But I didn't sub today, so I was unaffected by the drill. I make every effort to avoid subbing on PSAT Day or Shakeout Day -- two very sub-unfriendly days. I don't want to be the sub to make sure that the students are behaving during a standardized test or earthquake drill. I very nearly received a late call in my old district (the one where it's actually Day 45 today) for a Chemistry class -- but ten minutes later, the call was canceled.
So that's no earthquake drill for me today. Instead, we'll get back to the book from yesterday.
Chapter 2 of Ian Stewart's The Story of Mathematics, "The Logic of Shape: First Steps in Geometry," begins as follows:
"There are two main types of reasoning in mathematics: symbolic and visual. Symbolic reasoning originated in number notation, and we will shortly see how it led to the invention of algebra, in which symbols can represent general numbers ('the unknown') rather than specific ones ('7')."
But this chapter is all about the other type of reasoning, visual -- and that leads to Geometry. As this is a Geometry blog, I will enjoy describing all the neat things in this chapter.
The beginnings of Geometry go back to the oldest cultures. Stewart describes a geometric diagram found on an ancient Babylonian tablet:
"For example, the tablet catalogued as YBC 7289 shows a square and two diagonals. The sides of the square are marked with cuneiform numerals for 30."
The author explains that the Babylonians calculated the diagonal as 42; 25, 35. This equals 30sqrt(2), correct to two sexagesimal places. (In decimal, the Babylonian approximation is 42.42641 and the modern approximation is 42.42639.)
We now move on to the ancient Greeks. Stewart tells us about that famous mathematician Pythagoras and his followers:
"On the mystic side, they considered the number 1 to be the prime source of everything in the universe."
(In a side bar called "Harmony of the World," the author describes the Pythagorean musical intervals, namely the 2/1 octave, the 3/2 perfect fifth, and the 4/3 perfect fourth. I've already given many posts the "music" label, so I defer to those posts for a continued discussion of Pythagorean music.)
Stewart tells us that Pythagoras originally believed that all pairs of lengths had a common measure:
"It seems natural to imagine that enough subdivision would hit the number exactly; if so, all lengths would be rational. If this were true, it would make geometry much simpler, because any two lengths would be whole number multiples of a common (perhaps small) length, and so could be obtained by fitting lots of copies of this common length together."
But as his follower Hippasus of Metapontum discovered, the lengths 1 and sqrt(2) don't have such a common measure, because sqrt(2) is irrational. (I gave a proof back in Tuesday's post.) The author mentions the story that Pythagoras had Hippasus drown at sea, but he points out it's more likely that he was just expelled from the cult. So we now move two centuries ahead to the mathematician who first tamed irrationals:
"The Greek theory of irrationals was invented by Eudoxus around 370 BC. His idea is to represent any magnitude, rational or irrational, as the ratio of two lengths -- that is, in terms of a pair of lengths."
But as Stewart explains, working with ratios entails division -- and division was difficult given the Greek numeration system:
"Instead, Eudoxus found a cumbersome but precise method of comparison that could be performed within the conventions of Greek geometry."
This trick involves replacing division with multiplication. The author explains that this trick allowed Greek geometers to extend theorems about rational lengths to those about irrational lengths. (This is what Hung-Hsi Wu once called "the Fundamental Theorem of School Mathematics.)
Stewart now moves on to the most famous of all Greek geometers -- Euclid. He describes Euclid's work as an examination of the logic of spatial relationships:
"If a shape has certain properties, these may logically imply other properties. For example, if a triangle has all three sides equal -- an equilateral triangle -- then all three angles must be equal."
The author informs us that Euclid began with several common notions and postulates -- these are also known as axioms. He compares Euclid's axioms to the rules of a game:
"Anyone who wants to play that particular game must accept the rules; if they don't, they are free to play a different game."
From these axioms, Euclid proved many theorems. His first theorem, Book I Proposition 1, is the theorem that the U of Chicago text covered in Lesson 4-4 -- how to construct an equilateral triangle. I skipped that lesson yesterday due to the PSAT -- but take that, those who thought you'd escape 4-4! I am now suddenly discussing yesterday's skipped lesson in today's side-along reading post.
Anyway, Stewart continues to describe what Euclid achieved:
"He built a logical tower, which climbed higher and higher towards the sky, with the axioms as its foundations and logical deduction as the mortar that bound the bricks together."
