 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 Misildine. This video will constitute the entirety of lecture five, for which we're going to talk some more about finite geometries. In previous lectures, we introduced some finite geometries. We have three-point geometry. We have four-point geometry, which we've done previously. We have phanogeometry, which we talked a lot about in lecture four. If you're working through the homework connected to this course, then we've been exploring five-point geometry in the homework assignments as well. In lecture five, we want to introduce one more finite geometry. Again, this is just meant as a toy to help us better understand the axiomatics of geometry. This is what's known as Young's Geometry. In which case, to motivate Young's Geometry, I actually want to connect it to the phanogeometry we talked about in lecture four. Before doing that, though, let me put a diagram of Young's Geometry on the screen, also the axioms, which we will go through them in just a moment. But I want to remind the viewer here that axiom five of phano geometry, which in the previous video we had talked about, is a statement about parallel lines and that, in fact, they don't exist. Axiom five of phanogeometry, as we stated, was all lines intersect. So that is to say, parallel lines don't exist in phanogeometry. This statement was essential in constructing the unique model for phanogeometry. You'll do recall from lecture four, as we talked about phanogeometry, we used the axioms essentially constructed the model, up to isomorphism, what it had to be. And several times, we used the fact that there were no parallel lines. It was essential. Also, as we were discussing phanogeometry, we mentioned how all five axioms of phanogeometry are, in fact, independent. That is to say, no one axiom is a theorem of the other four axioms. And we did that by constructing models for which the specific axiom in question was negated, but the other four axioms were satisfied, thus proving the axiom was independent. We did that for axioms one two one two three four for phanogeometry, but we didn't do it for axiom five. This one that said that parallel lines don't exist because that was a little bit more complicated. And I actually promised that in this lecture, lecture five, we would use Young's geometry to prove such a thing. So let's go through the axioms of young geometry. You're going to see that the first four axioms are identical to the axioms of phanogeometry introduced in lecture four. So axiom one, there exists a line. Okay. And so with this model, you'll notice there's a bunch of lines and they're color coded to make it easier to read here. You have these three horizontal lines which are colored here in white. You have these three vertical lines which are colored here in blue. Then you have these three green lines that are slant or diagonal. You can think of as like, you have a northwest to southeast line going here. The other ones also have that idea. You have northwest to southeast like so and this one also northeast to southwest. Although it's the curvature, it's not exact, but that's the idea there. And then we also have these northeast to southwest red lines like so that you see there. So actually have 12 lines into these four color groups. The color will make more sense why we did it later on. It's not just for readability, but in particular, the model we have right here has 12 lines. So a line does exist. Okay. Axiom 2, there exist exactly three points on every line. And so you'll notice with our 12 lines, three points, three points, three points. You have three points, three points, three points for the horizontal and vertical lines. The slant ones, you have three points, three points, three points. And then lastly, three points, three points, and three points. Every line has three points. Okay. Axiom 3, not all points lie on the same line. And if you give me any line, I can find a point not on it. So given this line, here's a point, here's a point, here's a point. There's actually a lot of points not on the line, but we only need one, right? Similar constructs should be done for any of the horizontal lines. If you look at the vertical lines, I can find a point not on it. Similar arguments can be done for any of the vertical lines. If we take some of these slant lines like this red one, I can find points not on it. In particular, you'll notice that there are nine points in the geometry and every line has exactly three points. So in particular, every line has a point not on it. In fact, there's six points not on it, but we only need there to be at least one. And then Axiom 4 says that for each two distinct points, there exists a unique not line that contains both of them. So let's keep this point fixed for a moment. If I want this point or this point, there's a unique line that grabs those. If I want this point, you get that blue line. If you want this point, there's this line. If you want this point, this one, you have to use this red line right here. If you want this point, you get this one. If you want this point, how are you going to connect that one? There's a red line that connects those two together. And if you want this point, there's a green line that connects them. So I did this just for this point here in the corner. But if you'd replicated this process with