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hi everyone welcome back i’m dave today’s episode models of finite

incident planes uh hands-on we’re going to be working uh

through a couple of different models and i’m going to be taking you through it

step by step and before we get started i wanted to

mention that this episode is part of the

series incidence geometry tutorials with

step-by-step proofs so the link is below

in the description and i’ll be referring to other episodes and this video here

so let’s refresh our memory real quick of the instant axioms

so we have three incident axioms every two distinct points determines a

unique line every line has two points on it and there exists three non-collinear

points and so we’ve talked about all this in great detail in our last

um several videos and we’ve also proven these theorems here

um we proved each one of these in a separate episode and then we proved

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seven through 11 in an episode altogether and so now let’s get started with some

models now in each model that we’re going to

talk about though we’re going to talk about a parallel property

so these are the statements here that are the parallel properties here

we also talked about these in a previous episode the first one was called the

elliptic parallel property and then we have the euclidean parallel

property and the hyperbolic parallel property so basically a parallel

property starts with a line any line and any point not on the line so i’ll

just draw a quick line right here l and we have a point not on the line

and so these statements here which are all uh mutually exclusive right so

there’s no lines that go through p that are parallel to l

or there exists exactly one line through p that is parallel to l

or there exists more than one line parallel to l that passes through p

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so that’s the hyperbolic uh property right there number number three so

in the uh previous episode we talked about um parallelism

and if we said that if the euclidean parallel property then parallelism was

an equivalence relation so we talked about that in a previous episode and

um we also gave a first model in our in our previous episode about um a1 a2 and

a3 and instance geometry and it was a three point uh model and this is the

model here that we talked about and so this episode we’re going to come up with

some more models but let’s just very briefly remember what a model is

so we have axioms a1 a2 and a3 and anything that satisfies them we’re

going to call it instance geometry so this is a three point incidence geometry

and in this geometry there’s only three lines and there’s only three points so a

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is a point b is the point c is a point here’s line one here’s line two line

three and so we know the exact uh geometry right here it’s all listed

right here in this table now there’s some additional facts that

hold for this geometry that don’t hold in all incident geometries these three

right here hold in all three geometries and because these axioms hold

all 11 theorems that we proved hold for this model right here

but these are some additional things that hold these don’t hold in all

in all instance geometry every point is incident with two lines and you can see

that’s true and there are no parallel lines so the

elliptic parallel property holds so these two statements are

specific to this model but we can call it a model because we

verified that these three axioms hold so that’s what we did in our previous

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episode if if all that’s really quick for you you might want to check out that

episode that we did where i spent about 15 minutes

describing this model right here so what i want to show you right now is

another model so this is called the four point model the four point geometry

so we’re going to have four points a b c and d and we’re going to have six lines

and here are the lines right here so this right here describes the whole

geometry these are all the points and these are all the lines so a and b are

on line l and nothing else is on line l a and c are online

line two and nothing else is on line two so this describes our geometry right

here now before i can call it a geometry to be honest before i can call it an

incidence geometry we have to check that a1 a2 and a3 hold

so you know that has to be verified before we can call it a geometry so

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remember what a1 says a1 says given any two points

there’s a unique line going through them right so give me any of these two uh

give me any two of these points any any of these four a b c and d pick any two

that you want and there’s only going to be one line going through it for example

if you give me b and d there’s only one line going through b

and d and here it is right here so we have to verify that step by step by step

pick any two of these four and run through all the possible cases

what’s the line that goes through a b here it is what’s the line that goes

through a and c here it is what’s the line that goes

through a and d here it is and so on and so this is all the possibilities and so

we can verify that a1 holds simply by writing down all that information

similarly we can also write down the axiom a2 and verify that actually

made two holds so axiom a2 says for every line there exists at least two

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points on it that one’s actually much easier to verify because

they’re all given in in you know this format right here

pick a line for example number four you can see the exact two points on it

now accent 82 says there’s at least two points on them but in this geometry

there’s exactly two points on them excuse me all right and then axiom a3 says

that there exists three non-collinear points so let’s see if we can pick three

nonconlinear points and show that axiom e3 must also hold

well what about a b and c are they all on the same line

uh actually no they’re not so a b and c are non-collinear

in fact we could actually choose any of the three any three of the four

so axiom a3 not only holds but it holds lots of times there’s lots of

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non-collinear points so axiom one and two and three hold and

so yes this is a model of incidence geometry and it has four

points on it so we’re going to call it the four point geometry

now there may be some additional theorems or some additional

statements that hold true in this specific geometry that didn’t hold in

the three-point geometry in the three-point geometry remember that we had

the elliptic parallel property holding but then the four point geometry here

it’s the elliptic parallel property isn’t going to hold

um and we’ll see that here in a second so we’re going to see here

the an additional statement is each point has three lines on it so for example

now you know because this is a finite geometry when i say each point

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what i mean is i verified a i verified b i verified c and i verified d so let me

