Friday, November 8, 2019

Lesson 6-1: Transformations (Day 61)

This Sunday 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 three 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 three 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 a pi-based parody of a Christmas song, "Jingle Bells." This year, I'll post a video about pi and probability -- the latter being the topic of today's Ian Stewart chapter.



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, third Pi Day is the indeed the second day of a three-day weekend.

Today I subbed in a high school math class. So I'm definitely doing "A Day in the Life" today!

Let me say a few things about this class. First of all, it's my second visit to this class -- I described this class back in my Mole Day post on October 23rd. (No, I didn't mention Third Pi Day in class today, since I just told you that I wouldn't.)

And today is the first day of a two-day subbing assignment. The second day will be on Tuesday, as Monday is Veteran's Day. This means that not only is this our second straight three-day weekend, but its the second such long weekend in the middle of a multi-day assignment. Unlike last weekend though, there's no chance of subbing in another district right in the middle of this assignment, since all districts in California will be closed on Monday. (I know, having two straight long weekends is unusual unless it's for two holidays that are close, such as Lincoln's Birthday and President's Day. In this case, the two close holidays are Halloween and Veteran's Day.)

The other thing unusual about today is that for some reason, it's a minimum day. When I arrived at the school, no one could tell me exactly why it's a minimum day today. Our best guess is that it's the last day before a holiday. Minimum days often occur before long breaks such as Thanksgiving or even President's Day (when it's combined with Lincoln to form a four-day weekend), but not usually for a three-day weekend like Veteran's Day.

(Next year, Veteran's Day will fall on a Wednesday. Does this mean that there'll be a minimum day on Tuesday the 10th next year?)

So anyway, let's start "A Day in the Life":

8:00 -- If you recall from my October 23rd post, the day begins with second period Algebra II. This class is studying conic sections. The students have 32 questions where they must find the equation of a conic given some of its properties. They aren't expected to finish it today -- they'll have more time on Tuesday.

Here are some of the problems that I help them with. The TA who was here on October 23rd is absent today, and so I work on these problems alone:

3. Circle -- Center: (0, 4). Radius: 2sqrt(5).
5. Circle -- Center: (3, 13). Point on Circle: (8, 11).
7. Circle -- Endpoints of a Diameter: (9, -17) and (-3, -5).

I've made a big deal about how the Common Core now introduces circle equations in Geometry rather than Algebra II. The first question can be completed after Lesson 11-3 of our text. The second question can in principle be answered after 11-3 as well, since the Distance Formula is taught in Lesson 11-2 of our text. The third question also requires the Midpoint Formula, which our text covers right after circle equations, in Lesson 11-4.

Of course, equations of other conics are never taught in Geometry:

9. Ellipse -- (from a graph) Vertices: (-3, 8) and (-3, 4). Co-Vertices: (2, 2) and (-8, 2).
26. Parabola -- (from a graph) Vertex: (2, 1). Focus: (-3, 1)
27. Parabola -- Vertex: (6, -3). Focus: (7, -3).
28. Parabola -- Vertex: (-7, -1). Focus: (-7, -7/2).

The students work quietly on this assignment, and so I name this period the best class of the day.

8:40 -- Second period leaves and third period arrives. This is the second Algebra II class.

In this class, the students are slightly more talkative, but they also ask me more questions. So I know that they're working hard on these questions.

9:25 -- Third period leaves for snack -- the only break on this minimum day.

9:45 -- Fourth period arrives. This is the last Algebra II class.

As far as behavior is concerned, there is a cluster of four seats where a fifth student, a sixth student, and briefly a seventh student join them. It's difficult to tell whether they are actually discussing math or other stuff -- probably a little of both. I notice that while all three classes are honors, this is the only class that's considered to be IB-level. Yet on neither October 23rd nor today is this period the best class of the day.

In all three Algebra II classes, the regular teacher allows the students to check their answers using a QR code on their phones. But she warns me in her lesson plan that the QR codes for the parabolas use a different form from what she's taught them. For Questions 26-27, the QR code uses:

(y - k)^2 = 4c(x - h)

while what she's taught them is:

x = a(y - k)^2 + h (with a = 1/(4c)).

Throughout the three Algebra II classes, I switch back-and-forth between the various forms, which may have confused the students.

10:25 -- Fourth period leaves. Fifth period is the teacher's conference period.

11:15 -- Sixth period arrives. This is the first Algebra I class.

These students are taking notes on linear functions. Today's lesson is an introduction -- the main ideas are function notation and graphing linear functions by using a table.

Even though the first part on function notation has already been covered, I just have to go over it again in order to sneak in Sarah Carter's DIXI-ROYD mnemonic. Then I move on to graphing.

On the good list, I write down the names of students who fill in the table on the front board. In this class, I place two names on the bad list for generally disturbing others. Recall from my October 23rd post that students on the bad list will lose participation points and receive a detention.

As I promised in my October 23rd post, I greet the students as they enter the classroom. This is to make sure that they sit in their assigned seats.

For today's song incentive, I continue rapping "One Billion Is Big" as I've done all this week. With the antics of the two guys whose names I must write, they earn only two of the three verses.

11:55 -- Sixth period leaves and seventh period arrives. This is the last Algebra I class.

Even though there is some general talkativeness, no individual sticks out for the bad list. This is something I observed back on October 23rd -- sixth period has more disruptive individuals, while seventh period has more talkative groups. I don't write anyone on the bad list in this period, but on Mole Day the regular teacher had written an entire group on the board. This is something that I might need to do on Tuesday.