the other eight points, you would see the same type of thing happening here. There's a lot of symmetry in this geometry here. So in fact, every pair of points has a unique line that contains only one line there. And so what we see from this discussion is that this geometry right here, Young's geometry, satisfies the first four axioms of phano geometry. So if we forget the fifth axiom, these are both models of this geometric axiomatic system, axioms one through four, phano geometry, and Young's geometry. Where they disagree, of course, is axiom five. Axiom five of phano geometry said there are no parallel lines. But this geometry does have parallel lines. Look at this one right here. These white lines, which I've now colored yellow, these white lines are parallel lines. These blue lines are parallel lines. The red lines are likewise parallel lines. They don't intersect each other. Same thing for the green lines. And that actually is the reason behind the color coding here, is that different parallel families were given the same color. So if we just stop right there, since parallel lines do exist on Young's geometry, this improves that the phano axiom five is independent of axioms one, two, three, four. But for Young's geometry, we're actually going to give a stronger statement. We're not just going to say that parallel lines exist. We're actually going to give a very strong condition on what that existence means. What we're going to take here is axiom five, is that for each line L, and each point P not on L, there exists exactly one parallel line to L, which contains that point P. So what it means to be something like this, if this is our line L, and we take some point P off of it, there exists, so it's not just the parallel lines exist. I can guarantee that there's a parallel line to L that goes through P. That would also take care of this point right here, if these were our points P. But if point P was maybe something down here, then here's a line that does that as well. Same thing if these were those point P's. And it doesn't matter which line you take. If you take one of these curvy red lines like this, take a point off of it, there's a red line parallel to it, because all the red lines are parallel. If you grab this point, you get that one, this one, this one, it's pretty easy in that case. Or if you pick any of these three points, you can see that every point that's not on the first line is on exactly one other red line, which is going to be parallel to it. So this young geometry satisfies a very strong existence statement about parallel lines. We're guaranteed that for any line in a fixed point off of the line, you're going to get a parallel through that line. We don't just have this nebulous statement that parallel lines exist. I actually know conditions that guarantee where they'll be. They're passing through these points. And that's a very, very strong parallel axiom that we'll actually provide. We'll talk some more about this in the next lecture. We'll compare axiom five of phanogeometry with axiom five of young geometry, then provide actually another parallel alternative. And so we're taking a slightly different approach to this than we did with phanogeometry. With phanogeometry, I gave you the five axioms, and then we essentially constructed the only geometry, uptoisomorphism. We had this constructive proof. In this situation, we're not exactly going to prove that this is the only young geometry uptoisomorphism, but it is. This is another complete theory. Instead, what I want to do is prove theorems about young geometry. With this model is in mind here, right? But I'm not saying this is young geometry. I'm saying this is young geometry. Young geometry is these five axioms. This is a model of young geometry, so we know young geometry is consistent. But is there a different model that's not isomorphic? We don't know that yet. But as we go through the theorems of young geometry, we'll essentially start to see that this is the only picture. So I want to provide some of these theorems right now. So the first one, and I should also mention that if you're following along with this video, it might be a good idea to pause the video at this moment and copy down these axioms, as they won't be on the screen for much longer. It's an inconvenience of the technology, mind you. But if you're following along, you probably want to copy those because we're going to use these axioms throughout the rest of this video. All right. So let's look at some theorems of young geometry. So the first theorem in young geometry, again, these are all going to, in this video, all of these theorems are theorems of young geometry, the five axioms we've just had. So in young geometry, every point is on exactly four lines. And I want you to think about that for a moment. If we come back to the picture we had, this is a statement true for this one. Like if you pick any point right here, we have one, two, three, four lines that go through it. And they're all different colors, right? That's significant here. Why is that true for every young geometry? Because after all, we don't know that diagram is the only young geometry. Got to keep on saying it, but we don't have proof. And therefore, there should be some scrutiny of that right now. Now notice our statement says every point is on exactly four lines. If