 In our previous video for lecture 13, we defined the notion of what it means for two rays to be between one another. We also proved the crossbar theorem. So an immediate corollary of the crossbar theorem is this idea of correspondence of betweenness. That is, we have two notions of betweenness that are playing out right now. We have the notion of betweenness of points. That's how we started our conversation of the betweenness axioms. Betweenness of points was this undefined term. But then as we've developed the theory, we've come up with this betweenness of rays. And it turns out that these two notions of betweenness are really related to each other, this so-called correspondence of betweenness. And so what we claim here is that we have three rays, B A, B C, B D, in an order geometry. Then the ray B C is between B A and B D, if and only if there exist points on the rays, call them A prime, C prime and D prime, so that C prime is between A prime and D prime. So that is a betweenness statement of rays coincides with a betweenness statement of points. And it's a really short proof. It follows basically from the crossbar theorem and the between cross lima, which in some respect we can think of these two properties as almost converses of each other. So let's assume we have some type of betweenness of rays. So we get a picture that looks something like the following. So we have a common vertex B here, and then we have these rays given by A, B, and excuse me, not B, C, and D, like so. So we have these three rays. And so as illustrated, B C sits between B A and B D. So I claim that there exist points A prime, C prime and D prime. So we have a betweenness relation there. What I'm going to do is I'm going to use the crossbar between the points A and D. So in this situation, A prime is A and D is D prime, like so. By the crossbar theorem, there's going to be a point of intersection between the ray B C and the crossbar A D. And that intersection will call it C prime. And by the crossbar theorem, we know that point is going to be interior to the angle. And by construction, since C prime is on the interval A to D, then it's going to be that, okay, you have that C prime is between A and D, which of course we're calling those A prime and D prime. So let me use the first direction. So necessity follows directly from the crossbar theorem. But we can also go the other way around. So imagine we have our rays again, I deleted too much. Okay. So I deleted too much. And so suppose we have some type of line like so. So we have a point B right here, there's some point A prime, there's some point C prime here, there's some point D prime right here. And these are points on the rays B A, B C and B D, like so. So we're assuming that C prime is between A prime and D prime. Now this follows by the between cross lemma. So by the between cross lemma, because C prime is between A prime and D prime, that means that C prime is an interior point to the angle A prime B D prime. But as B A prime is the same ray as B A, and as the ray B D prime is the same as the ray B D, that means the angles, the angle A prime B D prime is equal to the angle A B C. In particular, that tells us that C prime belongs to the angle A B C sorry A that should be A B D, like so. So C prime belongs to that one. So in particular, the entire ray B C prime lives inside of that as we previously shown. But that's the same thing as just the ray B C. And therefore B C is between B A and B D because C is an interior point. So sufficiency is essentially an immediate consequence of the between cross lemma, although I want to provide all the details here. So because of the correspondence of betweenness of points in rays, so betweenness between points relates to betweenness of rays, betweenness of rays is going to inherit many of the same properties that the betweenness of points have. For example, the symmetric property. So because we can turn this around to become B prime C prime dash A prime, we can also turn this one around and say B D dash B C. There shouldn't be a prime there sorry. That's a ray dash B C dash B A. Because betweenness of points is symmetric, betweenness of rays will likewise be symmetric. It doesn't just stop there. We also get extension, right? The extension axiom becomes a property for betweenness here because we could extend this point, get some new point E prime. We could look at the ray there. Whoops, try that again. We could extend the ray there like so. And then we have now extended the rays. So we have this new ray B E prime. And so B A is going to be between that one, right? So extension works because we can extend the points. We can extend the rays. We also get density, right? What if we come up some new point F prime right here? By the density property, there's some point F prime that lives between A prime and C prime. And if I connect the dots here, then we're going to get a ray that sits between B A and B C. So density is also a consequence here because betweenness of points has density, betweenness of rays will likewise have density. What else can we talk about? The linearity rules, which talk about if we have certain betweenness statements, we can infer from them other betweenness statements. I'm not going to write those ones on the screen here, but the linearity rules are going to carry over because of this correspondence. Now, not everything's going to correspond perfectly. It is also true that trichotomy holds, well, we know trichotomy holds for three points that are collinear. Would it also be true that trichotomy holds for three rays that are concurrent so they have the same vertex? The answer is we have to have, it's almost true. We have to take a slightly weaker statement. And that's what we're going to prove right now. If we have a line that contains the point B in an order geometry, so again, I'm going to try to start prepping my picture down below. So we have a line, and this is our line L, and the point B lives on that line. Then I want you to consider the points A, C, and D so that they all reside in the same half plane. So you have A, you have C, and you have D. So A, C, D are all in the same half plane of the line L here. If you look at the rays associated to B, A, to B, C, and to B, D, so all three of these rays live in the same half plane of L, then one and only one of the three between this statements is true here. So either, as I have illustrated, either B, C sits between B, A, and B, D, or B, A is between B, C, and B, D, or B, D is between B, A, and B, C. So one of the three between the statements has to hold. So trichotomy holds if you live in a half plane. So before we prove this theorem, let me kind of mention to you what can go wrong, right? If you exit the half plane, you might get something like the following, something like this. So here's our B. So you have like A, C, and D like so. In this situation, because we're no longer in the same half plane, we might get some ambiguity about who is between who is B, C between B, A, and B, D, or is B, A between B, D, and B, C, or is B, D between B, A, and B, C. It's hard to keep track of these things, right? So if you take the entire plane, trichotomy falls apart because of rays because of this issue right here. But if you if you restrict it to just one half plane, that is if you have a flat angle, then trichotomy will be will be withheld. It will be