Monday, November 30, 2015

Lesson 6-1: Transformations (Day 61)

I wish you all a wonderful welcome back, and that you enjoyed your Thanksgiving break.

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. Of course, on this blog we've already covered reflections, rotations, and translations, since we jumped around.

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. (Actually, this is based on continuity, not betweenness, but these are closely related.)
  • 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.

As we work in Chapter 6 -- which counts as our Unit 5 -- we will revisit some of the transformations that's we've already studied. Of all the transformations covered in Chapter 6, the only one that we haven't studied yet is the glide reflection -- the most complicated of the isometries.

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 k, l, m, and n 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 pair k 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 mirror d 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

Even though we've already given a parallel postulate, this proof is valid in neutral geometry. Yes, the four mirrors are said to form a rectangle, but notice that the only perpendicular relationships needed are k and m, m and l, and l and n. So the four mirrors need only form a Lambert quadrilateral. We must make sure to choose the mirrors carefully as to avoid making the fourth angle a right angle. The following assignment will work:

k: the line with equation y = -k/2
l: the x-axis
m: the y-axis
n: the line with equation x = h/2

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 (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 (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?

Friday, November 27, 2015

The Common Core Debate, Black Friday Edition

I said that I'd post two blog entries during the Thanksgiving vacation. The first was yesterday, on Thanksgiving, and the second is today, the day after Thanksgiving -- also known as Black Friday.

Just as with Thanksgiving yesterday, today I discuss Black Friday's significance on the calendar. I've seen many theories describing the origin of the name "Black Friday." The prevailing theory is that the name dates back to the 1960's, when customers complained about the amount of traffic at the malls.

It's not too difficult to see how Black Friday developed into a shopping day. With Thanksgiving always on a Thursday, workers began taking Friday off in addition to the scheduled off days on Thursday, Saturday, and Sunday. And since Thanksgiving has long marked the start of the countdown to Christmas, it was often treated at the first shopping day of the holiday season. Closely related to Black Friday are Small Business Saturday, intended for non-chain retailers, and Cyber Monday, intended for online retailers. Notice that if Thanksgiving falls on its latest possible date (the 28th) then Small Business Saturday falls on the last day of November -- therefore "the Saturday after Thanksgiving" and "the last Saturday in November" are equivalent.

But Black Friday is often not the biggest sales day for retailers. The problem is that customers take up parking spaces at the malls, but when they enter the stores they are only window shopping rather than opening their wallets and purses to make purchases. Customers often want to wait until the latest possible day to make their Christmas purchases. Since Saturdays are the days when people went to neither work nor church, this means that the biggest sales day is often the last Saturday before Christmas -- often known as Super Saturday.

Notice that Super Saturday actually has a significance on the academic calendar. Since the last day of school before winter break in most districts is a Friday, the first full day of vacation when students don't have to go to school is Super Saturday.

Just as we've seen with Advent, Super Saturday is tricky to define when Christmas is on Sunday. The last Saturday before the holiday would be Christmas Eve, when stores close as early as 5 or 6 PM. On the other hand, a week earlier, stores may be open all the way to midnight -- and so it's more likely for stores to make more money when they are open six or seven extra hours. And so it's uncertain whether Super Saturday should be defined as the 17th or the 24th in such years.

About a decade ago, Black Friday overtook Super Saturday in sales when retailers started offering deep discounts on Black Friday. But then these Black Friday sales started creeping into Thanksgiving itself -- a phenomenon often known as Gray Thursday. (Don't forget that last year, I bought my computer on Gray Thursday.) With the early holiday sales being divided between Gray Thursday and Black Friday, it means that Super Saturday has regain its status as the biggest sales day.

It appears that the ten biggest shopping days of the year are usually (in no particular order):

-- Black Friday,
-- Four Saturdays, from Small Business Saturday to Super Saturday,
-- Four Sundays of Advent, and
-- The day after Christmas. This day is often known as Boxing Day in many countries, and it's Boxing Day, not Black Day, that's known as a day of deep discounts. Here in the U.S. the name "Boxing Day" is rare, but many stores nonetheless hold After Christmas sales that day.

Likewise Cyber Monday is often not the biggest online sales day. The equivalent of Super Saturday for online retailers is the last day that an item sent using free shipping can arrive by Christmas. This day doesn't follow a simple formula like "the third X in December." Instead it varies by retailer -- this year it's approximately Wednesday, December 16th.

