In an Affine Plane, All Lines Have the Same Number of Points

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 in an alpine plane all
lines have the same number of points we’re going to prove this fact right here
and uh let’s do some math let’s get started
so i’m going to begin by talking what an alfine plane is and by that i mean i’m
going to review very quickly in the previous episode of the series we talked
about what alpha planes are in great detail and so i just want to quickly review
now for an incident plane we have actions a1 a2 and a3
and for an alphine plane we have axiom a1 the same axiom up here
and we’re going to switch out a2 and we’re going to put axiom pa and then we
have the same axiom a3 now keep in mind that uh this episode is
part of the series incidence geometry tutorials with step by step proofs link
below for the full description i mean for the full series is below in the

00:01
description so you definitely want to check out um
the previous episodes and so we’re going to start on our new
new episode it’s going to be about proving
all lines have the same number of points um so here’s what incidence geometry is
or or an incident plane and an alpha plane here and the main difference is
the pa now in the previous episode we we proved that if we put pa here
then uh axiom a2 will be a theorem in uh this geometry here and we proved that
last time so we proved this theorem here in an
alpha plane you automatically have the clitorian parallel
property which i’m abbreviating is pa right here so we have that right here
and we have the incident axioms holding so we proved the axiom a2 has to hold if
you assume axiom a1 and then pa and then p and then uh

00:02
and then a3 yeah all right and so we also said that
um if we look back at the episode on parallelism if you have an incident plane
with with the euclidean parallel property then you get that parallelism
is an equivalence relation so definitely check out that previous episode also
and then the very last theorem that we did on the very last episode is we
approve this theorem right here and so yeah check out the previous episodes
all right now this series all started by proving these theorems here
and we proved in the last episode that these theorems not wholly only hold an
incidence geometry but they also hold the alpha planes
so good so we got um all that going now as far as the question goes do all lines
have the same number of points well it just depends upon where your lines are
in an incidence geometry when you have axiom a1 a2 and a3

00:03
then it’s very clear that this is not true
and the way the reason why i say that’s clear is because we’ve already given
several models one of the models we’ve given for incidence geometry was the
handshake plane and this was an example of a five-point geometry uh and we
called it the handshake plane it um had two
points on every line and so in this in this geometry right here the um
um every you know the the answer is yes there’s the same number of points on
every line there’s two points on every line line one through line ten but
we also saw examples of straight fan planes and here’s an
example of a straight fan plane and both of these geometry satisfies a1
a2 and a3 so if you’re an incident plane then the answer to this is no
there are some geometries where and here’s an example of one this line has

00:04
four points on it and these other lines only have two points on it so not every
line has the same number of points in this geometry so if you’re in instance
geometry and you have these three axioms holding it may happen that you have the
same number of points on every line or it may not happen that you have the same
number of points on every line however if we replace axiom a2 with
the euclidean parallel postulate or as i’m abbreviating it pa
then this cannot happen right here you you will have the case where um
all lines will have the same cardinality in other words uh every line has the
same number of points on it so it’s important to to realize that um um actually
i got a that should say uh i’ll find right there so let’s see if we can go
fix this right here all right here we go in an alphine plane

00:05
right if we put pa here now we have an alpha plane
uh in an alpha plane now we’re going to have this as a theorem and this is the
main subject for this episode right here in an alpha plane all lines have the
same cardinality and we want to prove this right here right now now in case
you don’t know what this word here means let me uh
talk about that for a second here so these theorems all say the same thing
in an alpha plane all lines have the same number of points
and enough i’m playing all lines have the same number of points
in an alpha plane all lines have the same number of points
okay so i i wrote it different ways depending upon
what kind of terminology you’re familiar with if you’re familiar with this word
right here well um i’m going to take that to mean this right here
for this episode so we have a bijection between any two lines and if you’re not

00:06
sure what that means well then in this episode i’m going to take that to mean
you have an onto and one-to-one function there’s some different terminologies
that you can use here but this is basically what i’m going to go with i’m
hoping that this third one here you have seen the word onto and you’ve seen the
word one-to-one before if not let me just give you an intuitive
impression of what this means so if i have a mapping
let’s say i have a mapping from let’s get a different marker here let’s
say this is a set a and this is a set b and i have a mapping
what does it mean to say on to so you know normally we’ll call this the
domain of f and when we map f we’re going to get some outputs
and these will be the outputs right here this is called the image of a and

