The section, titled "Uniqueness," begins by discussing Euclidean geometry. It discusses the five geometric postulates of Euclid, and states and proves the Uniqueness of Parallels Theorem -- also known as Playfair's Parallel Postulate. It then refers to non-Euclidean geometry, in which Playfair doesn't hold.
Notice that we've already covered this topic during the first semester. On this blog, we began by accepting Perpendicular to Parallels as our version of the Fifth Postulate. We used this postulate to prove Playfair, and then used Playfair to prove the Parallel Consequences. Therefore, we're already done with this part of Section 13-6.
I've briefly discussed the concept of non-Euclidean geometry a few times before. But I admit that this is still a very confusing concept. Last week, there was a question on my Mathematical Calendar -- you know, the one that Theoni Pappas publishes nearly every year -- about non-Euclidean geometry:
In elliptic geometry the number of lines passing through two distinct points is:
(5) infinitely many
Because this question appeared on February 5th, the intended answer must be choice (5). (I would've actually posted this last week on the 5th, except that I wanted to spend more time preparing a worksheet for my geometry student. This time, I had an entire three-day weekend to prepare the worksheet for today.)
But choice (5), under the usual interpretation, is wrong. There are not infinitely many lines passing through two points in the version of non-Euclidean geometry known as elliptic geometry! Usually, we model elliptic geometry using a sphere, with great circles as the lines. Now there are indeed infinitely many great circles passing through the North and South Poles -- these are the meridians.
Yet there are two problems here. The first is that there are infinitely many great circles passing through two points if and only if these points are antipodal -- that is, if these two points are directly opposite on the sphere, like the North and South Poles. If the points aren't antipodal, then there's only one great circle through the two points. The second problem is that elliptic geometry differs slightly from spherical geometry in that antipodal points count as a single point! And so technically, the correct answer is (1). In elliptic geometry, there is exactly one line passing through any two distinct points -- just like in Euclidean geometry!
Of the two types of non-Euclidean geometry, hyperbolic geometry is actually more straightforward than elliptic geometry -- despite it being much easier to visualize a model of the latter (a sphere) than the former. Hyperbolic geometry satisfies the first four postulates of Euclid, and the fifth fails -- there are infinitely many lines parallel to a given line through a given point.
But with elliptic geometry, it's more difficult to state which of Euclid's postulates fail. Nearly any of the postulates can be said to hold or fail in elliptic geometry (including the fifth!), based on how one interprets them.
For example, Euclid's first postulate can be interpreted as "two points determine a line." My argument above, in which (1) is the correct answer, implies that this postulate is true. But let's look at how the U of Chicago interprets this postulate in Section 13-6:
Postulates of Euclid:
1. Two points determine a line segment. [emphasis mine]
If a line is a great circle, then a segment is just an arc of that circle. And so, given two points on the sphere, there are at least two such arcs between the two points -- the short way and the long way around the sphere. This would make the first postulate false, and (2) the correct answer.
Postulates of Euclid:
2. A line segment can be extended indefinitely along a line.
This postulate could be considered false, since a great circle is finite. But one could consider continuing to go around the circle, forming arcs that are greater than 360 degrees. In this case, this would make the second postulate true. In fact, one could even imagine counting all of the arcs between two points that differ by multiples of 360 degrees as distinct segments. This is the only way to make (5) the correct answer -- perhaps this is what Pappas had in mind.
Notice my claim that even the fifth postulate can be considered true in elliptic geometry. Once again, let's look at how we can interpret the fifth postulate. If we take Playfair as stated in Section 13-6:
Uniqueness of Parallels Theorem (Playfair's Parallel Postulate):
Through a point not on a line, there is exactly one [line] parallel to the given line.
then this statement is clearly false in elliptic geometry, where there are no parallel lines. But let's look at how Dr. Franklin Mason states Playfair:
The Playfair Postulate (as stated by Dr. M):
Through a point not on a given line, there's at most one line parallel to the given line.
Those two extra words, at most, allow for the possibility of zero parallel lines. And so Playfair, as stated by Dr. M. is true in elliptic geometry!
Next, let's look at our version of the Fifth Postulate:
Fifth Postulate (as stated on this blog):
In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.
If l is perpendicular to m and m is parallel to n, then l is perpendicular to n.
Notice that there are no parallel lines, so there can be no m and n -- or can there? Recall that using our definition of parallel, any line is parallel to itself! Therefore, in elliptic geometry, we can say, "two lines are parallel if and only if they are identical" -- that is, "parallel" is just another word for "identical" in this geometry. (We've seen that in both elliptic and hyperbolic geometry, "similar" is just another word for "congruent.")
So the Fifth Postulate, in elliptic geometry, becomes:
If l is perpendicular to m and m is identical to n, then l is perpendicular to n.
or, since m is identical to n, we can substitute m for n, to get:
If l is perpendicular to m, then l is perpendicular to m.
