*Magic of Mathematics*:

*"The hours from seven 'til nearly midnight are normally quiet ones on the bridge.... Beginning almost exactly at seven o'clock.... it just looked as if everybody in Manhattan who owned a motorcar had decided to drive out on Long Island that evening."*

*This is the first page of the subsection "Complexity & the Present." Pappas opens with this excerpt from*

*The Law*by Robert M. Coates to illustrate how things take place for no apparent reason. Indeed, she continues:

"We have all experienced such events and usually attributed them to 'coincidence,' since there were no apparent indicators to predict otherwise."

There is only picture on this page, but it's quite large and takes up more than half the page. It depicts a huge crowd -- probably the people all trying to cross the Brooklyn Bridge at the same time.

Chapter 2 of George Szpiro's

*Poincare's Prize*is called "What Flies Know and Ants Don't." Here's how it begins:

"Standing at the command bridge of his converted cargo ship

*Santa Maria*, Christopher Columbus faced a problem. His men were close to a mutiny."

In this chapter, Szpiro introduces us to the idea of dimension before jumping into the statement of the Poincare conjecture. Again, this is why I considered waiting until Chapter 9 in the U of Chicago text (on the third dimension) before covering this chapter.

The mutiny on the

*Santa Maria*is over the fear of falling off the edge of the earth -- in other words, because the crew thinks that the earth is flat. But in fact, the shape of the earth has been known for thousands of years. As Szpiro explains:

"The philosopher Eratosthenes (276-194 BC), most famous for his prime number sieve [that we saw back in Ogilvy's book -- dw], had even calculated the earth's circumference. By planting vertical poles into the ground at the towns of Aswan and Alexandria and observing their shadows, he inferred that the circumference of the earth must have the length of 250,000 stadia [about 40,000 km]."

Of course, Columbus lands safely in the New World, and the world is not flat. Nonetheless, not everyone accepts the mainstream theory of the earth's shape:

"Members of the Flat Earth Society believe that the north pole is in the center of a disk and the edges consist of insurmountable, fifty-meter high ice mountains."

Here's a link to the Flat Earth Society, which holds this fringe view of the earth's shape:

https://www.tfes.org/

Also, apparently last weekend was the First Annual Flat Earth International Conference:

https://fe2017.com/

The reason for the chapter title is that from the perspective of an ant or fly crawling on its surface, a basketball is flat. In other words, a sphere's surface is a two-dimensional manifold. But unlike an ant, a fly can travel up and down in the air:

"What is the deeper view for the flies' broader view? The answer is that by flying away from the object, airborne animals are able to move in three dimensions."

In this chapter, Szpiro also introduces the reader to topology. He explains:

"Ball, bagel, and pretzel differ from another by the number of holes they have. So what else is new? Maybe this: In topology, all surfaces with the same number of holes are identical."

After providing some more examples of dimensions and topology, the author tells us the relationship between the dimension of an object and of its surface:

"We already pointed out that the surface of the three-dimensional earth is two-dimensional. In the same manner we may say that the surface of a two-dimensional disk is the one-dimensional circle that surrounds it; the surface of the one-dimensional line is the zero-dimensional endpoint. And -- to challenge your imagination -- let me point out that the surface of four-dimensional balls are three-dimensional spheres."

Yes, we will be reading about the fourth dimension. I wrote extensively about the fourth dimension in reading Rudy Rucker's book -- search for the label "Rudy Rucker" to find these old posts.

Szpiro concludes this chapter with a hint of what's to come:

"Can we always tell a ball from a bagel by inspecting its surface? In three dimensions it's easy -- at least for us and flies, if not for ants. But would we be able to tell a four-dimensional ball if we tripped over one? This is where Monsieur Henri Poincare comes in."

By the way, he explains that in topology, mathematicians "study which bodies can be transformed into one another by pulling, squeezing, twisting, and turning...but not by tearing or folding." All similarity transformations and isometries are legal -- including translations, today's lesson:

Lesson 6-2 of the U of Chicago text is called "Translations." In the modern Third Edition, we must backtrack to Lesson 4-4 to learn about translations.

This is what I wrote two years ago about this lesson:

To emphasize the coordinate plane, I've titled this lesson "Translations

*on the Coordinate Plane*" rather than just "Translations." Yesterday I warned you that another translation lesson was coming up soon, and as it turns out, that day is today.

I said that there are several ways to prove that the transformation mapping (

*x*,

*y*) to (

*x*+

*h*,

*y*+

*k*) really is a translation. First is a coordinate proof using the Slope and Distance Formulas, but I mentioned that we can't use those until after dilations in the second semester. Second is to manipulate the transformations until various mirrors cancel, but that requires arcane symbols such as:

T = r_U(

*n*) o r_U(

*m*)

and I said that it would only confuse students. Fortunately, there's a third way to complete the proof that I alluded to a few weeks back -- and I've decided to use this form of the proof in today's post.