00:07
b might be a larger set so b is the codomain is the codomain
and f a which are the outputs so you want to think outputs is called the
image of a is called the image of a the domain which is the inputs so
you want to think a is the domain or the inputs
and f of a are the outputs so we’re going to map from a into f of a and we
collect together all the outputs together into this set here it’s called
the image of a under f and b could be a larger set here so b is the domain
so to give you a concrete example of this we can take the real numbers here
and we can map it into the real numbers over here
so i’ll just draw the real numbers up here actually let me just

00:08
um i’ll just draw a number line here in fact why not just give a concrete
example let’s say sine x so now we’re going to graph
from the real numbers we can input any real number we want in here this is the
domain which is r all real numbers so i just
drew it like that and the domain here sorry and the outputs are between 1 and -1
right so there is the uh image of r under sine that’s the outputs right there so
um you can say that you know we can define f to be a function
f of x equals sine x and the x’s are inputs here and the range
which is is this part right here this is the range the range of f and

00:09
so the point is is that here the codomain here is all real numbers
and you know like two is so you can see this function right here is not onto
because there’s something that is not in the outputs like two and you know so
so the range doesn’t necessarily have to be equal to the co domain and if you
have something out here then the function is not one to one sorry it’s not onto
if the f of a is equal to the b then your function is going to be onto
so for example if i have something like x to the third
then this function will be on to because if we sketch it you can see that
no matter if we use a 1 we get an input no matter if we use a 2

00:10
we get an input no matter what we choose along the outputs or no longer
no matter what we choose in the co domain we’re going to get some input for it
and so that’s going to be the intuitive idea of onto so
i guess to be more formal let f be a function from a to b then f is
um said to be on to whenever whenever the following condition holds
no matter what you choose and be for all x and b
uh there exists uh something in a i’ll call it an a and a such that f of a
equals to the b equals to the output now
let’s change this to a y and change this to an x

00:11
and then that’ll look more familiar to you over here
okay so yeah this is just a definition of onto no matter what you choose over
here in b there exists something in a that mapped to that
so there exists an out so for everything in b it’s an output of some x
okay so that’s um on to there and then um
the one to one we’ll also get to here in a minute let’s go ahead and get started
on this proof here all right so let’s reword it now and i’m
going to choose particular letters here if l and m are lines and they could be
any lines at all then there existed on 2 and one to one
function between them and that will give us our bijection which will give us the
same number of points okay so here’s here goes the proof
so first off we’re going to assume that l is not equal to m that we have two

00:12
different lines if they’re the same line then we can just use the identity
function and then you know the identity function
is the bijection because you’re just mapping everything on the line every
point on the line you just map it back to itself and so whenever you do that
you just get the same the same point so go ahead and assume the lines are
different otherwise you’re you know if you only have one line in your geometry
well anyways i guess you can’t have that but anyways
let’s just assume the lines are different all right so by theorem one
l and m are um gonna have a unique point in common and that’s case one
or they’re actually going to be parallel to each other so i’m going to have two
cases in my proof here case one will be they have a unique
point in common so we’re going to have line l we’re going to have line m
and we’re going to have some point there in common to them
and then case two we’ll worry about if they’re parallel all right so we’ll do

00:13
case two last all right case one so here we go l and m are um
um non-parallel by theorem one we got our point o now
remember also we have axiom uh not axiom anymore but we have um a2 is holding
um in other words every line has got two points on it and so i can say i have uh
let’s make this come out a little bit further and a little bit further so
let’s say i have here point a and over here point a prime
and let me move it up a little bit and up a little bit more and this is a line m
and this is line l okay so every line’s got two points on
it and so m also has two points on it and these are not equal to each other
and the reason why is because a is on line l and a prime
is on line m and the lines intersect at only one place so we have two uh two
distinct points here a and a prime and these two are on line m and these two

00:14
are in line l okay and by axiom a1 we have a line
going through those two points there so let me draw the that line right there
and i’ll put a prime back here a prime right there
and i’ll just put an a right here all right that’s we got so far now we’re
trying to make a bijection between line l and line m
and i’m going to use this triangle and right here to do that
we’re not really using triangles it’s just it looks like a triangle the way i
draw it here but you know we haven’t said anything about triangles in this uh
and these geometries yet so all right here we go now to make a
bijection i need to pick a point on a line just an arbitrary point and i need
to map it to the other line or i need to make a function
right so to do that i’m going to pick x to be a point on line l just any point

