Wednesday, November 26, 2014

The Common Core Debate, Grades 4-7

"Give a man a fish and you feed him for a day," the ancient proverb goes. "Teach a man to fish and you feed him for a lifetime."

Last night, I tutored my geometry student. His school is one of those that observes a five-day Thanksgiving vacation, from Wednesday to Sunday. So naturally, he was not in a mood to study geometry on his first night of vacation. But still, he hadn't done as well on his last quiz, which covered the basics of proof, and so I need to help show him what he did wrong.

So I showed him some worksheets from the blog. I began with Section 3-3, "Justifying Conclusions," but as it turned out, what helped him was Section 4-4, "The First Theorem in Euclid's Elements." In fact, the proof he needed was very similar to Example 2, which I included on my worksheet. The only difference was that he needed to prove AC = BD rather than AB = CD.

Given: B is the midpoint of AC
           C is the midpoint of BD
Prove: AC = BD

Now naturally, my student would've preferred that I just give him a fish and tell him the missing steps in his proof. But instead, I taught him to be a fisherman and worked on finding the missing steps of the proof together. His teacher was expecting a paragraph proof -- not the two-column proof that the U of Chicago text and I provided. Here is a complete answer, by his teacher's standards:

We are given that B is the midpoint of AC and C is the midpoint of AD. By the definition of midpoint, AB is congruent to BC and BC is congruent to CD. By the Transitive Property, AB is congruent to CD. By the definition of congruent, AB = CD. By the Addition Property of Equality, AB + BC = BC + CD. By the Segment Addition Property, AB + BC = AC and BC + CD = BD. And substituting, we get AC = BD. QED

As it turns out, my student had forgotten the "by the definition of congruent" and the "substituting" steps of the proof. And so I taught my student to become a fisherman, and fish out the steps that often come up in geometry proofs. This is the context in which I want to discuss the Common Core Standards. Are the math standards for grades 4-7 the equivalent of teaching men to fish or just giving them fish?

Before we begin, you may be wondering, why did I choose grades 4-7 as this grade band? Well, back when I was in the third grade, our school had a policy for which students were allowed to play on the jungle gym, or "big toy," during afternoon recess for grades 1-4. Each grade was assigned a different day of the week on which they could play on the big toy -- first graders on Monday, second graders on Tuesday, third graders on Thursday, and fourth graders on Friday. (Wednesdays were common planning days -- students went home early and there was no afternoon recess.) But a strange thing happened -- there were many kids using the big toy on Mondays, Tuesdays, and Thursdays, but on Fridays it was almost empty. This was because fourth graders were "big kids" -- too mature to want to play on something as juvenile as a big toy. And so the teacher who was assigned to watch afternoon recess on Fridays allowed us third graders to have a second day on the big toy!

The academic work became tougher in the fourth grade as well. Up until then, our homework mainly consisted of spelling words, but in the fourth grade -- here in California, the standards require students to learn about our state's history that year -- we had our first extended project. We had to visit a mission set up by Spanish explorers and write a report about it. The report card shows that my grades dropped slightly that year -- I'd earned straight A's in second and third grades, but received a few B's during the fourth grade. (Of course, my math grade was still an A.)

Last week, I subbed at a middle school -- but I very nearly subbed at an elementary school. As it turned out, the only teachers at the middle school who needed a sub were coaches who were only out the last two periods of the day, so they wanted to have me go to an elementary school -- most likely the one that's right next door to the middle school. I observed the school's bell schedule, and noticed that in this district, fourth and fifth graders have a longer school day -- nearly half an hour more than students in grades K-3. And furthermore, there are even a few schools in the district that are only for fourth and fifth graders! So we see that in this district, there is a sharp distinction between the "little kids" of grades K-3 and the "big kids" of grades 4-5. (I ended up turning down the offer, since I'm not accustomed to working in elementary. I'm not used to having to deal with students having to line up for lunch, or supervising them for recess, or dealing with lessons for multiple subjects. Instead, I ended up subbing in a math class for sixth graders -- some of whom were only a few months older than the fifth graders I might have taught instead.)

Now on to the math. What math should students learn in grades 4-7? Well, there's one word that many preteens dread hearing during math class, long before hearing the word "algebra" -- and that word, of course, is "fractions." Dr. Hung-Hsi Wu not only wrote an extensive essay regarding Common Core Geometry. He also wrote about how fractions should be covered under the Core:

Why do we make students learn fractions? Well, all the math in grades 4-7 is supposed to culminate in the study of one of the most important sets in mathematics -- the set of rational numbers. This set is so important that mathematics use the bold letter Q to refer to the set of rational numbers. The letter Q stands for the German word Quotient, which means essentially the same thing that it means in English. The rational numbers are the quotients of integers -- that is, they are obtained by dividing the integers. The reason that Q is so important is that it is the simplest ordered field.  A field is basically a set in which one can add, subtract, multiply, and divide any two members of the set we want (with one exception -- we can't divide by zero).

There are four fields mentioned in the Common Core Standards. They are:

Q, the field of rational numbers (Grades 6-7)
R, the field of real numbers (Grades 8-Algebra I)
C, the field of complex numbers (Algebra II)
R(X), the field of rational functions (Algebra II)

This last field appears in the following standard:

(+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.

(Notice that the phrase "system analogous to the rational numbers" is just another way of saying "field" without using that word.)

But there's more to Q than simply being a field. The set Q is an ordered field -- by "order" we mean greater than and less than. In order to qualify as an ordered field, the field must have < and > inequalities that satisfy the four "Properties of Inequality and Operations" mentioned in Section 1-7 of the U of Chicago text. Clearly both Q and R are ordered fields. As it turns out, one can prove that R(X) is also an ordered field, but C isn't.

