I wish you all a wonderful welcome back, and that you enjoyed your Thanksgiving break.

Today we begin Chapter 6, which is on transformations -- the heart of Common Core Geometry. We see that the first lesson, Lesson 6-1, is simply a general introduction to transformations. Of course, on this blog we've already covered reflections, rotations, and translations, since we jumped around.

Today we begin Chapter 6, which is on transformations -- the heart of Common Core Geometry. We see that the first lesson, Lesson 6-1, is simply a general introduction to transformations. Of course, on this blog we've already covered reflections, rotations, and translations, since we jumped around.

This lesson begins with a definition of transformation. Once again, I omit the function notation T(

*P*) for transformations since I fear that they'll confuse students. But I do use prime notation. The best way to demonstrate transformations is on the coordinate plane, so I do use them.
The book uses N(S) to denote the number of elements of a set S, an example of function notation. I include this question in the review, since the number of elements in a set (cardinality) is such a basic concept for students to understand.

Interestingly enough, in college one learns about these special types of transformations:

- A transformation preserving only betweenness is called a
*homeomorphism.*(Actually, this is based on continuity, not betweenness, but these are closely related.) - Add in collinearity, and it becomes an
*affine transformation*. - Add in angle measure, and it becomes a
*similarity transformation*. - And finally, add in distance, and now we have an
*isometry*.

The final question on my worksheet brings back shades of Jen Silverman -- the distance between lines is constant if and only if they are parallel in Euclidean geometry. I mentioned Silverman's page earlier when we were getting ready to learn about parallel lines.

As we work in Chapter 6 -- which counts as our Unit 5 -- we will revisit some of the transformations that's we've already studied. Of all the transformations covered in Chapter 6, the only one that we haven't studied yet is the

*glide reflection*-- the most complicated of the isometries.One question on the worksheet involves the transformation mapping (

*x*,

*y*) to (

*x*+ 2,

*y*- 3). The students can see that this transformation is clearly a translation. But so far, we haven't completed a

*proof*that the transformation mapping (

*x*,

*y*) to (

*x*+

*h*,

*y*+

*k*) is a translation.

It's easy to show that the transformation mapping (

*x*,

*y*) to (

*x*+

*h*,

*y*) is a translation -- where we define translation as the composite of two reflections in parallel mirrors, with the direction of the translation being a common perpendicular of the mirrors. In this case, we can let the

*y*-axis and the line

*x*=

*h*/2 be the two mirrors, with their common perpendicular the

*x*-axis. Likewise, it's trivial to show that the mapping (

*x*,

*y*) to (

*x*,

*y*+

*k*) is a translation in the direction of the

*y*-axis. What we want to show is that the composite of these two translations is itself a translation.

It seems as if it should be

*trivial*to show that the composite of two translations is a translation, but in fact it isn't. After all, a translation is the composite of two reflections in parallel mirrors, so the composite of*two translations*is also the composite of*four reflections*. It's not obvious why the composite of these four reflections is also the composite of two reflections in parallel mirrors -- especially if these two new mirrors have nothing to do with the four original mirrors. Of course, at some point we'd like to say something about translation*vectors*-- for example, the composite of two translations is a new translation whose vector is the*sum*of the original two vectors. But this would need to be proved.To understand the proof, let's use some notation that appears in the U of Chicago text. If

*m*is a line, then we let r_

*m*denote the reflection in mirror

*m*. In the text, the letter

*m*appears as a subscript, but this is hard to reproduce in ASCII, so we use r_

*m*instead. Meanwhile a small circle o denotes the composite, and so we write r_

*n*o r_

*m*to denote the composite of two reflections -- first the reflection in mirror

*m*, then the reflection in mirror

*n*.

