Models of Finite Incidence Planes – Young’s Geometry

Video Series: Incidence Geometry (Tutorials with Step-by-Step Proofs)

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

hi everyone welcome back today’s uh episode models of finite incidence
planes we’re going to be looking at young’s geometry
and this episode is part of the series incidence geometry tutorials with
step-by-step proofs uh in this episode we’re going to continue giving some
models and so let’s briefly review what incidence geometry is so we’re going to
be looking at these three actions right here and we’re going to be calling them
an incidence geometry so for every two points there’s a unique line going
through them for every line there exists two points on it at least
and there exists three non-collinear points so those are our three axioms
anything that holds for these three axioms we’re going to call an instance
geometry and we’re going to continue looking at some models now in previous
episodes we proved these 11 theorems hold for any incidence geometry

we did these step by step in previous episodes and then we talked about three
additional statements that may or may not hold in fact we get we gave a model
three-point geometry where the elliptic property held
and the four-point geometry where the euclidean property held
and in five point geometry these were done in previous episodes
um and then in our last episode we talked about finals geometry and the
ellipt and we showed that the elliptic parallel property was true or we
discussed uh that is true all right and so today let’s
talk about young’s geometry what is young’s geometry so first of all it’s a
model for incidence geometry that means axioms a1 a2 and a3 hold
and what we’re gonna have is nine points or nine symbols each one of these are
going to represent points in our in our geometry so uh a through i
and we’re gonna have the meaning of incidents is going to be

any of these points as uh on a line and it consists of one point and two
other points and so here’s our um you know actual
concrete lines right here right they’re just spelled out for us all the way
and so we can understand the uh statements of our geometry much more
clearly and we can see what our points and we can see what our lines so we have
nine symbols and we have 12 lines and in order to call this an incidence
geometry we need to show that axiom a1 a2 and a3 hold
so let’s discuss this first um so i always think a2 is the easiest
um because what it means is for every uh for every line there’s at least two
points on it right so as we can see here every line has three exactly three
points on it so we definitely got a2 you know each line has at least two points

on it and you can just see that all right so just by observation
uh what about a3 so a3 says there exists three non-collinear points uh a b and d
what about them are they non-collinear let’s look for a line
and see if a b and d are on one line uh so a b and d anyone
and so as we check here for a b and d all in one line and we can see we don’t
have them on one line so they’re non clean year
now could there be other points that are non-clinical absolutely i’m sure it
would be easier to find three more but axiom e3 just says you have one pair uh
or one triple right these three are non-collinear so eight axiom a3 is done
um you know notice the difference between a2 and a3 a2 i had to check
every line had at least two points a3 is in existence it just says there exists

three you can name any three that you want and they’re non-linear
all right so now what about a1 here so a1 says
for every two points there’s only one line going through it
um and so when i have these two points here a and b uh a and b can no longer be
on any two any other line right because they’re on
this this line here so if i check any other line a and b better not be on it
so if i check all the other lines there’s no a and b on it
right this one’s got an a on it and this one’s got a b on it but a and b are on
both on this line and there’s no other line where a and b on it well that’s one
case what about a and c here’s a line with a and c on it are
there any other lines with a and c on it no
right so there’s a unique line so i’ll write the unique line right here it’s l1
and l1 and what about a and d so axiom a1 says we have to go through every two

every two distinct points every pair and find one and only one line that goes
through them what about a and d well we have l2 right here goes to a and d are
there any other lines that go through a and d no so it’s just l2
and so then i’ll check a and e and then a and f and
all the way to i right so a and g a and a and h and in each case we will check
how many lines go through it this should only be one let’s check a and i
how many lines go through a and i so then then here’s one line that goes
through a and i it’s l4 are there any other lines that go through a and i
always we check here none of them do so a1 is satisfied for this case and if
we check every single case then we will see that a1 holds all right

so you got to go through all the pairs there and this is a brute force technique
of just by looking at a concrete geometry and checking every single one
here um right and so then once we’ve gone through all the a’s and checked out
all the cases we can start with b and we can check bc bd be bf
and we can check all those cases and make sure there’s only one line that
goes through uh both of those um points there and then we’ll and then
we’ll uh finish the b so then we’ll do the c’s and d’s and e’s and f and all
the way and you have to check every case so i i’m assuming that you’re going to
go and check all the cases and make sure that this is right here
i’ve already checked all the cases i’m not going to put all the cases inside of
the episode here all right so now we can say that this is a
incidence geometry young’s geometry is an incidence geometry and we got the

point so we got the lines and we verified the axioms a1
we verified case by case by case you have to verify each and every case if it
fails even in one case then it’s not an incidence geometry all right so
we checked all cases here i checked all cases here
all right so now what can we say that’s additional
to young’s geometry not only is it an incidence geometry actually there’s a
lot of incidence geometries so some of the geometries um had
different properties for them for example three point four point five
point and fiona’s geometry right so they had different parallel properties right
so each one each geometry that you look at
who knows which parallel property might hold true so and young’s geometry here
so the first thing is the first property is each point is incident with exactly
four lines right so this is each point right so