I name seventh period as the best Algebra I class today due to the sixth period pair on the bad list.

12:40 -- Seventh period leaves, thus concluding this minimum day.

There are two changes that I wish to make before Tuesday. In Algebra II, my plan is to cover seven more problems -- mostly hyperbolas (since I don't go over them today), along with one each of the other types. For the parabolas, I want to clarify the correct form due to the QR code confusion.

For Algebra I, Tuesday's lesson will be on Chromebooks. But I wish to introduce a new song to perform in this class on Tuesday. This is a rare opportunity for me to prepare such a song -- most of the time I don't sub in math, and when I do, I enter the class not knowing what the lesson is. It's only fortunate that I have the "Solving Equations" song to fit many equation lessons. Unfortunately, I don't have any songs so far for the unit on graphing linear equations. (Eighth graders do graph some equations, but I never quite reached this unit at the old charter school before I left.) Now that I have a multi-day math assignment, I know exactly what lesson is planned for Tuesday.

Meanwhile, there's one song that I've been meaning to make a parody of. This is Elle King's "Ex's and Oh's." That title is practically begging to be parodied in a math class with "x's and y's" -- and such a parody would fit during a lesson on graphing functions on the xy-plane.

I'm aware that last year, I devoted several posts to music -- particularly, composing music in new scales based on EDL's and the Mocha computer emulator. So far, I haven't really composed music -- and when I do create a new song, it ends up being a parody rather than an original tune. As much as I'd like to use EDL's and Mocha, I can't help but want to make the Elle King parody -- it's set up perfectly with the multi-day subbing assignment during the linear functions unit.

Once again, the "x's and y's" parody should be so obvious that surely, someone has already created a version of this. I was able to find only one such parody -- it's created by "Brianna and Chase." I might as well link to it here:

My plan for this three-day weekend is to write down lyrics for this version. I might mix in some of my own lyrics as well, for two reasons. The first is that there may be parts of this video where I can't figure out the lyrics. The other is to sneak in lyrics that refer to the particular lesson that the students will actually learn on Tuesday -- graphing using intercepts.

I'll let you know how everything goes in my next post.

Chapter 18 of Ian Stewart's The Story of Mathematics is called "How Likely is That? The Rational Approach to Chance." Here's how it begins:

"The growth of mathematics in the 20th and early 21st centuries has been explosive. More new mathematics has been discovered in the last 100 years than in the whole of previous human history."

As you might have guessed from the title, this chapter is all about probability and statistics. I really need to post more about probability more on this Geometry blog, considering that probability is part of the California Common Core curriculum for Geometry. Stewart explains:

"All probabilities lie between 0 and 1, and the probability of an event represents the proportion of trials in which the event concerned happens. Back to that medieval question."

Here the author refers to questions that some people had in the Middle Ages about the probability of rolling certain numbers with a pair of dice.

Stewart discusses the idea of combinations (or binomial coefficients), which are found using nCr on the TI calculators:

"This argument shows that the number of ways to choose m objects from a total of n objects is  nCm = n(n - 1)...(n - m + 1)/(1 * 2 * 3 *...*m)."

These numbers also appear in Pascal's triangle. Stewart quotes Jacob Bernoulli as he discusses a certain problem in probability -- choosing pebbles from an urn with 3000 white and 2000 black ones:

"'Can you do this so often that it becomes ten times, one hundred times, one thousand times, etc., more probable...that the numbers of whites and blacks chosen are in the same 3:2 ratio as the pebbles in the urn, rather than a different ratio?'"

But there is a problem here with creating trials and inferring probabilities:

"How can we observe that the probability of all pebbles being white is very small? If we do that by lots of trials, we have to face the possibility that the result there is misleading, for the same reason; and the only way out seems to be even more trials to show that this event is itself highly unlikely."

Yes, circularity arises even if we try to define the probability of a "fair" coin -- whatever that means:

"Presumably, that heads and tails are equally likely. But the phrase 'equally likely' refers to probabilities."

The way to escape this circularity is by using axioms or postulates. Andrei Kolmogorov was one of the first to come up with axioms for a probability space:

"For a die, the set X consists of the numbers 1-6, and the set B contains every subset of X. The measure of any set Y in B is the number of members of Y, divided by 6."

Others who worked with probability and statistics was Francis Galton -- who wanted to study human heredity -- and Ysidor Edgeworth:

"Edgeworth lacked Galton's vision, but he was a far better technician, and he put Galton's ideas on a sound mathematical basis."

On that note, Stewart concludes the chapter as follows:

"Probability is now one of the most widely used mathematical techniques, employed in science and medicine to ensure that any deductions made from observations are significant, rather than apparent patterns resulting from chance associations."

The sidebars in this chapter don't contain any biographies of mathematicians -- instead, it's just the usual, what probability did for them and what probability does for us.

Because I spent so much time writing about today's class, I only briefly discuss probability (and this is a relatively short chapter anyway). But once again, the material is important, especially for California Geometry teachers.

Today I've posted a video of pi and probability, as well as a video of a song about graphing. If you want a song on probability, try Square One TV's "Ghost of a Chance."

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 of two 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 the sum of the original two vectors. But this would need to be proved.


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 klm, 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 klm, 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 klm, 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 klm, 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' and n' are parallel. This may look appealing, but the problem is that the original proof commuted l and m, 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 (xy) to (x + hy + 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 (xy) to (x hy 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 (xy) to (x + hy + k) is irrelevant. The other would be that if students really need to know that the mapping (xy) to (x + hy + 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.

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(xy) = (2xy) 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, so my next post will be on Tuesday.


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