Of course, there is one exception to this -- division by zero. If we're going to work hard to make 1 / 2 possible, why not 1 / 0?  Well, making 1 / 0 leads to contradictions. Indeed, there is a classic "proof" that 2 = 1:

Given: a = b
Prove: 2 = 1

Statements                    Reasons
1. a = b                         1. Given
2. a^2 = ab                   2. Multiplication Property of Equality
3. a^2 - b^2 = ab - b^2 3. Subtraction Property of Equality
4. (a+b)(a-b) = b(a-b)  4. Factoring (officially, Distributive Property)
5. a + b = b                   5. Division Property of Equality
6. b + b = b                   6. Substitution Property of Equality
7. 2b = b                       7. Combining like terms (officially, Distributive Property)
8. 2 = 1                         8. Division Property of Equality

But the proof is fallacious. In step 5, we divided by zero (disguised as a - b). Because of this contradiction, we instead conclude that division by zero is impossible. Therefore, in a field we may add, subtract, multiply, or divide any two elements, provided that we don't divide by zero.

Now Q is a field, but it's by no means the simplest field. Q is an infinite set, but the simplest field has only two members -- 0 and 1! In order to be a field, we must be able to add, subtract, multiply, and divide any two members (except division by zero), and here's how we do it in the field GF(2):

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0

0 - 0 = 0
0 - 1 = 1
1 - 0 = 1
1 - 1 = 0

0 * 0 = 0
0 * 1 = 0
1 * 0 = 0
1 * 1 = 1

0 / 1 = 0
1 / 1 = 1

Now, I bet that those preteens who hate fractions would love it if math classes would teach the field GF(2) instead of Q. Compared to QGF(2) is so easy. Everyone would get straight A's in math if this were taught instead of fractions! But that would be giving the students fish. We want them to learn how to be fishermen instead. The field GF(2) doesn't describe the real world. In real life, one fish plus one fish is more than one fish. We want 1 + 1 > 1, not 1 + 1 = 0. We want an ordered field, where addition works the way we expect it to. In short, we need Q, not GF(2).

When people complain about the Common Core Standards in grades 4-7, the two most common complaints are of the form:

  • Students are not receiving credit for correct answers.
  • Students are receiving credit for incorrect answers.
The issue here is on questions that require a student to explain how he or she obtained an answer. I notice what ends up happening is that a student gets an answer perfectly correct, but can't give an adequate explanation to the answer. If the student is in grades 4-7, what ends up happening is that the parents remember when the students earned high marks in earlier grades, then observe a significant drop -- sometimes as much as two or three letter grades -- after the transition to Common Core. They often say that their children loved math before Common Core, and now they hate it. On the other hand, these parents fear that other students can give incorrect answers to the same questions, yet come up with explanations so ingenious that they can receive credit.

What should happen when a student gets a question wrong? To many traditionalists, that student should be told "You're wrong!" as quickly and emphatically as possible. To say anything else is to give the student a fish. Students who aren't told that they are wrong -- the traditionalists fear -- will have misconceptions for years and will never become fishermen.

But there is an underlying assumption made by the traditionalists here. They assume that the students will meekly accept that they are wrong and learn the correct answer. But human nature tells us otherwise. In his work How to Win Friends & Influence People, nearly 80 years ago, Dale Carnegie writes that most people are resistant to being told that they are wrong. Rather than correcting themselves, they will become defensive. This is why one of Carnegie's principles is, "Show respect for the other person's opinions. Never say, 'You're wrong.'"

But now I can hear you saying that this is mathematics, not opinion. Any student who tries to claim that 2 + 2 = 5 doesn't have an opinion that any teacher is bound to respect, because that statement flies in the face of the fact that 2 + 2 = 4. I admit that I was as doubtful as you when I first read Carnegie's book. Yet still, simply telling the student that he or she is wrong and that 2 + 2 is really 4 won't result in the student magically accepting it. A student is more likely to say, "No, you're wrong, 2 + 2 = 5," than "You're right. 2 + 2 is really 4."

One Carnegie principle that can be especially helpful in a math class is, "Get the other person saying 'yes, yes' immediately." Here's an example of "yes, yes" in a math class:

Student: 2 + 2 = 5.
Teacher: Well, if you have 2 apples and add one more, you have 3 apples, right?
S: Yes.
T: And if you add one more to that, you have 4 apples, right?
S: Yes.
T: So now that you added 2 apples to 2 apples, how many apples do you have?
S: 4.

According to Carnegie, by getting the student to say "yes, yes," the teacher has led the student to the inescapable conclusion that 2 + 2 = 4, and this is more effective than getting into a one-on-one argument, student vs. teacher, with the teacher saying 2 + 2 = 4 and the student saying 2 + 2 = 5.

Even though nearly everyone is subject to the human nature of rejecting "You're wrong," I suspect that preteens -- and teens, of course -- are especially resistant to "You're wrong." It is because of this that while I support traditionalism in the lower grades, I cannot be a full traditionalist for fourth grade and above -- instead I support a mixture of traditionalist and progressive methods.

Sarah Hagan, a high school teacher from Oklahoma, discusses how her Algebra I students have trouble with overgeneralizing -- which is one of the most common errors made at this level. The students tell her that since two negatives make a positive, the sum of -3 and -5 must be +8. And -- despite what the traditionalists believe -- Hagan's telling the students that they are wrong and that the sum is -8 didn't lead to the students correcting themselves:

Because as soon as I start reteaching something that they have heard before, their minds shut down and start ignoring me.  I guess they are thinking, "I don't have to listen.  I already know this!"  But, the problem is that they don't know this.  They think that a negative exponent means that you need to change the fraction to its reciprocal to make the exponents positive.  In some cases, this works.  But, they are overgeneralizing.  They've been told that two negatives make a positive.  So, -3 + (-5) must be +8.  Again, they've taken a rule for multiplication and division and overgeneralized it.  And, don't even get me started on the order of operations.  No matter how many times I say that multiplication and division must be performed from left to right, I have a student who will argue with me that multiplication comes before division in PEMDAS so we must always do it first.

I repeat this for emphasis -- upon being told that they were wrong, Hagan's students either ignored or argued with their teacher. As Carnegie warned, the students certainly didn't correct themselves. I would point out that one easy way to get the students to realize that they are wrong without raising their defenses would be to have them add -3 to -5 on a calculator. As soon as they saw that -8 smiling back at them on their calculator screen, they'd know that they were wrong. But many traditionalists are also opposed to calculator use at any time prior to pre-calculus.

