Tuesday, November 10, 2015

The Parallel Tests: Lessons 3-4 and 5-6 (Day 53)

Today is a special day on the calendar, as is tomorrow. Let me repeat what I wrote last year about these two days (updated so that it refers to the 2015 calendar):

Here in California, not only is Veteran's Day a holiday on which all schools are closed, but also it must be observed on its actual calendar date -- November 11th -- no matter what day of the week that happens to be. As it turns out, this year, November 11th falls on a Wednesday.

Another thing that I want to point out is that today, November 10th, really is a holiday. No, it's not Admissions Day, but Pi Day!

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:


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.)

The only problem would be at a school where Veteran's Day must be November 11th. With the third Pi Day falling the day before a holiday and often part of a long weekend, more often than not there will be no school on November 10th. If that day falls on a Saturday or Sunday, then there will obviously be no school. Nor can there be school on Friday, November 10th, since then Veteran's Day would be on a Saturday and all Saturday holidays get pushed back to Friday. Finally, if November 10th falls on a Monday, as it did last year, some schools have a four-day weekend. So more often than not, there's no school on third Pi Day. Note that if it's a leap year, the 314th day will fall on the 9th -- a day on which there's more likely to be school than the 10th. But I have absolutely no idea what schools do when Veteran's Day falls on a Wednesday. (At least I did until this year -- and as we can see, both of my school districts take only the Wednesday off.)

Hmm, I wonder what I'd do if I taught an integrated math class -- where the students may be studying geometry around November 10th but algebra around March 14th. If possible, I'd try to schedule geometry near the original Pi Day of March 14th, so algebra would be before this.

(Actually having seen a few integrated texts since last year, I've seen that most do appear to save Geometry for the second half of the text. This would be convenient for Pi Day, though I point out that not every text places a chapter on circles at that point of the text.)

Both November 10th 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.

As I like to do on the other two Pi Days, let me post a pi video. This Numberphile video is especially relevant because it's about Pi and the Mandelbrot Set -- recall that I spent much of last month discussing Mandelbrot and fractals:

No, today's lesson has nothing to do with pi. Instead, we'll prove an important set of theorems -- the Parallel Tests. These theorems are spread out in the U of Chicago text between Lessons 3-4 and 5-6.

This is what I wrote last year in describing the Parallel Tests -- including why I call these theorems by the name "Parallel Tests":

But there are a few things to point out about this lesson. First of all, let's remind everyone where the name "Parallel Tests" comes from -- Dr. Franklin Mason. Dr. M then uses the name "Parallel Consequences" to refer to the statements of the form "if two lines are parallel, then..." -- in other words, they are the converses of the Parallel Tests.

But why should we distinguish between Parallel Tests and Parallel Consequences? For one thing, in geometry, a statement can be true even while its converse is false. "If a quadrilateral is a rhombus, then it is a parallelogram" is one such statement. But there are so many true statements with true converses -- the Parallel Tests among them -- that students fail to make the distinction. This is especially true when looking at the exercises typical in many geometry texts. An exercise based on a Parallel Consequence may give a diagram of two lines cut by a transversal, with one angle labeled with a degree value and another labeled with a variable, with the instruction:

If l | | m, then find the value of x.

And then, an exercise for a Parallel Test would be the same diagram, with the instruction:

Find the value of x that would make l | | m.

So there's no surprise that the students confuse a statement with its converse. Moreover, it turns out that in non-Euclidean hyperbolic geometry, the Parallel Tests are true, yet their converses are false (just like "if a quadrilateral is a rhombus, then it is a parallelogram")! So the Parallel Consequences all require a Parallel Postulate to prove, while the Parallel Tests do not.

In Dr. M's original formation of geometry, he uses triangle congruence to prove the Triangle Exterior Angle Inequality, then in turn uses the TEAI to prove the Alternate Interior Angles Test, and finally uses AIA to prove the other tests. But as I mentioned earlier, Dr. M eventually dropped the proof of TEAI and just included it as a postulate instead. Last year, I noticed that he'd taken this one step further, and dropped TEAI from Chapter 4 on parallels entirely. Dr. M simply had a Corresponding Angles Postulate, more in line with traditional texts.

...except that Dr. M has changed his Parallel Tests lesson yet again! He's returned to his original plan of using triangle congruence to prove TEAI. So if any of you complain that I keep changing my own lessons on this blog over and over again, just look at what Dr. M is doing! (Of course, many of my lessons are based in part on Dr. M's.)

There are a few things that I'd like to say about the sequence of proofs currently appearing on Dr. M's website (TEAI -> AIA -> other Parallel Tests). First, we compare Dr. M's proofs to what Dr. David Joyce writes about parallels on his website:

Chapter 7 [of the Prentice-Hall text -- dw] is on the theory of parallel lines. It begins by postulating that corresponding angles made by a transversal cutting two parallel lines are equal. One postulate is enough, but for some reason two others are also given: the converse to the first postulate, and Euclid's parallel postulate (actually Playfair's postulate). A proliferation of unnecessary postulates is not a good thing. One postulate should be selected, and the others made into theorems.