Now the author skips up to Book V Proposition 1. This theorem is a bit obscure, but it essentially states that 10'10" (ten feet ten inches) is ten times as much as 1'1":
"Is it trivialities dressed up as theorems? Mystic nonsense? Not at all. This material may seem obscure, but it leads up to the most profound part of the Elements: Eudoxus's techniques for dealing with irrational ratios. Nowadays mathematicians prefer to work with numbers, and because these are more familiar I will often interpret the Greek ideas in that language."
At this point Euclid describes two regular solids -- the dodecahedron and icosahedron (the two I left out from my September 30th post). These are related to the regular pentagon, and Euclid proved that the ratio of the hypotenuse to the side of a regular pentagon is what the Greek sage called "extreme and mean ratio," which we now call Phi, the golden ratio:
"Much of the material on number theory is not needed for the classification of regular solids -- so why did Euclid include this material?"
Stewart suspects it's because this material is also related to irrational numbers. Meanwhile, we move on to the greatest Greek mathematician of all -- Archimedes:
We learn that Archimedes both discovered and proved the formula for the volume of a sphere, using what is known as "exhaustion":
"This method has an important limitation: you have to know what the answer is before you have much chance of proving it."
It wasn't until 1906 when the Danish scholar Heiberg found a document that helped to determine how exactly Archimedes figured out the sphere volume formula:
"He discovered that the original document was a copy of several works by Archimedes, some of them previously unknown."
In the end, we found out that Archimedes used a forerunner of Cavalieri's method, similar to what we see in Lesson 10-8 of the U of Chicago text.
In my October 9th post, I wrote about the impossibility of angle trisection using only a straightedge and compass. Actually, Archimedes was able to trisect an angle if we add two extra marks to the straightedge (rather than use a pure unmarked ruler):
"Two other famous problems from that period are duplicating the cube (constructing a cube whose volume is twice that of a given cube) and squaring the circle (constructing a square with the same area as a given circle). These are also known to be impossible using ruler and compass."
At this point, Stewart describes the three conic sections:
- The ellipse, a closed oval curve obtained when the plane meets only one half of the cone. Circles are special ellipses.
- The hyperbola, a curve with two infinite branches, obtained when the plane meets both halves of the cone.
- The parabola, a transitional curve lying between ellipses and hyperbolas, in the sense that it is parallel to some line passing through the vertex of the cone and lying on the cone. A parabola has only one branch, but extends to infinity.
The author tells us that the foci of these conics have some special properties:
"Among their many properties, we single out just one: the distance from one focus of an ellipse, to any point, and back to the other focus is constant (equal to the diameter of the ellipse)."
Why did we suddenly jump from angle trisection to conics? Stewart explains:
"The Greeks knew how to trisect angles and duplicate the cube using conics. With the aid of other special curves, notably the quadratrix, they could also square the circle."
Anyway, the author informs us that the Greeks made two important contributions to the development of math, one of which was a systematic understanding of our favorite subject, Geometry:
"The second Greek contribution was the systematic use of logical deduction to make sure that what was being asserted could also be justified."
And indeed, that contribution lives on today:
"Every tower block, every suspension bridge, every football stadium is a tribute to the geometers of ancient Greece."
And indeed, that contribution lives on today:
"Every tower block, every suspension bridge, every football stadium is a tribute to the geometers of ancient Greece."
To that end, Stewart concludes the chapter as follows:
"In science, emphasis is placed on trying to prove that what you deeply believe to be the case is wrong. Ideas that survive stringent attempts to disprove them are more likely to be correct."
There are a few more sidebars remaining though in this chapter. One is on the last of the classical Greek mathematicians, Hypatia. (Stewart suggests that she was the first person to be executed by Christians as a "witch.") This is followed by other examples of what geometry does for us, including computer graphics and measurement. But his example -- the definition of the kilogram as the mass of a particular sphere -- has been obsolete for almost a year now. Yes, Metric Week is over, but I will point out that mass of "Big K" is no longer the definition of the kilogram as of last year:
Lesson 4-5 of the U of Chicago text is "The Perpendicular Bisector Theorem." (This theorem is part of Lesson 5-5 of the modern edition.) This is what I wrote last year about today's lesson:
Lesson 4-5 of the U of Chicago text covers the Perpendicular Bisector Theorem. This theorem is specifically mentioned in the Common Core Standards, so it's important that we prove it.