Now I want to write some math standards for grades 4-7, but what should they accomplish? Notice that the traditionalists often fear that their children's grades will drop significantly with the new Common Core Standards, but notice that they aren't actually opposed to huge grade drops. They just want the grades to drop for the right reasons. Students who can give all the correct answers, yet can't give the correct explanation that the teachers are looking for, don't deserve grade drops. But students who've been skating along by giving clever explanations and excuses without knowing the correct answers can and should have their grades drop fast.

But I say that it's neither those who know all the right answers, nor those who can give the best explanations, but those who can apply math to real-world situation, who deserve the rewards. So a student who can multiply 58 times 46 perfectly using the standard algorithm, but can't tell me the area of their backyard that measures 58 feet by 46 feet, deserves a grade drop. I'd much prefer the student who can tell me the area, but uses a nonstandard algorithm (such as the much-maligned "lattice method") to perform the multiplication. Explanations matter only to the extent that someone who explains why they multiplied 58 and 46 to find the area is more likely to know how to find area for years to come than someone who can't explain why. There's zero value in getting right answers if they answer the wrong question.

So what this all boils down to is, are the assessments -- that is, PARCC, SBAC, and worksheets created in the name of Common Core -- authentic enough that only those who can apply the math that they learned get the highest scores? I don't know -- but I suspect not, only because truly authentic assessments are extremely difficult to write. I want the standards that I propose here to lend themselves to such authentic assessment, but I will almost certainly fail.

So instead, I look online and read about the texts that homeschoolers -- especially parents who say that they took their students out of public schools specifically to avoid Common Core -- use to teach their preteens math. One text that's often mentioned is published by John Saxon. And I was able to purchase a Saxon text -- specifically, Saxon Math 65 (Second Edition) -- for $1, from the same library where I found the U of Chicago text.

There are 140 lessons in the Saxon text. Presumably, one lesson is to be given per day. There are 180 days in the school year, and we can assume that 40 days are given up to first/last day of school bureaucratic issues, assemblies, or testing (both teacher directed and state/Common Core). Here are a few randomly chosen lessons mentioned in the Saxon text:

Lesson 49 - Reading an Inch Ruler to the Nearest Fourth of an Inch
Lesson 56 - Making Equal Groups to Find an Average
Lesson 57 - Multiplying by Two-Digit Numbers
Lesson 65 - Dividing and Writing Quotients with Fractions
Lesson 75 - Writing Hundredths in Decimal Form
Lesson 82 - Adding and Subtracting Decimal Numbers, Part 1
Lesson 90 - Identifying Prime Numbers
Lesson 121 - Multiplying Decimal Numbers by 10, 100, and 1000
Lesson 126 - Probability and Chance
Lesson 131 - Dividing Decimal Numbers: Keep Dividing

We see a mixture of traditionalist and progressive lessons in this random cross-section. Lesson 57 is traditionalist as it focuses on the standard algorithm for multiplication, and Lessons 65 and 131 show long division with fractions and decimals. But other lessons are more progressive and focus on the practical uses of math. The title of Lesson 49 makes its real-life use explicit, and Lesson 126 focuses on probability, which has many real life uses (weather reports, games of chance, etc.). Lesson 56 sounds very progressive -- a traditionalist would just add up all the values and divide by the number of values to find the mean -- until we see that the making equal groups is just there to motivate the definition of mean.

Therefore, I recommend the Saxon text for teaching math to preteens. Notice that the title Math 65 indicates two grade levels -- the text is appropriate for below average sixth graders and above average fifth graders. At the school-age level, the Saxon books are numbered with a single grade, but at the preteen level they are numbered with two grades. At the teen-age level, Saxon originally based his texts on an integrated sequence (despite naming his texts Algebra 1, Algebra 2, Advanced Math), but in the Fourth Edition, the texts follow a traditional sequence, with Geometry thrown in.

But recommending a text isn't the same as giving standards. It's difficult to convert the Saxon lessons into standards because the text jumps around topics so often. (Indeed, for all the praise the Saxon texts receive, this is a frequent criticism of them.) Notice that Lessons 56 and 57 above aren't directly related at all. This is more striking when we see Lessons 55-58 in sequence:

Lesson 55 - Solving Two-Step Word Problems

Lesson 56 - Making Equal Groups to Find an Average
Lesson 57 - Multiplying by Two-Digit Numbers
Lesson 58 - Identifying Place Value Through Hundred-Millions

although 54 ("Finding Information to Solve Problems") is clearly related to 55, just as 59 ("Naming Numbers Through Hundred Millions") obviously follows after 58.

Then again, it's instructive to see how the Saxon lessons correlate with Common Core. Lesson 49 is actually a third grade standard in Common Core:

Generate measurement data by measuring lengths using rulers marked with halves and fourths of an inch. Show the data by making a line plot, where the horizontal scale is marked off in appropriate units— whole numbers, halves, or quarters.

Lesson 57 is on multiplying two-digit numbers. The standard algorithm for multiplying is given as a fifth grade standard:

Fluently multiply multi-digit whole numbers using the standard algorithm.

Lesson 121 is on multiplying decimals by 10, 100, or 1000. Many teachers give horror stories about students who are asked to multiply decimals by a power of ten and say "I don't know!" instead of "That's so easy -- just move the decimal point!" Indeed, I'd follow that up with, "I wish all decimal multiplication problems were by easy numbers like 10, 100, and 1000 instead of hard numbers like 7, 58, or 3.14.") Many traditionalists lament the Core's emphasis on ten and its multiples and powers, such as asking first graders to "make ten" when adding 8 + 6. What do I mean by "make ten"? It's best just to look at the relevant standard directly:

Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).

I tend to agree with the traditionalists here (as I usually do regarding the first grade). But multiplying decimals is definitely a proper use of ten and its powers. It's a fifth grade standard:

Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10.

As we expect, most of the lessons in Math 65 correspond to fifth and sixth grade standards. A few are introduced slightly earlier in the Core. But probability, Lesson 126, is a seventh grade standard:

Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around 1/2 indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event.