We begin by noting that the two transformations (

*x*,

*y*) -> (

*x*+

*h*,

*y*) and (

*x*,

*y*) -> (

*x*,

*y*+

*k*) are already proved to be translations. The first translation slides points along the

*x*-axis

*h*units, and the second slides points along the

*y*-axis

*k*units. But how are points

*not*on either axis transformed?

Now that we're in Euclidean geometry, we can prove that translations slide every point the same distance -- which is what we expect translations to do. We consider the points (0, 0), (

*a*, 0), (

*a*,

*b*), and (0,

*b*), and these points are the vertices of some quadrilateral. We know by definition, the side along the

*x*-axis has length

*a*and also the side along

*y*-axis has length

*b*. We also know that three of the angles are right angles -- the angle at (0, 0) since the axes are perpendicular, the angle at (

*a*, 0) because

*x*=

*a*is perpendicular to the

*x*-axis, and the angle at (0,

*b*) because

*y*=

*b*is perpendicular to the

*y*-axis. In Euclidean geometry, we know that the sum of the angles of a quadrilateral is 360, so if three of the angles are right angles, so is the fourth, and the quadrilateral is a rectangle.

But now we know the lengths of two sides of this rectangle,

*a*and

*b*, and we want to find the lengths of the other two sides. Of course, the other two sides must also have length

*a*and

*b*as opposite sides of a rectangle are congruent. Yet how do we know this? Most Geometry texts would now say that every rectangle is a parallelogram and the opposite sides of a parallelogram are congruent, therefore the opposite sides of a rectangle are congruent. But unfortunately, the U of Chicago text doesn't give the Parallelogram Tests and Consequences until the last part of Chapter 7.

But let's consider the Quadrilateral Hierarchy yet again. Now only is every rectangle a parallelogram, but every rectangle is an isosceles trapezoid in two different ways -- and we've indeed proved that the opposite sides (legs) of an isosceles trapezoid are congruent. Therefore, we really do know that the opposite sides of a rectangle are congruent.

What we've shown is that the distance from (

*a*, 0) to (

*a*,

*b*) is

*b*units, and that the distance from (0,

*b*) to (

*a*,

*b*) is

*a*units. So horizontal and vertical distance work exactly as we expect them too -- that is, we now know that the distance from (

*x*,

*y*) to (

*x*+

*h*,

*y*) is

*h*units, and the distance from (

*x*+

*h*,

*y*) to (

*x*+

*h*,

*y*+

*k*) is

*k*units.

So (

*x*,

*y*) -> (

*x*+

*h*,

*y*+

*k*) moves points

*h*units horizontally and then

*k*units vertically. But that still doesn't tell us that the composite of the horizontal and vertical translations is itself a translation. Let's instead try the following -- let

*P*(

*a*,

*b*) and

*Q*(

*c*,

*d*) be two points,

*P'*and

*Q'*be the images of

*P*and

*Q*under the first translation, and then

*P"*and

*Q"*be the images of

*P'*and

*Q'*under the second one.

Then as the first translation slides

*h*units,

*PP'*=

*QQ'*=

*h*, and as the second translation slides

*k*units,

*P'P"*=

*Q'Q"*=

*k*, and as all horizontal and vertical lines are perpendicular (by the same rectangle argument given earlier), both angles

*PP'P"*and

*QQ'Q"*have measure 90. Thus triangles

*PP'P"*and

*QQ'Q"*are congruent by SAS.

So by CPCTC,

*PP"*=

*QQ"*without requiring the Distance Formula. Also, we have that the angles

*P'PP"*and

*Q'QQ"*-- the angles

*PP"*and

*QQ"*make with the horizontal -- are congruent, meaning that

*PP"*and

*QQ"*are in the same direction. So the transformation maps every point the same distance in the same direction. Therefore the transformation is a translation. QED

Notice that this proof essentially assumes a sort of "Converse to the Two Reflections Theorem for Translations" -- the forward theorem asserts that if a transformation is a translation, then it moves every point the same distance and direction, and so the converse would say that if a transformation moves every point the same distance and direction, then it's a translation. But this converse is trivial to prove -- as soon as we have a point

*P*and its image

*P'*and say that every point is moved the same distance and direction as

*P*, then it's easy to find two mirrors to set up the translation. One of the mirrors can be placed at

*P*and the other at the midpoint of

*, both mirrors perpendicular to*~~PP'~~

*.*~~PP"~~

Also, notice we assume that just because the two angles are equal,

*P*and

*Q*must be moved in the same direction. (And technically, we assume that any horizontal line must be perpendicular to any vertical line.) Both of these can be proved using corresponding angles.

Still, this proof should be more intuitive than the other versions which require symbolic manipulation or formulas we haven't covered yet to prove.

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