There are many ways to prove the Perpendicular Bisector Theorem. The usual methods involve showing that two right triangles are congruent. But that's not how the U of Chicago proves it. Once we have reflections -- and we're expected to use reflections in Common Core, the proof becomes very nearly a triviality.
Here's the proof, based on the U of Chicago text but rewritten so that the column "Conclusions" and "Justifications" become "Statements" and "Reasons," and with "Given" as the first step (as I explained in one of the posts last week).
Lesson 4-5 of the U of Chicago text covers the Perpendicular Bisector Theorem. This theorem is specifically mentioned in the Common Core Standards, so it's important that we prove it.
There are many ways to prove the Perpendicular Bisector Theorem. The usual methods involve showing that two right triangles are congruent. But that's not how the U of Chicago proves it. Once we have reflections -- and we're expected to use reflections in Common Core, the proof becomes very nearly a triviality.
Here's the proof, based on the U of Chicago text but rewritten so that the column "Conclusions" and "Justifications" become "Statements" and "Reasons," and with "Given" as the first step (as I explained in one of the posts last week).
[2019 update: Yesterday's Lesson 4-4 was cancelled due to the PSAT, and I already mentioned Euclid's Book I Proposition 1 earlier in this post, but that doesn't mean that 4-4 doesn't matter. We notice that the U of Chicago text uses Book 1 Proposition 1 to introduce two-column proofs. With Lesson 4-4 omitted, the proof below becomes our first two-column proof of the school year. And Question 6 on today's worksheet is no longer "review," but new material.]
Given: P is on the perpendicular bisector m of segmentAB.
Prove: PA = PB
Proof:
Statements Reasons
1. m is the perp. bis. of AB 1. Given
2. m reflects A to B 2. Definition of reflection
3. P is on m 3. Given
4. m reflects P to P 4. Definition of reflection
5. PA = PB 5. Reflections preserve distance.
So the line m becomes our reflecting line -- that is, our mirror. Since m is the perpendicular bisector ofAB, the mirror image of A is exactly B. After all, that was exactly our definition of reflection! And since P is on the mirror, its image must be itself. Then the last step is the D of our ABCD properties that are preserved by reflections. The tricky part for teachers is that we're not used to thinking about the definition or properties of reflections as reasons in a proof. Well, it's time for us to start thinking that way!
Now the text writes:
"The Perpendicular Bisector Theorem has a surprising application. It can help locate the center of a circle."
This is the circumcenter of the triangle, one of our concurrency proofs. I mentioned last week that this is the easiest of the concurrency proofs. At first, it appears to be a straightforward application of the Perpendicular Bisector Theorem plus the Transitive Property of Equality. But there's a catch:
"If m and n intersect, it can be proven that this construction works."
Here m and n are the perpendicular bisectors ofAB and BC, respectively. But the text doesn't state how to prove that these two lines must intersect. As it turns out, the necessary and sufficient conditions for the lines to intersect is for the three points A, B, and C to be noncollinear. Well, that's no problem since right at the top of the page, it's stated that the three points are noncollinear -- and besides, we don't expect there to be a circle through three collinear points anyway. (And if this is part of a concurrency proof, then the three points are the vertices of a triangle, so they are clearly noncollinear.)
The Two Perpendiculars Theorem doesn't require Playfair, but the Perpendicular to Parallels Theorem does, so the above proof requires Playfair. As it turns out, in hyperbolic geometry, it's possible for three points to be noncollinear and yet no circle passes through them -- and so there's a triangle with neither a circumcenter nor a circumcircle!
And, if one has any lingering doubts that the existence of a circle through the three noncollinear points requires a Parallel Postulate, here's a link to Cut the Knot, one of the oldest mathematical websites still in existence. It was first created the year after I passed high school geometry as a student, and it has recently been redesigned:
http://www.cut-the-knot.org/triangle/pythpar/Fifth.shtml
Of the statements that require Euclid's Fifth Postulate to prove, we see listed at the above link:
"3. For any three noncollinear points, there exists a circle passing through them."
Now I didn't plan on giving Playfair's Postulate until Chapter 5. But here's something I noticed -- Playfair is used to prove the Parallel Consequences -- that is, the theorems that if parallel lines are cut by a transversal, then corresponding (or alternate interior) angles are congruent -- of which the Perpendicular to Parallels Theorem is a special case. But only that special case is needed to prove our circumcenter theorem. Indeed, I saw that the proof that the orthocenters are concurrent needs only that special case, and so does the Two Reflection Theorem for Translations.