Finally, notice that assigning two grade levels to this content can lead directly into tracking -- Math 65 is intended for sixth graders on a lower track and fifth graders on a higher track. I'd discussed tracking in a previous post and would like to continue that discussion, but not in this post since it's already bloated longer than I want it to be. The issue with tracking is, can one truly be fair to the students on the lower track?

The answer is, go back to the proverb at the beginning of this post. If one is denying fish to those on the lower track in order to teach them to become fishermen, then I support it, as long as one is genuinely teaching them to become fishermen. But if one is merely saying that he is teaching those on the lower track to be fishermen only as an excuse to deny them fish and let them starve, then I am opposed to it. And I believe that both of the above are occurring in various tracked schools. 

Happy Thanksgiving, everyone! My next post should be on my new computer -- hopefully it will be on December 1st, but it depends on when the Internet is connect to the new PC. Once again, I'll begin Day 70 on December 1st with Section 7-1 of the U of Chicago text, but if I must start late I'll cover Section 7-2 no matter what.

Friday, November 21, 2014

Activity: Drawing Triangles -- Based on Section 7-1 (Day 69)

Here it is, the last day before Thanksgiving vacation. This means that I should be taking all of next week off on this blog and wait to post Day 70 on Monday, December 1st.

But here's the problem. I've been wanting to purchase a new computer to replace my current PC. It is seven-and-a-half years old, and you know how computers fail when they are so old. Not only do I finally have the money to pay for a new computer thanks to my new job, but coming up is the perfect opportunity to buy it on a day when many retailers will have huge discounts -- Black Friday.

Now if I get the new computer, it's likely that I'll have to call our phone company to make an appointment to restore my Internet service -- and you know how things like that go. So, after taking a week off from this blog to celebrate Thanksgiving, I take another even more time off since I won't be able to post until my Internet is restored. And all of this will right when I'm supposed to post on Chapter 7, the chapter on triangle congruence. Even before Common Core, this is one of the most important chapters in the book, and now it's even more critical with the new standards.

Instead, what I was considering doing is treating Monday, Tuesday, and Wednesday as regular school days, even though my districts are closed those days. Then I could observe a full Thanksgiving week beginning on Turkey Day itself. I would start posting Chapter 7 next week and then continue it on Thursday, December 4th, is Day 73 -- by which point, I'd have a new computer with the connection to the Internet complete.

If the Internet is disconnected for much longer than I expect and I'm forced to skip days, no matter what happens, I will post Section 7-2, the Triangle Congruence Theorems SSS, SAS, and ASA. This is because this lesson contains these three theorems proved Common Core Style -- that is, using reflections and other transformations. Teachers can obtain lessons from the other sections after 7-2 from sources other than this blog. My duty here on this blog is to introduce geometry from the perspective of Common Core, and that means the mid-level proofs of SSS, SAS, and ASA. I don't mind skipping the higher-level proofs that one can read anywhere.

So instead, what I decided to do is take a chance and observe next week as Thanksgiving week, as scheduled in the schools. But since we finished Chapter 6 yesterday and said that I would post an activity today, why not let the activity be based on Section 7-1? After all, Section 7-1 of the U of Chicago text is on Drawing Triangles -- and what better activity is there to do than drawing? So Section 7-1 naturally lends itself to being an activity anyway. If I'm able to post on Monday, December 1st, then I might do Lesson 7-1 in more detail, but if I have to skip any days, then I can go straight to Lesson 7-2 without any trouble. After that, I'll cover the rest of Chapter 7 based on how many days I have to wait before my Internet is restored.

There are nine triangles to draw for this activity, based on Questions 11-19 from Section 7-1 of the U of Chicago text. The students are to draw a triangle ABC with the given information, and then conjecture whether all other accurately drawn triangles will be congruent to theirs -- so this can be done in groups. To add some fun to this pre-holiday activity, the students may cut out and glue their triangles to another sheet. They should label the triangles SS, AA, SA, SSS, ... based on the condition to which the triangle corresponds. Then they can compare their triangles with those of the other group in order to make their congruence conjectures -- possibly even before gluing them.

Note: SSS is the hardest one to draw, unless one uses a compass or possibly even strings cut to the correct lengths. But I don't necessarily want to make the students to a hard ruler-and-compass construction during a pre-holiday lesson. Most of the others could be drawn with just a protractor and a ruler.

This concludes the activity. During the Thanksgiving week, I won't post any school lessons, but I may post the my next Common Core pros and cons entry. In October I discussed Grades K-3, and this time I'll focus on Grades 4-7. I should be able to get that posted before the big purchase.

I won't post a worksheet for this activity -- instead I just give the conditions right here. This is why I'm buying a new computer next week -- one problem on my old computer is that so often, Windows Explorer locks up and I can't open up any new programs that aren't already open. So I can't access the word processor to type up the activity without turning the computer off and on. Rather than do that (since another problem is the infamous blue screen when I restart the computer), I will just post the nine questions right here:

For each triangle:
a. Accurately draw a triangle ABC with the given information.
b. Conjecture whether all other accurately drawn triangles will be congruent to yours.

1. SS condition: AB = 2", BC = 1 3/4".
2. AA condition: angle A = 70, angle B = 38.
3. SA condition: AB = 4 cm, angle A = 60.
4. SSS condition: AB = 4 cm, BC = 5 cm, AC = 6 cm.
5. AAA condition: angle A = 41, angle B = 100, angle C = 39.
6. SSA condition: AB = 2", BC = 1", angle A = 20.
7. SAS condition: AB = 2", BC = 1", angle B = 20.
8. ASA condition: angle A = 55, AB = 3", angle B = 90.
9. AAS condition: angle A = 40, angle B = 60, BC = 3 cm.

Thursday, November 20, 2014

Chapter 6 Test (Day 68)

Today I subbed in a sixth grade math class. They were basically working on graphing points and the four quadrants. (I almost ended up subbing elementary!)

Today I post the Chapter 6 Test -- the first test I am posting on a day other than Friday. The reason I am doing so is that the district whose calendar that I'm following takes off an entire week for Thanksgiving break. So I certainly want to end the chapter before the holiday. Also, my policy is that I don't post a test on the very last day before an extended vacation. This is in deference to schools that have a minimum day -- and therefore not enough time to give a test -- on the last day before an extended vacation. No, neither of the districts I work for will have a minimum day on Friday, but I've once been at schools that do. And so today is the day that I'll post the test, and tomorrow I will post some sort of activity instead.

