"Centuries prior to this, the Babylonians developed and improved upon the Sumerian sexagesimal system to develop a base 60 number system. But this section on number worlds is not about number systems but about types of numbers."
This is the third page of the section "Number Worlds." (Once again, we missed the first two pages due to the non-posting weekend.) Fortunately, Pappas tells us exactly what this section is about -- types of numbers.
Here she writes about negative numbers -- and their nonacceptance by Renaissance mathematicians:
"Let's take a glimpse at the first type of numbers -- the counting numbers. But the world of counting numbers were not enough to solve all the problems that were to evolve over the years. What would have been some reactions -- the problem is defective, there is no answer. Nicholas Chuquet (15th century) and Michael Stidel (16th century) referred to negative numbers as absurd. Even Blaise Pascal said 'I have known those who could not understand that to take four from zero there remains zero.'"
The numberhood of fractions -- and even irrational numbers -- was accepted centuries before the numberhood of negative numbers. The hierarchy of the sets (or "number worlds") -- N, Z, Q, R, C -- implies that the set Z of integers comes before the set Q of rationals. Instead, the historical hierarchy of sets should be something like N, Q+, R+, R.
Notice that the order in which the Common Core introduces the worlds is N (grades K-4), Q+ (grades 5-6), Q (grade 7), R (grade 8). This is closer to the historically correct order, and explains why the standards use term "rational numbers" when explaining the integer arithmetic rules:
CCSS.MATH.CONTENT.7.NS.A.1
Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.
On the other hand, irrational numbers appear after negative numbers in the Common Core. Blaise Pascal would have accepted sqrt(2) as a "true root" of the equation x^2 - 2 = 0, while -4 was a "false root" of the equation implied above, x + 4 = 0.
Before we get to today's lesson, let me make a few side comments. Today's Google Doodle celebrates a scientist/mathematician -- Max Born, the founder of quantum mechanics. As usual, I like to highlight scientific or mathematical Google Doodles on the blog. (Apparently his granddaughter is Olivia Newton-John. She's all about getting "physical" while he was about getting "physics-al.")
Actually, yesterday's Google Doodle also featured a scientist, Robert Koch, the discoverer of the tuberculosis bacterium. Both Germans were actually born on today's date, December 11th, but of course Google couldn't celebrate both of them on the same day.
Backtracking even further, there was also a Google Doodle for Dutch scientist Jan Ingenhousz, the discoverer of photosynthesis. It really did appear on his actual birthday, which was last Friday (the day after my own birthday). Wow -- it might have been interesting if I were still teaching a science class to mention these all these celebrated scientists in class!
Finally, some readers may have noticed that my home state of California has been in the news lately for all the wrong reasons -- relentless wildfires. In the San Fernando Valley, part of LAUSD had an impromptu five-day weekend from Wednesday to Sunday, but all schools in Los Angeles County are open today. UCLA, my alma mater, held only morning classes on Thursday and Friday -- and that was 10th week, the last week before finals. Meanwhile, Some schools in Ventura and Santa Barbara counties are closed today.
Lesson 7-6 of the U of Chicago text is called "Properties of Special Figures." (In the new Third Edition of the U of Chicago text, this is Lesson 7-7. Oh, and the title of the new Lesson 7-7 makes it clear that the "special figures" referred to here are parallelograms.)
I will post what I wrote two years ago on this lesson -- but it's edited heavily, since I made so many changes to the lesson that year -- I included some of Chapter 13 in that lesson. (Recall that just as I did two years ago, the new Third Edition also blows up the old Chapter 13 by moving all of its topics to other chapters. Ironically, Chapter 7 isn't one of them -- Lesson 7-6 of the new edition, on tessellations, appears in Chapter 8 of the old edition, and the other two extra lessons in the new edition don't appear in the old edition at all.)
Anyway, this is what I wrote two years ago about today's lesson:
The unit test will not be given until just before winter break, leaving us with an extra week to fill. So I asked myself, what else can I fit into this unit?
And so I decided just to post the parallelogram properties after all. This week we cover Lesson 7-6, on the Parallelogram Consequences, and Lesson 7-7, on the Parallelogram Tests.
We've already seen how useful the Parallelogram Consequences really are. The reason that they are delayed until 7-6 in the U of Chicago text is that they are best proved using triangle congruence, but triangle congruence doesn't appear until Chapter 7.
It would probably make more sense to cover parallelograms along with the other quadrilaterals in Chapter 5, but it's too late now. Fortunately, I have this opening in the schedule now to cover both Lessons 7-6 and 7-7.
