1. Introduction
2. Natural Point-Line-Plane Postulate
3. Definition of Antipodal Points
4. Spherical Geometry and Playfair
5. Spherical Geometry and AAA Congruence
6. Revisiting Birkhoff's Postulates
7. Pappas Problem of the Day
8. Traditionalists and Rich Problems
9. Feynman, Pages 1-70
10. Conclusion
Introduction
Before we begin, I address what's on most Southern Californians' minds lately -- earthquakes. On the morning of the Fourth of July, an 6.4 quake hit the town of Ridgecrest, in Kern County. (This is the same county as McFarland, with the two towns about 140 miles apart.) About a day and a half later, a 7.1 quake hit near the same area. We often consider a second earthquake in close succession to be an aftershock of the first quake -- but since the second earthquake was stronger, it might make more sense to consider the first quake to be a foreshock.
I was a young seventh grade student when I experienced my largest earthquake -- the 1994 Northridge earthquake. The quake jolted me awake -- it was Monday morning, but school was closed for Martin Luther King Day.
On Square One TV, there is a Mathnet episode called "The Case of the Ersatz Earthquake." Those of you who aren't Californian can watch it to learn more about earthquakes (including what those numbers 6.4 and 7.1 stand for). An unrelated episode, "The Case of the Map with a Gap," is set in a ghost town in Kern County, close to the epicenter of the earthquake.
This is the first of a series of summer posts about what I call "natural geometry." Here, "natural geometry" is intended to integrate Euclidean and spherical geometry in the same way that "neutral geometry" integrates Euclidean and hyperbolic geometry.
The purpose of these posts is to describe a possible high school Geometry course where we begin with Euclidean geometry as usual. Then right at the end of the year -- after state testing is complete -- we introduce the basics of spherical geometry. Thus our goal in these posts is to compare and contrast the two geometries -- find out which theorems and proofs the two systems have in common.
The starting point for our discussion is the first postulate in the U of Chicago text, from Lesson 1-7:
Point-Line-Plane Postulate:
(a) Unique line assumption: Through any two points, there is exactly one line.
(b) Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line. Given a plane in space, there exists a point in space not on the line.
(c) Number line assumption: Every line is a set of points that can be put in a one-to-one correspondence with the real numbers, with any point on it corresponding to 0 and any other point corresponding to 1.
(d) Distance assumption: On a number line, there is a unique distance between two points.
We see that this Point-Line-Plane Postulate holds in both Euclidean and hyperbolic geometry, but fails in spherical geometry. Thus our natural geometry requires a different postulate. In today's post, we'll start by determine how much of this postulate can be salvaged. Then we'll describe what we need to add to make a spherical version of this postulate.
There's already one part of this postulate that we want to drop right away -- part (b), which is the "dimension assumption." Our introduction of spherical geometry is two-dimensional -- we don't wish to mention 3D spherical geometry (that is, the surface of a 4D hypersphere). We can keep the first part about "given a line in a plane" (which becomes interpreted as "given a great circle on a sphere") and drop the second part about "given a plane in space."
Our natural geometry does not include hyperbolic geometry. But once again, sometimes we'll look at hyperbolic geometry to make sure that our Euclidean and spherical proofs are valid. This is what we needed to do with the concurrence of perpendicular bisectors proof in Lesson 4-5. That proof takes it for granted that the perpendicular bisectors intersect, yet in hyperbolic geometry it's possible for those lines to be parallel. That's when we realize that the correct Euclidean proof must in fact use a parallel postulate somehow.
Natural Point-Line-Plane Postulate
Let's look at the most important part of this postulate:
Point-Line-Plane Postulate:
(a) Unique line assumption: Through any two points, there is exactly one line.
As I wrote a few weeks ago, this statement fails in spherical geometry. That day, I already suggested a few changes we can make to make it hold on the sphere. We begin with:
(a') Through any two points, there is at least one line.
So far, we might have three, four, or even more points lying on multiple lines. So I suggested that we add the following:
(a") Through any three points, there is at most one line.
But after thinking about this for a while, we realize that we can be even more specific here. Notice that most of the time, through two points on the sphere there really is just one line. We'd have to be especially unlucky to choose two points which happen to be antipodal -- the only way to have two points on more than one line. So instead we might write:
(a") Suppose three points lie on a line. Then for at least one of those three points, that line is the only line that passes through that point and each of the other two points.
We notice that if we're given three points, at most two of those points can be antipodal -- so we can choose the point that isn't antipodal (to the other two). Suppose that a line l passes through three points, namely N, S, and P. If N and S are an antipodal pair, then we choose point P. Then (a") tells us that l is the only line containing both P and N. And it also tells us that l is the only line containing both P and S.
Of course, none of the three points need to be antipodal to each other at all. If we have three points P, Q, and R lying on line l, with none of them antipodal, then l is the only line through both P and Q, as well as the only line through both P and R, as well as the only line through both Q and R.
Statement (a") tells us that given any point, there is at most one point antipodal to it. After all, suppose we have three points N, S, and some other point S'. Then by (a"), at least one of those points share exactly one line with each of the other two points -- and that point would thus not be antipodal with either of the others.
But so far, we haven't shown that there is at least one point antipodal to any other points. In fact, we haven't claimed that any point has an antipodal point -- indeed, the statements written down so far hold in Euclidean geometry, where there are no antipodal points. So far, what we've written down is actually a natural Point-Line-Plane Postulate, holding in both Euclidean and spherical geometry.
So far, our natural Point-Line-Plane consists of (a'), (a"), and the first part of (b). I'm still trying to decide what to do with (c) and (d). With (c), we don't really want a correspondence between the points of a great circle and all the real numbers, just the ones between 0 and 360 (or 2pi, or tau, or whatever), since that's the correspondence used to determine the distance in part (d).
Of course, this correspondence on the sphere is intimately related to angle measure. Thus, I wonder whether we even need to define distance on a sphere -- instead, we define it in terms of angles. For example, we might define the length of a segment to be the third angle of a triangle with two right angles and that segment as the base. (Something similar might be done with area -- just define the area of a triangle to be the excess of the sum of its angles beyond 180.)
For now, I wish to focus on parts (a) and (b). So this is what natural Point-Line-Plane looks like now:
Natural Point-Line-Plane Postulate
(a') Through any two points, there is at least one line.
(a") Suppose three points lie on a line. Then for at least one of those three points, that line is the only line that passes through that point and each of the other two points.
(b') Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line.
Definition of Antipodal Points
So now we have natural Point-Line-Plane, which holds in both Euclidean and spherical geometry. It appears that all we need to do to make it spherical is to add something like:
- Antipodal points exist.
- Two points are antipodal if there exist two distinct lines passing through them.
- Two points are antipodal if there exist infinitely many distinct lines passing through them.
- Two points are antipodal if, given any other point, there exists a line passing through all three.
And once we have a definition of "antipodal," there are several ways of stating the postulate:
- There exists at least one pair of antipodal points.