we want to prove a statement about every point, what that really means is we're going to take a generic point P and improve something about that. So we know P is in the geometry, but we don't know anything else about it. If we put any further assumptions on P, then it might not be every point. It's going to be that point. We don't want that. It needs to be a generic point. So we have some point P right here. Now oftentimes as we try to prove a theorem, we sometimes reach a point where we have to make a claim. That is, we have to prove something else that's necessary for this proof, but hasn't been proven already. We could have presented this earlier as a lima or some proposition prior to this theorem. If it's worth using over and over again, that would be a lot better. But sometimes if it's simple enough proof and maybe not too noteworthy, we just put it here. We could have done it differently, but that's okay. It's more just a stylistic choice here. And so we have an important claim that we have to provide here. We claim that there exists a line, L, such that P is not on L. How do we know there's a line for which P is not on? Now if you think axiom three, that's not quite what axiom three says. Axiom three said not all points lie on the same line, but how do we know you don't have something like this where this is our point P and all lines are concurrent to the point P, right? In this situation, not all points and so like we could have other points like this because each line supposed to have three points. We could have some diagram like this potentially, right? In this situation, you'll notice that there's a line. Every line has three points, not all points lie on the same line, right? If you take, for example, this line right here, this point's not on it, this point's not on it. That's not what axiom three says. Axiom three says that not all points lie on the same line. How do we know not all lines lie on the same point? That's what we're trying to prove here. We're actually trying to get the dual of axiom three. We're trying to show that there's a line L for which P is not on it, so not all lines are concurrent. Now axiom one does guarantee that there is a line. So you have some line L prime right here. Now if P is not on L prime like the diagram suggests, then that's the line we're looking for. We call that L and we're done. So let's assume in fact that P is on L prime. So we get something weird like this going on. So P is a point on L prime. Now by axiom two, this line L prime does have three points. One of them is P. There's two other points yet unaccounted for. Let's call one of them A and so A is some point not equal to P but is also on L prime. Now axiom three, I mentioned it before right and said that this claim is not an immediate statement of axiom three, but axiom three does say there is a point not on L prime. Let's call that point B and then using axiom four. Axiom four tells us that given any pair of points, there's a line uniquely determined by them. So in particular using the pair A and B, there has to be a line that contains A and B and this is unique line. And I claim this line is the line L we're looking for. Now you'll notice that B is on L but it's not on L prime. So this point can't be P because P is on L prime. B is not. So B and P are different points. That's important to note here and I claim that P is not on L. L only contains one other point. What is that point? Could it be P? If it was P, right? If P and L both contained, excuse me, if L and L prime both contained P, that means L and L prime, they both have P, but they also both have A. So there would be two lines that contain P and A which would violate axiom four because axiom four says that between two distinct points, there is a unique line that contains both. We would have two at this point. Therefore, this line L does not contain P and thus that proves our claim. And I want you to note here what axioms we use. We use axiom one, axiom two, axiom three, and axiom four. So this idea that you can construct a line for which the point is not on is only a theorem of axioms one, two, three, four. We didn't use axiom five without parallel lines, which means this is also a theorem of phanogeometry. We had the exact same four axioms one, two, three, and four. All right. So now let's move to the main idea of this theorem here. So we want to prove that every point has exactly four lines on it. So let's return to our point P for a moment. We know there's some line L for which P is not on L by the claim we just did there before. The next thing to also mention is that this line L has exactly three points on it. That came from axiom two. So there's three points like so. And let's call these points A, B, and C. And these A, B, and C's are not necessarily the same A, B, and C that we were talking about in the previous claims proof. Now by axiom four, there exist unique lines that connect A to P, B to P, and C to P. So we get lines like this. And these lines necessarily are all distinct from each other because if any of these two lines overlap, that means there would be two lines that connect A and B together or A and C together or B and C together, which axiom four doesn't allow that. So we have these so far. And you'll notice that we've now counted one to three lines that are incident to P, three lines on P. We