It's funny how Black Friday's popularity has spread to other nations -- and of course those other countries don't observe Thanksgiving. I remember last year when I was tutoring some students born in Korea, and I asked them what they were looking forward to the most during the days off from school at the end of November. The Korean students' answer was not "Thanksgiving dinner," but "Black Friday shopping."

Let's get back to the Common Core. No, I'm not going to attempt to connect Common Core to Black Friday (although many Common Core opponents would argue that the purpose of the Core is to make money for businesses like Pearson, just as the purpose of Black Friday is to make money).

Instead, I will make a connection to science. This week marks the 100th anniversary of Albert Einstein's Theory of Relativity. Actually, it appears that Einstein first formulated his famous theory in 1905, but it took ten years until the predictions he made could be confirmed. Therefore 2005 was the centennial of his annus mirabilis, but 2015 is the centennial of the actual publishing.

And so it's only fitting that right around the time of the centennial celebration, the Next Generation Science Standards (i.e., Common Core for science) have been revealed:

Science, of course, is the school subject most closely allied with mathematics -- indeed, both science and math fall under the STEM umbrella. So even though this is a math blog, I want to have some discussion of the science standards.

But I begin with my own scientific studies. While I've always been a strong math student, my science grades have gone up and down throughout my academic career. In middle school, my science grades alternated between B's and C's. Despite this, I was allowed to take the Advanced General Science class as a freshman. In that class, though, I performed so well during the first quarter that the teacher was to recommend me for a higher class, Applied Biology/Chemistry, for the second quarter.

But I never stepped into the Applied Bio/Chem classroom. This is because just before the start of the second quarter, I transferred to a school in another district. My new school had a magnet program for which I might have qualified, except that students must apply to the magnet program from their middle schools in eighth grade -- and I will still in my old district as an eighth grader. So I was placed in one of the regular programs.

At this school, magnet students took traditionalist science classes like Biology and Physics, but regular students were placed into Integrated Science classes. (That's right -- just as there exist integrated math classes, science can also be integrated.) Officially, Integrated Science I, II, and III contain elements of Biology, Chemistry, and Physics. In California, two years of science are required for college admission. But since each year of Integrated Science is considered equivalent to a third of a year each in Biology, Chemistry, and Physics, Integrated Science I and II are insufficient for college, as they don't comprise a full year in any of the three disciplines. So two years of Integrated Science count as zero -- one must take I, II, and III in order for any of them to count.

In Science I and II, my grades hovered between A's and B's. But as I began Integrated Science III my junior year, I excelled once again. My teacher often noticed that I would finished my tests quickly and spend the rest of the time making them look neat, and then she'd grade them, and of course my answers were correct. And so she -- just as my freshman science teacher had two years earlier -- recommended me for a higher class the second quarter.

My teacher convinced the counselors to admit me to the magnet program as a junior, even though most students entered as freshmen. She had me take science in the room next door, which turned out to be a Chemistry class. But on my transcript, the class was still notated as Integrated Science III, because she knew that unless I took Science III, my first two years of science wouldn't count. Even though I could still take Chemistry as a junior and another class as a senior and still have two years of science for college, notating the class as Science III meant that all four years would count.

I did well in Chemistry, and the following year, I took AP Physics C. I believe that I earned a score of 4 on the AP exam. I went on to take some science classes in college, with mixed results. I earned an A- for my first quarter of physics, B- my second quarter, and C- my third quarter.

We might argue that since my best subject was math, I should perform much better in the physical sciences -- Chemistry and Physics -- that are heavy with calculations, and more poorly in classes like Biology where there were few calculations. In some ways, this was true, since there was much Bio in my Science I and II courses while Science III began with Chemistry. In California, seventh grade has always tended toward Life Science, which explains my mediocre grades in seventh grade science, but I fared no better in eighth grade which tended toward Physical Science. So the type of science only partially explains why I performed better in some science classes than others.

Anyway, let's take a look at the new Next Generation Standards. Notice that just like the Common Core High School Math Standards, the High School Science Standards aren't divided into courses. It's up to the states to make this division. Here's a link to my home state of California and how it is dividing the standards into courses:

After all this discussion about my Integrated Science courses in high school, notice that California is now recommending an integrated approach in middle school. Recall that 7th and 8th grade science had always been devoted to Life and Physical Science, respectively. More recently, the sixth grade science course was geared towards Earth Science -- according to the link above, this was back in 1998, by which time I had already passed sixth grade.