- There exists infinitely many pairs of antipodal points.
- For every point, there exists another point antipodal to it.
As it turns out, we only need the weakest definition and postulate:
- Two points are antipodal if there exist two distinct lines passing through them.
- There exists at least one pair of antipodal points.
These can be extended to the entire sphere by using transformations. Once again, we will have a Reflection Postulate available that applies to the sphere.
Let N and S be the antipodal pair guaranteed by the postulate. Of course, these letters were chosen to remind us of the North and South Poles of a sphere -- the ultimate antipodal points. By definition, at least two lines pass through these poles.
We begin by showing that there are many more lines passing through these poles. We start by letting T be any isometry. Now suppose T maps N to N -- that is, N is a fixed point of T. Then we ask, to what point does T map S?
Well, we know that there are two lines passing N and S, and since isometries preserve collinearity, both lines must pass through T(N) and T(S), which is sufficient for those images to be antipodal points of each other. Since T(N) = N, this means that T(S) must be antipodal to N -- and the point that's antipodal to N is exactly S. Thus T(S) = S -- that is, S is also a fixed point of T. In other words, any isometry that fixes one pole must necessarily fix the other pole.
So now if l is one of the lines passing through N and S, and P is any other point, we let T be an isometry that maps l to the line passing through N and P. (For example, T can be a reflection whose mirror is the angle bisector of l and NP.) By the above reasoning, S must also lie on the image of l. So the image of l passes through N, S, and P. Since P is arbitrary, this shows that there exists a line passing through N, S, and any other point you choose. As soon as we have two lines through the poles, we immediately have infinitely many such lines filling the entire sphere.
Now we wish to show that every point has another point antipodal to it. Let P be the point whose antipodal point we wish to find. Again, let T is an isometry, but this time we map N to P. (For example, T can be a reflection whose mirror is the perpendicular bisector ofNP.) Then we ask again, to what point does T map S?
Well, we know that there are two lines passing N and S, and since isometries preserve collinearity, both lines must pass through T(N) and T(S), which is sufficient for those images to be antipodal points of each other. Since T(N) = P, this means that T(S) must be antipodal to P -- which is exactly the point we're looking for. Since P is arbitrary, this shows that there exists a point antipodal to any point you choose. As soon as we have one antipodal pair, we immediately have infinitely many such pairs filling the entire sphere.
Both of these arguments rely on the fact that isometries, including reflections, preserve antipodal points -- if two points are antipodal, then so are their images under any isometry. (If we had defined a distance on the sphere, this would be obvious since antipodal points lie a hemisphere apart. But here we used only collinearity and didn't attempt to use distance at all.)
Spherical Geometry and Playfair
OK, so now we have a definition of antipodal points and a spherical postulate asserting that antipodal points exist. But now we ask, is this the best possible definition to add to natural geometry to form spherical geometry?
Recall that in our natural geometry, we wish to avoid vacuous postulates and theorems -- those that assert properties of objects that don't exist in a particular geometry. For example, instead of proving properties of rectangles (which exist in Euclidean, but not spherical, geometry), we prove the properties of a generalization of rectangles, such as equiangular, Saccheri, or Lambert quadrilaterals (which exist in both geometries).
I've mentioned in previous posts that, while Point-Line-Plane must be changed to accommodate spherical geometry, Playfair's Parallel Postulate remains, believe it or not. That's because Playfair is formally given as:
So now if l is one of the lines passing through N and S, and P is any other point, we let T be an isometry that maps l to the line passing through N and P. (For example, T can be a reflection whose mirror is the angle bisector of l and NP.) By the above reasoning, S must also lie on the image of l. So the image of l passes through N, S, and P. Since P is arbitrary, this shows that there exists a line passing through N, S, and any other point you choose. As soon as we have two lines through the poles, we immediately have infinitely many such lines filling the entire sphere.
Now we wish to show that every point has another point antipodal to it. Let P be the point whose antipodal point we wish to find. Again, let T is an isometry, but this time we map N to P. (For example, T can be a reflection whose mirror is the perpendicular bisector of
Well, we know that there are two lines passing N and S, and since isometries preserve collinearity, both lines must pass through T(N) and T(S), which is sufficient for those images to be antipodal points of each other. Since T(N) = P, this means that T(S) must be antipodal to P -- which is exactly the point we're looking for. Since P is arbitrary, this shows that there exists a point antipodal to any point you choose. As soon as we have one antipodal pair, we immediately have infinitely many such pairs filling the entire sphere.
Both of these arguments rely on the fact that isometries, including reflections, preserve antipodal points -- if two points are antipodal, then so are their images under any isometry. (If we had defined a distance on the sphere, this would be obvious since antipodal points lie a hemisphere apart. But here we used only collinearity and didn't attempt to use distance at all.)
Spherical Geometry and Playfair
OK, so now we have a definition of antipodal points and a spherical postulate asserting that antipodal points exist. But now we ask, is this the best possible definition to add to natural geometry to form spherical geometry?
Recall that in our natural geometry, we wish to avoid vacuous postulates and theorems -- those that assert properties of objects that don't exist in a particular geometry. For example, instead of proving properties of rectangles (which exist in Euclidean, but not spherical, geometry), we prove the properties of a generalization of rectangles, such as equiangular, Saccheri, or Lambert quadrilaterals (which exist in both geometries).
I've mentioned in previous posts that, while Point-Line-Plane must be changed to accommodate spherical geometry, Playfair's Parallel Postulate remains, believe it or not. That's because Playfair is formally given as:
- Through a point not on a line, there exists at most one line parallel to the given line.
Because of the words "at most," Playfair still holds in spherical geometry! But then again, is it useful to have Playfair in spherical geometry?
We'd think that if we postulate the existence of antipodal points, then we could conclude that the geometry is spherical. Then since the geometry is spherical, no parallel lines exist -- which means that there's no longer a need for Playfair to assert that there's at most one parallel line. This would make Playfair a wasted postulate -- it adds no new theorems to spherical geometry.
Let's think back to why Playfair is needed in Euclidean geometry in the first place -- it allows us to rule out hyperbolic geometry. In other words, Euclidean Point-Line-Plane is satisfied by both Euclidean and hyperbolic geometry. Adding Playfair rules out hyperbolic geometry, so that the resulting geometry must be Euclidean.
Thus, what we really need here is an alternate "Point-Line-Plane" postulate that holds in both spherical and hyperbolic (but not Euclidean) geometry. Then adding Playfair rules out hyperbolic geometry, so that the resulting geometry must be spherical.
So let's seek out statements that hold in spherical and hyperbolic, but not Euclidean, geometry. This is tricky, since spherical and hyperbolic geometry are sort of like opposite ends of a spectrum with Euclidean in the middle. We might try statements such as:
- The sum of the angles of a triangle is not 180.
- Rectangles do not exist.
