 Welcome back to our lecture series Math 3130, Modern Geometries for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Mistledine. At the end of lecture seven, which just as a reminder in lecture seven, we introduced the three parallel alternatives, the elliptic parallel postulate, the Euclidean parallel postulate, and the hyperbolic parallel postulate. The most famous of all of those three is certainly Euclidean parallel postulate. At the end of lecture seven, we define the idea of an affine geometry, which the definition is now here on the screen. An affine geometry was defined to be an incidence geometry which satisfies the Euclidean parallel postulate. So to be an incidence geometry, you satisfy the four axioms of incidence. We have line determination, which means for any pair of points, there exists a unique line determined by that pair. C can't see, which guarantees that every line has at least two distinct points on it. Point existence, which says there's at least three points in the geometry. Non-colinearity, which tells us that not all points lie on the same line. Again, those four axioms give us an incidence geometry. And then if we equip to the incidence axioms, one parallel axiom, particularly the Euclidean parallel postulate, this creates for us the idea of an affine geometry, which reminder, the Euclidean parallel postulate gives us the uniqueness of parallel lines. For any line, L, and any point P that's not on the line, there exists exactly one line parallel to L, which contains the point P right here. And so in this lecture I should say, because there'll be more than one video, we're gonna formalize the idea of affine geometry into an axiomatic system. And we're gonna see that the affine geometry's theorems that we'll do in this video in the next will generalize that of young geometry. Because we've seen affine geometries before, our so-called four-point geometry we introduced earlier in this lecture series is really the four-point affine geometry. And young geometry is really the nine-point affine geometry. And so many of the theorems we've proved about four-point and young geometry will be identical and some will generalize it, of course. So the first one here, we saw this when we studied young geometry here. We called it Proclus Lema. And I gave it a name then because really this was a general theorem for all of affine geometry. And I wanna make mention that affine geometry is a very important study of geometries. It's a very broad setting. After all, Euclidean geometry R2 is an affine geometry. Clearly there's more structure to Euclidean geometry than just the affine structure, incidence, and parallelism. But it's very important when we study Euclidean geometry we understand its affine structure. In fact, if one goes to linear algebra, linear algebra has tons of affine geometry all over the place. You use words like affine a lot. And when you talk about linear algebra, affine transformations, affine spans, affine combinations. Because honestly, if you take any field, which of course a field for those who might not have the abstract algebra background, this is a collection of numbers for which we can add to track multiple and divide. And those operations follow the usual rules such as associativity, commutivity, distributive laws. We have identities, we have inverses, all that stuff. So if you have a field, if you take the square of a field you can actually make an affine plane with it. And so essentially every two dimensional vector space is an affine plane and in higher dimensions as well are affine geometries as well. But of course in this lecture series we'll look only at planar geometry. And so in particular, if your field is a finite field you can get finite affine geometries. Our four point geometry is a result by taking the field of two elements and squaring it. Young's geometry can actually be given a coordinate system by taking the field of three elements and squaring that as well. So again, just want you to be aware that this affine geometry can be found everywhere. So back to Proclas Lima. I'll restate it because there's a good chance we don't remember it when we studied Young's geometry here but this is a theorem of any affine geometry. Suppose we have lines L, M and N and let's suppose that L is parallel to M and that M intersects in. Then we claim that L intersects in right here. And so we need to prove we're gonna prove this one right here. That is, if these two lines are parallel and those two lines intersect then we guarantee that these two lines intersect. And this is gonna lead directly to our transitivity parallelism which we're gonna do in just a second. So this proof is gonna be identical. I want you to compare it. Maybe you have notes on this but this is gonna be identical to our proof we did for Proclas Lima in Young's geometry. There is no part of the proof that we have to change because essentially the proof only comes from really Euclidean parallel posh. So there is of course one important exception here but let's start the proof here. Let P be the point of intersection between M and N. So of course we know by assumption that M and N intersect each other. So let me draw a sketch of that. So we have our two lines M, M, M and N. Excuse me, we're gonna label them. We'll call this one M, we'll call this one N and by assumption they intersect each other. So let's call that point P. We claim that this point of intersection is unique. This is a theorem of incidence geometry but now that I've written it I wonder do we even need uniqueness there? We'll see it in just a moment. Well, we know that in an M are parallel to, or intersect each other, excuse me. We know that L and M are parallel. That's what I of course was trying to say. So let's add L to our diagram right here. Let's label it L. So we know that L and M are parallel but N and M intersect at the point P. So we need to prove that L and N intersect each other. So we're gonna do this by proof by contradiction. Let's suppose that L is parallel to N. And so that's what we're gonna do here by way of contradiction. We're gonna assume that L is parallel to N, right? Why is that a problem here? Well, notice here that we have the line L like so. And we have the point P, which we know that P is not on L. How do we know that? Well, P is on M and L and M are parallel. So P can't be on L. So we have a line and we have a point not on the line by the Euclidean parallel postulate. I