If a circle's diameter is 3 times another's, then how many times greater is its area?
Let's call the diameters of the two circles d and D, so that D = 3d. Then the area of the smaller circle is pi(d/2)^2 and the area of the larger circle is pi(D/2)^2. The ratio of the areas is:
pi(D/2)^2/pi(d/2)^2
= pi(3d/2)^2/pi(d/2)^2
= (9pi d^2/4)/(pi d^2/4)
= 9
So the ratio of the two areas is nine -- and of course, today's date is the ninth.
Notice that the Fundamental Theorem of Similarity (Lesson 12-6) shows us that if lengths (such as diameters) in similar figures are in a ratio of k, then their areas are in a ratio of k^2. This gives us the solution to the Pappas problem as 3^2 = 9 directly without needing the circle area formula. But I decided to use the formula anyway, only because tomorrow is the 314th day of the year:
That's right -- tomorrow is the third Pi Day. In some years, the 314th day of the year is November 9th because of February 29th, but this year it is the usual November 10th. This is what I wrote two years ago about how I (inadvertently) celebrated third Pi Day in my classroom:
Hold on a minute! You probably thought that Pi Day was on March 14th -- and the date on which this blog was launched was Pi Approximation Day, July 22nd. So how can November 10th be yet another Pi Day?
Well, November 10th is the 314th day of the year. And so some people have declared the day to be a third Pi Day:
http://mathforum.org/kb/message.jspa?messageID=7605691
I like the idea of a third Pi Day, based on the ordinal date (where January 1 = 1, November 10 = 314, and December 31 = 365). As the author at the above link pointed out, the three Pi Days are nearly equally spaced throughout the year. So I can celebrate Pi Day every fourth month.
I wouldn't mention the third Pi Day in a classroom, unless I was at a school that was on a 4x4 block schedule, where a student may take math first semester and then have absolutely no math class in the second semester (when the original Pi Day occurs). The only chance a student has to celebrate Pi Day would be the November Pi Day. (Likewise, the second Pi Day -- July 22nd -- may be convenient for a summer school class.)
Both November 10th [or 9th] and March 14th suffer from falling near the ends of trimesters or quarters (depending on whether the school started in August or after Labor Day). Classes may be too busy with trimester or quarter tests to have any sort of Pi Day party.
At home, I like to celebrate and eat pie for all three Pi Days. The pie that I choose is the pie most associated with the season in which that Pi Day occurs. Today, I will eat either pumpkin or sweet potato pie due to its proximity to Thanksgiving. On Pi Approximation Day, I ate apple pie, since it occurs right after the Fourth of July, a date as American as apple pie. And for the original Pi Day in March, I eat cherry pie -- the National Cherry Blossom Festival usually occurs between a week and a month after Pi Day.
Two years ago, right in the middle of class, some eighth graders start singing the song "Thrift Shop" by Macklemore & Ryan Lewis for who knows what reason. So then I told them that there's a math parody of this song called "Pi Shop."
Unfortunately, I wasn't prepared to sing Kevin Lee's "Pi Shop" today -- and of course my mentioning of this song had nothing to do with Pumpkin Pi Day. Some of the students tried to guess the lyrics of "Pi Shop" -- for some reason, they thought it had something to do with buying fourteen pies!
Sadly, third Pi Day was the only Pi Day I actually spent in my class two years ago, since I ended up leaving just before the first (original) Pi Day. Perhaps I would have made a bigger deal about third Pi Day if I knew I'd be gone by first Pi Day -- but then again, back in November there was no reason for me to think that I wouldn't complete the school year.
For each of the three Pi Days I like to post some sort of video. Last year on Pumpkin Pi Day I posted "Oh Number Pi," a parody of "O Christmas Tree." This time, I post a pi-based parody of another Christmas song, "Jingle Bells":
Hold on a minute! You probably thought that Pi Day was on March 14th -- and the date on which this blog was launched was Pi Approximation Day, July 22nd. So how can November 10th be yet another Pi Day?
Well, November 10th is the 314th day of the year. And so some people have declared the day to be a third Pi Day:
http://mathforum.org/kb/message.jspa?messageID=7605691
I like the idea of a third Pi Day, based on the ordinal date (where January 1 = 1, November 10 = 314, and December 31 = 365). As the author at the above link pointed out, the three Pi Days are nearly equally spaced throughout the year. So I can celebrate Pi Day every fourth month.
I wouldn't mention the third Pi Day in a classroom, unless I was at a school that was on a 4x4 block schedule, where a student may take math first semester and then have absolutely no math class in the second semester (when the original Pi Day occurs). The only chance a student has to celebrate Pi Day would be the November Pi Day. (Likewise, the second Pi Day -- July 22nd -- may be convenient for a summer school class.)
Both November 10th [or 9th] and March 14th suffer from falling near the ends of trimesters or quarters (depending on whether the school started in August or after Labor Day). Classes may be too busy with trimester or quarter tests to have any sort of Pi Day party.