need a fourth line. Where's that fourth line going to come from? Well, that's where axiom five isn't coming to play here. We haven't used axiom five yet. Axiom five says that with respect to the line L and with respect to the point P, there is a line passing through P, which is parallel to L. Let's call that line M right here. And so we get this parallel line. So that now accounts for four lines. But one has to be careful. Is there a fifth line? We've now constructed using the axioms four distinct lines and these lines have to be distinct, right? Because notice that lines one, two, and three, they are intersecting line L. And so if M is parallel, it can't be these three lines. And we already argued why lines one, two, and three are distinct. So why can't there be another line? Why can't there be a fifth line? Well, if there was a fifth line passing through P, what can we say about it? Well, if it's parallel to L, then that would contradict axiom five, because axiom five says there's exactly one line parallel to L passing through P. So it can't be parallel. Therefore, it has to intersect L. L has only three points. So it has to intersect at one of the points A, B, or C. If it intersects at A, then we now have two lines that contain P and A that violates axiom four. But if it were to intersect at the point B, then we get two lines containing P and B. That's a contradiction. If they intersect at C, then if M and L intersect at C, then we have two lines that contain P and C, which gives us a contradiction. As all roads lead to contradiction there, that means there's not a fifth line, and that thus proves the statement right here that we do, in fact, have exactly four lines incidence to the point P. That proves our first theorem of young geometry. The next two theorems of young geometry are actually going to get names. This one's called Proclus Lemon. Why does it get a name? Because after today, we're really not going to study young geometry in any depth. The idea here is that young geometry is actually an example of a broader family of geometries, which we call affine geometry, which we'll study those in lecture seven here, for which Proclus Lemon, the next one, transitivity of parallelism, are actually theorems of affine geometry, and we'll use the exact same proof that we're using right now for young geometry. The previous theorem we just did is not a general theorem of affine geometry, although we will generalize it in that situation. That proof will require modification. These two that we're going to do right now on this page require no modification whatsoever. Proclus Lemon, for the current moment, we're going to state this as a theorem of young geometry, but in the future, this exact same proof can be used for general affine geometry. If lines L and M intersect and if the lines L and N are parallel, then I claim that M and N must intersect each other. I'm going to rewrite this using some symbolic notation a little bit. If L and M intersect, that means they're not parallel, and if we have that L is parallel to N, then we want to prove that M and N intersect, that is, they're not parallel. You just have a not parallel right there. That's what we're trying to prove right here. Since L and M intersect, that's our assumption, there's got to be some point of intersection between them. Let's have our two lines be right here. We'll call this one right here L and this one right here M, they intersect, and we'll call their point of intersection P. My diagram might be someone misleading here. I said they intersect. How do I know they intersect at a unique point? It turns out that using the axioms of young geometry, I can prove that the intersections are unique. I don't actually need that right here, but one should always be cautious about your diagrams. Your diagrams might be misleading. The proof is not the diagram. The proof is actually this friend right here. I'm not reading the proof because I want this to be much more approachable to you, the viewer, right? I want you to understand, instead of just being the dry proof, that illustration is to help to understand what's going on here, but the diagram I'm constructing is exactly following the proof that you see here on the screen. So getting back to the point here, P is the unique point of intersection, but I actually don't need that fact. I just need a point of intersection. What we're going to do is to prove that M and N intersect, which means they're not parallel. I'm actually going to prove by contradiction. I'm going to assume that M and N are parallel and get a contradiction because if N and M are parallel, we get this other line N right here for which I want you to note that with the line N by assumption, N and L are parallel. And then for the sake of contradiction, N and M are parallel. But note here, we have the line N, we have the point P, and there are two lines parallel to N that pass to the point P. This is a direct violation of axiom five. And therefore we get a contradiction, which means this assumption that N and M being parallel must be false and therefore M and N are not parallel, aka they intersect each other. Now in this proof, the only thing we used was axiom five. I said things like, oh, the point of intersection between L and M is unique. I get that from axiom four, but I didn't need that. And this theorem, Proclus