 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 Mislide. Lecture 7 has two main goals. The first of which is to finish developing the theory of incidence geometry, which we began in lecture 6. The second half is that we're going to introduce the so-called parallel axioms, the parallel postulates, parallel alternatives. That discussion will actually take place in the second video for lecture 7. This first one, again, is just developing the theory of incidence geometry. You which we'll recall in lecture 6, we introduced the four axioms, the four axioms that David Hilberg gave for incidence between lines and points. We gave them names line determination, which says that between any two points, there exists a unique line, so two points determine a line. The second one was secancy. Secancy says that every line has at least two points. Point existence gave us that there exists at least three points in the geometry, and the non-colonarity says that not all points are on the same line. Now, we developed a long list of theorems of incidence geometry using axioms of line determination, point existence, and non-colonarity. We haven't used the secancy axiom yet, and I promise you we do it in this video. Well, a slight JK on that one, that we're going to do it on the next theorem. This theorem probably could have got squeezed onto the previous video, but I didn't want to make it too long, so I wanted to break it where we are now, and this break is the same natural break that one would usually have in a face-to-face lecture that we do all of this in 50 minutes, just like so. So let's remind ourselves where we are. We've proven that non-colonial sets exist in incidence geometry, that I can give in two points, or one points, or no points. I can always form a set of three non-colonial points. What we're trying to do is develop the same idea for non-concurrent lines. That is, we don't want to have something like this where every line goes through the exact same point in our geometry. We don't want that. We want that the set of all lines is non-concurrent. We're working towards that. So the last thing we did in the previous video, just as a reminder, we proved that if there's a point, then there's a line not on that point. So this is like the dual arguments to the non-colonarity axiom. We've also proven that for each point, there exist at least two lines that are incidence to it. So what we want to do now is this first theorem, we're going to prove really line existence. We're going to prove that there exist at least three lines. The point existence axiom says there's at least three points. Why are there at least three lines? Well, basically we're going to come up with a set of three non-colonial points. We can't start with point existence because point existence says there's three points, but those points could be collinear and we only get one line. That would give us at least one line, but we want to show there's at least three lines. So by theorem 148 from the previous lecture, which told us that there exists a set of three non-colonial points, we have three points, call them P, Q and R, and then using these three points which we know are non-colonial line determination says there's a unique line between P and Q, there's a unique line between P and R, and there's a unique line between R and Q. None of these lines can be the same line because if any of them were the same line, then that means all three points would have to be on the same line, they'd be non-colonial. So that would violate theorem 148 in our lecture series. Theorem 148 required point existence, it required non-coloniality and it required line determination. This theorem also uses line determination, so we've used those three axioms and this proves the dual of point existence. This gives us line existence, there's at least three lines in our geometry. So, you know, a little bit of a fib, we didn't use secant C yet, but that'll be remedied in the next theorem. And this is the last major theorem. We're gonna do, there's like a short corollary that's gonna follow after this, which is our main target, but this is the last heavy lifting we have to do. If L is any line, then there exist lines M and N such that L, M and N don't have the same intersection. That is, they're non-concurrent. So basically the picture we're trying to find is the following. If you have a line L, there's gonna be other lines that form something like this, that they're not concurrent to it. And this is finally where secant C is gonna come into play. So we're starting off with a line. And I guess I should mention that this statement we're trying to prove right now is very similar to a statement we proved previously. In fact, this statement we're trying to prove is essentially the dual statement to theorem 147. One force theorem, theorem 147 we proved in the previous video says that if P is a point, then there exist points Q and R so that P, Q and R are non-colonial. We wanna do the same thing with lines and non-concurrency. All right? So imagine we have our line L like so. So this would be our line L. Now finally, finally, finally, we're gonna use the secant C axiom just because we postponed it doesn't mean it's not important. We developed all the previous theory without secant C but to get this, to be able to switch from points to lines we do need secant C here. So by secant C, we have that L contains at least two points on it. These points will be distinct. We'll call these points P and Q. Now by theorem 1.6, which in the previous lecture theorem 1.6 said if you have two points P and Q that are distinct, there exists a third point R such that the set P, Q and R are non-colonial. So in particular, R is not on L because P, Q and R are not collinear. Now using line determination, there exists lines between, a unique line between P and Q, excuse me, P and R and a unique line between Q and R like so. Now this feels like the set of points I'm looking for. So let's call these lines. Do I give them names? I think I do. There is a line M. M is the line that's between P and R and then N is the line between Q and R. So I'll stick with that using my, the same notation that theorem I hear. I claim that these lines are non-cocurrent. Is there a point that's on all three of these lines? Call that point S. So let's say that S is on the intersection of L with M with N. Is there such a point? Well, if there was a common point S incident to all three lines, then that would mean that S would have to be distinct from P, Q and R because as P, Q and R are not collinear, there's no line that contains, well, I should say, but these three lines do not contain, they don't all contain P, they all contain R, they don't contain Q. So S is a fourth point, that's important to note here. Then the lines L and M would have two distinct intersections, right? So if we look at L, P is on both of them and so is S. So we get that P and S, both are on the intersection of L and M. And as we proved with theorem 144, which is basically just line determination, that the intersection between two lines must be at most one point. So if P, since S is not P, there would have to be a second point of intersection, that's a contradiction. So there's no point that's on all three of these lines. So these three lines are non-concurrent. And then as a corollary of that theorem right there, there exist three non-concurrent lines. So the previous theorem says that if we have a line, I can construct two other lines which are not concurrent to it. So how do we get lines? Well, line existence says there is a line. So we have a line L and then the previous theorem is applied for which then we are done. And so this then gives us three lex, or three, excuse me, three theorems of instance geometry in this video. The previous video gave us, let's see, I'll just count them real quick. One, two, three, four, five, six, seven theorems. So plus three is 10. Between the two videos, we've now proven 10 theorems of instance geometry. And honestly speaking, that's about as far as we can go. There's a few other things you can prove for which for my students, I'll ask you to explore some of these things in the homework, but I really can't push the envelope much farther than we are right now because instance geometry is a very broad axiomatic system. And this is always the concern one has to take. Do I want diversity or do I want lots of theorems, right? Do you want lots of models or do you want lots of theorems? The more theorems you have, the fewer models you have, but the more models you have, the fewer theorems you have. When it comes to Euclidean geometry, we have tons of theorems, tons of things we can say about Euclidean geometry because there's only one Euclidean geometry up to isomorphism. But for instance geometry, there are so many, so many different instance geometries, but as a consequence, there's only so many things we can say about them. So in some essence, we've completed our theory of instance geometry because we've basically said, well, we can't. What are the broad theorems that cover everything about instance geometry? That's not to say that we're done talking about instance geometry. What that really means is that we're just gonna stack more things on top of it. We're gonna stack more axioms onto instance geometry to get more restrictive geometric structures, but all of those geometries will retain these 10 theorems we've developed right here and any other theorems of instance geometries we haven't talked about. So in the next video, like I said, we're gonna talk about parallel alternatives. So the incidence axioms don't say anything about do parallel lines exist or not. And when we look at phano geometry, it had no parallel lines. Young geometry did. So it appears that parallelism is independent of the incidence axioms. So if we want to say things about parallel lines, we are going to need another axiom that says something about parallel lines. And that's exactly the target of our next video.