But it doesn't seem right to choose statements like these as postulates. These statements just assert the nonexistence of familiar Euclidean objects, such as rectangles or triangles whose angles add up to the usual 180 degrees.
It seems much more satisfying to use something that positively, rather than negatively, distinguishes both spherical and hyperbolic geometry from the Euclidean variety. And at this point, there's only one example that I can think of.
Spherical Geometry and AAA Congruence
As a congruence statement, Angle-Angle-Angle indeed holds in both spherical and hyperbolic geometry, but not Euclidean geometry.
And so I wish to see what would happen if we accepted AAA as a postulate and add it to the natural version of Point-Line-Plane (which is satisfied by all three geometries). Then the resulting geometry could be either spherical or hyperbolic. If we then added Playfair, this should allow us to rule out hyperbolic geometry and leave spherical as the only possibility.
This isn't a complete proof yet, but let me at least get some of the ideas out there.
Theorem: AAA plus Playfair imply that no parallel lines exist.
Indirect Proof:
Assume that k | | l. Let A be any point lying on neither k nor l. Now draw any two transversals to both parallel lines through A. These intersect k at B and C respectively, and l at D and E respectively.
Now Playfair is equivalent to the Corresponding Angles Consequence -- the statement that if two lines are cut by a transversal, then corresponding angles are congruent. From this we conclude that Angle ABC = ADE and Angle ABC = AED. Since Angle A = A by the Reflexive Property, we observe that Triangles ABC and ADE are congruent by AAA.
(Notice that this doesn't work if A lies "between" the two parallel lines. In that case, just place A at another location so that the proof does work.)
But then AB = AD by CPCTC -- and since B and D lie on a transversal on the same side of A, this makes B and D the same point. Likewise C and E become the same point. In other words, we don't have two parallel lines k and l -- we only have one line. This is a contradiction, and so parallel lines can't exist. QED
The gap in this proof is that we assumed that Playfair implies the Parallel Line Consequences. We know that Playfair and the Parallel Line Consequences are equivalent in neutral geometry, but we don't know yet that they're equivalent in natural geometry. And we can't say "of course Playfair and Parallel Line Consequences are equivalent, since if there exist two parallel lines cut by a transversal then we must be in Euclidean geometry." This would be circular -- we're assuming that parallel lines don't exist on the sphere to prove that parallel lines don't exist on the sphere.
Of course, we could just adopt the Corresponding Angles Consequence as our parallel postulate rather than Playfair -- after all, this is essentially what the U of Chicago text does in Lesson 3-4 (where it is called "Parallel Lines Postulate"). But this is inelegant -- we'd have a postulate in natural geometry that begins "if two parallel lines are cut by a transversal" that thus would be vacuous unless the geometry is Euclidean. It also takes me away from my other goal on the blog -- transformations to prove the properties of parallel lines instead of just stating a postulate.
Then again, you might object that we're assuming AAA as a postulate in this proof. I've spent so many posts defending Common Core transformations and made a big deal how we can prove SAS, ASA, and SSS using transformations instead of assuming them as postulates. And now suddenly we're adopting AAA as a postulate!
Well, notice that the transformations used in Chapter 7 to prove SAS, ASA, and SSS can't be used to prove AAA at all. All of those proofs begin by performing an isometry on one of the triangles to make congruent sides coincide. Thus we must be given an S to make them work. None of them work in the AAA case (otherwise we'd be able to prove AAA in Euclidean geometry). And so there's nothing wrong with making AAA a postulate while still proving SAS, ASA, and SSS.
Of course, I'm not quite sure whether AAA belongs in a Point-Line-Postulate at all. We don't even have angle measures defined until we reach the Angle Measure Postulate of Chapter 3, so it doesn't quite make sense to have AAA there.
If we wish to avoid AAA, the only other possibilities I know of would be something like "dilations do not exist" (other than one with scale factor 1, or maybe -1) or "if two triangles are similar, then they are congruent." For now, we'll keep the following as our Spherical Point-Line-Plane postulate:
Spherical Point-Line-Plane Postulate
(a') Through any two points, there is at least one line.
(a") Suppose three points lie on a line. Then for at least one of those three points, that line is the only line that passes through that point and each of the other two points.
(b') Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line.
(c') AAA: If three angles of a triangle are congruent to three corresponding angles of another triangle, then the triangles are congruent.
Revisiting Birkhoff's Postulates
So far in this post, I've written so much about choosing the right postulates, and making sure that no postulates are wasted. Mathematicians prefer to prove as many theorems as possible from a small set of postulates. But what's best for mathematicians isn't necessarily best for high school students.
One of the most compact postulate sets I've ever seen is Birkhoff's Postulates. Apparently, all of plane Euclidean geometry can be derived using just four postulates. (Even Euclid himself had five!)
This week on her Mathematics Calendar 2019, Theoni Pappas wrote:
If we wish to avoid AAA, the only other possibilities I know of would be something like "dilations do not exist" (other than one with scale factor 1, or maybe -1) or "if two triangles are similar, then they are congruent." For now, we'll keep the following as our Spherical Point-Line-Plane postulate:
Spherical Point-Line-Plane Postulate
(a') Through any two points, there is at least one line.
(a") Suppose three points lie on a line. Then for at least one of those three points, that line is the only line that passes through that point and each of the other two points.
(b') Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line.
(c') AAA: If three angles of a triangle are congruent to three corresponding angles of another triangle, then the triangles are congruent.
Revisiting Birkhoff's Postulates
So far in this post, I've written so much about choosing the right postulates, and making sure that no postulates are wasted. Mathematicians prefer to prove as many theorems as possible from a small set of postulates. But what's best for mathematicians isn't necessarily best for high school students.
One of the most compact postulate sets I've ever seen is Birkhoff's Postulates. Apparently, all of plane Euclidean geometry can be derived using just four postulates. (Even Euclid himself had five!)
- Unique Line Assumption
- Ruler Postulate
- Protractor Postulate
- SAS Similarity
We know that the presence of a SAS Similarity Postulate forces the geometry to be Euclidean, since that's the only geometry where similar triangles exist (without the scale factor being 1). Thus we know that we can ultimately derive Playfair from SAS~.
Many modern texts include Ruler and Protractor Postulates. The Ruler Postulate corresponds to statements (c)-(d) of the U of Chicago's Point-Line-Plane Postulate, while the Protractor Postulate is essentially our Angle Measure Postulate.
Now imagine a high school Geometry course based on Birkhoff's Postulates. Well, his first three postulates allow us to derive all of Lessons 1-1 through 3-3, except for Lesson 1-9 on the Triangle Inequality (which our text states as a postulate). This past year, we reached Lesson 3-3 on the first day of October.
There's nothing else we can prove in Birkhoff's system without his fourth postulate, SAS~. Thus in a class based on Birkhoff, we'd be forced to teach similarity in October. Compare this to when I posted similarity lessons in the recent school year -- not until March!
