But this can be reduced to 14/43 = 1/3:

Nonary: 14/43 = 1/3, 28/86 = 2/6

I ran a short TI program to find more examples. The next odd base with cancellations is 15-mal (pentadecimal), which figures since both the base and its omega are composite. It's the first base to have six cancelling fractions:

Pentadecimal: 17/75, 27/76, 2c/c9, 2e/e9, 3c/ca, 4e/ec

And hexadecimal (used in computers) beats base 15. It's the first base with seven fractions:

Hexadecimal: 15/54, 19/96, 1f/f8, 2a/a8, 39/98, 3f/fc, 7f/fe

There's no need to go any further, since by this point we're beyond the human scale. I do point out that I was unable to find any prime base with a cancelling fraction, even though theoretically they should exist since they have composite omegas.

Returning to Ogilvy, his next curio involves perfect numbers:

6, 28, 496, 8128, 33550336, 8589869056, ...

As it turns out, when we repeatedly add the digits of any perfect number, the answer is always 1 -- except for the first perfect number 6. Ogilvy asks, why is the digital root always 1, and why is 6 the only exception?

To answer this, we recall that the perfect numbers are of the form 2^(p-1)(2^p - 1), p prime. We begin by working in mod 3, noting that if p is an odd prime then p - 1 is even:

2 == -1 (mod 3)
Therefore 2^(p-1) == 1 (mod 3)

So 2^(p - 1) = 3k + 1 for some k. Then we multiply by 2:

2^p = 6k + 2
2^p - 1 = 6k + 1
2^(p-1)(2^p - 1) = (3k + 1)(6k + 1)
                           = 18k^2 + 9k + 1 == (1 mod 9)
2^(p-1)(2^p - 1) - 1 == 0 (mod 9)

Since we're thinking about mod 9, we consider the omega rule for 9. Multiples of 9 have a digital root of 9, and so numbers with a digital root of 1 are == 1 mod 9. So we just proved that perfect numbers have a digital root of 1.

That is, except for the perfect number 6. The reason for the exception is that we assumed that p is an odd prime, while the perfect number 6 corresponds to the prime p = 2, which is not an odd prime.

In the next example, Ogilvy notes that the product of a two-digit number and its reversal (57 * 75) is never a square, except in the trivial case where a number is its own reversal (55). He tells us that there are non-trivial examples of square products of longer numbers and their reversals:

169 * 961 = 162409 = 403^2
1089 * 9801 = 10673289 = 3267^2

But in case, the number and its reversal are both squares themselves. Ogilvy conjectures that the product of a number and its reversal is never a square unless both factors are squares or palindromes.

The next example is definitely base-dependent. Our base 10, factors as 2 * 5, and so its possible to factor the first few powers of 10 without any zeros in the factors:

10^2 = 2^2 * 5^2 = 4 * 25
10^3 = 2^3 * 5^3 = 8 * 125

Ogilvy writes that the only other known powers of 10 that factor without zeros are 4, 5, 6, 7, 9, 18, and finally 33. But 10^8 doesn't work because 5^8 = 390625 has a zero. As the powers gain more digits, it's more likely that a zero will show up in either 2^n or 5^n, and so it's likely that there are no more powers of 10 that can factor without zeros.

The author continues looking at the digits of 2^n and 5^n:

2, 4, 8, 16, 32, 64, 128, 256, 512, ...

He notices that the final digits are periodic: 2-4-8-6. Moreover, the tens, hundreds, and further digits are also periodic, but they take longer to repeat. Ogilvy tells us that the digits in the nth place from the right repeat with period 4*5^(n - 1).

5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125, ...

Obviously all the units digits are 5, and all the tens digits are 2. Technically, this means that both are periodic with period 1. The hundreds digits are seen to repeat 1-6 (period 2), and as it turns out, the thousands digits repeat 3-5-8-0 (period 4). The general period is 1/2*2^(n - 1) for n > 1. That is,

5^(n - 1) controls the period length in the powers of 2
2^(n - 1) controls the period length in the powers of 5

Obviously, in other bases besides 10 = 2 * 5, the powers of the factors of the base follow somewhat different patterns.

