Models of Incidence Planes – Hand-Shake Planes

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

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

00:00
hi everyone welcome back i’m dave in this episode models of incidents planes
handshake planes um it’s part of the episode uh incidence
geometry tutorials with step-by-step proofs so let’s do some math
so first off um i’m going to review uh what the incident axioms are
just very briefly we talked about these in uh all the episodes so far
uh especially in the first episode but the first axiom says um for each um
for every point for every two distinct points there’s a unique line going
through them every line has at least two points on it
and actually made three says there exists three non-conlinear points
and if we have axioms 1 2 and 3 holding then we say that we have an incidence
geometry we’ve gone through each of these
theorems here step by step detail in the previous episodes

00:01
and we also talked about parallelism so we talked about these three parallel
properties and we also gave models uh different models
so in three point geometry we gave a model for that
and we we talked about how that has the elliptic parallel property
and we talked about four-point geometry and it had the euclidean parallel
property and we gave some more in some other episodes
and five-point geometry had the hyperbolic parallel property holding
so i definitely want to check out those episodes the link is below in the
description um and and so i just want to briefly re mention
what the three point geometry was the points where a b and c and we had three
lines and the four point geometry is we had
six lines and we had points a b and c and d we had four points and six lines
and in five point geometry we had points a b c d e and we have ten lines

00:02
now one thing in common for all three of these geometries we talked about how
axiom one and two and three hold for each of these geometries
and we also talked about how this had the um
elliptic and then euclidean and then the hyperbolic parallel property for each
one of these all right and so the topic for today is
the handshake plane handshake plane is an incidence plane in other words in
other words axioms one two and three are holding
um but every line uh has to have exactly
two points and so that’s why i wanted to kind of refresh your memory on these uh
ones right here because each each of these have
two points on their line now in previous episodes we also talked about the um
fianna’s geometry and young’s geometry and those are examples some geometries
where there were more than two points on each line
but in today’s episode uh three four and five are examples of handshake planes
so to be a handshake plane all you need to do is to have axioms one two and

00:03
three hold and then every line has also two points
on it so there’s an extra condition for that to hold
so in the three examples below above we see that the three four and five
uh point uh geometries are actually handshake geometries
and 3 1 satisfy the elliptic and the 4 1 the euclidean and the 5 point geometry
satisfied the hyperbolic parallel property here so here’s the theorem about the
handshake geometries if you have five or more points then it
must satisfy the hyperbolic parallel property
and we did a geometry the five point geometry and we saw that it satisfied
the hyperbolic parallel property if it has three or four points
then we saw that it didn’t satisfy this but if it has five or more points in
your geometry and it’s a handshake handshake geometry or handshake plane in
other words every line has exactly two points on it

00:04
um and so let’s see um why that’s true so if we take a line so remember what
the hyperbolic parallel property means uh let’s back up and not start the proof
yet what does this right here mean let’s just review that real quick
uh so if we take any line and any point not on the line any point
p not on the line there has to exist more than two or more lines through p
that are parallel to l so m and l are parallel and n and n and l are parallel
and they pass through p for any line for any point p
there exists at least two lines passing through p that are parallel to l
and that’s the hyperbolic parallel property so why does a hyperbolic
parallel property have to hold in a handshake plane with five or more points
well first off is we need to take an arbitrary line so suppose p is a line

00:05
and p is a point not on the line sorry so that’s how we set up our proof um
now there exist points two points on the line
that’s my axiom eight two every line has two points on it
so let’s um kind of follow along with the proof right here so we’re going to
take a line and a point nut on it and axiom 82 says we have two points on
every line in fact this is a handshake plane so it
has exactly two so i can say equals right here so this is um there’s exactly
two points on this line right here i use set notation here um other
notation that we’ve used in the past is a b with an arrow double-sided arrow
over it but because we’re in a handshake plane
we can say equals here we can we can say more than a and b are on the line
we can say that this line is equal to l and i just use this i think it’s more
intuitive at this point because it’s saying that this is the only two