Indeed, I explained that I like to introduce similarity late so that students who just barely passed Algebra I aren't bogged down with having to solve proportions first thing in Geometry. But once again, under Birkhoff, similarity and hence proportions must appear early.
Some proofs in the Birkhoff system are more complex that those in the U of Chicago, but one proof that's obviously simpler in Birkhoff is that of SAS Congruence. (Proof: if two sides and an included angle are congruent to the corresponding parts of another triangle, then the triangles are similar by the SAS~ Postulate with scale factor 1. QED) But despite the easier proof, we wouldn't want to teach this to high school students, since it forces them to learn similarity before congruence.
Some readers might argue that our spherical proof of "there exist no parallel lines" is just as bad as the proof of SAS Congruence in Birkhoff. Indeed, it might be worse, since it requires learning the properties of parallel lines just to prove that they don't exist.
But once again, in no high school course will we actually start with spherical geometry. My vision is for most of the class to be Euclidean geometry as usual. It's only at the end of the year -- after Euclidean geometry has been mastered -- will our proof of "there exist no parallel lines" work. (We say, this is why we can't have parallel lines on the sphere.)
In fact, the class will begin with a statement of the Euclidean Point-Line-Plane Postulate. It's only that throughout the course of the year, we (secretly) use the Natural Point-Line-Plane Postulate so that we can prove as many theorems that hold in both geometries. After these run out, then we use the full Euclidean PLP -- but at the end of the year, it's now easier to point out which theorems no longer hold in spherical geometry.
Pappas Problem of the Day
Today's Pappas problem is all about palindromes. There's also a Square One TV song on palindromes, but since this has nothing to do with Geometry, I won't post the problem here. Instead, I wish to discuss a Geometry problem from a few days ago.
This week on her Mathematics Calendar 2019, Theoni Pappas wrote:
Which statement is true?
1) All rectangles are squares.
2) All rectangles have at most one right angle.
3) All 5 statements are true.
4) All rectangles are rhombi.
5) No rectangles are trapezoids.
The intended correct answer is 5) -- and of course, this question was asked on the fifth (the day of the mainshock of the earthquake). But notice that it depends on the exclusive definition of "trapezoid." If we use the U of Chicago inclusive definition of "trapezoid" instead, then all rectangles are trapezoids (indeed isosceles trapezoids), and thus all five statements are false.
The reason I bring up Friday's Pappas question is because -- in today's spherical geometry post -- I ask, which of these statements hold in spherical geometry.
The answer, of course, is 3) -- all of them are vacuously true on the sphere. There are zero rectangles, and all zero of them are squares, rhombi, and non-trapezoids with at most one right angle.
Then again, only 5) holds in both Euclidean and spherical geometry. Therefore, we can only hope to prove 5) in natural geometry. In order to avoid issues with exclusive vs. inclusive definitions of "trapezoid," let's rewrite 5) as:
5) All rectangles are parallelograms.
Once again, our goal is to prove this statement in natural geometry (without the usual cop-out, "if a rectangle exist, then we must be in Euclidean geometry, so therefore...").
Actually, this statement is very easy to prove in neutral (Euclidean plus hyperbolic) geometry. In fact, the usual Euclidean proof suffices -- Lesson 5-2 of the U of Chicago text gives the proof in a single line: "Because two perpendiculars to the same line are parallel, every rectangle is a parallelogram." In hyperbolic geometry we'd call such lines "ultraparallel," but such lines are nonetheless parallel and thus can form the sides of a parallelogram.
Once again, this proof is valid in neutral geometry, but it's vacuously true unless, of course, the geometry is Euclidean. So we'd replace "rectangle" with a generalization that exists in both Euclidean and hyperbolic geometry. Using Lambert quadrilaterals here is easiest, since actually we need only three right angles to make the Euclidean proof work:
And it turns out that in neutral geometry, all Saccheri quads -- and, I believe, all equiangular quads as well -- are also pgrams. But these proofs are more complex.
But we're not working in neutral geometry, but in natural geometry. Unfortunately, the Euclidean proof no longer holds in natural geometry, since the line "two perpendiculars to the same line are parallel" fails on the sphere.
Of course, no parallelograms can exist on the sphere. So our statement "all rectangles are pgrams" seems even sillier in spherical than in hyperbolic geometry, where at least pgrams can exist. Clearly, generalizing "rectangles" to "Lambert quadrilaterals," etc., can't possibly lead to pgrams, since once again there are no pgrams on the sphere.
An interesting question is, just as we were able to generalize "rectangle," we might wish to generalize "parallelogram" as well. It might be possible to consider figures with the same symmetry as the Euclidean pgram -- namely 180-degree rotational symmetry. Thus the goal would be to prove that the equilateral quadrilateral has 180-degree rotational symmetry. (Here I already rule out Lambert and Saccheri quads since neither can have rotational symmetry -- else they'd have four right angles.)
If we return to the inclusive definition of "trapezoid," then a rectangle in Euclidean geometry is an isosceles trapezoid. Since there are no parallel lines on the sphere, there can be no trapezoids, whether isosceles or not. But consider the following description:
In Euclidean geometry, we can prove that such quadrilaterals must be isosceles trapezoids. Such quads also exist on the sphere -- but of course these quads are not trapezoids.
In natural geometry, it's trivial that all equiangular quads have two pairs of congruent adjacent angles, since all four angles are congruent. For Saccheri quads, we can prove that they all have a pair of congruent summit angles. No Lambert quad on the sphere can have two pairs of congruent adjacent angles (since then it would have four right angles).
Traditionalists and Rich Problems
There's been some traditionalist action over the past few days. It all starts on July 5th -- earthquake day -- when Barry Garelick posted the following:
https://traditionalmath.wordpress.com/2019/07/05/rich-problems-dept/
Here Garelick discusses one of his pet peeves -- "rich problems." He writes:
Why would we want problems such that "every student can be successful on at least part of it"? Oh, that's easy -- it's because students who don't feel that they can be successful on at least part of it are likely to leave it blank. As usual, Garelick forgets that students tend to leave hard problems blank.
For example: “What are the dimensions of a rectangle with a perimeter of 24 units?”
Let's try replacing this with a similar problem that is no longer open-ended:
Now suppose a student answers "ten units by two units." The teacher replies "Wrong!" The student is now subdued and decides simply to quit solving this problem, or any other on the p-set. Once again, our goal is to avoid labeling students "wrong" as long as possible. Once again, humans aren't Vulcans, since we often think emotionally, not logically. When students hear that they are wrong, many aren't motivated to correct themselves -- instead, they are no longer motivated to work at all.
Rather, the student’s understanding is viewed as “inauthentic” and “algorithmic” because the practice, repetition and imitation of procedures is merely “imitation of thinking”.
Hmm, I assume Garelick implies here that labeling a student's understanding as "inauthentic" is just as bad as labeling a student "wrong." Thus, in the name of consistency, should I be just as concerned with the "inauthentic" label as with the "wrong" label?