In the next example (which is base-independent), Ogilvy tries to find integers to solve:

(a^2 + b^2)(c^2 + d^2) = 2

The simplest solution is:

(2^2 + 4^2)/(1^2 + 3^2) = 2

The author gives us a more spectacular curio:

(3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2)/(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2) = 2

In the final example of the chapter, Ogilvy provides some more patterns:

                          1 +  2  =   3
                  4 +   5 +  6  =   7 +   8
          9 + 10 + 11 + 12 = 13 + 14 + 15
16 + 17 + 18 + 19 + 20 = 21 + 22 + 23 + 24

There's a similar pattern for sums of squares:

                                      3^2 +  4^2 =   5^2
                        10^2 + 11^2 + 12^2 = 13^2 + 14^2
            21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2
36^2 + 37^2 + 38^2 + 39^2 + 40^2 = 41^2 + 42^2 + 43^2 + 44^2

The author tells us that in the first sequence, the last term before the equal sign in the n'th row is given as n(n + 1), and in the second, the last term before equals is [2n(n + 1)]^2. So then we wonder whether there is a similar pattern for cubes with [3n(n + 1)]^2. But then he points out that the first equation of that sequence would be:

5^3 + 6^3 = 7^3

which contradicts Fermat's Last Theorem.

Ogilvy uses this as an excuse to write about FLT a little more. Even though the equation above is false, the two sides really are close:

5^3 + 6^3 = 341
          7^3 = 343

He asks, why is it that:

x^n + y^n = z^n

has infinitely many solutions for n = 1 and n = 2, yet no solutions for higher n? His answer is:

"This is the question that has baffled mathematicians for centuries."

That should be had baffled, of course, until Andrew Wiles came along.

The case n = 2 is associated with triangles -- right triangles, to be exact. In today's lesson, students learn about triangles and other polygons.

Lesson 2-7 of the U of Chicago text is called "Terms Associated with Polygons." (It appears as Lesson 2-6 in the modern edition of the text.)

This is what I wrote two years ago about today's lesson:

Lesson 2-7 of the U of Chicago text deals with polygons. Notice that this lesson consists almost entirely of definitions and examples. But this chapter was setting up for this lesson, since a polygon is defined (Lesson 2-5, Definitions) in terms of unions (Lesson 2-6, Unions and Intersections) of segments:

A polygon is the union of three or more segments in the same plane such that each segment intersects exactly two others, one at each of its endpoints.

It follows that this section will be very tough on -- but very important for -- English learners. I made sure that there is plenty of room for the students to include both examples and non-examples of polygons. The names of n-gons for various values of n -- given as a list in the text -- will be given in a chart on my worksheet.

The text moves on to define a polygonal region. Many people -- students and teachers alike -- often abuse the term polygon by using it to refer to both the polygon and the polygonal region (which contains both the polygon and its interior). Indeed, even this book does it -- when we reach the chapter on area. Technically, triangles don't have areas -- triangular regions have areas -- but nearly every textbook refers to the "area of a triangle," not the "area of a triangular region." Our text mentions polygonal regions to define the convexity of a polygon -- in particular, if the polygonal region is convex (that is, if any segment whose endpoints lie in the region lies entirely in the region), then the polygon itself is convex.

The text then proceeds to define equilateral, isosceles, and scalene triangles. A triangle hierarchy is shown -- probably to prepare students for the more complicated quadrilateral hierarchy in a later chapter.

Many math teachers who write blogs say that they sometimes show YouTube videos in class. Here is one that gives a song about the three types of triangle. It comes from a TV show from my youth -- a PBS show called "Square One TV." This show contains several songs that may be appropriate for various levels of math, but I don't believe that I've ever seen any teacher recommend them for the classroom. I suspect it's because a teacher has to be exactly the correct age to have been in the target demographic when the show first aired and therefore have fond memories of the show. So let me be the first to recommend this link:

Another song from Square One TV that's relevant to this lesson is "Shape Up." Notice that many geometric figures appear on the singer's head -- though not every shape appearing on her head is a polygon:

2017 Update: Today is an activity day. Two years ago, right after Lesson 2-7, I posted activities for the Daffynition Game and Jeopardy, and so I repeat those activities today. This is what I wrote two years ago about these activities:

And now I present my worksheet for the Daffynition Game. Remember that only one of these worksheets need to be given to each group -- in particular, to the scorekeeper in each group. The students write their guesses for Rounds 1-4 (or 5) on their own separate sheet of paper. I recommend that it be torn into strips so that they are harder to recognize. And the teacher provides the index cards, one for each student. Make sure that the students give back the index cards so you can reuse them for the next period. The students may keep their "guess cards," so there should be one for every student in every period.

The second page is for the Jeopardy game -- just as with the Daffynition game, there should be index cards, with the number of points on one side and the question (um, the answer, since the response is the question) on the other. In my class the questions were taped to the front board. In this version of the game, the categories correspond to the four lessons covered earlier this week. Of course, some lessons, such as Lessons 3-1 and 3-2 on angles, are tailor-made for Jeopardy, but unfortunately we haven't quite covered the lesson. Of course, we'll get there next week. I didn't include a Final Jeopardy Question, but here's a tricky one:

Final Jeopardy Category: Types of Polygons

If two points lie in the interior of this type of polygon, then the segment joining them lies in the interior.