00:15
at all and it’s not going to be o and a so i’ll draw it down here
so x is just any point on line l other than o and a
and i need to show how to map that function how to make a function how to
relate a point x over here to a point on m and i need to do that in a way
in which i get a unique point over here because if we take an x and we get lots
of points over here then we won’t have a function
so we need to get exactly one point over
here on this line m so how do we do that so you might say oh just draw a line
right well what gives us the right to do that the axioms okay so here we go so
notice that um we have this line right here through a prime and a and
we’re going to say there exists a unique line k through x

00:16
and it’s parallel so all we need is that x is not on this line right here
and so then by axiom um by p4 by p4 in fact i actually want to put that here
it says there exists so i want to say all right let’s see if that did it
so uh by p4 i left off my t there all right by axiom p4 there exists a
line i’m going to call it k that goes through this x here
so here’s this k right here line k and it goes right through the point now
notice i didn’t draw it hitting line m yet because i don’t know that it

00:17
hits line m yet but i know that here’s a line and here’s a point not on the line
and so there’s a line that goes through it that’s parallel so k is parallel to
this line right here by p4 i have this line k so by p4 there exists a k
that passes through x and it’s parallel to this line right here all right good
now what i’m going to do is i’m going to suppose there is no point here in other
words suppose k and m are parallel now we really don’t want that to to happen
because remember we’re trying to take an input on this line and get an output on
this line so i really want them to intersect so what i’m going to do is i’m
going to see if they don’t intersect what happens if k and m
are actually parallel what could go wrong with that so in
order to get them to intersect i’m going
to suppose they don’t intersect so let’s suppose they don’t intersect suppose

00:18
they’re parallel so by p4 again so we have that k is parallel to m and um
so k is parallel to m k is parallel to this one right here already
by the way that we constructed k k is parallel to that
so if we got two lines parallel and they both pass through um right so m [Music]
so uh the lines so let’s see here um m passes through a prime uh i got my
diagram here in the way uh since both lines lines m and these two
pass through a prime and are parallel to k so that tells us that both of these
lines right here are equal let me draw this again a little bit out
of the way here so this was the point here x

00:19
i’ll see if i can squeeze in here x here and here’s the line coming through here
k which is parallel to that line right there and we want them to be parallel
here i mean we don’t want them to be parallel to k and m so suppose they are
parallel but then the m and the and the k here they’re both parallel to
this line right here um if you assume this right here so both lines m
and these and this one right here pass through a prime and they’re parallel to k
right so definitely this one is parallel
to k because that’s how we constructed k
and if we assume that m is parallel to k
then they’re both parallel to k and they both pass through a prime so they have
to be equal to each other what’s so wrong with that well if if they’re um
you know if they’re equal to each other then um you know

00:20
a is on m but a is not on m and a was on l and m and l only intersected at o
so we cannot have these two lines equal to each other because that would put a
on m all right so this cannot happen by p four p four says this has to happen if
this if this holds um and so we can’t have that and so
these two right here cannot be parallel this cannot happen right here and so
what that means is uh by theorem one we’re going to get a k to intersect the m
at a unique point right here so this is line uh k right here coming through here
and k is going to come down through here and
it’s going to intersect this m right here and it’s going to do that right here
intersect m right here at some point right here and we’re going to call it x
prime and that’s going to be the output right there
so let’s just make sure that we understand the construction because we
have the nitty gritty over here which is making sure we got everything covered

00:21
logically but how does it actually work you know just intuitively right so we
got a line l we got a line m they intersect at one point because
we’re in case one they intersect at one point
now each of those lines have another point on them and we’re fixing those
points we say we got a prime we say we got a
so now we can start trying to map if i pick a point on l
then i have to get a line parallel through that point that’s parallel to
this one and that parallel line has that
parallel to these two lines are parallel so this line here actually has to
intersect down here that’s we showed right here it has to intersect down here
and we’re going to get a unique point right there and it and that uniqueness
comes from theorem one and so that point right there is x prime and
that’s going to be the output right there and so what we got is we got a
function we choose the point on l which we called