But the Perpendicular to Parallels Theorem is tricky to prove. And we've already seen what other texts do when a theorem is tricky to prove, yet useful to prove medium- or higher-level theorems later on. We just simply declare the theorem to be a postulate! So instead of giving Playfair in Chapter 5, I state the Perpendicular to Parallels Postulate. (Note -- this is what I did last year. I have yet to decide whether I will change this for this year or not.)
Notice that the Perpendicular to Parallels Postulate can then be used to prove full Playfair. A proof of Playfair is given in the text at Lesson 13-6. The only changes we need to make to that proof is making sure that angles 1, 2, and 3 are all right angles. The new Step 2 can read:
2. The blue line is the line passing through P perpendicular to line l, which uniquely exists by the Uniqueness of Perpendiculars Theorem (which we proved on this blog last week). So angle 1 measures 90 degrees. So, by the new Perpendicular to Parallels Postulate, since the blue line is perpendicular to l, it must be perpendicular to both x and y, as both are parallel to l. So angles 2 and 3 both measure 90 degrees.
Then Playfair is used to prove the full Parallel Consequences. Notice that like Dr. Franklin Mason, we plan on adding a new postulate. But unlike his Triangle Exterior Angle Inequality Postulate, my postulate can replace Playfair, while Dr. M's TEAI Postulate must be used in addition to Playfair.
The final thing I want to say about a possible Perpendicular to Parallels Postulate is that this postulate is worth adding if it will make things easier for the students. I believe that it will. Notice that the full Parallel Consequences require identifying corresponding angles, alternate interior angles, same-side interior angles, and so on, and students may have trouble remembering which is which. But the Perpendicular to Parallels Postulate simply states that in a plane, if m and n are parallel and l is perpendicular to m, then l is perpendicular to n -- no need to remember what alternate or same-side interior angles are! So, in the name of making things easier for the students, I just might include this postulate in Chapter 5.
But I won't include it right now. And so I'll skip that part of the lesson -- and throw out the questions like 4 and 5 that require it.
That makes this lesson rather thin. So we ask, is there anything else that can be included? Let's look at the Common Core Standard that requires the Perpendicular Bisector Theorem again:
CCSS.MATH.CONTENT.HSG.CO.C.9
Prove theorems about lines and angles. Theorems include [...] points on a perpendicular bisector of a line segment are exactly those equidistant from the segment's endpoints.
We notice a key word there -- exactly. It means that if we rewrote this statement as an if-then statement, then it would have to be written as a biconditional:
A point is on the perpendicular bisector of a segment if and only if it is equidistant from its endpoints.
That is -- we need the converse of the Perpendicular Bisector Theorem. For some strange reason, the U of Chicago text makes zero mention of the converse! As it turns out, we can prove the converse quite easily, but it requires a theorem that's still two sections away. Once we reach Lesson 4-7, then we can finally prove the Converse of the Perpendicular Bisector Theorem.
And so there's not much left in this section -- but one could say that the Perpendicular Bisector Theorem is so important that it nonetheless merits its own section. Notice that I kept Question 6, which is similar to the construction of the circumcircle, except it's given that lines e and f intersect at point C. So we don't need a Parallel Postulate to prove that they intersect.
Given: P is on the perpendicular bisector m of segment
Prove: PA = PB
Proof:
Statements Reasons
1. m is the perp. bis. of AB 1. Given
2. m reflects A to B 2. Definition of reflection
3. P is on m 3. Given
4. m reflects P to P 4. Definition of reflection
5. PA = PB 5. Reflections preserve distance.
So the line m becomes our reflecting line -- that is, our mirror. Since m is the perpendicular bisector of
Now the text writes:
"The Perpendicular Bisector Theorem has a surprising application. It can help locate the center of a circle."
This is the circumcenter of the triangle, one of our concurrency proofs. I mentioned last week that this is the easiest of the concurrency proofs. At first, it appears to be a straightforward application of the Perpendicular Bisector Theorem plus the Transitive Property of Equality. But there's a catch:
"If m and n intersect, it can be proven that this construction works."
Here m and n are the perpendicular bisectors of
The Two Perpendiculars Theorem doesn't require Playfair, but the Perpendicular to Parallels Theorem does, so the above proof requires Playfair. As it turns out, in hyperbolic geometry, it's possible for three points to be noncollinear and yet no circle passes through them -- and so there's a triangle with neither a circumcenter nor a circumcircle!