Some readers may notice how accommodating I am of having a test before Thanksgiving -- and that's after posting a test on Halloween. The difference is that unlike Thanksgiving, Halloween is not a holiday on which American schools typically close for even a day -- much less a full week -- nor have even a minimum day. (Of course, I did mention that some local school district apparently has staff development and two days off for students around Halloween.) To me, Halloween is a nighttime celebration, so my policy is no homework on Halloween -- as opposed to Thanksgiving, which is a week-long celebration starting as early as midday (if there's a minimum day) the Friday before.

Then again, the class I subbed in will be taking a test tomorrow. So perhaps I could have waited until tomorrow to post the test. Oh well, I already typed up this test today.

Traditionally schools had only two long breaks during the school year -- winter break near Christmas, and spring break near Easter. Those are in addition to the long summer break. But recently, more and more schools are having a third long break during the school year, Thanksgiving vacation. The first time that I heard of the phrase was in the old Peanuts comics -- and immortalized in the second Christmas special, "It's Christmas Time Again Charlie Brown." In the December 27, 1990, comic, Charles Schulz wrote the following exchange:

Peppermint Patty: Marcie, what book were we supposed to read during Thanksgiving vacation?
Marcie: This is Christmas vacation, sir..
Peppermint Patty: Christmas vacation?! How can I read something during Christmas vacation when I didn't read what I was supposed to read during Thanksgiving vacation?
Marcie: Duck, sir! Easter is coming!!

Now this comic came out when I was in the fourth grade -- and up until then, Thanksgiving was merely a four-day weekend, from Thursday to Sunday. Of course, Thanksgiving is on a Thursday, but schools have been also closing on Friday since before I was born. We discussed back on Veteran's Day how many students and teachers alike feel entitled to an extra day off when Veteran's Day falls on Tuesday or Thursday, so I can easily see how Black Friday and the four-day Thanksgiving weekend for schools first began.

Of course, since I was born, air travel has become more common, and families often live in different states on opposite coasts. Some news reports began to identify Wednesday, the day before Thanksgiving, as the biggest travel day of the year. Families that travel on Wednesday obviously can't send their children to school that day. So in the 1990's, some districts began to observe a five-day weekend from Wednesday to Sunday, including the largest district in the area, LAUSD, for a few years around this time. On my way home from subbing today, I drove past a school that still takes a five-day weekend, with Tuesday, the last day before the holiday, to be a minimum day.

The schools I attended as a student always held school the day before Thanksgiving, but for a few years, when I was in the sixth through ninth grades, a staff development day was held on the Monday after Thanksgiving (the day now called Cyber Monday, but this was back when the Internet was still in its infancy).

When I was a college student at UCLA, we had only the four-day weekend off. But I once read a Daily Bruin article suggest that the school take the entire week off. The following 2008 editorial published by the school paper (The Daily Bruin) points out that since UCLA began classes on a Thursday, called "Zero Week." So the suggestion is just to start classes the previous Monday so that the entire Thanksgiving week could be taken off. Naturally, travel is given as the top reason for having the whole week off:

As of today, UCLA still has only a four-day weekend for Thanksgiving. But shortly after that 2008 article was written, here in California, the budget crisis began. Many schools started having furlough days, and school years were fewer than 180 days. When districts chose which days to take off for the furlough days, the first three dates chosen were invariably the three days before Thanksgiving.

Then, of course, once funding for schools was restored, students and teachers alike decided that they liked having the entire week off. And so the Thanksgiving week stuck around at many schools districts, including LAUSD as well as both of the districts where I work.

In a way, the entire week off is the next logical step after a five-day weekend. Wednesday may be the biggest travel day of the year -- and so in order to get the jump on the crowds, flyers may leave on Tuesday instead. And once we take Tuesday off from school, we might as well take Monday off as well, since no one wants a lone day, a one-day week. And so the entire week is taken off.

But the week off still causes problems. Not everyone travels for Thanksgiving -- after all, someone has to host all of the big turkey dinners. And parents who don't travel may still put in full days of work Monday, Tuesday, and Wednesday, leaving their children who go to schools that take the full week off without daycare. So there is no holiday schedule that will satisfy everyone.

That's enough about Thanksgiving weekend -- now let's give the test. Here are the answers for the Chapter 6 Test, posted below:

1. a translation 2 inches to the left

2. a translation 2 inches to the right

3. a rotation with center O and magnitude 180 degrees

4. a translation 8 centimeters to the right

5. true

6. angles D and G

7. triangle DEF, triangle GHI

8. Reflexive Property of Congruence

9. definition of congruence

10. Isometries preserve distance.

11. translation

12. translation

13. glide reflection

14. glide reflection

15.-16. The trick is to reflect the hole H twice, over the walls in reverse order, and then aim the golf ball G towards the image point H". In #15, notice that y and w are parallel, so reflecting in both of them is equivalent to a translation twice the length of the course. In #16, notice that x and y are perpendicular, so reflecting in both of them is equivalent to a 180-degree rotation.

17. glide reflection (changing the sign of y is the reflection part, adding to x is the translation part)

18. translation

19. 7 (A Thanksgiving reference! These are the seven dates in November which could be turkey day!)

20. d (for dilation, of course!)

Wednesday, November 19, 2014

Review for Chapter 6 Test (Day 67)

Last night I tutored my geometry student. He was learning from Section 2-4 of the Glencoe text, which is titled "Deductive Reasoning." Even though Chapter 2 of both the Glencoe and U of Chicago texts cover logic, the worksheets that I covered back in the second chapter are very different from what the student learned today. In fact, the closest counterpart in the U of Chicago text is actually Section 13-1, "The Logic of Making Conclusions."

Both Section 2-4 of the Glencoe text and Section 13-1 of the U of Chicago text mention two laws that are important to logical thinking. Both texts call the first law the "Law of Detachment," which states that from p and p=>q, conclude q. This law was mentioned in my geometry text 20 years ago, except it was in the back rather than in the main text, and called by its Latin name, modus ponens.