At home, I like to celebrate and eat pie for all three Pi Days. The pie that I choose is the pie most associated with the season in which that Pi Day occurs. Today, I will eat either pumpkin or sweet potato pie due to its proximity to Thanksgiving. On Pi Approximation Day, I ate apple pie, since it occurs right after the Fourth of July, a date as American as apple pie. And for the original Pi Day in March, I eat cherry pie -- the National Cherry Blossom Festival usually occurs between a week and a month after Pi Day.
Two years ago, right in the middle of class, some eighth graders start singing the song "Thrift Shop" by Macklemore & Ryan Lewis for who knows what reason. So then I told them that there's a math parody of this song called "Pi Shop."
Unfortunately, I wasn't prepared to sing Kevin Lee's "Pi Shop" today -- and of course my mentioning of this song had nothing to do with Pumpkin Pi Day. Some of the students tried to guess the lyrics of "Pi Shop" -- for some reason, they thought it had something to do with buying fourteen pies!
Sadly, third Pi Day was the only Pi Day I actually spent in my class two years ago, since I ended up leaving just before the first (original) Pi Day. Perhaps I would have made a bigger deal about third Pi Day if I knew I'd be gone by first Pi Day -- but then again, back in November there was no reason for me to think that I wouldn't complete the school year.
For each of the three Pi Days I like to post some sort of video. Last year on Pumpkin Pi Day I posted "Oh Number Pi," a parody of "O Christmas Tree." This time, I post a pi-based parody of another Christmas song, "Jingle Bells":
As I've mentioned before, third Pi Day falls close to Veteran's Day on November 11th. This means that more often than not, third Pi Day falls on a three-day weekend when schools are closed. This year, schools observe Vets Day on Monday, November 12th, so third Pi Day is the indeed the first day of a three-day weekend.
Meanwhile, today I subbed in a middle school special ed class. This is my second visit to this class -- the first was back on September 5th (the first day I subbed this school year). As usual, you can refer to my September 5th post to learn more about this class.
This is a self-contained class where the same six or seven students (all boys) stay in our classroom for most of the day. The middle school rotation today starts with third period, which is history -- these students are also reading My Brother Sam Is Dead.
Sixth period is the teacher's conference -- but I'm assigned another class this period. As it turns out, it's a math class. This is the same math teacher I co-taught with on Monday, except that sixth period is an eighth grade class. The students are now taking a quiz on slope. (It's a quiz, so Conversation Silent is posted on the front wall.) I notice that in addition to Sarah Carter's Interactive Notebook and Slope Dude ideas, this teacher has another way to remember slope at least in the two special cases. The slope of a vertical line is Undefined -- and the capital letter U contains two vertical lines. The slope of a horizontal line is Zero -- and the capital letter Z contains two horizontal lines.
The main behavior problem for this special ed class occurs after lunch. Actually, during lunch the aides decide today to allow the students to eat lunch and play on Chromebooks in the room. The problem is that when lunchtime ends, the students refuse to do any tasks. This includes voting (and you thought Election Day was over!) for ASB President, VP, and other offices, as well as the two classes that rotate today into the after lunch blocks -- science and math.
During science, the aides must send out three boys (nearly half the class!) to the office. One student is using profanity, a second student walks outside to use his cell phone, and the third is playing "Keep Away" with the second student. Because of this, one aide decides that she won't even bother trying to teach the math lesson today. It's a shame, since I'm a math teacher. (Things would have worked out yesterday when the rotation starts with second period math, but not today. Then again, that's why the school has rotations, so that the same class isn't always first or last.)
Actually, one student (the cusser) doesn't stay for math -- instead he goes to a general ed seventh grade math class -- the same one I co-taught on Monday. Of course, I already know that this resident teacher is out, and so is the co-teacher (the one I subbed for on Monday -- both teachers are at meetings today). Because there are subs in that class, the aide want to protect those subs from having to deal with the cusser, so she tells him to stay in our class and make up the quiz next week. I actually help this student review for his quiz. (It's on the distributive property, if you recall from Monday.)
So that's all the math I taught today. Back on September 5th, I wrote about how I could have used some of the Number Talks techniques to help these kids out, but then math was cancelled. Here is a sample of some the problems the regular teacher left on the board (one of each type):
1. 42 * 26 =
2. 5^3 =
3. 8/16 = (reduce the fraction)
4. 932/8 = (long division)
Cathy Humphreys has chapters for multiplication, division, and fractions, but not exponents -- but then again, students find 5^3 simply by calculating 5 * 5 * 5 or 25 * 5, so that's multiplication.
Speaking of Number Talks, you might notice that this is another traditionalists' post. You guessed it -- yesterday I wrote that the traditionalists haven't blogged anything major lately, so of course today they post something major:
“When students take Algebra I matters, but many students do not have early access.”
So says a recently released report by the U.S. Dept. of Education. It also states:
The Department is encouraging both access to and enrollment in STEM courses. Both aspects are important because, as we will see through the story, even where access to Algebra I classes are available students do not necessarily enroll in them.