00:06
points on the line and it’s because it’s a handshake plane right so
we know exactly what our line l is and we know p is not on l
so um now there exists at least five points and we only have three down here
so there has to be two other points somewhere q and r somewhere
um but actually q cannot be on l because every point has got exactly two lines
and same thing with r r cannot be on l so they’re somewhere off l um
so they’re not going to be equal to p a b because there’s three points
already and we know we have at least five so i got this q and r somewhere so
i can say draw a q here and say an r over here and
we can make the lines remember we can make a line through any two distinct
points let me put q up here and so i can draw a line through p and q

00:07
this will be line m and notice that line m here is parallel because
it does not have a or b on it so here’s line m and this line is parallel because
there’s only two points on it in other words i i don’t need to keep drawing it
right there’s just two points on it i draw it like like maybe there’s a whole
bunch of points on it but that’s really just to just to see a line it’s really
made up of these two points right here and same thing with n um i can draw
and if i draw it like this it looks like
there’s a point there but there’s not so let me just put the r over here
just so it doesn’t break our brain brains here all right
so there’s a line through pnr which just consists of p and r because we’re in a
handshake plane only two points on that line only two points on that line so
here’s two lines that pass through p and they’re both parallel to l they’re

00:08
parallel to l because they don’t have any points in common there’s only two
points on this line and two points on this line and all four both points are
different from each other all right so there’s our proof right
there take any line l take any point p on it i can find two lines that are
passing through p and parallel to l so that’s an interesting thing there
it always has the hyperbolic parallel property holding okay so um one more thing
if we actually have a finite number of points
then we can actually count the number of lines
and here’s the number of the lines so right so if your handshake plane has say
20 points in it then we can actually count the number of
lines in the geometry it’ll be 20 over 2 which will be 10 10 times 19
so if it had 20 points there would be let’s see 190 lines

00:09
um and that’s just by plugging in 20 here 20 points and 190 lines so
rather than using letters you might use numbers as your points for example one
two three four here are all my points 19 20. so those
are my points right there and i can start building my lines what my lines
look like one two would be a line and i’d call that line say one
line two would be made up of say one three line three would be made up of say
line one and four and i could go all the way to say line twenty
which should be made up of one and twenty well let’s see here we didn’t quite

00:10
match that so um the first line is line 21 so one and two so maybe line 19
there we go there would be my first 19 lines
made up of these two points these two points and these two points and then
line 20 we would start at two so two and we would do two and one again
or two and two we would do two and three and 21 would be two and four
and so on until we got to 190 lines right there just as a
so you can see how that works if we choose a smaller end we can actually
write them all out for example if we choose five
then well that’s five point geometry you already saw all the lines for that
if you choose n equals seven then you could say here are my points
and how many lines would we have seven minus one would be six six over

00:11
two is three so we’d have three times seven which would be 21 lines
and so we could write all those lines out the first line would be 1 2
and then the next line would be 1 3 1 4 dot 1 7 and then 2 3 2 4
all the way to the 21st line so each of these would be lines
and you can name them however you wanted to name the lines
line one line two line three so on in any case let’s see why this is true
as long as you have a finite number of points these are going to be the number
of lines and handshake geometry with endpoints has so here’s how we came up
with this right here this formula right here because if you add up all these
numbers right here you get this number right here
so this is a nice form because it’s just simply multiply and divide and this is
adding up a whole bunch of things but in fact they’re actually equal to each
other if you’re not sure why these are equal to each other

00:12
check out my series uh on mathematical induction where we actually go through
this and prove this is true here um but what we’re going to do in this uh
proof here is understand the left side and then we’ll leave to the
reader or the viewer here how these two things are equal to each other right so
so this is important right here sorry so we can connect the first point to the
other n minus one points so let’s do say an example with say
six points because we already did five so
um this would be six over three six over two would be three
three times five said be fifteen lines so six points and fifteen lines
so let’s see why that’s true right so again we list out all of our points
one two three four five six and now i’m going to start making lines so
i’m going gonna pick this one right here and in fact let me put my points over