Once again, my concern is with students who leave problems blank. I know that calling them "wrong" leads to students giving up and leaving problems blank. Does calling them "inauthentic" also lead to students just giving up? (The honest answer here is, "I don't know.")
Garelick concludes his post with what he considers a rich problem from his Golden Age text:
“If a+1 = b, then which is true? a>b, a<2b, or a<b?”
OK, I'll concede that this is an interesting problem. It turns out that a > b is never true, while a < b is always true. We can substitute to solve the last inequality:
a < 2b
a < 2(a + 1)
a < 2a + 2
-a < 2
a > -2
Students might assume that a < 2b is always true, because they fail to consider negative numbers.
OK, here's what I consider to be a rich problem, but don't expect to see this in a Geometry course anytime soon (nor should it appear there):
Let's see what Garelick's commenters have to say. We begin with Chester Draws, who describes an activity that he assigned this week:
Chester Draws:
I set my students the task of designing Aztec pyramids with 1,500 m^3 volume yesterday, as a bit of end of week fun. Am I meant to take in each of the 25 answers and mark them for the next lesson? Because they won’t know if they were successful otherwise,nor where they made mistakes. But if I do that regularly, then I’m marking two hours for every lesson I teach. That’s simply not sustainable.
(You might wonder what class Draws was teaching the first week in July -- summer school? Well, I notice that Draws uses the word "Maths" later in this comment, implying that he isn't American. In Great Britain, there are more breaks -- "half-term" -- during the year with a shorter summer, and so school extends deep into July. Meanwhile, in Australia it's currently winter. There is probably a break of two weeks between terms at some point in July, but not this week. At any rate, since Draws isn't American, he isn't affected by the Common Core Standards.)
Here Draws criticizes open-ended "rich problems" because they make it difficult for the students to know whether they were successful or where they made mistakes. But then again, that's the point -- the "rich problems" are supposed to protect students who would refuse to work after being told that they are "wrong."
And of course, we can't leave SteveH out of the fun here:
SteveH:
This is the classic ed-speak I heard about differentiated instruction (learning) when my son was young. This wasn’t just for math. The onus is on the child (parents at home) for anything more than NON-STEM CCSS proficiency. I’ve always thought that this was just a K-8 excuse for full inclusion academic age tracking. Somehow, those things magically disappear in high school due to the coming reality of college and real life.
There's that "full inclusion" again, which we already know that the traditionalists hate. It's also notable that SteveH follows "full inclusion" with "academic age tracking." Here SteveH knows that his opponents profess to be against "tracking." Thus he rubs it in that the current status quo is indeed "tracking," except that it's tracking by age (the idea that all five-year-olds should start kindergarten, all six-year-olds should start first grade, and so on). Of course, the opponents of "tracking" would counter that tracking by age is preferable to what amounts to tracking by race (or income).
Feynman, Pages 1-70
It's time for us to begin our summer side-along graphic novel, Jim Ottaviani's Feynman. By the way, my copy of his newest book, Hawking, has arrived.
Once again, I like the idea behind Ottaviani's books -- depicting mathematicians and scientists as heroes, not nerds. I hope that books such as these can inspire students to learn math and science -- and indeed, this is why I'm introducing a new blogging label for these books, "heroes."
The book opens with a lecture from "some fool physicist" in 1964:
"So you see," he says, "nature has always looked like a horrible mess...but as we go along, we discover patterns and come up with theories. The result? The mess I'm going to show you is smaller than the mess I would've had to show you ten years ago.
"And we had a bigger mess at the beginning of the 20th century, before quantum mechanics. Laws of motion, heat, magnetism, electricity, light, X-rays...and on and on and on. Classical physics couldn't handle any of it right. An absurd situation! But along with the smaller mess, I'm also going to show you that different theories of physics are very similar. Maybe that's just because of our limited imagination. We try fitting every new phenomenon in the framework we already have! Some when some fool physicist gives a lecture on physics and says: 'This is the way it works, and look how wonderfully similar all our theories are," ...maybe it's not because they really are similar. Maybe it's because we physicists have only been able to think of the same damn thing over and over again."
That fool physicist, of course, was Richard Feynman. Today's section takes us through Feynman's youth, from his youth in 1923 up to his work during World War II in 1944.
As a five-year-old child, Ritty, as he was called, was always inquisitive:
Ritty: Dad?
Mr. Feynman: What, Ritty?
Ritty: Why does it keep moving?
Mr. Feynman: Well, the reason the ball keeps rolling is because it has "inertia."
Ritty: What's inertia?
Mr. Feynman: That's what scientists call the reason the ball keeps moving, but it's just a name. Nobody really knows what it means.
The scene suddenly jumps to 1986, when an elderly Richard is reflecting on his life:
"If we know the rules," he says, "we consider that we 'understand' the world. That was a lot of years ago and things haven't changed much since I was a kid in Far Rockaway [in Queens, New York]. We still don't have them all. Maybe someday."
The young Ritty tried to study Calculus on his own. (Hey, SteveH, you can't blame his inability to learn Calculus in school on the Common Core!)
Librarian: This is too advanced. Who is this for? You can't read this.
Ritty: It's...for my father.
"That's one of the few lies I told," he narrates, "because I was always a very good boy. So...my father told me wonderful things about the world, but the nearest thing to a scientist for me was my dentist. And he had a crony, Mr. Leseur, who was teaching science here at that time. And he talked Mr. Leseur into letting me visit Far Rockaway High when I was still in grammar school. And here I met Mr. Johnson. And I'd ask him to give me some experiment to do, and he'd haul out all kinds of apparatus...magnets, tubes, and pipes."
Eventually, the young Richard is admitted to MIT. While there, he tries to join a fraternity, but he has some problems with the hazing:
"...I fought like hell. They eventually took us way out of town; we had to find our way back. But first we had to get free. They tied everybody up. All but one guy, that is."
Other guy: Whatsa matter?
Richard: Yeah, you look awful.
"They didn't want to do it to him, but not because he fought back."
It was because Richard's friend had escaped the Holocaust. Meanwhile, Richard is attending classes at MIT, but his father wants him to break up with his girlfriend Arline. Unfortunately, the elder Feynman is a product of his times -- he warns his son about women's brains:
Richard: What about them?
Mr. Feynman: Well, science. She wants to be an astronomer, and women can't do that. Richy, be careful with those [plates]!"
"Very, very different from now," he narrates. "But regarding me and Arline, they didn't need me to worry; MIT was the perfect school. I loved it there, and settled pretty quickly on a field that suited me well, I think." (Since math was too theoretical and electrical engineering was too practical, he settled on physics.) "The only problem was this rule about taking other courses [to satisfy the college general ed requirements]. You know, to get more 'culture.' Once I escaped because they had astronomy listed in humanities. At the time, I was interested in science and nothing else."
His days at MIT continue:
"I also worked on other stuff, including my senior thesis and a paper on cosmic rays with Prof. Vallarta. When I finished..."