And today's Lesson 7-6 also includes the Center of a Regular Polygon Theorem, which the U of Chicago text proves using induction. This fits with the Putnam-based lesson that I posted last week.
Oh, and by the way, I found the following page about the Fibonacci sequence and generalizations:
http://mrob.com/pub/seq/linrec2.html
(That's right -- this is my second straight week on the same Putnam problem.) It shows why numbers of the Fibonacci sequence have other such numbers as factors -- it's because the Fibonacci numbers can be written as polynomials that can be factored to get other Fibonacci polynomials.
So how does Dr. Merryfield solve Putnam problems like 2014 B1 or 2015 A2? As it turns out, he uses a proof technique called mathematical induction. Most proofs in a high school geometry course aren't proved used mathematical induction. Indeed, only one proof in the U of Chicago text is proved this way -- and it just happens to be the Center of a Regular Polygon Theorem! Furthermore, Dr. Wu uses induction in his proofs on similar triangles, so this is a powerful proof technique indeed. (On the other hand, Wu simply defines a regular polygon as an equilateral polygon with a circle through its vertices, so he would have no need to prove the following theorem at all.)
We've already seen how useful the Parallelogram Consequences really are. The reason that they are delayed until 7-6 in the U of Chicago text is that they are best proved using triangle congruence, but triangle congruence doesn't appear until Chapter 7.
It would probably make more sense to cover parallelograms along with the other quadrilaterals in Chapter 5, but it's too late now. Fortunately, I have this opening in the schedule now to cover both Lessons 7-6 and 7-7.
And today's Lesson 7-6 also includes the Center of a Regular Polygon Theorem, which the U of Chicago text proves using induction. This fits with the Putnam-based lesson that I posted last week.
Oh, and by the way, I found the following page about the Fibonacci sequence and generalizations:
http://mrob.com/pub/seq/linrec2.html
(That's right -- this is my second straight week on the same Putnam problem.) It shows why numbers of the Fibonacci sequence have other such numbers as factors -- it's because the Fibonacci numbers can be written as polynomials that can be factored to get other Fibonacci polynomials.
So how does Dr. Merryfield solve Putnam problems like 2014 B1 or 2015 A2? As it turns out, he uses a proof technique called mathematical induction. Most proofs in a high school geometry course aren't proved used mathematical induction. Indeed, only one proof in the U of Chicago text is proved this way -- and it just happens to be the Center of a Regular Polygon Theorem! Furthermore, Dr. Wu uses induction in his proofs on similar triangles, so this is a powerful proof technique indeed. (On the other hand, Wu simply defines a regular polygon as an equilateral polygon with a circle through its vertices, so he would have no need to prove the following theorem at all.)
[2017 update: I retain this discussion from two years ago about old Putnam problems in order to demonstrate mathematical induction. But actually, this year's Putnam problem 2017 A1 is also an example of induction. For example, the statement that all numbers ending in 2 or 7 are sexy is proved via induction -- the initial case is that 2 is sexy, and Merryfield proved the lemma that if n is sexy, then n + 5 is sexy. Also, with the Fibonacci link above, notice that some of the properties of the sequence that I mentioned on Black Friday are provable via induction.]
So here is the proof of the Center of a Regular Polygon Theorem as given by the U of Chicago -- in paragraph form, just as printed in the text (rather than converted to two columns as I usually do):
Proof:
Analyze: Since the theorem is known to be true for regular polygons of 3 and 4 sides, the cases that need to be dealt with have 5 or more sides. What is done is to show that the circle through three consecutive vertices of the regular polygon contains the next vertex. Then that fourth vertex can be used with two others to obtain the fifth, and so on, as many times as needed.
Given: regular polygon ABCD...
Prove: There is a point O equidistant from A, B, C, D, ...
Draw: ABCD...
Write: Let O be the center of the circle containing A, B, and C. Then OA = OB = OC. Since AB = BC by the definition of regular polygon, OABC is a kite with symmetry diagonalOB. Thus ray BO bisects angle ABC. Let x be the common measure of angles ABO and OBC. Since triangle OBC is isosceles, angle OCB must have the same measure as angle OBC, namely x. Now the measure of the angles of the regular polygon are equal to 2x, so angle OCD has measure x also. Then triangles OCB and OCD are congruent by the SAS Congruence Theorem, and so by CPCTC, OC = OD. QED
Now the "Analyze" part of this proof contains the induction. If the first three vertices lie on the circle, then so does the fourth. If the fourth vertex lies on the circle, then so does the fifth. If the fifth vertex lies on the circle, then so does the sixth. If the nth vertex lies on the circle, then so does the (n+1)st. I point out that this is induction -- from n to n+1.