00:13
here one two three four five six so my geometry has got those six points and
now i’m going to start making lines and the way i’m going to pick the lines i’m
going to choose the first point right here which is just called a one and i’m
going to connect it to the other and how
many others are there there’s five right
so if there’s n then there’s n minus one points to connect to so i connect this
one to a two there’s my first line one to a three one to a four one to a five
one two is six then i move on to the next one right here which is two
now how many can i connect two to well it’s not n minus one anymore now
it’s n minus 2. so that’s the next line in our proof you
know once you check all the ones off then you disregard the 1 now you start at 2
and then you connect 2 2 3 2 4 2 5 2 6. and so we had here how many did
we have here we had 5 we have n minus 1 and then here we have n minus 2

00:14
and then how we could have here n minus 3 and we’re going to have n minus 4
and then we’re gonna have n minus five so the last step right here will be the
last line to consider is the one between the nth minus one and the nth point
so the last one would be the five and the sixth right there’s six
points in the six so the in myth one and the six so the last line after we made
all the lines out the last line would be that let’s say the fifteenth line
would be five would be made up of the numbers five and six so that would be
the last line in my list of lines so i would start out with l1
uh which would be one and two and then the last line after we made uh
you know all the ones in here the last one would be five and six there
all right so um again that’s how we count them all up you know the this one
only has one then we’d have two and the last one give us the five right so we
add up all those numbers right there and

00:15
this is just a nicer easier way to write all of this out right here okay so
again if you’re stuck at this equal sign right here why this is equal to that
check out my series on mathematical deduction where i prove a lot of results
like that right there all right and so this is a nice handy
way of calculating the lines for example if there’s 10 points i can easily
determine this right here if there’s 10 points i can say that it’ll be 10 over 2
which will be 5 5 times the 10 minus 1 which is 9 so it’ll be 45 lines
this is so handshake planes are kind of nice
if you know this is a finite number of points say 10 i can immediately tell you
how many lines there will be all right um i was just wondering about
this right here so is there an incidence geometry with infinitely many points so

00:16
we just talked about handshake with endpoints could we make up one with
infinitely many points so let me know what you think in the
comments and to see if we can do that um but let’s just throw some ideas down
here and see what you think about them so let’s take for our points just the
natural numbers one two three and so on um yeah
so let’s just take all those just infinitely many numbers here um
can we make a handshake geometry out of this
can we um so let’s call this our script p for our points
and for our lines let’s use a script l and our lines will be made up of
these two points so we’ll make this a handshake geometry so we’ll say it’ll be
made up of x and y where x and y are natural numbers so just two numbers

00:17
and then of course make them not equal to each other
and so these will be the lines so for example 1 2 is a line or 35 48 is a line
35 48’s another line and we can just keep doing this so all we need to do is
to choose a number from here and another number from here and make sure they’re
not equal and then put those two numbers together
and put them on a line and there’s a line
so script l is just made up of all these little two sets
and there’s infinitely many lines there’s infinitely many points this is a
handshake geometry how do we know that well first of all it satisfies axioms a1
a2 a3 why well because if i were to choose say two one
right there they’re both in here two’s in here and one’s in here and they’re
not equal but these are the same set though right they’re the same line so

00:18
there’s only going to be one line going through um
two lines two two numbers or two points um and every line will have two points
on them because i’m choosing two things not equals so each one will have two
points and it’s easy to find three non-collinear points for example
um points one two and three are non-collinear there’s no line that
can go through all three of them because
lines only are made up of two numbers at all
so yeah axioms a1 a2 and a3 hold it’s a handshake geometry it’s got more than
five points so we know it has to be hyperbolic um
and so yeah let me know what you think in the comments so for instance you know
we’re just saying we have points we have lines
and for a point to be incident just means it’s inside the
um you know set like for example one is incident with this line and two is
incident with this line and 35 is incident with this line

00:19
in any case um that’s a lot of fun so let’s do um some more um
geometry right here check it out let’s do some more math

About The Author
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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.

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