The first thing we know about reflections is the Flip-Flop Theorem, which tells us that if a reflection maps

*F*to

*F'*, then it maps

*F'*to

*F*. That is, if r_

*m*(

*F*) =

*F'*, then r_

*m*(

*F'*) =

*F*. Therefore the composite of the reflection with itself must map every point to

*itself*. This transformation is often called the

*identity*transformation, and we can write it using the symbol I. So we write:

r_

*m*o r_

*m*= I

Since I is a transformation in its own right, we can compose it with other transformations. Of course, this is trivial -- the composite of I and any other transformation is the other transformation:

I o r_

*m*= r_

*m*

r_

*n*o I = r_

*n*

r_

*n*o I o r_

*m*= r_

*n*o r_

*m*

*Notice that composition of transformations is associative, but not commutative -- so in general, we have that r_*

*m*o r_

*n*is not the same as r_

*n*o r_

*m*. In fact, we can see how r_

*m*o r_

*n*and r_

*n*o r_

*m*are related by finding

*their*composite:

r_

*m*o r_

*n*o r_

*n*o r_

*m*= r_

*m*o I o r_

*m*

= r_

*m*o r_

*m*

= I

So r_

*m*o r_

*n*is the transformation which, when composed with r_

*n*o r_

*m*, yields the identity. This transformation has a special name -- the

*inverse*transformation. As it turns out, the inverse of any reflection is itself. But the inverse of a translation is a translation in the opposite direction, and the inverse of the counterclockwise rotation with magnitude theta is, quite obviously, the clockwise rotation with magnitude theta with the same center.

Notice that r_

*n*o r_

*m*, being the composite of two reflections, can't itself be a reflection -- it must be either a translation or rotation. So r_

*n*o r_

*m*can almost never equal its inverse r_

*m*o r_

*n*. But there is a special case -- notice that the inverse of a 180-degree rotation counterclockwise is the 180-degree rotation clockwise, but these are in fact that same rotation, since +180 and -180 differ by 360. So if r_

*n*o r_

*m*equals a 180-degree rotation, then r_

*m*and r_

*n*commute after all.

The last thing we need in our proof that the composite of two translations is a translation will be both of Two Reflections Theorems (one for Translations, the other for Rotations). The Two Reflections Theorem for Translations tells us that the direction and distance of a translation depend only on the common perpendicular and the distance between the mirrors. So if

*k*,

*l*,

*m*, and

*n*are all parallel mirrors, and the distance from

*k*to

*l*equals the distance from

*m*to

*n*, then r_

*n*o r_

*m*= r_

*l*o r_

*k*. And the same happens for rotations -- if

*k*,

*l*,

*m*, and

*n*are all concurrent mirrors, and the angle from

*k*to

*l*equals the angle from

*m*to

*n*, then r_

*n*o r_

*m*= r_

*l*o r_

*k*.

Notice that in some ways, we are actually performing a transformation

*on the mirrors*-- that is, given two mirrors

*m*and

*n*, we wish to transform them to

*m'*and

*n'*such that r_

*n*o r_

*m*= r_

*n'*o r_

*m'*. Or if we want to get very formal, if we have some transformation T such that:

T = r_

*n*o r_

*m*

*then we wish to find another transformation U such that:*

T = r_U(

*n*) o r_U(

*m*)

and the point is that the transformation U has nothing to do with the transformation T. Indeed, if T is a translation, then U can be

*any*translation. Likewise, if T is a rotation, then U can be

*any*rotation with the same center as T.

So now let's prove a simple case, that the composite of two translations is a translation. Our simple case will be when the two translations are in the same direction (or in opposite directions). This means that the mirrors

*k*,

*l*,

*m*, and

*n*are all parallel, and we wish to prove that:

r_

*n*o r_

*m*o r_

*l*o r_

*k*

*is a translation. Since all the mirrors are parallel, in particular*

*m*| |

*n*. By the Two Reflections Theorem for Translations, we're allowed to slide the mirrors themselves. We can slide mirrors