Lemma, is really just a consequence of axiom five. In fact, one can prove with the appropriate settings that Proclus Lemma is logically equivalent to axiom five. We've seen that axiom five implies Proclus Lemma. But again, under the right structure, I can prove to you that Proclus Lemma implies axiom five as well, that they're logically equivalent to each other. Now related to Proclus Lemma is our next theorem which is called the Transitivity of Parallelism. Why does it get its name? Well, let's read the statement. In Young's Geometry, two lines parallel to the same line are parallel to each other. So what this is telling us here is that the parallel relation is transitive in Young's Geometry. If these two lines are parallel to the same line, they're parallel to each other. So if M is parallel to N and if L is parallel to N, then this gets us that M is parallel to L. Now that's not exactly how transitivity works. Transitivity, typically, you have to switch something like this around for which you get that N is parallel to L. So if M is parallel to N and N is parallel to L, then M is parallel to L. That's transitivity. But if you have symmetry, then these two statements are equivalent to each other for which as we define parallelism as a relation, it's always symmetric. And you'll also recall that we buy, it seems sort of artificial, but we actually defined the parallel relationship that aligns parallel to itself, so it's reflexive. And so we always have that parallel, the parallel relation is reflexive and symmetric. But in Young's Geometry, it's also transitive, which is why we call this the transitivity of parallelism. So in Young's Geometry, parallel lines actually form an equivalence relation. And therefore, there are equivalence classes to parallel lines. Before proving this, let's go back to the diagram of Young's Geometry, which we see now on the screen. So you'll notice here that I color coded these parallel lines on purpose. Those are the equivalence classes. We have the white lines. Those are a parallel equivalence family. We have the blue lines, the vertical ones. This is a parallel equivalence class. We have the red lines. That's a parallel class. We have the green lines. So this color coding actually was suggestive of the general observation we have on Young's Geometry. Because again, this is a model of Young's Geometry. We haven't yet proven, we're not going in this lecture, but we haven't proven that Young's Geometry is complete. There could be other models. But nonetheless, every model of Young's Geometry must exhibit this equivalence relation of parallelism. This model does, in fact, as it's the only model, all models will. But again, that's beside the point. So let's get back to the proof of this thing. Why is parallelism transitive? So let's go with the assumptions here. So let's assume we have some lines, L and M, which are both parallel to N. So we're assuming we're going to have N over here, something like this. Here's our N. We're going to have our L and our M. So let me label these things. This is L. This is M. And this one right here is our N. And so by assumption, we've assumed L is parallel to N. We've assumed that M is parallel to N. And what we want, what we want right now is that L is parallel to M. That's what we're going for. So we're going to, again, proceed by contradiction here. Let's suppose this isn't the case. So if L and M are not parallel, that means at some point they get to intersect each other. So they have a point of intersection, call it P, but it won't really make much of a difference because we can actually apply Proclaslima at this situation. So you'll notice that for the sake of contradiction, we're assuming that L intersects M. So we have an if right there. We also have, by assumption, that L is parallel to N. Then by Proclaslima, we should get that M is intersecting N, which is a contradiction. So this is an immediate contradiction to Proclaslima, which means that our assumption was bad. So the lines L and M don't intersect each other, so in fact, they must have been parallel, thus giving us what we want. So transitivity of parallelism followed from Proclaslima, followed from Axiom 5. Those are the only axioms of young jump we've used to prove this. And like I said before, these directions can actually be reversed. The transitivity of parallelism is logically equivalent to Proclaslima, but that's not a topic we're going to talk about in this video right now. All right, so now let's return to a theorem that is true for young geometry, perhaps not the general family of affine geometries, which I haven't even defined what that is yet, because the last two theorems are only consequences of Axiom 5. Let me give you one that is unique for young geometry. In young geometry, we claim that every line has exactly two lines that are parallel to it. In our diagram, I drew it so that parallel families were color-coded, and those color coding always came in groups of three. So if I have one line fixed, there was two lines parallel to it. We saw that in the model. Why is that true for every young geometry? Because again, we can't use the fact that it's only one model. We don't know that. We haven't proven that. These are the things that help one prove such a thing. So every line is parallel to exactly two other lines. That's what we want to prove