Here Barry Garelick, our favorite traditionalist, is trying to push eighth grade Algebra I again. Here he's linking to a Department of Ed article. Let's skip to the concluding paragraph:
We know that a strong STEM education is a path to successful career and we know that the need for STEM knowledge and skills will continue to grow in the future. [5] Taking Algebra I before high school, such as in 8thgrade, can set students up for a strong foundation of STEM education and open the door for various college and career options.
Notice that phrase "open the door." Traditionalists like to push eighth grade Algebra I because they believe that it opens the doors that lead to STEM.
Once again, Garelick criticizes Jo Boaler, who wrote the foreword to our Number Talks book:
Once again, Garelick criticizes Jo Boaler, who wrote the foreword to our Number Talks book:
But such words do not matter. Algebra continues to be the forbidden fruit of education, reserved for those whose parents can afford to have their kids learn it outside of school–or have enough clout to get their kids in to 8th grade algebra programs when they are offered. As I wrote about here, the San Francisco school district did away with algebra in 8th grade. Jo Boaler and Alan Schoenfeld wrote an article in the San Francisco Chronicle, lauding this decision, and stating:
They (i.e., San Francisco USD) found a unique balance that is now seen as a national model. They decided to challenge students earlier with depth and rigor in middle school. All students in the district take Common Core Math 6, 7 and 8, a robust foundation that allows them to be more successful in advanced math courses in high school
Let's see what our favorite commenter, SteveH, has to say:
SteveH:
Exactly!. When our non-urban middle school finally (!) got rid of CMP with no real algebra in 8th grade, they brought in Glencoe Pre-Algebra and Algebra I courses in 7th and 8th grades. This happened because parents complained that the school did not offer a proper curriculum sequence to advanced placement in math and languages in high school. This is still a problem because our lower schools use Everyday Math. Students and parents still feel like they have been hit in the head with a brick when they don’t get into Pre-Algebra in 7th grade even though they have “distinguished” grades on their yearly math assessment.
SteveH:
Exactly!. When our non-urban middle school finally (!) got rid of CMP with no real algebra in 8th grade, they brought in Glencoe Pre-Algebra and Algebra I courses in 7th and 8th grades. This happened because parents complained that the school did not offer a proper curriculum sequence to advanced placement in math and languages in high school. This is still a problem because our lower schools use Everyday Math. Students and parents still feel like they have been hit in the head with a brick when they don’t get into Pre-Algebra in 7th grade even though they have “distinguished” grades on their yearly math assessment.
Once again, SteveH refers to the U of Chicago elementary text (Everyday Math). His argument, of course, is that the U of Chicago doesn't prepare students for seventh grade Pre-Algebra and thus eighth grade Algebra I.
SteveH:
This is also not just about STEM. Back when I taught college math and CS (in a math department), we saw many students who had to change majors because they couldn’t handle the required math classes. This even happened to nursing students who couldn’t handle a proper statistics course. What is needed is a complete list of majors and what level of math is needed. This is not just a STEM issue. Schools need to offer remediation/advancement opportunities for each grade. When kids in 7th grade say they want to become marine biologists, they need to know exactly what is expected of them in school.
I already mentioned that nursing example in yesterday's traditionalists' post -- this is one of SteveH's favorite examples. Marine biology is a new example of his, though. But what happens if the seventh grader who want to be marine biologists already know that they hate math?
Wayne:
Bad as it is to deny students who are prepared for algebra the right to enroll in it, a much worse problem is such a large percentage who should be prepared and are not because of the same misconceptions in the earlier grades. Harmful as this is to children from affluent, educated communities, many of these students have compensating opportunities. Students from underprivileged communities have no such and the results are disastrous.
Here I assume that "compensating opportunities" means "private tutors," since they like to mention these often.
Here's the question I have for the traditionalists -- suppose we were to try taking SteveH's and Wayne's advice. We tell young tween math students how much math is needed for their desired future career paths. Will that make them want to work hard on traditional p-sets? Will this result in not only more eighth graders taking Algebra I, but actually receiving good grades in that class? In other words, if we were to tell Anthony (a student of Cathy Humphreys), who didn't want to learn fractions, that he must master them to become a nurse or marine biologist, would that make him want to work harder on a traditional fractions p-set? The answer is -- I don't know.
Lesson 6-1 of the U of Chicago text is called "Transformations." There is no corresponding lesson in the modern Third Edition -- transformations are spread out in Chapters 3 through 6, with no separate introductory lesson.
This is what I wrote last year about this lesson:
It's easy to show that the transformation mapping (x, y) to (x + h, y) is a translation -- where we define translation as the composite of two reflections in parallel mirrors, with the direction of the translation being a common perpendicular of the mirrors. In this case, we can let the y-axis and the line x = h/2 be the two mirrors, with their common perpendicular the x-axis. Likewise, it's trivial to show that the mapping (x, y) to (x, y + k) is a translation in the direction of the y-axis. What we want to show is that the composite of these two translations is itself a translation.
I wasn't able to come up with a general proof that the composite of two translations is a translation, but I was able to prove it in certain cases. This includes the case where one of the translations is horizontal and the other is vertical -- which is the only case that really matters.