show what i mean so let’s pick a first a so point a all right

has three lines on it let’s see here’s one here’s two here’s three

and ar is not on any of these so that works for a now what about point b

does p have three lines on it let’s see one two three

and none of the other lines have b on it all right so it works for b and you see

i went through everything here a b c d e oh sorry not a d not an e in this one

let’s just check d real quick what are the three lines d is on one

two three all right so that works for d also might as well check c one two three

it’s exactly on three lines so this statement right here

is proven just simply by going through and listing all the evidence and because

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it’s finite we can actually go through each and every point and verify

now it’s important to remember that this statement is true

in four-point geometry but this statement was not derived by the axioms

a1 a2 and a3 so this is not true in every incidence geometry but we verified

it as true in four point geometry we can also verify that each distinct

line has one line parallel to it exactly one line parallel to it so let’s see

that this is true so it says each distinct line so now i need to check l1 l2 l3

l4 i need to check each of these six cases is it true

um that each distinct line has exactly one one line parallel to so let’s see

line l1 so here’s a line and this is not parallel to it these are not parallel

because they have point a in common uh l3 is not parallel to l1

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l4 is not parallel l5 is not parallel but l6 is parallel

so we haven’t verified each distinct line but we have verified that for l1 it

has only one parallel line to it so that works so now let’s go check l2 l2

so these two are not parallel because they have a in common

these two are not parallel because they have a these two are not parallel ah

these two are parallel and these two are not parallel so again

for l2 there’s exactly one line parallel to it namely l5

so l2 works also and so we can go and check each line

one at a time in fact let’s just skip to the end in l6 here i claim that it has

exactly one line parallel to it what is it what’s the one line parallel to l6

well it’s l1 isn’t it and so that also holds and so you can go

and verify all of these and so we can check each one of these by brute force

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and we can verify this statement is true now again this is not a true statement

in every incident acts in in every incidence geometry but it is a true

statement in four point geometry and we can either try to prove it’s true

in this geometry or we could just list all the evidence out by brute force

all right and then the last one is the euclidean parallel property holds

this one may take a lot more work and the reason why is because

it requires a line and a point this one just required us to verify every point

so we had four cases this step statement two

said every distinct line so it had six cases to verify

but the euclidean parallel property requires us to take any line

and any point not on the line and so if i pick a line that’s already

got two points on it and so how many points will be off the

line because we have four points uh total

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so we’ll have to verify for each of the six lines and then each line

i’ll have to pick uh the other four points that are not on it for

example this would be one case here this could be a and b

and then this could be c or it could be d and so those would be two different

cases so we’ll have to check line one and the point c

that will be in one case and line one and d

and then for each of the lines there’s going to be 12 cases in total for l2

which has a and c on it it doesn’t have b on it so that’ll be a different case

and then l2 and then we’ll have a d case it’s all together we’ll have 12 cases

and we can by brute force go and verify the equilibrium parallel property so

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let’s see one case so for line one and point c not on line one there has to

exist a unique line that passes through c that is parallel to l

and that line is l6 so l6 is parallel to l1 and it passes through c

so the evidence right here is l6 right here and so there’s the unique line right

here so for each of these lines and points

we’re going to have a unique line here and that’s exactly what we need for the

euclidean parallel property so let’s just go and check one more case here how

about l 2 and d so here’s a line and a point not on l2

and i’m saying that there’s a unique line that passes through d and it’s

parallel so what is that line that line is uh l2 is uh it has to have d in it so

let’s go with uh l5 here right so l5 here

because l2 and l5 are parallel they have no points in common and this right here

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passes through d so for each line and point here in all

12 cases you’ll come up with a unique line and once you go through all 12

cases then you’ll see that the euclidean parallel property holds all right so

there’s some nice geometry right there the four point geometry now

again to have to make sure this is a geometry we specify that

we verify that a1 a2 and a3 hold and so then we have the word geometry

and then once we know it’s an incidence geometry then we can try to look for

some more patterns and and so far for three-point geometry

the elliptic parallel property held in four point geometry the euclidean

parallel property holds and so now let’s look at five point geometry

so five point geometry says we have five points and now we’re going

to call them a b c d and e those are our five points

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and in this geometry we’re going to have 10 lines

and here are our lines right here here’s all the lines

again each line has just got two points on it and that’s it

now you might say why not choose three line or sorry why not choose three

points on each of the lines and why not have seven lines

well the number of points and the number

of lines and the number of points on the line are all determined by these three

axioms here so you have to verify the axioms you

can’t just write down anything random and it’ll work

you have to make sure the axioms hold so you might ask are there any other

five-point geometries and that would be a really good question

and you have to just play with it and see and find the geometries so you have to

have all three axioms axiom two says there has to be at least two points on

any line so that’s the bare minimum right here i got two points on every line

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and a three here says not all points can be on one line