Well, Dr. M's website actually meets Dr. Joyce's ideal -- he only uses one postulate, Playfair's. In fact, it was by reading Dr. M's website when I first realized that one doesn't need any parallel postulates to prove the Parallel Tests -- these theorems are all neutral. Dr. M then uses Playfair to convert each Parallel Test into its converse, a Parallel Consequence.

I'm trying as hard as I can to meet the Joyce-Dr. M ideal of having only one parallel postulate -- which explains why I'm insisting on remaining in neutral geometry even as we start learning about the Parallel Tests. But it's tough to do this and use translations to derive the Parallel Tests, because much of what we know about translations is not neutral.

Dr. M uses TEAI to derive the Alternate Interior Angles Test. But it's also possible to use TEAI to prove the Corresponding Angles Test instead. We look at the proof on Dr. M's website, where he draws alternate interior angles 1 and 2, with angle 1 exterior to the triangle formed by the two lines and their transversal and angle 2 an interior angle of the triangle. But we could have easily used angle 3 instead -- where angles 1 and 3 are vertical. Now angles 2 and 3 are corresponding angles, while angle 3 is still exterior to the triangle. In either case we derive the same contradiction -- the exterior angle 1 or 3 is greater than the remote interior angle 2, yet it's given that they are congruent. So we can go TEAI -> CAT -> AIAT rather than TEAI -> AIAT -> CAT. (I've actually once seen a text that goes TEAI -> CAT -- actually, CAT is stated as a theorem early in the text, but not proved until late in the text when TEAI is given.)

Here's the thing about TEAI -- I associate this theorem with Dr. M so strongly that I almost always call it by its initials TEAI, just as Dr. M does, rather than Triangle Exterior Angle Inequality (while I only seldom call other theorems by their initials, such as ITT for Isosceles Triangle Theorem). The TEAI theorem is Euclid's Proposition 16, which is the first proposition that is provable in neutral geometry, but not spherical geometry.

It's because of this that many theorems that hold in neutral but not spherical geometry tend to be provable using TEAI. The most obvious are the Unequal Sides and Unequal Angles Theorems, which even the U of Chicago text derives from TEAI. A less obvious example is AAS Congruence. Even Dr. M proves AAS using ASA plus Triangle Sum, but we've seen before that Euclid proves AAS using his Proposition 16, TEAI.

Recall that I've chosen to avoid TEAI. So I've come up with another proof of AAS. It's still neutral because it only requires the Corresponding Angles Test and not full Triangle Sum. Let me post the proof that I wrote a few weeks ago:

AAS Congruence Theorem:
If, in two triangles, two angles and a non-included side of one are congruent respectively to two angles and the corresponding non-included side of the other, then the triangles are congruent.

Given AB = DE, Angle A = FDE, Angle C = DFE. We begin, as in the other proofs, by performing the isometry that maps AB to DE. And as in the proofs of ASA and SAS, the Side-Switching Theorem implies that using line DE as a mirror, Ray DC is mapped to Ray DF. So we know that the image of C is somewhere on Ray DF, but we don't know yet that C' is exactly F.

So let's try an indirect proof -- assume that C' and F are not the same point. This may be tough to visualize, so try drawing a picture. We already know that C' lies on Ray DF, so we can draw C' to be any point on Ray DF other than F. It doesn't matter whether C' is between D and F or on the opposite side of F from D -- both will lead to the same contradiction.

After we label the two angles known to be congruent, DC'E and DFE, we notice something about the diagram we've drawn. We see that lines C'E and FE are in fact two lines cut by the transversal DF, and the two angles DC'E and DFE turn out to be corresponding angles that are congruent. Thus, by the Corresponding Angles Test, lines C'E and FE are parallel! (Recall that the Parallel Tests are neutral, unlike the Parallel Consequences.) And so we have two parallel lines that intersect at E, a blatant contradiction. So the assumption that C' is not F must be false, and so C' is exactly F. QED

But if TEAI is so handy, why do I avoid it so much? I've always assumed that the proof of TEAI was too difficult for the students to understand, which is why Dr. M. first made TEAI a postulate, then dropped it altogether. But now that Dr. M has brought the TEAI theorem back, it almost makes me wonder whether I should include it in my lessons as well.

We could include TEAI in a black box unit where we prove all of the congruence theorems, including AAS from TEAI. If we were to use Dr. M's proofs of SSS and HL which require the Isosceles Triangle Theorem, then the next unit -- which would focus on applications of triangle congruence -- could be all about quadrilaterals, since we would have already proved both the Isosceles Triangle Theorem and its converse.

Ah, that's right -- quadrilaterals. As it turns out, the last theorem in the Chapter 5 of the text that we can prove in neutral geometry is the full Quadrilateral Hierarchy Theorem. This may seem incredible since without a parallel postulate, we can't prove that all of the quadrilaterals in the hierarchy (such as rectangles and squares) even exist!