To understand the proof, let's use some notation that appears in the U of Chicago text. If m is a line, then we let r_m denote the reflection in mirror m. In the text, the letter m appears as a subscript, but this is hard to reproduce in ASCII, so we use r_m instead. Meanwhile a small circle o denotes the composite, and so we write r_n o r_m to denote the composite of two reflections -- first the reflection in mirror m, then the reflection in mirror n.
The first thing we know about reflections is the Flip-Flop Theorem, which tells us that if a reflection maps F to F', then it maps F' to F. That is, if r_m(F) = F', then r_m(F') = F. Therefore the composite of the reflection with itself must map every point to itself. This transformation is often called the identity transformation, and we can write it using the symbol I. So we write:
r_m o r_m = I
Since I is a transformation in its own right, we can compose it with other transformations. Of course, this is trivial -- the composite of I and any other transformation is the other transformation:
I o r_m = r_m
r_n o I = r_n
r_n o I o r_m = r_n o r_m
Notice that composition of transformations is associative, but not commutative -- so in general, we have that r_m o r_n is not the same as r_n o r_m. In fact, we can see how r_m o r_n and r_n o r_m are related by finding their composite:
r_m o r_n o r_n o r_m = r_m o I o r_m
= r_m o r_m
= I
So r_m o r_n is the transformation which, when composed with r_n o r_m, yields the identity. This transformation has a special name -- the inverse transformation. As it turns out, the inverse of any reflection is itself. But the inverse of a translation is a translation in the opposite direction, and the inverse of the counterclockwise rotation with magnitude theta is, quite obviously, the clockwise rotation with magnitude theta with the same center.
Notice that r_n o r_m, being the composite of two reflections, can't itself be a reflection -- it must be either a translation or rotation. So r_n o r_m can almost never equal its inverse r_m o r_n. But there is a special case -- notice that the inverse of a 180-degree rotation counterclockwise is the 180-degree rotation clockwise, but these are in fact that same rotation, since +180 and -180 differ by 360. So if r_n o r_m equals a 180-degree rotation, then r_m and r_n commute after all.
The last thing we need in our proof that the composite of two translations is a translation will be both of Two Reflections Theorems (one for Translations, the other for Rotations). The Two Reflections Theorem for Translations tells us that the direction and distance of a translation depend only on the common perpendicular and the distance between the mirrors. So if k, l, m, andn are all parallel mirrors, and the distance from k to l equals the distance from m to n, then r_n o r_m = r_l o r_k. And the same happens for rotations -- if k, l, m, and n are all concurrent mirrors, and the angle from k to l equals the angle from m to n, then r_n o r_m = r_l o r_k.
Notice that in some ways, we are actually performing a transformation on the mirrors -- that is, given two mirrors m and n, we wish to transform them to m' and n' such that r_n o r_m = r_n' o r_m'. Or if we want to get very formal, if we have some transformation T such that:
T = r_n o r_m
then we wish to find another transformation U such that:
T = r_U(n) o r_U(m)
and the point is that the transformation U has nothing to do with the transformation T. Indeed, if T is a translation, then U can be any translation. Likewise, if T is a rotation, then U can be any rotation with the same center as T.
So now let's prove a simple case, that the composite of two translations is a translation. Our simple case will be when the two translations are in the same direction (or in opposite directions). This means that the mirrors k, l, m, and n are all parallel, and we wish to prove that:
r_n o r_m o r_l o r_k
is a translation. Since all the mirrors are parallel, in particular m | | n. By the Two Reflections Theorem for Translations, we're allowed to slide the mirrors themselves. We can slide mirrors m and n to two new mirrors, m' and n', such that m' | | n' with the same distance between them:
r_n o r_m o r_l o r_k = r_n' o r_m' o r_l o r_k
How does this even help us at all? That's easy -- we slide m until its image is exactly l! Then we have:
r_n o r_m o r_l o r_k = r_n' o r_m' o r_l o r_k
= r_n' o r_l o r_l o r_k
= r_n' o I o r_k
= r_n' o r_k
And so we've done it -- we've reduced four mirrors to two. And we know that n is parallel to its translation image n' and n is parallel to k as all the original mirrors were parallel. So k | | n' -- that is, r_n' o r_k is the composite of reflections in parallel mirrors. Therefore it is a translation. QED
So now we see our trick to reduce four mirrors to two -- we transform the mirrors a pair at a time, using the Two Reflections Theorems, until two of the mirrors coincide. Then the composite of a reflection with itself is the identity, which disappears, leaving two mirrors behind. But when we transform the mirrors, we must be careful to transform the correct mirrors. When we have:
r_n o r_m o r_l o r_k
we may transform k and l together, or l and m together, or m and n together. But we can't transform, say, k and m together, or l and n, or k and n, since these aren't listed consecutively in the composite.
Let's finally prove that the composite of a horizontal and a vertical translation is a translation. We begin by writing:
r_n o r_m o r_l o r_k
where k and l are horizontal mirrors (for the vertical translation), and m and n are vertical mirrors (for the horizontal translation). We notice that k, l, m, and n form the sides of a rectangle.
