 In this video I want to talk about four point geometry. So in our previous video we talked about three point geometry and we came up with an axiomatic system that produced exactly what we call three point geometry. This one is just another toy example. It's meant to be an example of a finite geometry. It's not the only geometry out there but I want to just give you three axioms, what we call the axioms of four point geometry and then we're going to prove some things about them, okay? So the first axiom here is that there exist exactly four points in four point geometry, which is why we give it the name. There's exactly four points there. Axiom two, this one we saw before. So with three point geometry clearly the axiom there was there was three points, now we have four points. Axiom two, this is actually identical to one of the axioms we had for three point geometry. It was also axiom two in that situation. It's verbatim, the same thing here. For each two distinct points, there exists a unique line that contains both of them. Now with three point geometry, we actually have four axioms. With four point geometry, we actually have three axioms, somewhat of an ironic twist there. But our third axiom is gonna tell us that each line is on exactly two points. And so looking here at this diagram, I claim this is a model of four point geometry which proves that then we have a consistent axiomatic system. The circles as usual represent points. So you see that we have exactly four points. So axiom one is satisfied. Axiom two says that if we take any pair of points, there's only one line that's incident to both of them. There's exactly one line. So if we look at the possible pairings, here's a pair of points. This is the only line that touches both of them. Another pairing would be this one. This is the only line that touches both. This is our last pair. And this would be the unique line that touches them. If we look at this pair, I should say it's the last pair. It's the last pair with the first point there. With this pair, we have that line. With this pair, we have that line. And then this is finally the last pairing. We have that one right here. And so between any two points, there's exactly one line that touches both of them. And I should also mention that axiom three, each line has exactly two points on it. So with this line, you have two points. With this line, you have two points. With this line, you have two points. With this line, you have two points. With this line, we have two points. And then the last line here, this one we have two points there as well. So this does in fact satisfy axiom three. I should mention that as we draw this diagram here, you notice that the lines in the middle cross, we are not suggesting that even though the lines cross, that there's a point of intersection. Our geometry only has four points, which are namely these four points right here. So even though our diagram has a crossing on it, there's no point there. So these two lines that you see crossing a mineral are actually parallel lines because they don't intersect anywhere. Now, if you wanted to, we could have redrawn the diagram so that you don't see any artificial crossings. So we could have drawn something like this. This is another way of drawing the diagram so that we don't see any crossing line segments. So we don't have the misconception that the lines intersect when they really don't. But I should also mention that some people don't like this diagram here because you're like, is this line straight? Because after all, it looks kind of wiggly. Middle here, I should mention that we don't have a definition of straight so far. We can talk about lines and geometry, but what does straight mean? I mean, if you had to define what straight is, what is straight? For which you could argue that straight is this, it's what a line does, it's the shape of a line, which is somewhat of a circular definition, right? Lines are straight and therefore straight things are lines. So the fact that this line is curved in the sort of usual Euclidean sense doesn't mean it's not straight in this four-point geometry sense because after all, lines are straight. But again, if you don't like that, we could have also drawn the diagram in the following manner. We could think of it as the vertices of a tetrahedron. Probably draw these ones as dashed lines. This one also would be a dashed line. Let me fix that. Probably draw it like this. So we could think of it as the vertices of a tetrahedron, but then of course you're kind of wondering like, well, it's not planar, right? It should be flat like a plane. But if you get, I go back into the topics of, well, what's a plane, what's flat mean? So these are all isomorphic models. And in fact, we'll see very quickly that all models of four-point geometry are isomorphic to this one. This one's a simple one to draw, so we're gonna stick with that. So let us prove some theorems about four-point geometry using these three axioms. And we're gonna see that essentially the proofs of these theorems are gonna lead to us that this is the picture we have to have. This is actually gonna provide to us that these three axioms form a complete system. There's only one model of isomorphism. And it's the one that was on the screen a moment ago. It's now disappeared, of course. So the first theorem, this is a theorem of just four-point geometry. I don't claim this is a theorem for general geometry, just four-point geometry. If two distinct lines intersect, then the intersection is exactly one point. So notice here, if, right? It's important to point out here that this geometry, unlike three-point geometry, it does have parallel lines. Notice that here's two parallel lines, here's two parallel lines. And like I mentioned earlier, these two lines that cross actually are parallel lines because there's no point that is on both of them. So it's important that we say the word if right here. If two line, distinct lines intersect, then