But now middle school science will all be integrated. Meanwhile, notice that high school science courses are left to local decision. There is another link back to the Next Generation website which shows options for both integrated and traditionalist courses, as well as the division of the standards into three and four courses (Biology, Chemistry, Physics, and Earth Science).

What is my own opinion of this adoption? I've said that one of the strongest arguments in favor of Integrated Math is that this is how most other nations arrange their math classes. So, we can ask ourselves, how do other countries arrange their science courses? I did some research and decided to look up how another country does science -- I chose Great Britain. (And yes, that's another country to which Black Friday has spread, even though Thanksgiving isn't celebrated in the country that the Pilgrims were fleeing from.)

I found out that the British definitely integrate science in middle school (which they call "Key Stage 3"), but what they do in high school ("Key Stage 4"), I found a bit confusing. Apparently, there's no such thing as taking "Biology" one year and "Chemistry" the next, but just "Science." Yet the GCSE exams -- the big tests that students take at the end of Key Stage 4 (JK Rowling got the inspiration for Harry Potter's OWL exams from the GCSE's and their predecessors) -- give separate tests for Biology, Chemistry, and Physics. On the other hand, there is only one math (or as they call it, "Maths") exam -- not separate tests for algebra and geometry.

So if we really want to follow other countries, we should have Integrated Science classes yet traditionalist standardized tests. (Note: I still have yet to hear how testing under the Next Generation Science Standards will work, except that there will be testing once each in elementary, middle, and high school, as is currently done.)

Not quite related to the Next Generation Science Standards is another issue surrounding the science curriculum -- the concept of "Physics First." This is the idea that freshmen should start the science curriculum with Physics, and then sophomores take Chemistry, and then Biology for juniors, rather than the reverse order. To me, this idea has merit. Much of Physics applies to Chemistry, and likewise much of Chemistry applies to Biology. And much of math applies to Physics -- so students can see why it's so important to learn math. The same is true with a subject like Statistics -- I believe that Stats and Data Analysis are more relevant for most people than much of algebra, and so teachers can demonstrate how math can be useful by showing its applications in Physics and Statistics.

But many traditionalists oppose this rearrangement. They'd argue that much of both Physics and Statistics are dependent on higher algebra and Calculus, and so students shouldn't see either Physics or Stats until after having completed Calculus. My problem is that students may never get through Calculus if they see algebra and Calculus as useless subjects. Of course, a freshman Physics course should start with the basics -- you can still save Einstein's Theory for AP Physics as a senior.

Thus ends my second holiday post. I will resume regular posting on Monday, November 30th.

Thursday, November 26, 2015

The Common Core Debate, Thanksgiving Edition

Happy Thanksgiving! Last year, I didn't post on Turkey Day itself, because that was when I purchased my new computer. But this year, I decided that the first of my two special holiday posts will indeed be today. As I often do during vacation periods, I'll go back to the Common Core Debate and discuss issues related to its implementation.

But first, as you already know, calendars fascinate me. And so on this Thanksgiving Day, we may ask, why is today Thanksgiving? And I don't simply mean, "because today's the fourth Thursday in November" -- instead I mean, why is Thanksgiving observed on the fourth Thursday in November?

(This post has been tagged "calendar" due to my long discussion of the Thanksgiving date, as well as "traditionalists" for the Common Core debate. I was considering tagging this "How to Fix Common Core," but officially this post is not a part of that series.)

Of course, we already know that Thanksgiving commemorates the feast observed by the Pilgrims and Native Americans back in 1621. But here's the thing -- it's unlikely that the first Thanksgiving was observed in late November, and it's unknown whether or not it was a Thursday. Indeed, it's possible that the feast was closer to Canadian Thanksgiving in October, than any date in November.

One theory is that the first Thanksgiving feast was actually on another holiday -- a little-known Christian holiday called Michaelmas:

Notice that the name "Michaelmas" is pronounced with a short i, just like "Christmas," even though the holidays are named after St. Michael and Christ, respectively, each with a long i. If you recall six months ago when I was discussing the Waldorf schools, notice that one holiday often celebrated in Waldorf schools is Michaelmas.