want you to be aware of that. Then there should be a unique line parallel to L that passes through P. But we have them count them one and two. That's a violation of the Euclidean parallel postulate. So we get a contradiction there, which then means we must get the opposite of what we assumed. So we're gonna get that L is not parallel to N, which means L and N intersect each other. So this theorem follows really just from the Euclidean parallel postulate. Like I said, why we do have uniqueness in the intersection, we never actually needed that. It was a theorem of instance geometry. We mentioned it also for Young's geometry, but really it's a superfluous fact. All that matters here is the Euclidean parallel postulate. If the assumptions of Proclus lima were satisfied but the conclusion was not, then we would have multiple parallel lines, which is in violation of the Euclidean parallel postulate. So this is a theorem for any affine geometry because it really just needs the Euclidean parallel postulate, which Young geometry satisfied, four point geometry satisfied and thus general affine geometry satisfies Proclus lima. In fact, I'm not gonna prove this in this video. You can actually prove that N incidence geometry, Proclus lima is logically equivalent to the Euclidean parallel postulate, although EPP is what we will assume as the axiom and we prove Proclus lima from it. Related to this is the transitivity of parallelism that I referenced earlier, right? This is something we explicitly proved for Young geometry, but the proof there is transferable immediately to affine geometry. We don't have to change the proof whatsoever because this is a generalization of Young geometry. This is generalization of four point geometry just affine geometry. We have the four incidence axioms, we have Euclidean parallel postulate, that's all we need. Now transitivity of parallelism, remember this tells us that if two lines are parallel to the same line, they're parallel to each other. So we're trying to prove something like the following. If L is parallel to M and if let's say L is parallel to N, then we have that M is parallel to N. Now to write transitivity, we typically have to write this a little bit differently. We might say something like this, let's swap these elements right there. So we'd say that M is parallel to L and then L is parallel to N, then M is parallel to N. That's how transitivity of the relation is typically written. Now because parallelism is a symmetric relationship always you can swap the order of that, no big deal. And also by construction, parallelism is a reflexive relation. So if a parallelism is a, if it's a equivalence relation depends entirely on whether it's transitive and it turns out it'll be transitive only in affine geometry. Transitivity of parallelism is also something logically equivalent to the Euclidean parallel postulate inside of incidence geometry. We will prove it as a consequence of Proclus lima. Now, how do we do that? Well, let's take the assumptions here. Let's assume that M and N are both parallel to the line L like we have assumed right here. And then let's suppose the contradiction there approved by contradiction, we'll say not M parallel to N. So let's say they intersect each other. So let's draw our picture. Well, our picture is gonna look exactly like we saw a moment ago. Since M and N intersect each other, they would cross like that. L is parallel to both of them like so. And so what's happening here? We have our M, we have our N, we have our L. We see that because L and M are parallel and because M and N intersect each other then Proclus lima applies which with our numbering system that was 148 that then gives us that L and M must intersect which then contradicts the assumption here. So if we didn't have transitivity of parallelism we would contradict Proclus lima which is already been established. So therefore M and M have to be parallel to each other. That's an important consequence here. So look at these theorems here, the Proclus lima, transitivity of parallelism. These are the exact same proofs we used with Young's geometry because it was really a theorem affine geometry. We didn't have to change the proof whatsoever. Now in some situations, well basically this, everything that's true about affine geometry must be true for Young's geometry. It must be true for four point geometry because those are affine geometries. But there are some theorems that will be more, will be more specific to those geometries. Like for example, Young's geometry only has nine points. Every line in Young geometry has three points. Now that theorem won't be true for every affine geometry because in four point geometry lines only had two points. But what we can do is generalize those theorems to any affine geometry. So like we observed, four point geometry, every line had two points. For Young geometry, every line had three points. In general affine geometry, we can generalize that principle to the following. All lines contain the same number of points, which of course in four point geometry that number was always two. In Young geometry, that number was always three. This happens in general. And this is actually a very beautiful proof. One of the most important proofs I'd say for affine geometry, that all lines contain the same number of points. All right, so what we're gonna do to prove this is we have to take two arbitrary lines. So we're gonna take two lines, L and M. And what we're gonna then do is then prove, since these two lines were chosen arbitrarily, we're gonna prove that these lines have the exact same number of points. We can do that by establishing a bijection between the points on L with the points on M. A bijection is gonna be a one-to-one onto map, an injective and surjective map. This will establish a one-to-one correspondence between the points of L and M, which then will say that the set of points on L, compared to the set of points on M is the same number. This will even apply if we have an infinite number of points. So if one line has a cannibly many points, then the other line has to also have cannably many and one line has uncannable many points, then the other one also has to be uncannable. This even applies to infinite sets as well. So how are we gonna do this? Well, we have to take a starter point. So the line L has some point on it, we're gonna call that point P. It doesn't matter which point you choose, but by secance theme