Every induction proof begins with an initial step, or "base case." In this proof, the base case is that the first three points lie on a circle. This is true because any three noncollinear points lie on a circle -- mentioned in Section 4-5 of the U of Chicago. The induction step allows us to prove that one more point at a time is on the circle, until all of the vertices of the regular polygon are on the circle.
We have seen how powerful a proof by induction can be. We have proved the Center of a Regular Polygon Theorem. But now we wish to prove another related theorem -- one that is specifically mentioned in the Common Core Standards:
What we wish to derive from the Center of a Regular Polygon Theorem is that we can rotate this polygon a certain number of degrees -- about that aformentioned center, of course -- or reflect it over any angle bisector or perpendicular bisector. But the U of Chicago text, unfortunately, doesn't give us a Regular Polygon Symmetry Theorem or anything like that.
I'm of two minds on this issue. One way would be to take this theorem and use it to prove that when rotating about O, the image of one of those isosceles triangles with vertex O and base one side of the polygon is another such triangle. The other way is to do Dr. Wu's trick -- he defines regular polygon so that it's vertices are already on the circle. Then we can perform rotations on the entire circle. (Rotations are easier to see, but it's preferable to do reflections because a rotation is the composite of two reflections.) Notice that the number of degrees of the rotation depends on the number of sides. In particular, for a regular n-gon we must rotate it 360/n degrees, or any multiple thereof.
[2017 update: The modern Third Edition of the text actually mentions rotation symmetry. This section tells us that a parallelogram has 2-fold rotation symmetry, and the statements about regular polygons actually appear in Lesson 6-8 of the new edition.]
So here is the proof of the Center of a Regular Polygon Theorem as given by the U of Chicago -- in paragraph form, just as printed in the text (rather than converted to two columns as I usually do):
Proof:
Analyze: Since the theorem is known to be true for regular polygons of 3 and 4 sides, the cases that need to be dealt with have 5 or more sides. What is done is to show that the circle through three consecutive vertices of the regular polygon contains the next vertex. Then that fourth vertex can be used with two others to obtain the fifth, and so on, as many times as needed.
Given: regular polygon ABCD...
Prove: There is a point O equidistant from A, B, C, D, ...
Draw: ABCD...
Write: Let O be the center of the circle containing A, B, and C. Then OA = OB = OC. Since AB = BC by the definition of regular polygon, OABC is a kite with symmetry diagonal
Now the "Analyze" part of this proof contains the induction. If the first three vertices lie on the circle, then so does the fourth. If the fourth vertex lies on the circle, then so does the fifth. If the fifth vertex lies on the circle, then so does the sixth. If the nth vertex lies on the circle, then so does the (n+1)st. I point out that this is induction -- from n to n+1.
Every induction proof begins with an initial step, or "base case." In this proof, the base case is that the first three points lie on a circle. This is true because any three noncollinear points lie on a circle -- mentioned in Section 4-5 of the U of Chicago. The induction step allows us to prove that one more point at a time is on the circle, until all of the vertices of the regular polygon are on the circle.
We have seen how powerful a proof by induction can be. We have proved the Center of a Regular Polygon Theorem. But now we wish to prove another related theorem -- one that is specifically mentioned in the Common Core Standards:
CCSS.Math.Content.HSG.CO.A.3
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.
What we wish to derive from the Center of a Regular Polygon Theorem is that we can rotate this polygon a certain number of degrees -- about that aformentioned center, of course -- or reflect it over any angle bisector or perpendicular bisector. But the U of Chicago text, unfortunately, doesn't give us a Regular Polygon Symmetry Theorem or anything like that.
I'm of two minds on this issue. One way would be to take this theorem and use it to prove that when rotating about O, the image of one of those isosceles triangles with vertex O and base one side of the polygon is another such triangle. The other way is to do Dr. Wu's trick -- he defines regular polygon so that it's vertices are already on the circle. Then we can perform rotations on the entire circle. (Rotations are easier to see, but it's preferable to do reflections because a rotation is the composite of two reflections.) Notice that the number of degrees of the rotation depends on the number of sides. In particular, for a regular n-gon we must rotate it 360/n degrees, or any multiple thereof.
[2017 update: The modern Third Edition of the text actually mentions rotation symmetry. This section tells us that a parallelogram has 2-fold rotation symmetry, and the statements about regular polygons actually appear in Lesson 6-8 of the new edition.]
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