Richard (typing): "The scattering of cosmic rays by the stars of a galaxy"...by...
Prof. Vallarta: Since I'm the senior scientist, the convention is that my name goes first.
"That seemed funny to me, but okay...Later, Heisenberg edited a book on cosmic rays, and he commented on every worthwhile paper. Ours had no place in it! But at the very end of the book, he mentioned our theory, saying, 'Mugga wugga blah blah blah blah this and that according to Vallarta and Feynman. I saw Vallarta a few years later."
Richard: Did you see the book by Heisenberg?
Prof. Vallarta: Yeah, you're the last word in cosmic rays. That's what I get for pulling rank!
Richard: Ha!
As a grad student, Feynman gets to meet other famous scientists, including Einstein. He also works with Wolfgang Pauli on quantum theory, but he is understandably distracted.
"I didn't solve the quantum version either, and I worked on it for years. So maybe Pauli's 'this, that, and the other thing' was right. I don't know...I wasn't paying attention. When I came home one time Arline had developed a bump on her neck. Sometime later she ran a temperature and the bump got bigger -- or maybe smaller -- and the hospital said she had typhoid fever. Right away I researched it. The next time I visited, I asked the doctor how the Wydell test came out."
Doctor: Negative.
Richard: What? How can that be! Why all these gowns, then? Maybe she doesn't have typhoid fever!
"The result of that was the doctor talked to her parents, who told me not to interfere."
Future Father-in-Law: ...after all, he's the doctor. You're only her fiance.
"Anyway, after a little while she seemed to get better, the swelling went down, and the fever left. But later it started again."
Feynman soon figures out that his young bride has tuberculosis. Even though she has only seven years or fewer to live, Richard marries Arline anyway. By 1944, he is working on the top-secret Manhattan Project with Hans Bethe, in Albuquerque, New Mexico. On the last page for today:
"So the next time he visited, he called me over first. We argued for hours."
Bethe: How about so and so?
Richard: That sounds a little bit better, but it's got this damn fool idea in it.
"And that's how it went. I was dumb that way, but it works with the right kind of person."
And then the caption reads, "Military guys are not the right kind of person." This last page contains a funny scene where in order to keep the mission top-secret, Richard arrives at the training site at 10:07, 10:12, 10:14, and finally 10:18:
Military guy: What the hell is going on Mr. Fine-eh-man? Yah keep goin' out, but I never see yah come back in!
And this is where our reading ends for today, so we'll figure out Feynman's magic trick later on. It might be that the training site satisfies spherical geometry, and so what appears to be a straight line is really a circle that takes him outside the site again...
Conclusion
Today's post is all about the natural and spherical Point-Line-Plane Postulate:
Natural Point-Line-Plane Postulate
(a') Through any two points, there is at least one line.
(a") Suppose three points lie on a line. Then for at least one of those three points, that line is the only line that passes through that point and each of the other two points.
(b') Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line.
Spherical Point-Line-Plane Postulate
(a') Through any two points, there is at least one line.
(a") Suppose three points lie on a line. Then for at least one of those three points, that line is the only line that passes through that point and each of the other two points.
(b') Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line.
(c') AAA: If three angles of a triangle are congruent to three corresponding angles of another triangle, then the triangles are congruent.
But in an actual class, we'd mention only the traditional Euclidean PLP Postulate:
Point-Line-Plane Postulate:
(a) Unique line assumption: Through any two points, there is exactly one line.
(b) Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line. Given a plane in space, there exists a point in space not on the line.
(c) Number line assumption: Every line is a set of points that can be put in a one-to-one correspondence with the real numbers, with any point on it corresponding to 0 and any other point corresponding to 1.
(d) Distance assumption: On a number line, there is a unique distance between two points.
Indeed, natural geometry is like our own Manhattan Project -- we state Euclidean PLP, but then use only Natural PLP as long as possible. Then at the end of class, we reveal spherical geometry
Let me address what's on most Americans' minds today -- the Women's World Cup of soccer. I congratulate the Americans for beating the Dutch and winning a second straight championship!
1) All rectangles are squares.
2) All rectangles have at most one right angle.
3) All 5 statements are true.
4) All rectangles are rhombi.
5) No rectangles are trapezoids.
The intended correct answer is 5) -- and of course, this question was asked on the fifth (the day of the mainshock of the earthquake). But notice that it depends on the exclusive definition of "trapezoid." If we use the U of Chicago inclusive definition of "trapezoid" instead, then all rectangles are trapezoids (indeed isosceles trapezoids), and thus all five statements are false.
The reason I bring up Friday's Pappas question is because -- in today's spherical geometry post -- I ask, which of these statements hold in spherical geometry.
The answer, of course, is 3) -- all of them are vacuously true on the sphere. There are zero rectangles, and all zero of them are squares, rhombi, and non-trapezoids with at most one right angle.
Then again, only 5) holds in both Euclidean and spherical geometry. Therefore, we can only hope to prove 5) in natural geometry. In order to avoid issues with exclusive vs. inclusive definitions of "trapezoid," let's rewrite 5) as:
5) All rectangles are parallelograms.
Once again, our goal is to prove this statement in natural geometry (without the usual cop-out, "if a rectangle exist, then we must be in Euclidean geometry, so therefore...").
Actually, this statement is very easy to prove in neutral (Euclidean plus hyperbolic) geometry. In fact, the usual Euclidean proof suffices -- Lesson 5-2 of the U of Chicago text gives the proof in a single line: "Because two perpendiculars to the same line are parallel, every rectangle is a parallelogram." In hyperbolic geometry we'd call such lines "ultraparallel," but such lines are nonetheless parallel and thus can form the sides of a parallelogram.
Once again, this proof is valid in neutral geometry, but it's vacuously true unless, of course, the geometry is Euclidean. So we'd replace "rectangle" with a generalization that exists in both Euclidean and hyperbolic geometry. Using Lambert quadrilaterals here is easiest, since actually we need only three right angles to make the Euclidean proof work:
- All Lambert quadrilaterals are parallelograms (neutral geometry).
And it turns out that in neutral geometry, all Saccheri quads -- and, I believe, all equiangular quads as well -- are also pgrams. But these proofs are more complex.
But we're not working in neutral geometry, but in natural geometry. Unfortunately, the Euclidean proof no longer holds in natural geometry, since the line "two perpendiculars to the same line are parallel" fails on the sphere.
Of course, no parallelograms can exist on the sphere. So our statement "all rectangles are pgrams" seems even sillier in spherical than in hyperbolic geometry, where at least pgrams can exist. Clearly, generalizing "rectangles" to "Lambert quadrilaterals," etc., can't possibly lead to pgrams, since once again there are no pgrams on the sphere.
An interesting question is, just as we were able to generalize "rectangle," we might wish to generalize "parallelogram" as well. It might be possible to consider figures with the same symmetry as the Euclidean pgram -- namely 180-degree rotational symmetry. Thus the goal would be to prove that the equilateral quadrilateral has 180-degree rotational symmetry. (Here I already rule out Lambert and Saccheri quads since neither can have rotational symmetry -- else they'd have four right angles.)