Just as Christmas falls near the winter solstice, Michaelmas falls near the autumn equinox. To be exact, Michaelmas is on September 29th. It's known that the first Thanksgiving was a three-day feast, so if it began on Michaelmas, it would have ended on October 1st. But notice that these are dates in the old-style Julian calendar, since the Pilgrims came from England, and the English didn't convert to the new-style Gregorian calendar until just before the American Revolution. There was a ten-day difference between the two calendars, and so the last day of the feast would have been October 11th on our modern calendar.

Notice that October 11th, 1621, on the Gregorian calendar was a Monday -- indeed, it was the second Monday in October. And Canadian Thanksgiving also falls on the second Monday in October -- so our neighbors to the north have a three-day weekend on Saturday, Sunday, and Monday, and these three days correspond exactly to the three days of the feast according to the Michaelmas theory!

Another theory is that the first Thanksgiving was not the Christian festival of Michaelmas, but instead the Jewish holiday of Sukkot, or the Feast of Tabernacles. Here's a link to this theory:

We know that the Jewish calendar is lunisolar, and so we must consult lunar tables. Sukkot falls on the 15th day of the month of Tishrei, right after Rosh Hashanah. It turns out that Rosh Hashanah in 1621 fell on September 6th Julian (September 16th Gregorian), and so Sukkot fell two weeks later, on September 20th Julian (September 30th Gregorian). This date is still much closer to the Canadian date than the American date, although Americans can take solace in the fact that at least September 30th Gregorian was a Thursday.

As it turns out, the only thing the Pilgrims actually did in November was not in 1621, but in 1620 -- they landed at Plymouth Rock on November 11th Julian (November 21st Gregorian), which is just before our Thanksgiving. So we could argue that American Thanksgiving actually commemorates the landing of the Mayflower, while only Canadian Thanksgiving commemorates the big feast.

But we still haven't figured out why Thanksgiving must be on a Thursday. We've mentioned it's possible that the feast began on a Thursday, according to the Sukkot theory. But it just as easily could have begun on a Saturday and lasted until Monday, according to the Michaelmas theory. And besides, most holidays either fall on an exact date (Christmas, Independence Day, Veterans Day) or observed on a Monday (Presidents Day, Memorial Day, Labor Day). Thanksgiving is an outlier in that it always falls on a Thursday. Why would anyone choose a Thursday for a holiday?

We know that President Lincoln first formalized Thanksgiving as the last Thursday in November. But there were earlier presidents who made Thanksgiving Proclamations, often on Thursday. For example, President Washington proclaimed Thursday, November 26th, 1789, a day of Thanksgiving, so the Thursday tradition was established well before Lincoln. But there is a relationship between Lincoln's Thanksgiving Thursday and another Christian holiday -- perhaps only slightly better known than Michaelmas -- called Advent.

Advent is defined as four Sundays before Christmas, and therefore serves somewhat as a countdown to Christmas. (To see what Sundays have to do with Thursdays, read on.) Notice that if Christmas falls on a Sunday, then Advent is a full four weeks earlier. But if Christmas falls on a Monday, then Christmas Eve is the fourth Sunday of Advent, and so the first Sunday of Advent is only three weeks and one day before Christmas.

Now suppose Christmas fell on a Monday. Then Advent is three weeks before Christmas Eve, which works out to be December 3rd, so this is the latest possible Advent. Notice that the previous Thursday would be the last day of November -- the latest possible Thanksgiving under Lincoln.

In other words, we conclude that "last Thursday in November" and "last Thursday before Advent" are in fact equivalent, and Thursday is the only day of the week for which "last X in November" and "last X before Advent" are equivalent. So under Lincoln's Thanksgiving date, Thanksgiving is three days before the Sunday when Christians begin counting down to Christmas. Under my Advent theory, Thanksgiving and Christmas have always been linked, decades before the Macy's Thanksgiving Day Parade featuring Santa Claus at the end.

Of course, notice that the Advent theory only explains why presidents proclaimed the last Thursday in November to be Thanksgiving. It has nothing to do with anything the Pilgrims did. In fact, the Pilgrims didn't even celebrate Christmas (much less Advent), since it's not mentioned in the Bible (as opposed to St. Michael and Tabernacles). It's said that the Pilgrims began construction of their common house on December 25th Julian (January 4th Gregorian), so it was anything but a holiday.