what are the cases that we have to check
to see that this statement is true so we have to check point a b c d e f g
h and i and for each one of these we have to
come through and show exactly four lines so let’s count what lines um are is are
what lines is a on right so one two three four so l1 through l4 1 l2 l3 el4
uh is a on any more of them just physically check just observe right a’s
not on any more of them so it’s exactly 4. what about b is b on any of them um
one let’s write it down a one and then l five and then l six l seven
and then l8 no no no uh and we got our four lines right there

we got our four lines right there so we have to check all the way through to i
let’s check i all right i’m saying this right here
statement is saying exactly four lines so what are the four lines that i is on
one two three and four and so we just write them down right
here now four this is the evidence right here one l6
and then usually evidence doesn’t add up to a proof unless you’ve checked all
cases each and every single case and so i’m going to write dot dot here
because i have confidence in you that you’ll sit down and do this l4 l6
l8 and l12 pause the video and fill in these steps right here for us
all right the next uh thing that i noticed is that
for every two distinct points there’s exactly five lines that are not incident

with these points so what are the cases we have to check
for every two distinct points right so we’re gonna have to check the
case a and b two distinct points right and i’m saying there’s exactly five
lines that don’t go through either of these so let’s see here
no this one goes to a a a b b b and here
we go one two three four five these five
lines right here do not go through a nor b so l eight through l
so exactly five lines right there one two three four five so what about ac
is there exactly five lines that do not go through these points let’s check a
and c it’s got a a uh here’s one of them so let’s write down here l5 and

a and c i’m looking at so l6 and l7 let’s write those down l6 l7 um l8’s got c
c c and then l 11 and l 12. and there’s five lines right there that
do not go through a and c so you have to check every single case here
all the way to f and g what are the five lines
that do not go through f and g right so l1 l2 goes to g um l3
l3 and then l4 goes through f and l5 and then l6 and that goes to g and then l8

and then goes g f f g and so there we go those are the five
lines there one three five six and eight those are the three fives uh three the
five lines for that case but we gotta check every case in here to make sure
that this line that this statement 2 is true alternatively
you could try to prove a1 a2 and a3 and try to prove step number two
however you will fail there are other incidence geometries
that satisfy a1 a2 a3 in fact every incidence geometry has to satisfy these
but this property right here does not hold for every incidence geometry
so you cannot prove statement two from these uh statements right here you
would have to make some additional axioms and you would have to actually
verify those axioms with these points here and these lines

all right step three the euclidean parallel property holds so what that says is
through every line any line at all in any point not on the line
there has to be one unique line that goes through p that’s parallel and so
in order to check the euclidean parallel property there are a lot of cases
because you got to check every line and for each of those lines for all 12 lines
you also have to check every point that’s not on the line so let’s make
sure we have an understanding of what all the cases would be
so again we have to check every line so we would start off with say check line
one check line one and we need a point not on the line now
line one has three points but how many points are in our geometry we got nine
so that means we’ve already got three so that means we need
to check six cases one two three four five six six cases there so a case for d

and e and f right and then g and then h and then line one and then i so
i’m checking line one and a point not on it
and as you can see for line one there’s going to be six cases we have to check
so six cases and then how many cases do we have total
right because we got 12 lines so we’re going to have 12 times six cases total
cases so as you can see we’re starting to get to the point where
uh this is not necessarily fun to go through all these brute force cases and
check and so um you know imagine how we check
things when we have infinitely many points right we’re not going to do
things by brute force we’re going to have other methods for doing this and
i’m going to show you those methods in an upcoming episode of upcoming episodes

so this euclidean parallel property you have to go through and test twelve times
six total cases and we will find out uh that it works
let’s just look at one case for examples these you can go through all of them on
your own so l one and d right l1 is a line and d is not on it and what we’re
going to say is the euclidean parallel property says is so now this is l1 and
this is d and there has to be one line through d
exactly one line that’s what euclidean parallel property says one line through
d that’s parallel to l one so let’s check uh here’s line l1
so to be parallel means it cannot have an a b and c on it it has to go through d
so this goes through d but it has an a on it so that doesn’t count
uh that doesn’t go through d doesn’t go through d
all right here this one goes through d but it has a point b on it so that one
doesn’t count uh are there any other lines that go through d
here’s a line that goes through d right here

and it has a c in common so that one doesn’t count
um here’s a line here d that goes through d
and it is parallel so this would be the line l1 through here or sorry l11
and it’s the only one l12 doesn’t go through d right so you
know this case right here works because of l1 so i’ll just draw an arrow l1
right here so l1 is the unique line that passes through d that’s parallel to l1
and we go through all of these cases and we find the unique line that passes
through the point not on the line and then we’ll verify the euclidean
parallel property holds for young’s geometry
all right well that’s it for uh this episode i look forward to seeing you in
the next one click right here for the next episode i’ll see you there

About The Author
Dave White Background Blue Shirt Squiggles Smile

David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

Let's do some math together, today!

Contact us today.



About Us


© 2022

Privacy Policy | Terms of Service