the intersection is exactly one point. I'm gonna position this in such a way that we can actually see the axioms and the theorem simultaneously here. So how are we gonna do this? So we have to assume the if, right? If you prove it conditional, we assume the hypothesis and then have to prove the conclusion as well. So by assumption, the two lines have at least one point of intersection. And so our lines, maybe do something like this, there's one point of intersection on them by assumption because they intersect. Now, if there were more than one, that means there's at least two points of intersection. So we get a picture that looks something like this. Now in this situation, this would mean that these two points, let's call them x and y are here, the two points x and y would then have two lines which are incident to both of them. That's in violation of axiom two because given any two distinct points, such as x and y, there's a unique line that contains both of them. We now have two lines. And so this theorem is essentially just a theorem. It's just the contrapositive of axiom two. This is a theorem of four point geometry, but it was also a theorem for three point geometry because axiom two was valid in that geometry as well. The difference in three point geometry we removed the if statement and just said the intersection is exactly one point because in three point geometry, I actually made a statement that guarantees all lines intersect. That is in three point geometry, there were no parallel lines. We don't have that statement in four point geometry. So in four point geometry, if lines intersect, it has to be at a unique point. This type of picture that you see still on the screen here is a no, no, can't happen. Well, another theorem for four point geometry says that there's exactly six lines by assumption we know there are four points, but how many lines are there? Well, the model has six, but maybe other models could have different numbers. It turns out six is the right number. And it's basically, you have to draw the picture that we have on the screen before the model because by axiom one, there's four points. So we can position our points, something like this. Then we know that by axiom two, each pair corresponds to exactly one line. So there's one line that grabs both of them. So if you look at the pairings, there's this one, there's this one, there's this one, there's this one, and then the other pairings will be something like this. So if you think of the six lines, there's six possible pairs. That is to say, how many ways can you pair together, how many ways can you pair together the points? You're gonna get four choose two, which gives us six. That's where this number six is coming from. But wait a second, maybe some of these lines are actually the same line, right? Maybe there's one line that grabs these three points, right? Cause axiom two just says that given any pair, there's only one line that's incident to that pair of points. But it doesn't mean that every line necessarily has to have two points. Oh, but wait, axiom three told us exactly that, that each line is on exactly two points. So you can't have a line with three points, that's in violation to axiom three. So something like this green line on the screen is not a possibility. When you combine axioms two and axiom threes together, what that tells us is that every line is exactly two points and since every pair of points gives us a unique line, that pairs of points is identical to lines. There's no difference between them. And since there's four points, four choose two gives us six lines. So that gives us our second theorem here of four point geometry. There's two more that I wanted to prove here. So the first one, excuse me, of the remaining two, so this is our third theorem of four point geometry. Each point has exactly three lines incident to it. Well, remember for each, for by axiom two, each point X has a line containing it with each of the other points of geometry which by axiom one, there's four points total. So if we have one point in mind here, here's X. Well by axiom one, there are four points total, one of them is X, so there's three other points that are not X. By axiom two, given each of these pairs, there is a line between it. So given the four points, there's gotta be at least three distinct lines that are incident to point X here. And we know, of course, that these are gonna be distinct lines because they do harbor different points like so. Remember, no line contains, well every line contains exactly two points. So if they contain X and then one other point, that determines the line. These are three distinct lines. But why can't there be another line, right? What if there was some fourth line that passes through X? Well, the issue there is that by axiom three, this new line would have to contain another point. But we've already accommodated for all of those. One of them's X, what's the other point? It would have to be this one, this one, or this one by the fact there's only four points. So there can't be another line. Thus, we get that this point X is incident to exactly three lines. Now, as this point was chosen arbitrarily, this gives us all of the points. Now, one thing before we do the last theorem I kind of wanna point out to you is that when we were proving things about the FIFOs that turned out to be three point geometry, we had to do everything perfectly abstractly, just in words, just in logic and statements and such. But as we're making proofs about these geometries, they naturally lead to these illustrations that you're seeing right here. Now, I want you to be aware that these illustrations are not the proof. Proof by illustration is not a valid technique because in fact, our illustrations can be misleading. The point of the drawings here is to help provide intuition behind the logic that's happening here on the left side of the screen. The diagram just helps us understand what's happening over here. And this