Nowadays Thanksgiving is celebrated on the fourth, not last, Thanksgiving in November (so it isn't always the Thursday before Advent). I mentioned on the blog last year that this is due to President FDR ("Franksgiving"), since FDR wanted to extend the Christmas shopping season to boost the Depression economy. By FDR's day, Thanksgiving was certainly known as the start of the countdown to Christmas.

That's enough about the calendar -- let's move on to Common Core. But since it's still Thanksgiving, let's talk about the state that the Pilgrims landed in -- Massachusetts.

As it turns out, Common Core has sort of a setback in the Bay State. Massachusetts has rejected the PARCC exam, to replace it with a hybrid of PARCC and its pre-Common Core test, the MCAS. I see that there are several articles at the Boston Globe that discuss this, but here's the one I found with the most comments:

Judging by the comments, not many people agree with the PARCC/MCAS compromise. That's the problem with compromises -- they end up infuriating everyone. Here are the three types of comments that I found:

-- Pro-Core advocates (including the editorial board) say stick with PARCC because it measures college and career readiness and allows comparisons between states.
-- Anti-Core advocates say go back to the pre-Common Core MCAS. If the Core and PARCC are bad enough that people would want to see an alternative in the first place, just cut out PARCC completely.
-- Anti-testing advocates say eliminate state standardized testing completely. They were opposed to the MCAS long before there ever was a Common Core.

I like compromises because each side has merit, and hopefully a compromise can incorporate the best of what each side has to offer. I myself have had problems with many PARCC questions as documented right here on the blog, so I wouldn't mind throwing some of them out. But we can keep enough PARCC questions to allow state-by-state comparison.

Then again, I admit that there could be a better way to achieve a compromise. Notice that comparison to students in other states is most relevant to high school students. After all, these are the ones who are applying to colleges which draw students from many states, including two well-known colleges right in Massachusetts (Harvard and MIT). There is less of a need for elementary and middle school students to be compared to their counterparts in other states -- and besides, much of what is in the Common Core for the lowest grades is not age-appropriate (as discussed often here on the blog).

And so a great compromise would be for the (old pre-Core) MCAS to be given to eighth grade and below, and the PARCC to be given to high school students. Since the MCAS was traditionally administered to grades 3-8 and 10, this plan would keep MCAS in grades 3-8. PARCC tests, meanwhile, are given to every grade 9-11, so they would stay that way.

Oh, and there's one more thing. Any eighth graders enrolled in Algebra I should take the PARCC Algebra I test (and the MCAS test for ELA) They certainly shouldn't have to double-test in math -- the MCAS because they're in 8th grade and the PARCC because they're in Algebra I. We've seen how California almost came up with a super-hard eighth grade course in order to incorporate both the 8th grade standards for SBAC and the Algebra I standards -- we avoid this by not requiring 8th graders in Algebra I to take the 8th grade math test. If, by chance, there are any seventh graders enrolled in Algebra I, they shouldn't have to take the 7th grade MCAS either, but only PARCC Algebra I.

Unfortunately, Massachusetts will do the exact opposite during the current year of transition from PARCC to the new MCAS. It will be the high school students taking the MCAS this year, and grades 3-8 will still take the PARCC!

Last year, just before Thanksgiving, I wrote a blog entry discussing how Common Core affects instruction in grades 4-7. This is some (but not all) of what I wrote last year:

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.)

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:

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.

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.'"

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.

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.

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.

Last year, I thought that perhaps requiring students in this grade span (Grades 4-7) to explain their answers, as on the PARCC test, would help them avoid making these kinds of mistakes. But after having seen traditionalists post so many bad PARCC and other Common Core questions requiring ridiculous explanations, I'm now starting to lean towards pre-Common Core assessments -- which would include the old MCAS -- for this grade span.

And so this is why my recommendation for the state of Plymouth Rock would be to revert to the old MCAS for grades 3-8, and keep PARCC for high school only.

I spent the rest of last year's post discussing a text often favored by traditionalists -- Saxon Math. I don't post that again, but I will have something to say about Saxon Math in the upcoming weeks.

Thus concludes this post. I hope you enjoy the rest of your Thanksgiving!