Now we begin transforming the mirrors. First, we notice that l and m are perpendicular (since l is horizontal, while m is vertical). As it turns out, reflections in perpendicular mirrors commute -- this is because by the Two Reflections Theorem for Rotations, the angle of rotation is double the angle between the mirrors, and since the angle between the mirrors is 90, the rotation angle is 180, which means that they commute. If you prefer, we could say that instead of commuting, we're actually rotating the two mirrors 90 degrees. The rotation image of l is m, and the rotation image of m is l. In either case, we obtain:
r_n o r_m o r_l o r_k = r_n o r_l o r_m o r_k
Now we rotate the perpendicular pairs k and m together, and l and n together. We rotate the pairk and m until the image of m is concurrent with l and n, and then we rotate the pair l and n until the image of l is concurrent with k and m:
r_n o r_m o r_l o r_k = r_n o r_l o r_m o r_k
= r_n' o r_l' o r_m' o r_k'
Now l' and m' are the same line -- the diagonal of the original rectangle. Let's call this new mirrord to emphasize that it's the diagonal of the rectangle:
r_n o r_m o r_l o r_k = r_n o r_l o r_m o r_k
= r_n' o r_d o r_d o r_k'
= r_n' o I o r_k'
= r_n' o r_k'
Even though d disappears, notice that d is still significant. Since both rotations preserved angle measures, we have that k' and n' are both perpendicular to the diagonal d. So, by Two Perpendiculars Theorem, k' | | n'. So we have the composite of reflections in parallel mirrors -- a translation. QED
But unfortunately, attempting to generalize this proof to the case when the directions of the two translations aren't perpendicular fails even in Euclidean geometry. Notice that the four mirrors now form merely a parallelogram, not a rectangle. But we know that in Euclidean geometry, the angles from k to m and from l to n are congruent because they're opposite angles of this parallelogram. After performing the rotations, the congruent angles have rotated into alternate interior angles formed by the lines k' and n' and the transversal d, which would mean that k' andn' are parallel. This may look appealing, but the problem is that the original proof commuted l andm, which is invalid unless l and m are perpendicular.
I'm not sure how to fix the proof that the composite of two translations is a translation, but luckily, we only need the perpendicular case to prove that mapping (x, y) to (x + h, y + k) is a translation.
Now here's the thing -- at some point this week, I wish to have students perform translations on a coordinate plane. I could give students this proof that mapping (x, y) to (x + h, y + k) is a translation, but look at how confusing this proof appears! It's highly symbolic and may be too abstract for students -- just look at lines like:
T = r_U(n) o r_U(m)
What in the world does this mean? It means that the lines which are the images of one transformation, U, are the mirrors (not the preimages, but the mirrors) of another transformation, T. If a student was to forget that U is a transformation, he or she might think that U is a line, and so r_U(m) would mean the image of the line m reflected in the mirror U. Notice that on a printed page rather than ASCII, it would be obvious that the subscript is all of U(m), not just the letter U.
As we've discussed so often recently, traditionalists like symbolic manipulation. If they could see the symbolic manipulation involved with this, they might realize that transformational geometry isn't just hand-waving but is actually rigorous mathematics. But high school students will be confused if I were to give a proof as symbolic as this one, so I'd replace the symbols with words -- yet as soon as we did this, it will appear to the traditionalists that transformations are mere hand-waving yet again.
We might point out that verbal descriptions might be more understandable to high school students, while symbols for transformations might be better suited at the college-level. But this is when a traditionalist might say, fine, then let's save transformations for college-level math and teach high school students only math for which they can understand the symbols! In particular, a traditionalist would look at this proof and say one of two things. The first would be that students shouldn't be learning about translations anyway, so the mapping (x, y) to (x + h, y + k) is irrelevant. The other would be that if students really need to know that the mapping (x, y) to (x + h, y + k) is a translation, then they should prove it by using the Slope and Distance Formulas and not by playing with mirrors. (Our problem is that we can't use the Slope and Distance Formulas until we've reached dilations and similarity.)
Indeed, I've noticed that some college geometry texts use the notation A-B-C to denote a very simple concept -- namely that point B is between A and C. Often I wondered why no high school Geometry text ever uses this A-B-C notation, but now I realize why -- the A-B-C notation is considered too abstract for high school students to understand.
None of this has anything to do with today's lesson, so I'll make the decision regarding how to present the proof when we get there. Meanwhile, if traditionalists really want to "privilege the symbol," why don't they begin by enforcing A-B-C notation in high school Geometry and see how far they get?
Meanwhile, today is an activity day. I've never posted an activity for Lesson 6-1 before. This is because in the past two years, Veteran's Day has fallen on either Friday or Saturday (so that in either case, the day that schools were closed was Friday). This year, the holiday is on a Monday, which forces Lesson 6-1 to move from Monday to Friday. This is the first new worksheet that I must create for this school year.