The second law has different names in the two texts. The Glencoe text calls it the "Law of Syllogism," and states that it is sort of like a transitive property for logic. Therefore, the U of Chicago calls it the "Law of Transitivity." From p=>q and q=>r, conclude p=>r. But two British logicians, Bertrand Russell and A.N. Whitehead, used the name "Syllogism" about a hundred years ago.

Even though there are many laws in logic, these two are by far the most important. In fact, the Metamath website -- a site that seeks to write two-column proofs for many math theorems, calls the Law of Syllogism the most commonly used assertion, followed by Detachment (modus ponens):

My student had a little trouble at first distinguishing the two, but by the end of our session he was getting the hang of it. I can see why Glencoe would place this before beginning proofs. I'm not sure why the U of Chicago waits until Chapter 13 to present this information. That one may want to delay indirect proofs until Chapter 13 is understandable, but I don't know why this lesson is in Chapter 13 while converses -- a related logical concept -- appear in Chapter 2. I had no worksheet of mine to show my student, since I'm nowhere near writing a worksheet for Lesson 13-1 yet.

Instead of Chapter 13, let's think about Chapter 6 instead. We have just about finished Chapter 6, and so let's get ready for a test. The following worksheet is intended as a review for the upcoming test on the material of Chapter 6. As usual, I base my problems on the SPUR section of the U of Chicago text, and this review worksheet may contain repeats from the Chapter 6 quiz.

Tuesday, November 18, 2014

Section 6-7: Corresponding Parts in Congruent Figures (Day 66)

Tonight I will tutor my geometry student.

Section 6-7 of the U of Chicago text covers the Corresponding Parts in Congruent Figures Theorem, which the text abbreviates as CPCF. But a special case of this theorem is more widely known -- corresponding parts in congruent triangles are congruent, or CPCTC.

When I was young, a local PBS station aired a show called Homework Hotline. After school, middle and high school students would call in their homework questions in math and English, and some would be chosen to have their questions answered on the air by special teachers. Even when I was in elementary school, I often followed the geometry proofs that were called in, and more often than not, there were triangle congruence two-column proofs where the Reason for a step was often CPCTC. So this was where I saw the abbreviation CPCTC for the first time. (By the time I reached high school, a few calculus problems were called in to the show. Nowadays, with the advent of the Internet, the show has become obsolete.)

Here's a link to an old LA Times article about Homework Hotline:

When I reached geometry, our text usually either wrote out "corresponding parts in congruent triangles are congruent," or abbreviated as "corr. parts of cong. tri. are cong.," probably with a symbol for congruent and possibly for triangle as well. But our teacher used the abbreviation CPCTC. Now most texts use the abbreviation CPCTC -- except the U of Chicago, that is. It's the only text where I see the abbreviation CPCF instead.

Dr. Franklin Mason, meanwhile, has changed his online text several times. In his latest version, Dr. M uses the abbreviation CPCTE, "corresponding parts of congruent triangles are equal."

Well, I'm going to use CPCTC in my worksheets, despite their being based on a text that uses the abbreviation CPCF instead, because CPCTC is so well known.

Once again, it all goes back to what is most easily understood by the students. Using CPCTC would confuse students if they often had to prove congruence of figures other than triangles. But as we all know, in practice the vast majority of figures to be proved congruent are triangles. In this case, using CPCF is far more confusing. Why should students had to learn the abbreviation CPCF -- especially if they have already seen CPCTC before (possibly by transferring from another class that uses a text with CPCTC, or possibly even in the eighth grade math course) -- for the sole purpose of proving the congruence of non-triangles, which they'd rarely do anyway?

So it's settled. On my worksheet, I only use CPCTC.

Notice that for many texts, CPCTC is a definition -- it's the meaning half of the old definition of congruent polygons (those having all segments and angles congruent). But for us, it's truly a theorem, as it follows from the fact that isometries preserve distance and angle measure.

In this lesson, students basically learn what CPCTC is. Of course, they won't actually use it in any proofs until Chapter 7. Notice that my lesson begins with the same example as the text, where we begin with a triangle and then translate it, reflect it, and then rotate it. I begin the same way, except I do it in a different order -- translation, then rotation, and then reflection. This is because when given two congruent triangles and one wants to map one to the other, the most intuitive way to do so is to translate it so that one pair of vertices coincides, then rotate it so that one whole side coincides, and if that's not enough to make the entire triangles coincide, then one final reflection will do the trick. This is, in fact, how Dr. M proves SAS. (This is also why my triangle images overlap -- I want to think in terms of mapping a preimage onto a target image, one isometry at a time.)

This also leads to another question. We've made a big deal about how some results require the Fifth Postulate and how others don't. Technically speaking, SSS, SAS, and ASA are all true in all three types of geometry -- Euclidean, hyperbolic, and spherical. (AAS and HL, on the other hand, do require the Fifth Postulate. This is why Dr. M introduces SSS, SAS, and ASA in his Chapter 3, but he had to wait until Chapter 4 to give AAS or HL) So we could have taught SSS, SAS, and ASA much earlier, before we covered the Fifth Postulate. But not only does this modify the U of Chicago order, but it means that we can't use translations, whose properties derive from the Fifth Postulate. Of course, we could just use reflections instead, but translations are much easier to understand. When we want to map triangle ABC onto a congruent triangle DEF, it's much easier to begin by translating A to D, rather than reflecting over the perpendicular bisector of AD. This is why we're waiting so late before introducing SSS, SAS, and ASA.

Another issue that comes up is the definition of the word "corresponding." Notice that by using isometries, it's now plain what "corresponding" parts are. Corresponding parts are the preimage and image of some isometry. Unfortunately, we use the word "corresponding angles" to mean two different things in geometry. When two lines are cut by a transversal and, "corresponding angles" are congruent, the lines are parallel, but when two triangles are congruent, "corresponding angles" (and sides) are congruent as well. The phrase "corresponding angles" has two different meanings here! Of course, one could unify the two definitions by noting that the corresponding angles at a transversal are the preimage and image under some isometry. I tried this earlier, remember? It turns out that the necessary isometry is a translation, whose properties depend on the Fifth Postulate, yet we want the Corresponding Angles Test (not the Consequence, but the Test) to be proved without using any Fifth Postulate at all.