held up. That's what I'm trying to say here. And so because of that, that's why we restrict to a half plane. And so now let's go through the details of this. Why does trichotomy hold? So there are a few cases that we have to consider. So first suppose that A and D are on opposite sides of the line B, C. So the ray B, C is already illustrated on the screen. So that gives us the line. So let's say that A and D are on opposite sides of this line B, C, just like I have illustrated right here. That means that the line segment A, D intersects the line B, C at some point C prime, which will label that. And it must be that C prime sits between A and D. Okay? So by the illustration here, it feels like C prime should be on the ray B, C. But what if it wasn't? It intersects the line B, C. So if it's not on that ray, then that means it's actually would then be on the opposite ray. So what if C prime is actually down here? So we get something weird like this happening. So by way of contradiction, suppose that C prime is on the opposite ray, negative B, C. Well, if that's the case, then C prime is not in the half plane that's bounded by L and it contains C. And that's because the point B sits between C and C prime in that situation. Okay? Because these two rays are on opposite sides of the line L. So we would get that C and C prime are on opposite sides. Now what we said earlier, right? By plane separation, it's going to be that A and C prime are on opposite sides of L. Well, why is that? Well, if we go through the details of this, like we just said a moment ago, C and C prime, they're in different half planes. So they're on opposite sides. What about A and C? A and C like so. They're on the A and C are on the same side of the line that L is, that's by assumption. So that's where plane separation comes into play. Since A and C are on the same side and since C and C prime are on opposite sides, that would then imply that A and C prime are on opposite sides of the line L. So that would then give us some point of intersection right here, which we're going to call that point P. So P is the intersection between the line segment A, C prime, and L. I noticed I'm going to get rid of this C prime that's already on the screen because it could be misleading. We're assuming it's somewhere else right now. All right. So likewise, D and C prime are on opposite sides of the line L. How do we know that? Well, it's basically the same argument here, right? So by similar reasoning, the line segment C prime D is going to have to intersect L somewhere. We'll call that point Q, right? Our picture is kind of curious. It looks like our line is a circle. Now if lines can be circles, clearly this would be acceptable. But what's going to happen here? Okay. Well, notice that the line segment AD, when you intersect with L, it'll contain the points, it'll contain the point Q and P. Well, why is that? Well, the line segment, so notice my construction, C prime is between A and D. So if I take A, C prime, the interval, and at union with C prime D, the interval, this is going to give you the interval, the line segment AD. So P is a point of intersection here and Q is intersection here. So P and Q are both on AD and they're also both on L, which that seems a little problematic, right? Because wait, are P and Q the same point? Or are they different points? That's very interesting here. The problem is that P and Q are actually different points. Why is that? Well, that's because C prime sits between P and Q. Where did that come from? This actually comes from the linearity rules, right? So we're going to use the fact that C prime is between A and D. We also know that P is between A and C prime. Okay. And we also know, of course, that C prime is between Q and D. So if you follow the linearity rules, I'm not going to provide all the details of this right now. But if you follow the linearity rules, you would then have to infer from these three between the statements that C prime is between P and Q. So if there's a point between P and Q, then they cannot be the same point. But wait a second. The intersection between a line segment and a line is two different points. Well, the line segment can be extended to a line. So we're saying we have two distinct intersections between two lines. That's a contradiction to line determination. All right. So this is the contradiction. What was the assumption? Well, we assume that C prime was on the opposite ray, negative BC. So we then have to assume the opposite. C actually belongs to the ray BC. So let me clean up my picture a little bit. Here we go. So we then, because of our contradiction, we now have the following picture that C prime does, in fact, live on the ray BC. And in that situation, because C prime lives on the ray BC, this really tells us that the ray BC is the same as the ray BC prime, like so. And so because C prime is an interior point of the angle, that means C is an interior point of the angle. And therefore BC is between the rays BA and BD. So that was under the assumption, of course, that A and D were on opposite sides of the line BC. All right. So consider the last possibility where A and D are on the same side of BC, and A and C are on the same side of BD. So if I reschedule the picture, I'll get something like the following. Here's our line L again. Here's our point B. Now to draw the picture this time, I kind of have to do the following. Here's A, here's C, here's D, something like this. So if we take the line BC, which would be illustrated over here, then we can allow for A and D to be on the same side of that line. But if we were to consider the line DB, then A and C are on the same side of that line. So we get something like this. So this diagram does seem to match up with our assumptions here, but let's analyze it here. Since A and D are on the same side of the line BC, that means A belongs to the open half plane determined by the line BC and on the same side as D. But then our other assumption was that A and C are on the same side of BD. So that means that A belongs to the open half plane bounded by the line BD on the same side as C. So A lives inside the intersection of these two open half planes, but the intersection of these two open half planes is exactly the angle, the interior of angle, CBD. So A is an interior angle to CBD, and that's exactly what it means like to have BA between the rays BC and BD. So just like I've illustrated right here. All right? So the uniqueness of the between the statements, so we've now proven that we've exhausted all the possibilities here. And so we've shown that one of the rays was always between the others. Now the uniqueness of the between the statements comes from the correspondence between between this of rays and between this of the points, right? Because if we have some betweenness between the, if we have some type of betweenness of the rays, then we can draw our line in which case we have the three points on this line as it intersects the rays, like so, for which in that situation there's only one between the statement between the points. And so correspondence of betweenness of points with rays will then give us that this will be unique. And so therefore we've established that in fact there is a, there is an analog between betweenness of rays with betweenness of points and they just do in fact, once they correspond with each other.