But let's think about it. In the hierarchy, rectangle is below parallelogram, which means that every rectangle is a parallelogram. In if-then form, we would say "if a figure is a rectangle, then that figure is a parallelogram." In non-Euclidean geometry, this conditional is vacuously true -- that is, it's true exactly because there are no rectangles! There are zero rectangles, and all zero are parallelograms! In fact, we can prove that all unicorns are parallelograms, because there exists zero unicorns, and all zero are parallelograms.

So now we can prove in neutral geometry that every rectangle is a parallelogram. The proof begins like this:

Given: ABCD is a rectangle.
Prove: ABCD is a parallelogram.

In non-Euclidean geometry, we can't prove that a rectangle exists -- but we don't have to, because it's given that a rectangle exists. As soon as a rectangle exists, we're in Euclidean geometry -- but that doesn't change the fact that the proof is neutral, since we didn't use any non-neutral theorems to prove that the rectangle exists (as "Given" is not a theorem). We then use the same proof that the U of Chicago text uses to prove the theorem -- it follows trivially from the Two Perpendiculars Theorem.

But here's a part of the Quadrilateral Hierarchy Theorem that's less trivial to prove -- namely that a rectangle is an isosceles trapezoid. Again we can begin:

Given: ABCD is a rectangle.
Prove: ABCD is an isosceles trapezoid.

In the U of Chicago text, this proof is trivial because there an isosceles trapezoid is defined to be a trapezoid with congruent base angles. We know that a rectangle is a trapezoid, because every rectangle is a parallelogram and every parallelogram is a trapezoid. And we know that its base angles are congruent, because its base angles are right angles.

But here on the blog, an isosceles trapezoid is defined to be a quadrilateral with congruent base angles and congruent legs. Again, the congruence of the base angles is trivial, but the congruence of the legs is not so trivial. Here is the proof written in full:

Statements                             Reasons
1. ABCD is a rectangle            1. Given
2. ADC, BCD, ABC, BAD       2. Definition of rectangle (meaning)
    are all right angles
3. Angle ADC = BCD,             3. All right angles are congruent.
    Angle ABC = BAD
4. Let N be midpoint of DC      4. Every segment has exactly one midpoint.
5. DN = CN                            5. Definition of midpoint (meaning)
6. M on AB with MN perp. DC 6. Uniqueness of Perpendiculars Theorem
7. Angle DNM = CNM            7. All right angles are congruent,
8. MN = MN                           8. Reflexive Property of Congruence
9. Triangle DNM = CNM         9. SAS Congruence Theorem (steps 5, 7, 8)
10. DM = CM, MDN = MCN   10. CPCTC
11. Angle ADM = BCM           11. Subtraction Property of Equality (step 3a minus 10b)
12. Triangle ADM = BCM        12, AAS Congruence Theorem (steps 3b, 11, 10a)
13. AD = BC                           13. CPCTC
14. ABCD isosceles trapezoid   14. Definition of isosceles trapezoid (sufficient)

Notice that nowhere did we actually use the fact that all four angles of the rectangle are 90 degrees -- to get to step 3, we only needed that the base angles are congruent to each other, as are the summit angles (and we didn't even need the first pair to be congruent to the second). And so we've just proved that any quadrilateral with two pairs of adjacent congruent angles is an isosceles trapezoid. This may be more satisfying to a neutral geometer than our proof that a rectangle is an isosceles trapezoid. So instead of merely proving that things that might not exist in neutral geometry -- such as rectangles or unicorns -- are isosceles trapezoids, we showed that objects that provably exist -- quadrilaterals with two pairs of adjacent congruent angles -- are isosceles trapezoids.

It may seem awkward to save the proof that rectangles are isosceles trapezoids for after the Parallel Tests (since AAS was needed in the proof). Then again, even in the U of Chicago text we needed Parallel Tests to show one last relationship in the hierarchy -- that every rhombus is a parallelogram.

The proof in the U of Chicago text is neutral because it only uses the Alternate Interior Angles Test and not any Parallel Consequences. Actually, we didn't need all four sides to be congruent to each other, but only that they are congruent in pairs, AB = CD and BC = DA. Therefore, just like the rectangle-isosceles trapezoid proof, this one generalizes. We've in fact proved that if the opposite sides of a quadrilateral are congruent, then it is a parallelogram. This is a very well-known Parallelogram Test, and in fact I gave the proof of the same from Lesson 7-7.

We have therefore given a neutral proof of the entire Quadrilateral Hierarchy Theorem.

On today's worksheet, I post a worksheet that contains the following proofs:

-- Corresponding Angles Test
-- Alternate Interior Angles Test
-- Every rhombus is a parallelogram.
-- AAS Congruence Theorem
-- Every rectangle is an isosceles trapezoid.

Enjoy your Veteran's Day Holiday. My next post will be Thursday, November 12th.

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