If we return to the inclusive definition of "trapezoid," then a rectangle in Euclidean geometry is an isosceles trapezoid. Since there are no parallel lines on the sphere, there can be no trapezoids, whether isosceles or not. But consider the following description:
- a quadrilateral with two pairs of congruent adjacent angles
In Euclidean geometry, we can prove that such quadrilaterals must be isosceles trapezoids. Such quads also exist on the sphere -- but of course these quads are not trapezoids.
In natural geometry, it's trivial that all equiangular quads have two pairs of congruent adjacent angles, since all four angles are congruent. For Saccheri quads, we can prove that they all have a pair of congruent summit angles. No Lambert quad on the sphere can have two pairs of congruent adjacent angles (since then it would have four right angles).
Traditionalists and Rich Problems
There's been some traditionalist action over the past few days. It all starts on July 5th -- earthquake day -- when Barry Garelick posted the following:
https://traditionalmath.wordpress.com/2019/07/05/rich-problems-dept/
Here Garelick discusses one of his pet peeves -- "rich problems." He writes:
One definition is: “A problem that has multiple entry points and has various levels of cognitive demands. Every student can be successful on at least part of it.”
My definition differs a bit: “One-off, open-ended, ill-posed problems that supposedly lead students to apply/transfer prior knowledge to new or novel problems that don’t generalize.”
Why would we want problems such that "every student can be successful on at least part of it"? Oh, that's easy -- it's because students who don't feel that they can be successful on at least part of it are likely to leave it blank. As usual, Garelick forgets that students tend to leave hard problems blank.
For example: “What are the dimensions of a rectangle with a perimeter of 24 units?”
Let's try replacing this with a similar problem that is no longer open-ended:
- What are the dimensions of a square with a perimeter of 24 units?
Now suppose a student answers "ten units by two units." The teacher replies "Wrong!" The student is now subdued and decides simply to quit solving this problem, or any other on the p-set. Once again, our goal is to avoid labeling students "wrong" as long as possible. Once again, humans aren't Vulcans, since we often think emotionally, not logically. When students hear that they are wrong, many aren't motivated to correct themselves -- instead, they are no longer motivated to work at all.
Rather, the student’s understanding is viewed as “inauthentic” and “algorithmic” because the practice, repetition and imitation of procedures is merely “imitation of thinking”.
Hmm, I assume Garelick implies here that labeling a student's understanding as "inauthentic" is just as bad as labeling a student "wrong." Thus, in the name of consistency, should I be just as concerned with the "inauthentic" label as with the "wrong" label?
Once again, my concern is with students who leave problems blank. I know that calling them "wrong" leads to students giving up and leaving problems blank. Does calling them "inauthentic" also lead to students just giving up? (The honest answer here is, "I don't know.")
Garelick concludes his post with what he considers a rich problem from his Golden Age text:
“If a+1 = b, then which is true? a>b, a<2b, or a<b?”
OK, I'll concede that this is an interesting problem. It turns out that a > b is never true, while a < b is always true. We can substitute to solve the last inequality:
a < 2b
a < 2(a + 1)
a < 2a + 2
-a < 2
a > -2
Students might assume that a < 2b is always true, because they fail to consider negative numbers.
OK, here's what I consider to be a rich problem, but don't expect to see this in a Geometry course anytime soon (nor should it appear there):
- What are the dimensions of a Saccheri quadrilateral (spherical) with a perimeter of 24 units?
Let's see what Garelick's commenters have to say. We begin with Chester Draws, who describes an activity that he assigned this week:
Chester Draws:
I set my students the task of designing Aztec pyramids with 1,500 m^3 volume yesterday, as a bit of end of week fun. Am I meant to take in each of the 25 answers and mark them for the next lesson? Because they won’t know if they were successful otherwise,nor where they made mistakes. But if I do that regularly, then I’m marking two hours for every lesson I teach. That’s simply not sustainable.
(You might wonder what class Draws was teaching the first week in July -- summer school? Well, I notice that Draws uses the word "Maths" later in this comment, implying that he isn't American. In Great Britain, there are more breaks -- "half-term" -- during the year with a shorter summer, and so school extends deep into July. Meanwhile, in Australia it's currently winter. There is probably a break of two weeks between terms at some point in July, but not this week. At any rate, since Draws isn't American, he isn't affected by the Common Core Standards.)
Here Draws criticizes open-ended "rich problems" because they make it difficult for the students to know whether they were successful or where they made mistakes. But then again, that's the point -- the "rich problems" are supposed to protect students who would refuse to work after being told that they are "wrong."
And of course, we can't leave SteveH out of the fun here:
SteveH:
This is the classic ed-speak I heard about differentiated instruction (learning) when my son was young. This wasn’t just for math. The onus is on the child (parents at home) for anything more than NON-STEM CCSS proficiency. I’ve always thought that this was just a K-8 excuse for full inclusion academic age tracking. Somehow, those things magically disappear in high school due to the coming reality of college and real life.
There's that "full inclusion" again, which we already know that the traditionalists hate. It's also notable that SteveH follows "full inclusion" with "academic age tracking." Here SteveH knows that his opponents profess to be against "tracking." Thus he rubs it in that the current status quo is indeed "tracking," except that it's tracking by age (the idea that all five-year-olds should start kindergarten, all six-year-olds should start first grade, and so on). Of course, the opponents of "tracking" would counter that tracking by age is preferable to what amounts to tracking by race (or income).
Feynman, Pages 1-70
It's time for us to begin our summer side-along graphic novel, Jim Ottaviani's Feynman. By the way, my copy of his newest book, Hawking, has arrived.
Once again, I like the idea behind Ottaviani's books -- depicting mathematicians and scientists as heroes, not nerds. I hope that books such as these can inspire students to learn math and science -- and indeed, this is why I'm introducing a new blogging label for these books, "heroes."
The book opens with a lecture from "some fool physicist" in 1964:
"So you see," he says, "nature has always looked like a horrible mess...but as we go along, we discover patterns and come up with theories. The result? The mess I'm going to show you is smaller than the mess I would've had to show you ten years ago.
"And we had a bigger mess at the beginning of the 20th century, before quantum mechanics. Laws of motion, heat, magnetism, electricity, light, X-rays...and on and on and on. Classical physics couldn't handle any of it right. An absurd situation! But along with the smaller mess, I'm also going to show you that different theories of physics are very similar. Maybe that's just because of our limited imagination. We try fitting every new phenomenon in the framework we already have! Some when some fool physicist gives a lecture on physics and says: 'This is the way it works, and look how wonderfully similar all our theories are," ...maybe it's not because they really are similar. Maybe it's because we physicists have only been able to think of the same damn thing over and over again."