So unfortunately, we're stuck with two unrelated definitions of "corresponding angles."

Monday, November 17, 2014

Section 6-6: Isometries (Day 65)

Section 6-5 of the U of Chicago text deals with isometries. In particular, this lesson points out that in addition to reflections, rotations, and translations, there is a fourth, often-forgotten isometry, known as the glide reflection.

What, exactly, is a glide reflection? Well, here's how the U of Chicago defines it:

Let r be the reflection in line m and T be any translation with nonzero magnitude and direction parallel to m. Then G, the composite of T and r, is a glide reflection.

Just as reflections, rotations, and translations have nicknames -- "flips," "slides," and "turns," respectively -- glide reflections have the nickname "walks." The U of Chicago gives the example of the isometry mapping the right footprint to the left footprint while walking as a glide reflection. Another name for glide reflection is "transflection," since it is the composite of a reflection and a translation.

Why do we need to worry about glide reflections? Well, we can look at the types of isometries that exist in different geometries. First of all, in 1D geometry, there are only two isometries -- reflections and translations. You can't perform a rotation on a one-dimensional line.

For two dimensions, the number of isometries depends on whether we are using Euclidean, hyperbolic, or spherical (elliptic) geometry. In particular, we have:

  • spherical: rotation, reflection, glide reflection
  • Euclidean: translation, rotation, reflection, glide reflection
  • hyperbolic: horolation, translation, rotation, reflection, glide reflection
Recall that a translation is defined to be the composite of reflections in parallel lines -- so we need to know how parallel lines work in order to know what isometries there are. In Euclidean geometry there is one type of parallel, but in hyperbolic geometry there are two types of parallel -- horoparallel and ultraparallel -- and so there are two different isometries, horolation and translation, depending on whether the mirrors are horoparallel or ultraparallel, respectively. But in spherical geometry, there are no types of parallel lines, so there are no translations at all (much less horolations).

Horolations, translations, and rotations all preserve orientation, since they are the composite of two reflections (the first mirror switching orientation and the second switching it back). But what seems counter-intuitive is that there are only two isometries that reverse orientation in all three types of geometry -- reflections and glide reflections. In hyperbolic geometry, the composite of a reflection and a translation is a glide reflection, but what about the composite of a horolation and a reflection? I suspect what happens is that if an isometry is the composite of a horolation and a reflection, one can find a translation and a (different) reflection whose composite is also that isometry, so it is really a glide reflection after all. (One cannot use the Glide Reflection Theorem mentioned in this chapter to find the translation or reflection, as its proof requires Playfair and so fails in hyperbolic geometry.)

In spherical geometry, the situation is every worse. There are no translations in spherical geometry, so how can there be glide reflections? Well, this link explains it well:

Glide-Reflections: Like Euclidean geometry, the combination of a reflection and a translation is a new kind of symmetry. We saw above that translations on the sphere are really rotations, and hence a glide-reflection could also be called a rotation-reflection.

Recall how I wrote that what appears to be a translation on the sphere is really a rotation.

Returning to Euclidean geometry, we see the above discusses the composite of a rotation and a reflection, so what happens when we compose these in Euclidean geometry? For that matter, what is the composite of a rotation and a translation? As it turns out, neither of these produces any new isometries in two dimensions. (Three dimensions is another matter.)

The reason for this is that in plane geometry, three mirrors suffice. That is, not only is any isometry the composite of just reflections, but that they are the composite of at most three reflections. The U of Chicago alludes to this fact, and shows a hierarchy where all isometries are the composite of one, two, or three reflections. The text doesn't prove this, but we can prove it here easily.

First, suppose we have three noncollinear points, A, B, and C, and another point P. Now suppose we let P' be any point other than P such that AP = AP', BP = BP', and CP = CP'. So, by the Converse of the Perpendicular Bisector Theorem, A, B, and C are all on the perpendicular bisector of PP' -- that is, they are collinear, a contradiction since these points are assumed to be noncollinear. Therefore, the conclusion is that there is no point other than P such that AP = AP'BP = BP', and CP = CP' -- that is, if P' is a point with the same distances to A, B, and C as P, then P' is the same point as P.

Now we take any isometry that maps three noncollinear points A, B, C to D, E, F. Now we begin by performing a sequence of reflections. The first mirror is the perpendicular bisector of AD. This reflection maps A to D, and B, C to new points B', C'. The second mirror is the perpendicular bisector of B'E. Notice that D is on this line, since DE = AB  = A'B' = DB' (isometries preserve distance and the Converse of the PBT). So D reflects to itself, B' reflects to E, and C' reflects to some C". Now the third mirror is the perpendicular bisector of C"F. By similar reasoning, both D and E are already on this line, and so D and E reflect to themselves and C" reflects to F.

So the composite of three reflections maps A, B, C to D, E, F. Now let O be any point in the plane, P its image under the original isometry, and P' its image under the three reflections. Since isometries preserve distance, AP = AP'BP = BP', and CP = CP'. Then P' is the same point as P, and since this is true for any point in the plane, the composite of three reflections is the original isometry. Thus any isometry is the composite of at most three reflections. QED

The "at most" part is there because not all three reflections are necessary. If A and D are the same point then the first reflection isn't necessary, and if B' and E are the same point then the second reflection isn't necessary. There's actually a 50-50 chance that the third mirror isn't necessary -- since D and E are on the reflecting line and reflections preserve angle measure, by the Two Sides of Line Assumption part of the Angle Measure Postulate, C" must be one of only two points, one of which is F and the other is its reflection across line DE.