right now. So what we're going to do is we're going to start off with a line. We're going to call that line L. We want to prove that line L has two lines parallel to it, exactly two. Now, because L is a line, by Axiom 3, there exists a point P that is not on L. So we get something like this. Here's our point P. Likewise, by Axiom 2, line L has three points on it. In particular, it has at least one point, Q, that is on the line. I'm going to put that point right here, Q. Now, since the point P is not on L, by Axiom 5, there exists a line that passes through P that is parallel to L. This gives us our first line, which is parallel to L. We need to come up with another one. Now, using Axiom 4, we know that there exists a unique line determined by P and Q. There's some line that'll go through P and Q, and it's the only line that does that. Let's call this line M for a moment. Now, by Axiom 2, M contains three points. P and Q are two of those points. There has to be a third point, which we're going to call that point R. Now, using Axiom 5 one more time, since R, let's make a comment here, R is not on L. Because if R was on L, we would have two lines that contain Q and R, which violates Axiom 4. So R is not on Q. R is not on L. Axiom 5 applies here. There is a line passing through R that is parallel to L. These lines right here, line 1 and line 2, I guess I did them in the other order, but it doesn't matter. These lines are parallel to L, and now we've come up with two lines that are parallel to L. Now, pot in here for a moment, I should mention that we have constructed at least two lines that are parallel to L. Could there be a third line parallel to L? That's kind of interesting here, right? So why can't there be another line? We have to consider that as a possibility. And so let's consider this other line, this other line right here, that's parallel to L. Let's give it the name N for a moment. Now, so by assumption, we're assuming that L is parallel to M. Excuse me, to N. That's what I meant to say. This new line we've constructed, but we also know that line L intersects M. So by Proclaslima, we then have that this line N must intersect the line M. That was theorem 1315 in our numbering systems, Proclaslima. So there's got to be some point of intersection right here, but what is that point, all right? That point could be Q, but if that point was Q, then N and L don't intersect each other. So what if it was something like R or something like P, right? We'll get to those in a second, but let's consider R for a moment. What if it were to intersect at the point R? Well, if it intersected R, remember this line is parallel to L and passes through R. Then this line N is parallel to L and passes through R. That would be a violation of Axiom 5, that we would have multiple parallel lines passing to the same point. So I guess it's a contradiction. The line can't pass through R, but could it pass through P? You get the exact same problem. So it turns out that this line cannot exist, and therefore we have only three parallel lines total, including L itself. So in particular, there are exactly two other lines parallel to L, thus proving the theorem for Young's Geometry here. Now this last term I'm only going to make in passing the statement. This is something that my students can prove. They will prove in the homework. And if you're not a student watching this video, then by all means, I still leave it as an exercise to you. Can we prove that Young's Geometry has exactly nine points and exactly 12 lines? The proof here is essentially finishing off that the diagram we started with is the only model for Young's Geometry. We have all the ingredients, the theorems we've developed so far, using those like Proclus lima, the transitivity parallelism, and then the other two theorems of Young's Geometry, which you'd call Young's theorem one and then them two, right? Using those without a situation which we can prove that there are exactly nine points and 12 lines. And the proof of that, much like we did with Phano geometry, will essentially prove that the model we had before is the only model of disomorphism. We talked about Phano geometry. I was very constructive. I constructed the geometry from scratch. We were much more abstract in this one. We haven't done that yet. I'm leaving that actually to you, the viewer, to produce that theorem. But just so you know, as we conclude this discussion on Young's Geometry, that that model is the only one. There's nine points and 12 lines. So I do appreciate you watching this longer video, but there's really no way of breaking it up. That made it a little bit more sense than what we have right here. Next time, we're going to talk about the axioms of incidence geometry, because all of these finite geometries we've talked about, three-point geometry, four-point geometry, Phano, Young geometry, five-point geometry from the homework, these are all examples of a common geometric theory for which we want to extrapolate from that the general theory, which we call incidence geometry. Check that out next time. If you like this video, learn something about Young geometry, give it a like. If you want to see more videos like this in the future, or get other updates on this math channel, subscribe. And as always, if you have any questions, please post them in the comments below.