That fool physicist, of course, was Richard Feynman. Today's section takes us through Feynman's youth, from his youth in 1923 up to his work during World War II in 1944.
As a five-year-old child, Ritty, as he was called, was always inquisitive:
Ritty: Dad?
Mr. Feynman: What, Ritty?
Ritty: Why does it keep moving?
Mr. Feynman: Well, the reason the ball keeps rolling is because it has "inertia."
Ritty: What's inertia?
Mr. Feynman: That's what scientists call the reason the ball keeps moving, but it's just a name. Nobody really knows what it means.
The scene suddenly jumps to 1986, when an elderly Richard is reflecting on his life:
"If we know the rules," he says, "we consider that we 'understand' the world. That was a lot of years ago and things haven't changed much since I was a kid in Far Rockaway [in Queens, New York]. We still don't have them all. Maybe someday."
The young Ritty tried to study Calculus on his own. (Hey, SteveH, you can't blame his inability to learn Calculus in school on the Common Core!)
Librarian: This is too advanced. Who is this for? You can't read this.
Ritty: It's...for my father.
"That's one of the few lies I told," he narrates, "because I was always a very good boy. So...my father told me wonderful things about the world, but the nearest thing to a scientist for me was my dentist. And he had a crony, Mr. Leseur, who was teaching science here at that time. And he talked Mr. Leseur into letting me visit Far Rockaway High when I was still in grammar school. And here I met Mr. Johnson. And I'd ask him to give me some experiment to do, and he'd haul out all kinds of apparatus...magnets, tubes, and pipes."
Eventually, the young Richard is admitted to MIT. While there, he tries to join a fraternity, but he has some problems with the hazing:
"...I fought like hell. They eventually took us way out of town; we had to find our way back. But first we had to get free. They tied everybody up. All but one guy, that is."
Other guy: Whatsa matter?
Richard: Yeah, you look awful.
"They didn't want to do it to him, but not because he fought back."
It was because Richard's friend had escaped the Holocaust. Meanwhile, Richard is attending classes at MIT, but his father wants him to break up with his girlfriend Arline. Unfortunately, the elder Feynman is a product of his times -- he warns his son about women's brains:
Richard: What about them?
Mr. Feynman: Well, science. She wants to be an astronomer, and women can't do that. Richy, be careful with those [plates]!"
"Very, very different from now," he narrates. "But regarding me and Arline, they didn't need me to worry; MIT was the perfect school. I loved it there, and settled pretty quickly on a field that suited me well, I think." (Since math was too theoretical and electrical engineering was too practical, he settled on physics.) "The only problem was this rule about taking other courses [to satisfy the college general ed requirements]. You know, to get more 'culture.' Once I escaped because they had astronomy listed in humanities. At the time, I was interested in science and nothing else."
His days at MIT continue:
"I also worked on other stuff, including my senior thesis and a paper on cosmic rays with Prof. Vallarta. When I finished..."
Richard (typing): "The scattering of cosmic rays by the stars of a galaxy"...by...
Prof. Vallarta: Since I'm the senior scientist, the convention is that my name goes first.
"That seemed funny to me, but okay...Later, Heisenberg edited a book on cosmic rays, and he commented on every worthwhile paper. Ours had no place in it! But at the very end of the book, he mentioned our theory, saying, 'Mugga wugga blah blah blah blah this and that according to Vallarta and Feynman. I saw Vallarta a few years later."
Richard: Did you see the book by Heisenberg?
Prof. Vallarta: Yeah, you're the last word in cosmic rays. That's what I get for pulling rank!
Richard: Ha!
As a grad student, Feynman gets to meet other famous scientists, including Einstein. He also works with Wolfgang Pauli on quantum theory, but he is understandably distracted.
"I didn't solve the quantum version either, and I worked on it for years. So maybe Pauli's 'this, that, and the other thing' was right. I don't know...I wasn't paying attention. When I came home one time Arline had developed a bump on her neck. Sometime later she ran a temperature and the bump got bigger -- or maybe smaller -- and the hospital said she had typhoid fever. Right away I researched it. The next time I visited, I asked the doctor how the Wydell test came out."
Doctor: Negative.
Richard: What? How can that be! Why all these gowns, then? Maybe she doesn't have typhoid fever!
"The result of that was the doctor talked to her parents, who told me not to interfere."
Future Father-in-Law: ...after all, he's the doctor. You're only her fiance.
"Anyway, after a little while she seemed to get better, the swelling went down, and the fever left. But later it started again."
Feynman soon figures out that his young bride has tuberculosis. Even though she has only seven years or fewer to live, Richard marries Arline anyway. By 1944, he is working on the top-secret Manhattan Project with Hans Bethe, in Albuquerque, New Mexico. On the last page for today:
"So the next time he visited, he called me over first. We argued for hours."
Bethe: How about so and so?
Richard: That sounds a little bit better, but it's got this damn fool idea in it.
"And that's how it went. I was dumb that way, but it works with the right kind of person."
And then the caption reads, "Military guys are not the right kind of person." This last page contains a funny scene where in order to keep the mission top-secret, Richard arrives at the training site at 10:07, 10:12, 10:14, and finally 10:18:
Military guy: What the hell is going on Mr. Fine-eh-man? Yah keep goin' out, but I never see yah come back in!
And this is where our reading ends for today, so we'll figure out Feynman's magic trick later on. It might be that the training site satisfies spherical geometry, and so what appears to be a straight line is really a circle that takes him outside the site again...
Conclusion
Today's post is all about the natural and spherical Point-Line-Plane Postulate:
Natural Point-Line-Plane Postulate
(a') Through any two points, there is at least one line.
(a") Suppose three points lie on a line. Then for at least one of those three points, that line is the only line that passes through that point and each of the other two points.
(b') Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line.
Spherical Point-Line-Plane Postulate
(a') Through any two points, there is at least one line.
(a") Suppose three points lie on a line. Then for at least one of those three points, that line is the only line that passes through that point and each of the other two points.
(b') Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line.
(c') AAA: If three angles of a triangle are congruent to three corresponding angles of another triangle, then the triangles are congruent.
But in an actual class, we'd mention only the traditional Euclidean PLP Postulate:
Point-Line-Plane Postulate:
(a) Unique line assumption: Through any two points, there is exactly one line.
(b) Dimension assumption: Given a line in a plane, there exists a point in the plane not on the line. Given a plane in space, there exists a point in space not on the line.
(c) Number line assumption: Every line is a set of points that can be put in a one-to-one correspondence with the real numbers, with any point on it corresponding to 0 and any other point corresponding to 1.
(d) Distance assumption: On a number line, there is a unique distance between two points.
Indeed, natural geometry is like our own Manhattan Project -- we state Euclidean PLP, but then use only Natural PLP as long as possible. Then at the end of class, we reveal spherical geometry
Let me address what's on most Americans' minds today -- the Women's World Cup of soccer. I congratulate the Americans for beating the Dutch and winning a second straight championship!
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