is why geometry is actually a really great place to have a second semester of proof writing. After you've done a course equivalent to math 3120 at SUU then you can actually take a class like math 3130, geometry for which we're proving stuff. The class is about writing proofs. This is a proofs class, but we can use the intuition of geometry to help us understand the proofs because in another place like topology, analysis, algebra or something, we might not have prior knowledge. We might not have much intuition and therefore geometry is one of the best places to develop our proof writing better. And so let's look at one last theorem here. And when you look at all four of these theorems together, you essentially have proven that the only model to four point geometry is the one we started with earlier, but let's look at this next one here. In four point geometry, each distinct line has exactly one line parallel to it. In fact, if L is a line and P is a point not on the line, then there exists exactly one parallel line, two L, which contains P. So we're actually gonna prove that second statement, the in fact statement, because if we get that, that proves the previous statement right there. So we're gonna take some line L and we're gonna take some point P that's not on L, something like this. And so we wanna prove what our goal here is, we're gonna prove there is a line that goes through it and it's parallel to L and it's gonna be the only one there is. And in particular, we'll say why there's exactly one line parallel to it, but we'll get to that in a moment. Well, okay, by axiom three, the line L contains two points. And I guess that before we continue on, I should say here with this hypothetical statement, if L is a line such and such, if P is such and such, this is a conditional statement, if then, right? If then, we assume the hypothesis, then we prove the conclusion. So we don't have to worry about whether the hypothesis is true or not, we assume it's true. Maybe it's a vacuous statement for the geometry, doesn't matter, we assume the hypothesis and then prove the conclusion from there. All right, so like I said, so we have this line L, we have this point P and we know that P is not on L. But because L is aligned by axiom three, there are exactly two points on that, we'll call them X and Y. And we know X and Y are distinct from P. We have that X doesn't equal P, we have that Y doesn't equal P, because if they were, that would apply P was on L because by construction, X and Y are on L. But by assumption, P is not on L. So we get that, okay, if X and Y have to be distinct because if they are the same, sorry, they're distinct from P because if they were the same, then they would be on L, which they're not. All right, so then by axiom one, we know that there's a fourth point in this geometry, let's call it Q, right? L only contains two points and P is not one of those two points. So if we counter for three of the four points in the geometry, this gotta be a fourth point, call it Q. And so it's different from P as well. By axiom two, there exists a line that contains both P and Q. Let's call that line M, like so. And so now M is the conjectured line to be parallel to L. Our diagram seems to suggest that, but is it parallel? Is there a point that's on L and M? If M intersected L, right? You know, we have a couple of possibilities. L intersect M, whoops, L intersect M. We've learned previously our first theorem that if they intersect, it's gotta be a single point. So our intersection is either X or the intersection could be Y or the intersection is empty, which of course in the last case that would imply they're parallel. So what about the first situation? What if X was on with the intersection between L and M? Well, that means X would then be on M itself for which X is not P, X is not Q, and M could only have two points. So X can't be on there, the same thing for Y as well. So by construction, we then see that M and L are parallel lines. They don't intersect each other whatsoever. And so given any line, and it doesn't matter which point you choose off of the line, we started with P, but if we actually started with Q, we could have done a similar construction here. And so this, in fact, proves the statement that each line has exactly one line that is parallel to it. And basically what you do is there's two points off of the line. You take the unique line that contains those two points off of the line, it'll give you your parallel line. So if we come back to our picture, yeah, this one will work here. We've said it already that if you have this line, here's its unique parallel line. If you take this line, it has its unique parallel line right here. And when you take this line, you have its unique parallel line right there as well. And so when you put all of these theorems together, it forces that our geometry has to look like this thing right here. And so we do in fact have a complete axiomatic system here. This is the only model of disomorphism that satisfies these three axioms because all of the theorems we've proven force us towards this picture right here. So this is what we call four point geometry. More specifically, we should probably call this four point affine geometry. We'll define what affine geometry means in a future lecture. Don't worry about it right now. But that brings us to the end of lecture three, which we've talked about three point geometry and four point affine geometry. In lectures four and five, we'll do some more examples of finite geometries. These are meant just to be toys, just to help us practice proving things about lines and points for these abstract geometries. So thanks for watching. If you learned anything, please hit the like button to these videos. If you have any questions, post them in the comments below and I'll be glad to answer them. And if you wanna see more videos like this in the future, please subscribe to the channel. I'll hopefully see you next time, everyone, bye.