This is a tautology -- that is, it's automatically true! If we had to prove it in two-column format, it would contain only one step -- the Given step!
Even the fifth postulate as written in Section 13-6 may be true:
Postulates of Euclid:
5. If two lines are cut by a transversal, and the interior angles on the same side of the transversal have a total measure of less than 180, then the lines will intersect on that side of the transversal.
In elliptic geometry, the conclusion is automatically true, as any two lines will intersect. And if the conclusion of a conditional is true, the conditional itself is true. The only problem is the phrase "on that side." Any two lines will intersect at two antipodal points -- one on each side. But in elliptic geometry antipodal points are in fact the same point. Technically, we can't actually define "side of a line" because of this is elliptic geometry, but we can in spherical geometry. Each line (great circle) divides the sphere into two hemispheres -- the two sides of that line.
This is what makes elliptic geometry different from hyperbolic geometry. In hyperbolic geometry, the first four postulates are true and the fifth is false, but in elliptic geometry, we can't even say which postulates are true until we interpret them. Thus the so-called "neutral geometry" -- the theorems of geometry that only require the first four postulates to prove -- end up proving the theorems of both Euclidean and hyperbolic geometry, but not elliptic or spherical geometry.
During the extra time after the PARCC and SBAC exams, I hope to provide some actual worksheets on non-Euclidean geometry. My plan is to focus on spherical (not elliptic) geometry, since this is the easiest to understand and motivate by considering the shape of the earth. In my interpretation, the second through fifth postulates will be true, but the first postulate will be false. My plan is to replace our Point-Line-Plane postulate with a replacement.
Stay tuned for more details on our spherical geometry in the upcoming months. But tying this back to the inequalities of Chapter 13, we ask, are these true in non-Euclidean geometry? As it turns out, the Triangle Exterior Angle Inequality (TEAI) holds in hyperbolic, but not spherical, geometry. (Of course, the Exterior Angle Equality holds only in Euclidean geometry.) Because of this, all of the inequalities that we proved last week and this week fail in spherical geometry, although they do hold in certain cases -- for example, the Triangle Inequality holds provided that we always measure the short way around the sphere and never the long way.
(Recall that Dr. M originally proved TEAI using only triangle congruence, not TEAE. He has since changed his site so that TEAI is proved simply using TEAE, just like Glencoe and U of Chicago.)
Now let's get back to Section 7-8 of the U of Chicago text and the SAS Inequality in preparation for my tutoring session with my geometry student tonight. As I mentioned before, this section gives a proof of the SAS Inequality using the Triangle Inequality. As usual, I have decided to convert the proof to two-column format. The figure accompanying the proof gives two triangles, ABC and XYZ.
SAS Inequality Theorem:
If two sides of a triangle are congruent to two sides of a second triangle, and the measure of the included angle of the first triangle is less that the measure of the included angle of the second, then the third side of the first triangle is shorter than the third side of the second.
Given: AB = XY, BC = YZ, angle B < angle Y
Prove: AC < XZ.
1. AB = XY, etc. 1. Given
2. exists isometry T s.t. A'B' is 2. Definition of congruent
XY, C' same side of
3. C'Y = ZY 3. Isometries preserve distance
4. Let m symmetry line C'YZ 4. Isosceles Triangle Symmetry Theorem
5. m perp. bis.
6. QC' = QZ 6. Perpendicular Bisector Theorem
7. A'C' < A'Q + QC' 7. Triangle Inequality
8. AC < XQ + QZ 8. Substitution
9. XQ + QZ = XZ 9. Betweenness Theorem (Segment Addition)
10. AC < XZ 10. Substitution
We see that the proof is similar to that of SAS Congruence Theorem, except that this isometry puts C' on the same side of
Both Dr. M and Glencoe state a converse to SAS Inequality -- Glencoe calls it SSS Inequality. Dr. M hints at this proof -- we can use the same strategy that we used to derive Unequal Angles Theorem from its converse, Unequal Sides Theorem. We prove it indirectly:
SSS Inequality Theorem:
If two sides of a triangle are congruent to two sides of a second triangle, and the third side of the first triangle is shorter than the third side of the second, then the measure of the included angle of the first triangle is less that the measure of the included angle of the second.
Given: AB = XY, BC = YZ, AC < XZ.
Prove: angle B < angle Y
Assume not. Then angle B is either less than or equal to angle Y.
Case 1: angle B = angle Y. Then triangles ABC and XYZ are congruent by SAS Congruence, and so AC = XZ, a contradiction.
Case 2: angle B > angle Y. Then AC > XZ by SAS Inequality, a contradiction.
In either case we have a contradiction of AC < XZ. Therefore angle B < angle Y. QED
For Euclid, the SAS Inequality is his Proposition 24. Dr. M's first proof is based on Euclid:
and the converse, the SSS Inequality, is Euclid's Proposition 25: