 Hi, my name's Adam. I'm a geometry student of Dr. Missildine's at SUU and today we're going to be talking about something called phanogeometry Whenever you have a new geometry you want to take a look at the axioms and really internalize those because those are your new Rules defining your geometric universe, right? So the axioms we have for phanogeometry Axiom one there exists at least one line axiom two there exists exactly three points on every line Axiom three not all points lie on the same line Axiom four for each two distinct points there exists a unique line containing both of them an axiom five all lines intersect Or at least there are no parallel lines All right, so let's take a look at what theorems we can prove in this geometry So here's the axioms up in the corner for reference the first thing we're going to prove in this geometry is that two distinct lines intersect in exactly one point Okay, so let's go ahead and let L and M Be two distinct lines All right L M By axiom five we know that all lines intersect so there's at least one point of intersection for L and M We're gonna put that right there. We call it P All right, but we have contradiction let's suppose there's another point of intersection because trying to prove There's exactly one so let's say there's two of them, okay? Might look something like this We'll call this one Q so L and M intersect at both P and Q However, this violates axiom four Which says that for each two distinct points there exists a unique line containing both of them because for these two distinct points P and Q There exist two distinct lines L and M that contain both of them so this can't happen Therefore there exists at least one Intersection of two distinct lines and there exists no more than one Intersection of two distinct lines so there's exactly one All right, so theorem one has been proven now. Let's prove theorem two there exists exactly seven points in seven lines now to do this We're gonna construct the geometry Okay, we're gonna do it axiom by axiom So the first axiom says there exists at least one line So let's draw a line and we're trying to show that there are seven points in seven lines So let's number the lines this line will be L one by axiom to there exists exactly three points on every line so let's Put three points on this line. We'll call them a B See okay by axiom three not all points lie on the same line So there must be some point up here Let's put it up here. We'll call it D It's not on L one. Okay Axiom four for each two distinct points there exists a unique line containing both of them So there must be a line L two that has a and D L three that has B and D and L four that has C and D Let's write those in L two L three Now for So the question is how do we know that L two and L three aren't the same line? Well, if L two and L three were the same line then they would both have a and D on them All right, that would be two distinct lines. I both have a and D and I would violate theorem one It would also violate axiom four so that can't happen. So these lines are all distinct We can by similar reasoning show that L two is my L four and L three is my L four Okay, let's go back to axiom two and there exists exactly three points on every line so There must be a third point on this sign on this sign and on this sign And we'll call them E F and G So how do we know That E and B are not the same point. Why couldn't L two be a BD? Well, if L two is a BD then L one and L two would both have AB on them and that would violate theorem one which says that two distinct lines intersecting exactly one point Because that shows two points where L one and L two intersect so E and B can't be the same Point likewise E and F can't be the same point because L two and L three would intersect at two points because they both intersected D Already and you can show the same thing by similar reasoning for all of the cases F and B G and B G and F Any case? All right, so therefore these points are on these lines L two L three and L four All right, but axiom four. We know that for each two distinct points there exists a unique line So there must be a unique line That has a and F right So axiom five says all lines intersect so this line L five which is Constructed must intersect L four at some point. So they have to share a point is what intersection means So it can't be D or else L two and L five would have two distinct points That they intersect that and that would violate theorem one and likewise It can't be C So it must be G because this line L four only has these three points on it So it's almost a method of exhaustion to show that it has to be G By similar reasoning there has to be a line from C to F and you can show With that exhaustion method method that that has to go to E Because it has to intersect L two and we'll call this line L six. All right So let's take stock so far. We have six lines We just wrote the sixth one and we have seven points So the question is where is the seventh line? We're trying to prove there exactly seven lines Well by axiom four that has to be a unique line With E and B All right, but this line has to intersect L four as well therefore It can't be at D or C or that would either conflict with L two L one So it has to be a G as well We don't have any notion of straight lines. So let's just draw this Weird curvy line connecting those three points and fits all the axioms are three points on it You know, it doesn't violate the fourth one. It doesn't violate a theorem. So this is the last point and it's a little lopsided circle But it'll work for us All right, so so far we've showed that there are at least seven points a v c d e f g and At least seven lines L one to L seven right this last one. We just drew is L seven So how do we prove that there are no more than seven of each? Well, let's do it by contradiction. Let's say there's another point out here H That we haven't explored yet Okay by axiom four there has to be a line With a and h on it Okay This line has to intersect L four at some point. All right But this can't happen because a and d already have a line a and g already have a line a and c already have a line So the only other option is to say, okay, this That must this third point here must be a new point in the geometry But if that happens and violate axiom five because This l eight that we just constructed Doesn't intersect l four among other lines. It doesn't intersect l seven, you know It doesn't intersect l six. So this line can't exist. Sorry. We just showed by contradiction that There has to be only seven lines and seven points And so let's erase this all right Therefore there exists exactly seven points and seven lines and this model that we just constructed Is the only model of phanogeometry up to isomorphism you can stretch points and lines and Move them around relabel them, but it's going to be the same exact construction There isn't another way to construct this and up to isomorphism means that This is the only Option this is the only construction So that is phanogeometry. Thanks for watching