Friday, November 20, 2015

Activity: Transformations -- Based on Lessons 6-1 and 12-1 (Day 60)

Here it is, the last day before Thanksgiving vacation. This means that I'm taking all of next week off on this blog and wait to post Day 61 on Monday, November 30th.

This is what I wrote last year in explaining why many schools take an entire week off for Turkey Day:

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 (yes, the Peanuts are still on my mind after watching the movie a few weeks ago) -- 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 last year, I drove past a school that took a five-day weekend, with Tuesday, the last day before the holiday, to be a minimum day (but this year that district switched to taking the whole week off).

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.

Notice that today is Day 60 on the school calendar. So we are now exactly one-third of the way through the year -- that is, one trimester. So we notice that even though winter break doesn't divide the semesters in this district, Thanksgiving break does divide the trimesters. And now that I'm thinking about it, I'm wondering whether this wasn't an intentional decision by the district administrators -- the first day of school was chosen by counting backward 60 school days from Thanksgiving, so that there would one whole trimester completed by the holiday.

Recall that trimesters are common in elementary schools and rare in high schools. They used to be common in middle schools, but now more and more middle schools are switching to semesters. My own district, therefore, uses trimesters only at the elementary level.

Also, recall that my district originally wanted to end the first semester by Christmas, but the union fought the idea since teachers didn't want to start school early enough in August. So I wouldn't be surprised if the compromise Middle Start calendar was created simply by saying that if we can't end the first semester by Christmas, let's at least end the first trimester by Thanksgiving.

A rule of thumb is that a trimester is approximately three months long. We observe that in my district, the first day of school was on August 26th and Thanksgiving is November 26th. So we conclude that if we start school exactly three months before Turkey Day -- on the fourth Wednesday in August -- then there will be exactly 60 days of school before the holiday break. The first trimester then consists of 12 weeks (as sixty divided by five is twelve), with three extra days to make up for the three holidays Labor Day, Columbus Day, and Veterans Day.

I posted the Chapter 5 Test yesterday in deference to schools that have a minimum day as the last day before the holiday. My own district doesn't have a minimum day before Thanksgiving, but I ended up subbing in a German class, and the students in that class took a test yesterday -- today they are taking notes on a video. On the other hand, one of the students in the German class is taking a Precalculus test today, on basic trig. He says that he's ready for the test, but he scared me a little when he told me that he had memorized the entire unit square -- when he should have said unit circle! Another student had to take a test today for his Integrated Math I class. Well, that shows the difference between the foreign language department and the math department!

And so today becomes an activity day instead. Last year, I posted a preview activity for Chapter 7, since under last year's Early Start Calendar, that was the first chapter after Thanksgiving. This year, Chapter 6 is ostensibly the first chapter after the break, but we've already covered two lessons from that chapter on the blog. But we haven't officially covered Lesson 6-1 yet.

Actually, Lesson 6-1 is a great lesson to cover. It's all about general transformations, with many of these transformations being done on the coordinate plane. Since the most transformation questions on the Common Core tests take place on the coordinate plane, I really want to emphasize them.

Many of the questions from this section of the U of Chicago text have students draw a figure on the plane, transform its vertices, and then identify the transformation as either a reflection, slide, turn, or size change. (The terms "slide" and "turn" are used since this in the text precedes Lessons 6-2 and 6-3 on translations and rotations.)

The previous page in the text gives a strange transformation, S(x, y) = (2x, x + y). This transformation is clearly not an isometry, as we can see that if the preimage is a house, the image is still a house, but a distorted house, not one where anyone would want to live.

What sort of transformation is S(xy) = (2xx + y)? Well, as it turns out, the transformation that maps (x, y) to (x, x + y) is the transformation I mentioned yesterday -- a transvection or shear. Its fixed line is the y-axis and its shear factor is 1.

Yesterday, I briefly mentioned that transvections preserve area, and we can use this fact to show that two triangles with the same base and height must have the same area, but I never gave the proof. So here's a paragraph proof:

Given: AB = DE, drop perpendiculars C to AB (meeting at M) and F to DE (meeting at N), CM = FN.
Prove: Triangles ABC and DEF have the same area.

We know that some isometry maps AB to DE, and so we begin our proof the same way that we started our proof of the SAS (and other) Congruence Theorem. As usual, if C and F are on opposite sides of DE, reflect C over DE. But the image of C isn't necessarily F. What we hope to show is that some transvection maps C to F.