so those two are pretty uh nice um to verify but then you also have to a

have to have a1 through each point is only one line so this line

has a and b on it so no other line can have both a and b on it so this has a

and c and a and e and a and d but because this one has a and b on it

no other line can have a b on it so when you start trying to find these

geometries that satisfies all three of these you know that’s going to determine

what’s possible and what isn’t possible to be called the geometry

now this geometry right here once you verify that axioms a1 a2 and a3 hold we

can call it a geometry and we’ll have some very um

specific statements that will hold for this geometry right here

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so um we can check that the axioms are true they actually hold

and we’re going to call this five point geometry

and here are some additional statements that hold for this particular geometry

for this particular model so each point has four lines on it

so let’s see if we can verify that now remember this is case by case it’s a

finite geometry so we can do this by brute force

each point has four points on it now it says each point so we got to verify each

point and there’s five cases to check so we’ve got to check all five cases that

this is true let’s check for case a does a have only four lines on it not

three not five exactly four so here’s one two three

four so that held and once we check all the cases then we’ll know this statement

is correct let’s check another point let’s check point b one two

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three four so again that holds and in most geometries you’re going to

work with day-to-day that aren’t these finite models they’re going to have

infinitely many points and so we’re not going to be able to go

point by point by point but in this model we can

c how about d all right let’s just check d real quick one two three oops

one two three four there it is so each each one is you have to verify

each one of these go through there and verify each one

all right another statement that holds for five point geometry is

each distinct line has three lines parallel to it

so this says each just each line right so now we got to check all 10 cases l1 l2

all the way to l10 let’s just check one of the cases here

just so that you can understand l2 has three lines parallel to it so let’s

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see what those three lines are so l2 has b and c

so this is claiming that there are three lines that don’t have b and c on it

this one does has b on it this one doesn’t have c on it oh here’s one so l4

and l2 are parallel so l4 is one of them um and then l5 no l5 is not parallel

they both have c on it um l6 they both have b on it l7 they both have b l8

l8 and l2 are parallel so here’s another one l8 and what about l9

l9 they both have c on it and then l10 so there are the three

so we verified for the case l2 it actually has three lines parallel to it

one two three so we have to go through each one of these cases and find the

three and then we’ll know that this statement right here is true

and then the last thing to check is that

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this actually is a geometry in incidence geometry

where axioms a1 a2 and 83 hold and the hyperbolic parallel property holds

so i’ll just verify this for one case um well actually how many cases will we

have we’ll probably have a lot of cases let’s

see if we can count all the cases really quick

so we have to go through and verify for each of the lines so i’ll pick l1 first

and then for each point not on the line so this has got two points on it

so you know if we take away five minus 2 we got 3 left so we’ll have l1 and

c d and e are not on it so l1c l1d l1 e and this will be true for each line

there’ll be three cases so all together there’ll be 30 cases

so in order to prove the statement right here we’ll need to check all 30 cases

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if all 30 cases work then all that evidence will add up to a proof

all right so let’s just check this case right here l1 and c so here’s l1

at every line every point not on the line i have to find more than one line that

goes through c and is parallel so i’m looking for a line

that goes through c and is parallel so this line goes through c and it’s

parallel so l3 works so i’ll put that evidence over here l3

uh what’s another one it has to go through c

what’s another line that goes through c this one goes through c but these are

not parallel to each other they both have a on it

goes through c goes through c all right this one goes through c and is

parallel to it so l9 so there we go we have more than one

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in fact we have two l3 and l9 are both parallel to l1 and they pass through c

so here’s an example of uh the hyperbolic parallel property here’s a instance

instantiation of it forget that word anyways check all 30

cases just like we check this one case right here and you’ll see that the

hyperbolic parallel property holds all right so what we’ve done so far is

we’ve found three point geometry four point geometry and five point geometry

and in each one of those three point geometry the elliptic

parallel property held in four point geometry the euclidean

parallel property held and in five point geometry the

uh hyperbolic parallel property held and so what we said is what we found is is

that if you have axioms a1 a2 and a3 in other words if you have an incidence

geometry you could have any one of these three

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holding of course you can’t have them more than one holding because they’re

mutually exclusive but just because you have these three

holding then you can you may have this one you may have this one you may have

this one these are independent of each other

because we found models where this one is true and this one and these two are

not true and and so on and so what we have here is that

these show that all of these uh three um statements are independent from each

other um the three parallel properties are all

independent from the incident axioms all right

so uh now let’s talk about some more models click right here

right here and we’ll get started and i’ll see you in that episode