Section 6-2 of the U of Chicago text covers translations, the last of the three major transformations. In some ways, the slide is the easiest for the students to understand.
I've mentioned earlier the reasons for saving translations for last. Translations, unlike reflections or rotations, are dependent on the Fifth Postulate. The U of Chicago's flawed proof of the Two Reflection Theorem for Translations omits reference to the Fifth Postulate, but rewriting this proof in two-column form makes the need for the Fifth Postulate evident:
Definition of translation:
A translation (or slide) is the composite of two reflections over parallel lines.
Two Reflection Theorem for Translations:
If m | | l, the composite of the reflection in l followed by the reflection in m is a translation that slides figures twice the distance between l and m, in the direction from l to m perpendicular to those lines.
Given: j | | k
A reflected over j is A'
A' reflected over k is A"
Prove: Line AA" perpendicular to both j and k
AA" is double the distance between j and k
(For the proof, let x be the distance from A' to j and y be the distance from A' to k.)
1. A reflected over j is A' 1. Given
A' reflected over k is A"
5. Lines AA', A'A" identical 5. (Our) definition of parallel
6. A, A', A" collinear 6. Definition of collinear
7. Line AA" perp. j and k 7. Substitution
8. dist. from j to k is x + y 8. Definition of distance between parallel lines (yesterday's exercises)
9. AA' = 2x, A'A" = 2y 9. Definition of distance from point to line & Reflections preserve dist.
10. AA" = 2(x + y) 10. Betweenness Theorem & Distributive Property
The text states that the proof is similar if A is located in a different location relative to j and k -- it turns out that the distance between the parallel lines may be x - y or y - x rather than x + y. But our emphasis is on line 3. We can't use the Two Perpendiculars Theorem -- despite what the text writes -- unless the two lines are perpendicular to the same line. Here
It's interesting to see what translations look like in other forms of geometry. In one-dimensional geometry, there are only two isometries -- reflections and translations. Think about it, if you have a line, you can only slide the line or flip it over. You can't turn the line without a second dimension.
A translation is in hyperbolic geometry is as composite of two reflections over ultraparallel lines -- that is, lines with a common perpendicular. This makes sense since, according to the Two Reflection Theorem for Translations, the direction of a translation is given by that common perpendicular.
In spherical geometry, there are no parallel lines and therefore no translations! This may seem strange, since on the spherical earth one can apparently slide (translate) objects. But as it turns out, what appears to be a translation is really a rotation (whose center is very, very far away). This is true even if the "translation" is along a great circle. Recall that transformations must map every point in the entire plane (or sphere in spherical geometry), not just a single object. Sliding an object east or west along the equator (a great circle) is actually a rotation centered at the poles -- they are the fixed points of the rotation (in elliptic geometry, the poles are a single point), and the degree of the rotation is simply the change in longitude of the object. True translations don't have a fixed point.
According to the text, every translation has a direction and a magnitude. So now some people may be expecting to see the word vector appear here -- since vectors also have direction and magnitude. But instead of using vectors to define translations, the U of Chicago text uses translations to define vectors (in Section 14-5)!