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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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