We see that both CM and FN are perpendicular to DE, and they are congruent. And so CMNF is once again a Saccheri quadrilateral (a rectangle, since by now we're in Euclidean geometry), and since a rectangle is a trapezoid, CF is parallel to DE, so CF is an invariant line of any transvection with a fixed line DE. And so now we've found our transvection -- we choose a transvection with fixed line DE and shear factor CF/CM. This transvection maps C to F.

So DEF is the image of ABC under the composite of an isometry and a transvection. Since both isometries and transvections preserve area, ABC and DEF have the same area. QED

Then we see that the preimage and image in the U of Chicago text don't have the same area. This is because S doesn't map (x, y) to (x, x + y), but rather to (2x, x + y). So S is not a pure transvection, but rather the composite of a transvection and some other transformation. This second transformation merely doubles the x-coordinate. It can't be a dilation because a dilation would have to double both of the coordinates. Many Precalculus texts would call this a "horizontal stretch." And while transvection preserves area, horizontal stretch by a scale factor of 2 obviously doubles area. And so the image of the figure under S has exactly twice the area of the preimage.

By the way, the statement that transvections preserve area only works in Euclidean geometry. We can see eventually that in non-Euclidean geometry, area is deeply related to angle measure, and since transvections don't preserve angle measures, they can't preserve area either.

I was wondering whether it's possible to teach transvections in our Geometry classes. Under the Common Core, we're already teaching reflections, translations, and rotations for congruence, as well as dilations for similarity. So the natural step would be to teach transvections -- maybe even as part of teaching area, as transvections preserve area. The first unit of the second semester will be on dilations and so the next unit -- when it's time to teach area -- could be on transvections.

But I won't -- the Common Core doesn't require anyone to teach transvections. It may be possible to have an activity, similar to the one in the Japanese classroom we were discussing yesterday, where students can see how because of translations, triangles with the same base and height will also have the same area, just before showing them the formula. But the formula should still be proved using the Area Postulate and decomposition, as is done traditionally.

That's enough about a later activity -- what about today's activity? Here's another tradition that I've seen in math classes (besides taking a test) -- the day before Thanksgiving break, the students are going to graph points on a coordinate plane in the shape of a turkey.

Now I'm not an artist, so I couldn't come up with the points for the turkey's shape myself. So I decided to search for a turkey on the Internet. Although there are many turkeys around, I actually found one that not only involves graphing, but also transformations! The following worksheet comes from a school district in Fort Bend, Texas (which isn't even a Common Core state):

And after all of that discussion on transvections, the transformation turns out to be an ordinary dilation, with a scale factor of one-half. And after billing this as a preview of Chapter 6, the activity actually fits better with Lesson 12-1 of the U of Chicago text on dilations. Well, it's still a preview, but just a preview of the second semester, not next week.

This worksheet is somewhat tricky. Many of the points on Tom Turkey's preimage have odd coordinates, meaning that a dilation of scale factor 1/2 maps these to fractional points. And a few of the points on Tom's preimage even have fractional coordinates already. But at least all of the points are in the first quadrant (or the axes bounding that quadrant), so that we don't have to worry about sign errors.

It's possible to modify this worksheet to cover other transformations. On the coordinate plane, translations are the easiest. Reflections and rotations usually involve negative signs (unless the mirror isn't an axis or the rotation center isn't the origin), and so we'd need graph paper with a window of -30 to 30 on at least one of the axes. It's even possible to have a transvection -- it might be fun to see what Tom looks like sheared. I recommend using only 1 as the shear factor. Mapping (x, y) to (x, x + y) is easy but involves large coordinates. We could map (x, y) to (x, x - y) instead -- this avoids large coordinates but includes negative coordinates.

Notice that the Tom's waddle is a simple parallelogram. If you want students to perform different transformations, you can have them practice on the waddle. (One of the four possible transvections maps the waddle to a square.)

This activity may be a bit long, and so if your school has a minimum day, you can divide the class into partners -- one partner draws the preimage and the other the image. Both partners should work on calculating the image points first.

Thus ends this post and the first trimester. As I usually do during holiday periods, I plan on making two special posts during Thanksgiving break. The next school day post will be on November 30th.