00:22
x we chose that point and we got the output x prime and we’re also going to
define the mapping so that o ghost o because o is on both lines right so it’s
just like a fixed point there and then a goes to uh f of a
sorry a goes to a prime right because we’re we’re mapping from l to m so you
start on something on line l so we start on a uh a on line l and it maps to a
prime so we got those two and then we got the all the other points we can get
by using this axiom here and remember how do we get this line
we got this line because there’s by p4 parallel to this line is going to be a
unique line so there’s only one line there and then they intersect right here
at a unique point and that was by theorem 1 there
so now we have our function defined here’s how we define our function right
here and we actually have a process for actually finding this one right here
we can take any point and we’ll just draw the line parallel through that

00:23
point that’s parallel to our two fixed points up here we keep those a and a
prime fixed and then we just draw a parallel line it
has to intersect we get a unique point and that unique point is the output
all right and so there is the beginning of case one we have a function now case
one we still have to show that this function is on two and we still have to
show this function as one to one so let’s see if we can do that now
so here we go to show that f is subjective
so now i need to take a point on line m and i’m going to call that point here c
so now we don’t know anything about x or anything like that and so
we’re still keeping the same setup though
so now i’m going to pick a point on this line
so to define the function i picked an arbitrary x and i mapped it over here
and got x prime but to show surjective i need to take something potentially in

00:24
the codomain something that the function
may or may not map to we’re going to see
and then we need to come up with a point over here that does map to it
so we’re going to pick an arbitrary point over here c on line m
now if c is o or if c is a prime then we’re done so so what we’re trying
to do is we’re trying to find the input where this is the output and we picked
arbitrary c we picked arbitrary c on line m
and we’re and we’re thinking about the cases what if c was o well then we’re
done we already know map we already know what maps to c because o would map to c
if c was o um what about c is a prime if c is a
prime well we already know that a maps to a prime so we would be done again
so the last case is well what if c is not uh o or a prime
okay so now we’re going to see what happens so now i’m just putting a c down
here it’s not o it’s not a prime we already know what happens what what gets

00:25
mapped to that so now i pick an arbitrary c down here
not o not a prime now what in the world could map to it okay so we need to find
something that over here that maps to it all right so by p4 again right so by p4
so because we have a line and we have a point not on the line
and so now i’m going to get a unique line parallel it’s parallel to this line
here through here so let’s just draw it like that so these lines are parallel
and so by p4 there exists a line i’m going to call it k prime
uh it passes through c and it’s parallel to this one right here
so play ferrous axiom or the euclidean parallel posture whatever you want to
call it gives us this line k prime here okay so
um we’re going to hope that that k prime will give us the input that we need to
get the c is the output so we’re still trying to find that what that question

00:26
mark is so we need these two lines to intersect we need k prime and l to
intersect right here so what we’re going to do is suppose
they don’t intersect and hopefully we’ll come up with a problem with this so
let’s suppose they don’t intersect let’s put that up here i like to i like
to keep that down here so f of question mark equals c
all right so what if they don’t intersect so let me just
drop that away there what if k prime and l don’t intersect what if they’re what
if they’re actually parallel well if they’re parallel then what we
know is that this line which uh passes through a and this line which passes
through a and a is not on k prime and both of these are parallel to k prime
so by p4 there’s only one line that does that
so these are two lines that pass through a and both are parallel to k prime
there’s only one line so they have to be equal to each other now this cannot

00:27
happen though because remember a prime is not on l a prime is on m
right so this cannot happen so you know this is going to happen right here k
prime and l are not parallel in other words they have to intersect at a unique
point by theorem one so k prime cannot be parallel to l and so k prime
intersects um and that says intersects l at a single point which we’re going to
call c prime right here so c prime right there um and so now we got is
f of c prime equals c so we found the input that’s going to come from this
output and that input exists using p4 right there so f is in fact subjective
okay so that proves the case for subjective right there we found that
point right there we got its existence uh by by p4 all right and so now

00:28
what we have here is it’s uh subjective so now we’re going to try to show that
it is actually one to one so um let’s um update our diagram over here
okay so we still have l and m and they’re intersecting we’re still in
case one they intersect at a unique point and we got two other points on
those lines and we got a line through that point through those two points
and so everything is saying the same we still have our same mapping we pick an x
over here and we can map it to an x prime but now the question is
if we take b and c on l like let’s put a b here and a c here
let me spread it apart a little bit let’s put a b here and a c here
and what if they map to the same that’s an o
and what if they map to the same point here d
but if they map to the same point can you see my difference between my o and d