we know it has at least one point. Likewise, there has to be a point on M. And I should also mention that this point P is not on M, okay? Because it could be possible that L and M actually intersect each other. We don't claim that they're parallel lines, right? It could be that they intersect each other, but intersections and incidence geometry are unique. So there's only one point of intersection between the two lines if there's even one. Maybe they are parallel lines. But by secance theme there's a second point, then that second point can't be on both lines. So we do know we have a point on L that's not on M. And likewise, there has to be a point Q on M that's not on P by the same reasoning. So we have a point on one line, but not the other and on one line out the other. So we have these two points P and Q. So so far, of course, how we've done this, we've used secance to get two points online. We use line determination to guarantee that the intersection of lines is unique. We're gonna use line determination again, because we're going to form the line that connects P and Q. Now let me draw that again. Connect the lines P and Q. We're gonna call this line little n. Like so, it's the line uniquely determined by P and Q. Then for each point on L, and I'm gonna write this as an X right at the moment, because I want you to think of this as a variable, right? Because for every point on L, it's called X, and we don't know how many points there could be on L here. Like in four point jumps there's only two points and young there's three. But in Euclidean geometry, we're gonna have uncountably many points. These are all affine geometries, doesn't matter. Just take, consider any other point on L. By the Euclidean parallel postulate, there exists a unique line that's parallel to n that passes through X. And so consider that line for a moment here. The unique parallel line that contains X that is parallel to n. Let me make it look a little bit more parallel, at least parallel in the Euclidean sense. So we get something like that. We're gonna call this line n sub X. This is a line parallel to n, but passes through the point X, all right? Such a line is guaranteed as, by the Euclidean parallel postulate, it's the only one there is. And I should mention that it will, the line nX will intersect L, but that intersection will be unique. And so the only point that's shared between nX and L is gonna be this point X, okay? So now when you look at this over here, my diagram seems to suggest that the line nX should intersect M. But how do we know that? The thing is we're not drawing a picture of Euclidean geometry, although my diagram looks Euclidean at the moment. How do I know this intersection actually exists? Because when we draw a four point geometry, after all, we draw it like this and we have four points, one, two, three, four. And while there's an intersection in the diagram, that intersection actually has no point there. How do we not know we have the similar thing going on here? It's just, it's a misinformation from the diagram or something like that. Well, we have to have an argument here. There is in fact a point of intersection here and this follows by Proclus lima because notice what we have here. Whoops. We have that the lines nX and n are parallel to each other. We have that n and L intersect each other. Excuse me, that's not the one we wanna do. We have that n and m intersect each other. The direction I was going would show that L and nX intersect, but you already know that. But let's see, nX and n are parallel, but n and m intersect. Then by Proclus lima, we have to then have that nX intersects M somewhere. We are gonna call that point Y sub X like so. So there's some point of intersection and again as intersections are unique, there's only one point on nX and on M. And so we can in an unambiguous manner identify this point as YX. And notice we gave it the subscript Y sub X. And the reason we're doing that is we're connecting a point on L to this point Y sub X right there. So there's an identification going on. In all reality, this is going to be a function relationship. So we've established a function from the points on L to the points on M and it follows the rule that X is going to map to Y sub X by this construction. I'll note of course that if you take F of P, this is identified with the point Q because there's a unique line parallel to n that passes through P and that's n itself because n is considered parallel to itself. So just you're aware Q is the point Y sub P. I've always continued to call it Q right here. So we have this identification and I want you to be aware this is a well-defined map because given any point here, there's only one point on M it's gonna be identified to. And likewise, it's also going to be a one to one map. If you take any two points over here, there's only one X, I guess what I'm trying to say, if you take different Y coordinates, you're gonna have two different lines parallel to n associated to it. And so just give you two different X coordinates. It's gonna be a one to one map. It's also onto, how do I know it's onto? Well, we can kind of work the other way around, right? If we take the line n and we take any point on M then by EPP there's a line parallel to n that contains that point. By Proclas Lima, that point will intersect here somewhere on L and then that point will map to the starter point. So this map is in fact, bijective. You can also construct the inverse of this map by working backwards. Notice we started with points on X. We could have started points on M, excuse me, we started points X on L. We could have started with points on M and worked backwards. It has an inverse, that also proves it's a bijection. So that's the important thing here. This map, F, is in fact a bijection. It's one to one and onto. Which means that the cardinality of the domain is equal to the cardinality of the co-domain. For which what do I mean by the cardinality of a line here? I'm talking about the points on the line. And so this establishes that the number of points on L is equal to the number of points on M. It's a nice, cute little argument but this is one of the most fundamental arguments for affine geometry. All lines have the exact same number of points. Young's geometry had three points per line. Four point geometry had two points per line. And we're gonna see in the next video what are the consequences of this very important observation.