So let's look at this activity -- sorry, traditionalists! As usual, our activity is based on the Exploration Question in the U of Chicago text. Here is the Lesson 6-1 Exploration question:
24. Explore the transformation with rule T(x, y) = (2x, y) by finding images of common figures under T. (Hint: Use points in all quadrants.)
The resulting transformation would be categorized under "affine transformation" above. And this particular transformation is called a shear or a transvection. Like reflections, transvections have an entire line of fixed points (in this case the y-axis). But unlike reflections, every line parallel to the axis of the transvection is invariant.
Monday is Veteran's Day Observed, so my next post will be on Tuesday.
SteveH:
This is also not just about STEM. Back when I taught college math and CS (in a math department), we saw many students who had to change majors because they couldn’t handle the required math classes. This even happened to nursing students who couldn’t handle a proper statistics course. What is needed is a complete list of majors and what level of math is needed. This is not just a STEM issue. Schools need to offer remediation/advancement opportunities for each grade. When kids in 7th grade say they want to become marine biologists, they need to know exactly what is expected of them in school.
I already mentioned that nursing example in yesterday's traditionalists' post -- this is one of SteveH's favorite examples. Marine biology is a new example of his, though. But what happens if the seventh grader who want to be marine biologists already know that they hate math?
Wayne:
Bad as it is to deny students who are prepared for algebra the right to enroll in it, a much worse problem is such a large percentage who should be prepared and are not because of the same misconceptions in the earlier grades. Harmful as this is to children from affluent, educated communities, many of these students have compensating opportunities. Students from underprivileged communities have no such and the results are disastrous.
Here I assume that "compensating opportunities" means "private tutors," since they like to mention these often.
Here's the question I have for the traditionalists -- suppose we were to try taking SteveH's and Wayne's advice. We tell young tween math students how much math is needed for their desired future career paths. Will that make them want to work hard on traditional p-sets? Will this result in not only more eighth graders taking Algebra I, but actually receiving good grades in that class? In other words, if we were to tell Anthony (a student of Cathy Humphreys), who didn't want to learn fractions, that he must master them to become a nurse or marine biologist, would that make him want to work harder on a traditional fractions p-set? The answer is -- I don't know.
Lesson 6-1 of the U of Chicago text is called "Transformations." There is no corresponding lesson in the modern Third Edition -- transformations are spread out in Chapters 3 through 6, with no separate introductory lesson.
This is what I wrote last year about this lesson:
Today we begin Chapter 6, which is on transformations -- the heart of Common Core Geometry. We see that the first lesson, Lesson 6-1, is simply a general introduction to transformations.
This lesson begins with a definition of transformation. Once again, I omit the function notation T(P) for transformations since I fear that they'll confuse students. But I do use prime notation. The best way to demonstrate transformations is on the coordinate plane, so I do use them.
The book uses N(S) to denote the number of elements of a set S, an example of function notation. I include this question in the review, since the number of elements in a set (cardinality) is such a basic concept for students to understand.
Interestingly enough, in college one learns about these special types of transformations:
- A transformation preserving only betweenness is called a homeomorphism.
- Add in collinearity, and it becomes an affine transformation.
- Add in angle measure, and it becomes a similarity transformation.
- And finally, add in distance, and now we have an isometry.
The final question on my worksheet brings back shades of Jen Silverman -- the distance between lines is constant if and only if they are parallel in Euclidean geometry. I mentioned Silverman's page earlier when we were getting ready to learn about parallel lines.
One question on the worksheet involves the transformation mapping (x, y) to (x + 2, y - 3). The students can see that this transformation is clearly a translation. But so far, we haven't completed a proof that the transformation mapping (x, y) to (x + h, y + k) is a translation.
It's easy to show that the transformation mapping (x, y) to (x + h, y) is a translation -- where we define translation as the composite of two reflections in parallel mirrors, with the direction of the translation being a common perpendicular of the mirrors. In this case, we can let the y-axis and the line x = h/2 be the two mirrors, with their common perpendicular the x-axis. Likewise, it's trivial to show that the mapping (x, y) to (x, y + k) is a translation in the direction of the y-axis. What we want to show is that the composite of these two translations is itself a translation.
It seems as if it should be trivial to show that the composite of two translations is a translation, but in fact it isn't. After all, a translation is the composite of two reflections in parallel mirrors, so the composite oftwo translations is also the composite of four reflections. It's not obvious why the composite of these four reflections is also the composite of two reflections in parallel mirrors -- especially if these two new mirrors have nothing to do with the four original mirrors. Of course, at some point we'd like to say something about translation vectors -- for example, the composite of two translations is a new translation whose vector is thesum of the original two vectors. But this would need to be proved.
To understand the proof, let's use some notation that appears in the U of Chicago text. If m is a line, then we let r_m denote the reflection in mirror m. In the text, the letter m appears as a subscript, but this is hard to reproduce in ASCII, so we use r_m instead. Meanwhile a small circle o denotes the composite, and so we write r_n o r_m to denote the composite of two reflections -- first the reflection in mirror m, then the reflection in mirror n.
The first thing we know about reflections is the Flip-Flop Theorem, which tells us that if a reflection maps F to F', then it maps F' to F. That is, if r_m(F) = F', then r_m(F') = F. Therefore the composite of the reflection with itself must map every point to itself. This transformation is often called the identity transformation, and we can write it using the symbol I. So we write:
r_m o r_m = I
Since I is a transformation in its own right, we can compose it with other transformations. Of course, this is trivial -- the composite of I and any other transformation is the other transformation:
I o r_m = r_m
r_n o I = r_n
r_n o I o r_m = r_n o r_m
Notice that composition of transformations is associative, but not commutative -- so in general, we have that r_m o r_n is not the same as r_n o r_m. In fact, we can see how r_m o r_n and r_n o r_m are related by finding their composite:
r_m o r_n o r_n o r_m = r_m o I o r_m
= r_m o r_m
= I
So r_m o r_n is the transformation which, when composed with r_n o r_m, yields the identity. This transformation has a special name -- the inverse transformation. As it turns out, the inverse of any reflection is itself. But the inverse of a translation is a translation in the opposite direction, and the inverse of the counterclockwise rotation with magnitude theta is, quite obviously, the clockwise rotation with magnitude theta with the same center.
Notice that r_n o r_m, being the composite of two reflections, can't itself be a reflection -- it must be either a translation or rotation. So r_n o r_m can almost never equal its inverse r_m o r_n. But there is a special case -- notice that the inverse of a 180-degree rotation counterclockwise is the 180-degree rotation clockwise, but these are in fact that same rotation, since +180 and -180 differ by 360. So if r_n o r_m equals a 180-degree rotation, then r_m and r_n commute after all.
The last thing we need in our proof that the composite of two translations is a translation will be both of Two Reflections Theorems (one for Translations, the other for Rotations). The Two Reflections Theorem for Translations tells us that the direction and distance of a translation depend only on the common perpendicular and the distance between the mirrors. So if k, l, m, andn are all parallel mirrors, and the distance from k to l equals the distance from m to n, then r_n o r_m = r_l o r_k. And the same happens for rotations -- if k, l, m, and n are all concurrent mirrors, and the angle from k to l equals the angle from m to n, then r_n o r_m = r_l o r_k.
Notice that in some ways, we are actually performing a transformation on the mirrors -- that is, given two mirrors m and n, we wish to transform them to m' and n' such that r_n o r_m = r_n' o r_m'. Or if we want to get very formal, if we have some transformation T such that:
T = r_n o r_m
then we wish to find another transformation U such that:
T = r_U(n) o r_U(m)
and the point is that the transformation U has nothing to do with the transformation T. Indeed, if T is a translation, then U can be any translation. Likewise, if T is a rotation, then U can be any rotation with the same center as T.
So now let's prove a simple case, that the composite of two translations is a translation. Our simple case will be when the two translations are in the same direction (or in opposite directions). This means that the mirrors k, l, m, and n are all parallel, and we wish to prove that:
r_n o r_m o r_l o r_k
is a translation. Since all the mirrors are parallel, in particular m | | n. By the Two Reflections Theorem for Translations, we're allowed to slide the mirrors themselves. We can slide mirrors m and n to two new mirrors, m' and n', such that m' | | n' with the same distance between them:
r_n o r_m o r_l o r_k = r_n' o r_m' o r_l o r_k
How does this even help us at all? That's easy -- we slide m until its image is exactly l! Then we have:
r_n o r_m o r_l o r_k = r_n' o r_m' o r_l o r_k
= r_n' o r_l o r_l o r_k
= r_n' o I o r_k
= r_n' o r_k
And so we've done it -- we've reduced four mirrors to two. And we know that n is parallel to its translation image n' and n is parallel to k as all the original mirrors were parallel. So k | | n' -- that is, r_n' o r_k is the composite of reflections in parallel mirrors. Therefore it is a translation. QED
So now we see our trick to reduce four mirrors to two -- we transform the mirrors a pair at a time, using the Two Reflections Theorems, until two of the mirrors coincide. Then the composite of a reflection with itself is the identity, which disappears, leaving two mirrors behind. But when we transform the mirrors, we must be careful to transform the correct mirrors. When we have:
r_n o r_m o r_l o r_k
we may transform k and l together, or l and m together, or m and n together. But we can't transform, say, k and m together, or l and n, or k and n, since these aren't listed consecutively in the composite.
Let's finally prove that the composite of a horizontal and a vertical translation is a translation. We begin by writing:
r_n o r_m o r_l o r_k
where k and l are horizontal mirrors (for the vertical translation), and m and n are vertical mirrors (for the horizontal translation). We notice that k, l, m, and n form the sides of a rectangle.
Now we begin transforming the mirrors. First, we notice that l and m are perpendicular (since l is horizontal, while m is vertical). As it turns out, reflections in perpendicular mirrors commute -- this is because by the Two Reflections Theorem for Rotations, the angle of rotation is double the angle between the mirrors, and since the angle between the mirrors is 90, the rotation angle is 180, which means that they commute. If you prefer, we could say that instead of commuting, we're actually rotating the two mirrors 90 degrees. The rotation image of l is m, and the rotation image of m is l. In either case, we obtain:
r_n o r_m o r_l o r_k = r_n o r_l o r_m o r_k
Now we rotate the perpendicular pairs k and m together, and l and n together. We rotate the pairk and m until the image of m is concurrent with l and n, and then we rotate the pair l and n until the image of l is concurrent with k and m:
r_n o r_m o r_l o r_k = r_n o r_l o r_m o r_k
= r_n' o r_l' o r_m' o r_k'
Now l' and m' are the same line -- the diagonal of the original rectangle. Let's call this new mirrord to emphasize that it's the diagonal of the rectangle:
r_n o r_m o r_l o r_k = r_n o r_l o r_m o r_k
= r_n' o r_d o r_d o r_k'
= r_n' o I o r_k'
= r_n' o r_k'
Even though d disappears, notice that d is still significant. Since both rotations preserved angle measures, we have that k' and n' are both perpendicular to the diagonal d. So, by Two Perpendiculars Theorem, k' | | n'. So we have the composite of reflections in parallel mirrors -- a translation. QED
But unfortunately, attempting to generalize this proof to the case when the directions of the two translations aren't perpendicular fails even in Euclidean geometry. Notice that the four mirrors now form merely a parallelogram, not a rectangle. But we know that in Euclidean geometry, the angles from k to m and from l to n are congruent because they're opposite angles of this parallelogram. After performing the rotations, the congruent angles have rotated into alternate interior angles formed by the lines k' and n' and the transversal d, which would mean that k' andn' are parallel. This may look appealing, but the problem is that the original proof commuted l andm, which is invalid unless l and m are perpendicular.
I'm not sure how to fix the proof that the composite of two translations is a translation, but luckily, we only need the perpendicular case to prove that mapping (x, y) to (x + h, y + k) is a translation.
Now here's the thing -- at some point this week, I wish to have students perform translations on a coordinate plane. I could give students this proof that mapping (x, y) to (x + h, y + k) is a translation, but look at how confusing this proof appears! It's highly symbolic and may be too abstract for students -- just look at lines like:
T = r_U(n) o r_U(m)
What in the world does this mean? It means that the lines which are the images of one transformation, U, are the mirrors (not the preimages, but the mirrors) of another transformation, T. If a student was to forget that U is a transformation, he or she might think that U is a line, and so r_U(m) would mean the image of the line m reflected in the mirror U. Notice that on a printed page rather than ASCII, it would be obvious that the subscript is all of U(m), not just the letter U.
As we've discussed so often recently, traditionalists like symbolic manipulation. If they could see the symbolic manipulation involved with this, they might realize that transformational geometry isn't just hand-waving but is actually rigorous mathematics. But high school students will be confused if I were to give a proof as symbolic as this one, so I'd replace the symbols with words -- yet as soon as we did this, it will appear to the traditionalists that transformations are mere hand-waving yet again.
We might point out that verbal descriptions might be more understandable to high school students, while symbols for transformations might be better suited at the college-level. But this is when a traditionalist might say, fine, then let's save transformations for college-level math and teach high school students only math for which they can understand the symbols! In particular, a traditionalist would look at this proof and say one of two things. The first would be that students shouldn't be learning about translations anyway, so the mapping (x, y) to (x + h, y + k) is irrelevant. The other would be that if students really need to know that the mapping (x, y) to (x + h, y + k) is a translation, then they should prove it by using the Slope and Distance Formulas and not by playing with mirrors. (Our problem is that we can't use the Slope and Distance Formulas until we've reached dilations and similarity.)
Indeed, I've noticed that some college geometry texts use the notation A-B-C to denote a very simple concept -- namely that point B is between A and C. Often I wondered why no high school Geometry text ever uses this A-B-C notation, but now I realize why -- the A-B-C notation is considered too abstract for high school students to understand.
None of this has anything to do with today's lesson, so I'll make the decision regarding how to present the proof when we get there. Meanwhile, if traditionalists really want to "privilege the symbol," why don't they begin by enforcing A-B-C notation in high school Geometry and see how far they get?
Meanwhile, today is an activity day. I've never posted an activity for Lesson 6-1 before. This is because in the past two years, Veteran's Day has fallen on either Friday or Saturday (so that in either case, the day that schools were closed was Friday). This year, the holiday is on a Monday, which forces Lesson 6-1 to move from Monday to Friday. This is the first new worksheet that I must create for this school year.
So let's look at this activity -- sorry, traditionalists! As usual, our activity is based on the Exploration Question in the U of Chicago text. Here is the Lesson 6-1 Exploration question:
24. Explore the transformation with rule T(x, y) = (2x, y) by finding images of common figures under T. (Hint: Use points in all quadrants.)
The resulting transformation would be categorized under "affine transformation" above. And this particular transformation is called a shear or a transvection. Like reflections, transvections have an entire line of fixed points (in this case the y-axis). But unlike reflections, every line parallel to the axis of the transvection is invariant.
Monday is Veteran's Day Observed, so my next post will be on Tuesday.
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