And, if one has any lingering doubts that the existence of a circle through the three noncollinear points requires a Parallel Postulate, here's a link to Cut the Knot, one of the oldest mathematical websites still in existence. It was first created the year after I passed high school geometry as a student, and it has recently been redesigned:
http://www.cut-the-knot.org/triangle/pythpar/Fifth.shtml
Of the statements that require Euclid's Fifth Postulate to prove, we see listed at the above link:
"3. For any three noncollinear points, there exists a circle passing through them."
Now I didn't plan on giving Playfair's Postulate until Chapter 5. But here's something I noticed -- Playfair is used to prove the Parallel Consequences -- that is, the theorems that if parallel lines are cut by a transversal, then corresponding (or alternate interior) angles are congruent -- of which the Perpendicular to Parallels Theorem is a special case. But only that special case is needed to prove our circumcenter theorem. Indeed, I saw that the proof that the orthocenters are concurrent needs only that special case, and so does the Two Reflection Theorem for Translations.
But the Perpendicular to Parallels Theorem is tricky to prove. And we've already seen what other texts do when a theorem is tricky to prove, yet useful to prove medium- or higher-level theorems later on. We just simply declare the theorem to be a postulate! So instead of giving Playfair in Chapter 5, I state the Perpendicular to Parallels Postulate. (Note -- this is what I did last year. I have yet to decide whether I will change this for this year or not.)
Notice that the Perpendicular to Parallels Postulate can then be used to prove full Playfair. A proof of Playfair is given in the text at Lesson 13-6. The only changes we need to make to that proof is making sure that angles 1, 2, and 3 are all right angles. The new Step 2 can read:
2. The blue line is the line passing through P perpendicular to line l, which uniquely exists by the Uniqueness of Perpendiculars Theorem (which we proved on this blog last week). So angle 1 measures 90 degrees. So, by the new Perpendicular to Parallels Postulate, since the blue line is perpendicular to l, it must be perpendicular to both x and y, as both are parallel to l. So angles 2 and 3 both measure 90 degrees.
Then Playfair is used to prove the full Parallel Consequences. Notice that like Dr. Franklin Mason, we plan on adding a new postulate. But unlike his Triangle Exterior Angle Inequality Postulate, my postulate can replace Playfair, while Dr. M's TEAI Postulate must be used in addition to Playfair.
The final thing I want to say about a possible Perpendicular to Parallels Postulate is that this postulate is worth adding if it will make things easier for the students. I believe that it will. Notice that the full Parallel Consequences require identifying corresponding angles, alternate interior angles, same-side interior angles, and so on, and students may have trouble remembering which is which. But the Perpendicular to Parallels Postulate simply states that in a plane, if m and n are parallel and l is perpendicular to m, then l is perpendicular to n -- no need to remember what alternate or same-side interior angles are! So, in the name of making things easier for the students, I just might include this postulate in Chapter 5.
But I won't include it right now. And so I'll skip that part of the lesson -- and throw out the questions like 4 and 5 that require it.
That makes this lesson rather thin. So we ask, is there anything else that can be included? Let's look at the Common Core Standard that requires the Perpendicular Bisector Theorem again:
CCSS.MATH.CONTENT.HSG.CO.C.9
Prove theorems about lines and angles. Theorems include [...] points on a perpendicular bisector of a line segment are exactly those equidistant from the segment's endpoints.
We notice a key word there -- exactly. It means that if we rewrote this statement as an if-then statement, then it would have to be written as a biconditional:
A point is on the perpendicular bisector of a segment if and only if it is equidistant from its endpoints.
That is -- we need the converse of the Perpendicular Bisector Theorem. For some strange reason, the U of Chicago text makes zero mention of the converse! As it turns out, we can prove the converse quite easily, but it requires a theorem that's still two sections away. Once we reach Lesson 4-7, then we can finally prove the Converse of the Perpendicular Bisector Theorem.
And so there's not much left in this section -- but one could say that the Perpendicular Bisector Theorem is so important that it nonetheless merits its own section. Notice that I kept Question 6, which is similar to the construction of the circumcircle, except it's given that lines e and f intersect at point C. So we don't need a Parallel Postulate to prove that they intersect.
[2019 update: The concurrency of the altitudes of a triangle, an important proof for Common Core, is implied in that aforementioned Question 6.]
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