*m*and

*n*to two new mirrors,

*m'*and

*n'*, such that

*m'*| |

*n'*with the same distance between them:

r_

*n*o r_

*m*o r_

*l*o r_

*k*= r_

*n'*o r_

*m'*o r_

*l*o r_

*k*

*How does this even help us at all? That's easy -- we slide*

*m*until its image is exactly

*l*! Then we have:

r_

*n*o r_

*m*o r_

*l*o r_

*k*= r_

*n'*o r_

*m'*o r_

*l*o r_

*k*

= r_

*n'*o r_

*l*o r_

*l*o r_

*k*

= r_

*n'*o I o r_

*k*

= r_

*n'*o r_

*k*

*And so we've done it -- we've reduced four mirrors to two. And we know that*

*n*is parallel to its translation image

*n'*and

*n*is parallel to

*k*as all the original mirrors were parallel. So

*k*| |

*n*' -- that is, r_

*n'*o r_

*k*is the composite of reflections in parallel mirrors. Therefore it is a translation. QED

So now we see our trick to reduce four mirrors to two -- we transform the mirrors a pair at a time, using the Two Reflections Theorems, until two of the mirrors coincide. Then the composite of a reflection with itself is the identity, which disappears, leaving two mirrors behind. But when we transform the mirrors, we must be careful to transform the

*correct*mirrors. When we have:

r_

*n*o r_

*m*o r_

*l*o r_

*k*

*we may transform*

*k*and

*l*together, or

*l*and

*m*together, or

*m*and

*n*together. But we can't transform, say,

*k*and

*m*together, or

*l*and

*n*, or

*k*and

*n*, since these aren't listed consecutively in the composite.

Let's finally prove that the composite of a horizontal and a vertical translation is a translation. We begin by writing:

r_

*n*o r_

*m*o r_

*l*o r_

*k*

*where*

*k*and

*l*are horizontal mirrors (for the vertical translation), and

*m*and

*n*are vertical mirrors (for the horizontal translation). We notice that

*k*,

*l*,

*m*, and

*n*form the sides of a rectangle.

Now we begin transforming the mirrors. First, we notice that

*l*and

*m*are perpendicular (since

*l*is horizontal, while

*m*is vertical). As it turns out, reflections in perpendicular mirrors commute -- this is because by the Two Reflections Theorem for Rotations, the angle of rotation is double the angle between the mirrors, and since the angle between the mirrors is 90, the rotation angle is 180, which means that they commute. If you prefer, we could say that instead of commuting, we're actually

*rotating*the two mirrors 90 degrees. The rotation image of

*l*is

*m*, and the rotation image of

*m*is

*l*. In either case, we obtain:

r_

*n*o r_

*m*o r_

*l*o r_

*k*= r_

*n*o r_

*l*o r_

*m*o r_

*k*

*Now we rotate the perpendicular pairs*

*k*and

*m*together, and

*l*and

*n*together. We rotate the pair

*k*and

*m*until the image of

*m*is concurrent with

*l*and

*n*, and then we rotate the pair

*l*and

*n*until the image of

*l*is concurrent with

*k*and

*m*:

r_

*n*o r_

*m*o r_

*l*o r_

*k*= r_

*n*o r_

*l*o r_

*m*o r_

*k*

= r_

*n'*o r_

*l'*o r_

*m'*o r_

*k'*

*Now*

*l'*and

*m'*are the same line -- the diagonal of the original rectangle. Let's call this new mirror

*d*to emphasize that it's the diagonal of the rectangle:

r_

*n*o r_

*m*o r_

*l*o r_

*k*= r_

*n*o r_

*l*o r_

*m*o r_

*k*

= r_

*n'*o r_

*d*o r_

*d*o r_

*k'*

= r_

*n'*o I o r_

*k'*

= r_

*n'*o r_

*k'*

*Even though*

*d*disappears, notice that

*d*is still significant. Since both rotations preserved angle measures, we have that

*k'*and

*n'*are both perpendicular to the diagonal

*d*. So, by Two Perpendiculars Theorem,

*k'*| |

*n'*. So we have the composite of reflections in parallel mirrors -- a translation. QED

