 Welcome back to our lecture series math 3130 modern geometries for students at Southern Utah University. As usual I'll be a professor today Dr. Andrew Missildine. In lecture 11 we're going to continue our discussion of the betweenness axioms which we introduced in lecture 10. Now you'll recall that we said that an incidence geometry which satisfies the four axioms of betweenness is what we called an ordered geometry. Why is it called ordered? What does that order have to do? And so in lecture 11 we want to get through the very technical argument that every line in an order geometry can be given an ordering. Now before we do that in this video we're going to prove the idea of linear density. That is if you take any two points on a line you could always find a bunch of other points related to them with well between this relationship. So more specifically if we have a line L and we have two distinct points on that line call them B and D and this of course is an ordered geometry. Then we can always find three other points A, C and E on the line L such that the point A has B between A and D. The point C will be between B and D and the point E will satisfy the relationship that D is between B and E. So let me try to draw an illustration to illustrate what we're thinking of right here. We have our line L, we have some points B and D like so. Here's L, here's B, here's D. So the linear density theorem tells us we can always have a line, excuse me, we always have a point A so that B is between A and D. We always have a point E so that D is between B and E and we always will have a point C such that C is between B and D. Okay? Now the existence of some of these points are quite trivial. Let's start with A for example. By using the extension axiom which was one of the between this axiom we get that there's going to be a point on the line such that A, there's a point A on the line such that in particular D dash B dash A holds. We also have by of course the symmetric axiom that this is the same thing as saying A dash B dash D which is what we're trying to start off with. So the existence of the point A is given it basically immediately by extension but we're also using the symmetric axiom there. The existence of the point E also follows immediately from extension so we have this point B dash D dash E. Alright so A and E are pretty easy to come by. Why do we have the point C? That one's a lot harder to explain so that's going to be the bulk of this argument right here. So to find the point C we are going to use the extension axiom repeatedly but we have to be a little more clever about it. So to start off with we're going to have to find a point F that lives off of the line L. Now the existence of this point is a consequence of incidence axioms. Given any line in the geometry we can always find a point off of that line called that point F. And then consider the line, let me try that again, let's consider the line that goes between F and B like so. For convenience we will call this line M. Okay? Well apply the extension axiom to the line M right here. There has to be a point, let's call it G, that extends the line from B in the direction of F. Again, extension right there. And so then we want to consider the line that connects G to the point D. Kind of missed that one. Let me try that one more time. By line determination and axiom of incidence there is a line that is determined by the points D and G. We're going to call that line N. Again just for convenience here. And then by using extension one more time there is a point, let's call it H, that lives on the line N like so. Okay? And so really what we're done here is of course we've constructed these lines and these points and I want you to focus on this idea right here. We've constructed this triangle, the triangle G D B. That's what we're playing around with right here. And some of my labels a little bit hard to see but remember here's F here is, oh we didn't actually write its name on the screen did we? So this was our point G, not G, H right here on our graph. Okay? And so the main objective here is then to consider the line segment that connects F and H together like so. Consider this line, the line between F and H. In particular it's the segment that we care about but we really we can think of it as a line right? And so when we look at the diagram does appear to be some type of point of intersection, right? But we can't just assume by the diagram because the lines cross they have to have an intersection there. That's what we're trying to prove right here. The whole point of the diagram is just to help us keep track of what all these points and lines mean, what are the between this relations, what are the incidence relations. But in all reality the proof doesn't come from the illustration. The proof here is going to follow from posh's axiom, another axiom of between this. So let's see how that happens here. Okay? I'm going to scooch my picture up just a little bit so we can see more of the proof here. And so scooch it just a tiny bit down here. Alright, so what's the argument here? So we're going to consider this line determined by F and H right here. And so we know that F is the unique point of intersection between the line FH with the line M, right? Because we have this line right here, the line M. We have this line FH, which I haven't really given a name just FH. But the unique point of intersection is going to be the point F. Because we know from incidence geometry because of line determination that when two lines intersect, they intersect a unique location. Okay? And so the intersection between these two red lines has to be F. In particular, the line FH doesn't contain G nor does it contain B. Because if it did, there would be nonunique intersections. So B and G are not on the line FH. Okay? We can also say a similar thing about the point D right here. Because after all, we know that the line FH is going to intersect the line N at a unique location. That unique location is going to be H right here. And so again, by line determination, lines intersect a unique point. Therefore, the line FH does not intersect the point D. Why are we bothering talking about this? But it's because of Poch's axiom here. By Poch's axiom, we have that we have to consider a triangle. We have to consider a line that doesn't intersect any of the vertices of the triangle. Okay? That's the big assumption of Poch's axiom. Then we also know that this line FH intersects one side of the triangle. I'm going to zoom out a little bit so we can still see the picture but we can also read the proof. Okay? So we know that the line FH intersects one of the sides of the triangle. It intersects the side BG in particular. So by Poch's axiom, since the line doesn't go through any of the vertices of the triangle and since it intersects one side of the triangle, it has to be true that FH intersects the segment BD or it intersects the segment GD. Now, if it intersects the segment BD, which is actually what we want, that's because the intersection, since it's not a B, it's not a D, it would have to be at some point between BD. That's the point C that we're looking for. Okay? But Poch's axiom does have two possibilities. So why can't it be GD? Why can't it intersect something over here? But the reason for that is the line FH we already know intersects in and it happens at H. Lines only intersect at two different lines only intersect at one point. So if there was a point of intersection between D and G, that would be a second point of intersection between the lines in with FH. Well, unless, of course, this point is H itself, but by construction H is not between G and D. H is actually what we do have is that D is between H and G. And by trichotomy, we can't have multiple between this relationship. Notice how all of the axioms between this are coming out. We've used extension a lot. We just talked about trichotomy. We used the co-linearity and symmetric axiom and then Poch's axiom is going to finish the day for us right here, right? And so I want you to be aware that this original construction, these extensions were used exactly so that we had control over the between these relationships. We started off with a point F that was off the line, then we extended it, then we connected the dots, then we extended it again. And so again, this was so that we can construct a triangle that forces the intersection to be not here but over here. And so that then concludes the theorem here. By Poch's axiom, since the intersection of F H on N can't lie between G D, this would mean that the line F H must intersect the segment B D and it happens at a point between B and D and that then gives us the point C. An important consequence, of course, of the density theorem is that in order geometry all lines have infinitely many points. We already knew that by extension because we can continue to extend, extend, extend in both direction. But the reason it's called dense is that it's not that we just make the longer the lines longer and longer longer is that between any two points we can find another point and between them we can find another point and between them we can find another point and we can find another point. So between any two points on the line, we have infinitely many points and that's what density is measuring right here.