# Models of Finite Incidence Planes – Hands-on

(D4M) — Here is the video transcript for this video.

00:00
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
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
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