Even though we've already given a parallel postulate, this proof is valid in neutral geometry. Yes, the four mirrors are said to form a rectangle, but notice that the only perpendicular relationships needed are

*k*and

*m*,

*m*and

*l*, and

*l*and

*n*. So the four mirrors need only form a Lambert quadrilateral. We must make sure to choose the mirrors carefully as to avoid making the fourth angle a right angle. The following assignment will work:

*k*: the line with equation

*y*= -

*k*/2

*l*: the

*x*-axis

*m*: the

*y*-axis

*n*: the line with equation

*x*=

*h*/2

But unfortunately, attempting to generalize this proof to the case when the directions of the two translations aren't perpendicular fails even in Euclidean geometry. Notice that the four mirrors now form merely a parallelogram, not a rectangle. But we know that in Euclidean geometry, the angles from

*k*to

*m*and from

*l*to

*n*are congruent because they're opposite angles of this parallelogram. After performing the rotations, the congruent angles have rotated into alternate interior angles formed by the lines

*k'*and

*n'*and the transversal

*d*, which would mean that

*k'*and

*n'*are parallel. This may look appealing, but the problem is that the original proof commuted

*l*and

*m*, which is invalid unless

*l*and

*m*are perpendicular.

I'm not sure how to fix the proof that the composite of two translations is a translation, but luckily, we only need the perpendicular case to prove that mapping (

*x*,

*y*) to (

*x*+

*h*,

*y*+

*k*) is a translation.

Now here's the thing -- at some point this week, I wish to have students perform translations on a coordinate plane. I could give students this proof that mapping (

*x*,

*y*) to (

*x*+

*h*,

*y*+

*k*) is a translation, but look at how confusing this proof appears! It's highly symbolic and may be too abstract for students -- just look at lines like:

T = r_U(

*n*) o r_U(

*m*)

What in the world does this mean? It means that the lines which are the

*images*of one transformation, U, are the

*mirrors*(not the

*preimages*, but the

*mirrors*) of another transformation, T. If a student was to forget that U is a transformation, he or she might think that U is a line, and so r_U(

*m*) would mean the image of the line

*m*reflected in the mirror U. Notice that on a printed page rather than ASCII, it would be obvious that the subscript is all of U(

*m*), not just the letter U.

As we've discussed so often recently, traditionalists like symbolic manipulation. If they could see the symbolic manipulation involved with this, they might realize that transformational geometry isn't just hand-waving but is actually rigorous mathematics. But high school students will be confused if I were to give a proof as symbolic as this one, so I'd replace the symbols with words -- yet as soon as we did this, it will appear to the traditionalists that transformations are mere hand-waving yet again.

We might point out that verbal descriptions might be more understandable to high school students, while symbols for transformations might be better suited at the college-level. But this is when a traditionalist might say,

*fine*, then let's save transformations for college-level math and teach high school students only math for which they

*can*understand the symbols! In particular, a traditionalist would look at this proof and say one of two things. The first would be that students shouldn't be learning about translations anyway, so the mapping (

*x*,

*y*) to (

*x*+

*h*,

*y*+

*k*) is irrelevant. The other would be that if students really need to know that the mapping (

*x*,

*y*) to (

*x*+

*h*,

*y*+

*k*) is a translation, then they should prove it by using the Slope and Distance Formulas and not by playing with mirrors. (Our problem is that we can't use the Slope and Distance Formulas until we've reached dilations and similarity.)

Indeed, I've noticed that some college geometry texts use the notation

*A*-

*B*-

*C*to denote a very simple concept -- namely that point

*B*is

*between*

*A*and

*C*. Often I wondered why no high school Geometry text ever uses this

*A*-

*B*-

*C*notation, but now I realize why -- the

*A*-

*B*-

*C*notation is considered too abstract for high school students to understand.

None of this has anything to do with today's lesson, so I'll make the decision regarding how to present the proof when we get there. Meanwhile, if traditionalists really want to "privilege the symbol," why don't they begin by enforcing

*A*-

*B*-

*C*notation in high school Geometry and see how far they get?