And so the hierarchy listed in the U of Chicago is complete. Three mirrors suffice, and all of the possibilities are listed in the chart. In particular, if one were to take two congruent triangles and toss them into the air, when they land, the first can be mapped to the second by one of the four possible isometries -- reflection, rotation, translation, or glide reflection. Chances are that it will be either a rotation or a glide reflection. Think about it -- there's a 50-50 chance that the figures will have the same orientation, which would mean that it's either a translation or a rotation. But we have to be extremely lucky for it to be a translation -- that would mean that the triangles are "facing the same way," with corresponding sides exactly parallel. If the sides deviated even by, say 1 degree, from being parallel, a translation is impossible and it must be a rotation -- the magnitude of that rotation would in fact be that same 1 degree. A rotation is said to have more degrees of freedom than a translation, which is why a rotation is much more likely.

Similarly, a glide reflection has more degrees of freedom than a mere reflection, in case the two triangles have opposite orientation. Therefore, a random isometry will almost surely be either a rotation or a glide reflection. Notice that degrees of freedom and dimension are very closely related -- that's why the only two isometries possible in 1D, reflections and translations, have fewer degrees of freedom than the two isometries that require 2D.

I once tutored a geometry student who had a worksheet on glide reflections. The student had to use a coordinate plane to perform the glide reflections, which were given as the composite of a reflection and a translation. But the problem was that on the worksheet, the direction of the translation wasn't always parallel to the reflecting line! In fact, in one of the problems the translation was perpendicular to the reflecting line. That would mean that the resulting composite wasn't truly a glide reflection at all, but just a mere reflection!

Friday, November 14, 2014

The Concurrency Theorems (Day 64)

I wanted to give these concurrency theorems during Chapter 5. But that unexpected week off trying to align my blog's calendar to my district's calendar ruined my Chapter 5 plans. So I put all of the concurrency theorems right here, since I can afford to put them here without ruining Chapter 6.

As it turns out, Hung-Hsi Wu proves all four of these concurrency theorems, as opposed to the U of Chicago, which only discusses one of them. We know enough to prove two of them using results from Chapter 5 and before -- the circumcenter and orthocenter -- the other two are more difficult and must wait until Chapter 7. Therefore, most of this post will be cut-and-paste from the proofs given by Wu.

We begin with the circumcenter, the intersection of the perpendicular bisectors. This is the only concurrency theorem mentioned in the U of Chicago. (The theorem numbering is Wu's.) The first part of Wu's proof is nearly identical to the U of Chicago's:

Theorem 13.
(iThe perpendicular bisectors of the three sides of a triangle meet at
a point, called the
circumcenter of the triangle. (iiThere is a unique circle that
passes through the vertices of a triangle, and the center of this circle
(the circumcircle
of the triangleis the circumcenter.
Let the triangle be ABC, as shown, and let Mbe the midpoints of BC and
AC, respectively. Also let the perpendicular bisectors of BCAC be m and n,
respectively. Let
m and n meet at O.
lies on the perpendicular bisector of BCOB OC (Perpendicular Bisector Theorem
). Similarly,
OC OA. Together, we have OB OA OC
is equidistant from Aand C. In particular, is equidistant from and .
By the Converse of the Perpendicular Bisector Theorem, lies on the perpendicular bisector of AB
already lies on the perpendicular bisectors of BC and AC, this proves the rst
part of the theorem.
To prove the second part, we have to rst prove that there is a circle with center at
that passes through the vertices AB, and C, and that any such circle must
coincide with this circle. Consider the circle
with center and radius OAK
must pass through
and C, on account of OB OA OC, so we have proved
the existence of such a circle. Next,we have to show that any other circle
Aand must coincide with K. Let the center of Kbe O0. Since Ois
by definition equidistant from
and COlies on the perpendicular bisector of BC
(Converse to PBT), and therefore lies on
m. For exactly the same reason, O0
must also lie on
n, the perpendicular bisector of AC. Therefore Ois the point of
intersection of
m and n, which is O. This shows O0. Since Kpasses through A,
the radius of
Kis also OA. Hence Kbecause they have the same center and
the same radius. The proof of the theorem is complete.

But then, as I pointed out earlier, part of this proof depends on some Parallel Postulate. I want, in fact, to use our version of the Fifth Postulate, since all that's needed for the proof was the part about Perpendicular to Parallels. But unfortunately, I couldn't find a way to avoid an indirect proof. (Wu, in a Remark, also refers to this fact that m and n don't necessarily intersect at any point, much less O, without a Parallel Postulate. In non-Euclidean hyperbolic geometry, this theorem is in fact false.)

Assume that m and n don't intersect -- that is, that they are parallel. It's given that m is perpendicular to BC and n is perpendicular to AC. So, by the Fifth Postulate, both BC and AC are perpendicular to m, so, by the Two Perpendiculars Theorem, BC | | AC. But C is on each line. So lines BC and AC are identical, and the three points are collinear. But A, B, and C are the vertices of a triangle, so they can't be collinear. This is a contradiction. Therefore m and n intersect. QED

Notice the similarity between this and the proof of the Two Reflection Theorem for Translations. I point out that the key difference is that in the latter, we were trying to prove that three points are collinear, whereas here, we use the fact that three points aren't collinear to derive a contradiction in an indirect proof.

Now for the other proof -- the orthocenter proof. I will cut and paste Wu's proof. In his proof, he ends up showing that the orthocenter of a triangle is the circumcenter of a larger triangle!

Theorem 14.
The three altitudes of a triangle meet at a point, called the orthocenter
of the triangle.
Let the altitudes of triangle ABC be ADBE and CF. Through each vertex,
draw a line parallel to the opposite side, resulting in a triangle which we denote by
triangle A0B0C0
By construction, the quadrilaterals
AC0BCABCBare parallelograms. Therefore,
by the Parallelogram Theorem (opposite sides are congruent)
C0A BC AB0. Thus is the midpoint of
C0B0. Moreover, since AD perpendicular BC and BC | | C0B0, we also know that AD perpendicular C0B0
(by our Fifth Postulate). It follows that
Line AD is the perpendicular bisector of C0B0Similarly,
Line FC and Line BE are perpendicular bisectors of A0Band C0A0, respectively.
By Theorem 13,
Line ADLine FC and Line BE meet at the circumcenter of triangle A0B0C0. The proof
is complete.

I provide no exercises for this lesson/activity -- since the proofs are difficult enough on their own