00:29
there’s a d all right so what if they what if they
map to the same point what if remember we took an x and we mapped it to an x
prime well what if b and c both get mapped to d in other words
uh uh assume that there exist points b and c
where they both you know the inputs give us the same output so this would be an
example of not being one to one because two went to 1
and so we’re going to find a problem with this all right now we have the line b
and b prime and c and c prime um so let’s see here um

00:30
what is the b prime and the c prime so what we have here is
the function is going to take b and go to a b prime
and the function is going to take c and go to a c prime
and if we’re under the assumption that these are equal to each other
then b prime equals c prime equals d so if that happens then this line right
here and this line right here they’re both parallel to this line right
here because remember that’s how we’re getting our outputs we draw a line
parallel to this line and it has to intersect over here and it intersects at
a b prime and then we draw this line right here through here and that’s how
we get the c prime and it’s parallel to this line right here
so and what if these two points are the same point here but in
in any case these two lines are both parallel to this one right here and if

00:31
these are equal right here if these two points are equal
then they both pass through the same point so both of these lines would have
to be equal by p4 and the line that we’re calling it equal to is k prime
so if these are equal to each other then we would just have one point down here
which we’re going to call d so here’s going to be b prime
we get output from the function c prime output from the function and that’s d
and so now what we got is the line through cd and bd
have to be the same line because they’re
both parallel to this line and they both pass through d
so by p4 they’re the same line right there
and we’re just going to call that line k prime
all right so by theorem 1 k prime and l have to intersect at one point
meaning b and c have to be the same point so this line right here which

00:32
looks like i drew it like two lines but you know that can’t happen by p4 those
lines have to be the same line which means because that line has to hit
l at only one place so that would force b and c to be equal to each other also
so that gives us the one to one right there
of the function in other words if you try to do two and one theorem one is
going to come back and say no no uh c and b have to be the same point you’re
only gonna be able to have one to one so we’re gonna have a one one-to-one
function there all right and so now we have case two
what if in fact they’re not going to intersect at all
so how would we do that so what we did so far was we had line l we had line m
and we had them intersecting and we were able to come up with two
other points here and we were able to map take a point any point and draw a
parallel and we get an output over here now what if l and m are parallel though

00:33
now we don’t have this o to to come up with these other two
points and and to come up with that so what if they’re just flat out parallel
so now how am i going to map between these two lines here
how are we going to map between these two parallel lines so here we go so um
we’re going to say o is on line l right l’s got two points i’ll just call
one of them o and o prime is on m right we know this line’s got two points
i’ll just pick one of them and we know this line’s got two points i’ll just
pick one of them all right so it follows that now these
two lines are not the same because because they’re parallel lines so
there’s no points in common so if o’s on l o is not on m so these two points
cannot be the same right so now what i have here is a line going through these

00:34
two here now so so what we have here is l and the line through o and o prime
and l i’ll put l first and then we also have m
and the line through o and o prime so what we have are two cases here we have
this line l and the line through o and o prime and
they intersect at a point don’t they so by case one if you have two lines
that intersect at a unique point there’s going to be a big bijection
between them and that and we’re going to call that bijection f so i don’t have a
bijection between l and m yet but i do have a bijection between l and and and
this line right here in fact let’s just call that line k so
k is the line through o and o prime let’s just call it k
so there’s a bijection between l and k because they intersect at one point
remember case one is we found a bijection between two lines that

00:35
intersect at a unique point well l and k
intersect at a unique point so there’s a bijection between l and k now
m and k also intersect at a unique point so there’s a bijection between m and k
so there’s a bijection which we’re going to call it g between k and m
and so we’re going to put those bijections together
so this is a composition first we do f we map
so what we’re going to do is we’re going to take a point on l
and f goes first right so we’re going to take a point on l and then we’re going
to map it to a point on k but we already have a bijection between
k and m so now i’ll take that point and map it to m and so the composition is
just l to m and so here’s the composition function
and it’s known it’s well known that if you take um
a composition of bijections you have a bijection so this right here is a

00:36
bijection so this function right here just maps from l to k and then k to m
and so you can think of it as just from l to m and we have a bijection there
all right and so that does the proof right there
so let’s do some more geometry right here yeah let’s do some more

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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