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hi everyone welcome back i’m dave in this episode uh finite alpine planes

fun and fun facts and interesting information

is part of the series incidence geometry tutorials with step by step proofs

uh let’s get started let’s do some math all right so first uh let’s uh quickly

uh review what an alphine plane is so in uh previous episodes we talked a

lot about what an incident plane is or sometimes we call it an incidence

geometry but we’re only working with points and lines here

so collectively we’ll call the whole thing an incident plane if we want

but we have these three axioms holding and we proved

lots of theorems with these three axioms here now an alpha plane uh we said in a

previous episode is we’re going to take a2 and we’re going to replace it with pa

and remember pa is the euclidean parallel postulate or sorry the

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euclidean parallel property and um so yeah we’re going to replace a2 with

pa and we did this in the last episode and what we said was we then get the uh

in an alpha plane the incident axioms hold in other words if you put pa here

as one of your three axioms then you can go prove that a2 holds must hold and we

prove that and so we call that a theorem so every alpha plane is already an

incident plane but an alpha plane you can do more in because you have this

stronger axiom here so in an alpha i’m playing the instant axioms hold

in an alpha plane we also proved this in the previous episode

and then we also showed in the very last

episode that in an alpha plane all lines have the same number of points so i

recommend checking out the uh the full series the link is below in the

description and let’s continue on uh now just one

more thing before we get started though that in an alpha plane uh the incident

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axioms hold which means we can prove all of these three all these 11 theorems

and we already did that for incident axioms so these actually hold also in an

alphine plane right because remember in an alpha plane the incident axioms hold

and we proved all of these using the incident axioms all right so

um i wanted to review one more thing before we get to the finite alpha plane

and that is the notion of a pencil now in a previous episode uh we also talked

about parallelism we defined this equivalence relation right here

um or actually we defined this relation right here and we talked about how it’s

reflexive and symmetric and then um in the episode on parallelism

uh we showed that if the euclidean parallel postulate holds then this

relation right here is actually an equivalence relation

and so in an alpha plane since we have pa holding which is the euclidean

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parallel property and so since that’s holding in the alpha plane

so parallelism is an equivalence relation and we’re going to use that and

since we’re going to be in a finite alpha plane we’ll be able to talk about

this equivalence relation here and the equivalence classes there so um a pencil

is going to be the equivalence classes right so pencil is a line together with

all the lines that are parallel to it so in the in the

you know nomenclature of equivalence relations it’s called an equivalence

class of parallel lines so that’s what a pencil is so if you have a line in in

one in a geometry say line l then the pencil

would be all the lines it would be l and all the lines parallel to l

that would be a pencil think of it as a really thick um

yeah so that’s a pencil all right so i thought we would do a quick example

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real quick um just find the pencils for this four point alpha geometry we talked

about how this is a four point geometry in a previous episode in other words if

my points are uh a b c and d and these are my lines

then we talked about how axiom a1 a2 and a3 hold

for four point geometry and we also noticed that in this geometry the

euclidean parallel property also holds so once we know all that then we know

this is actually an alphine plane and so we can actually or this is a

concrete model of one and so we can talk about the pencils we

can talk about this parallelism it’s an equivalence relation and what we can do

is we can actually go find all the pencils so um

pencil we need to find the line and then find all the lines parallel to it so

i’ll denote the pencils by say p1 p2 p3 so here’s our first pencil right here so

the first pencil remember pencil is made up of a bunch of lines

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right so i’m going to choose line a b first so a b here

and now all the lines that are parallel to a b

so help me find all the lines that are parallel to a b here so that’s not

parallel to a b because it has an a on it so not parallel not not not all right

so c d is a parallel line c d is parallel to a b so

this is the line a b in all the lines parallel to a b so this is a pencil and

this is our first pencil so pencil two could be ac line ac

and all the lines parallel to ac so what are the lines parallel to ac

well if you look the only other line parallel to ac is in fact bd

so here’s a second pencil b2 let’s find p3

so let’s find another line let’s say a d

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and what are all the lines incident with a d or sorry parallel with a d

and so the line right next to it bc is and if we check there are no more lines

parallel to ad and so let’s find uh p4 let’s say p4 is over here

and our next line will be say bc but actually we already did bc so

uh p3 right here we’re gonna get p are we gonna get p3 again right so p bc and

all the lines parallel to bc and ad is the only other line so this this would be

the pencil for bc this is which is equal to the pencil of

a d now what about l5 if i choose l5 is bd which is right here and so this is

the line bd and all the lines parallel to it and then the last one would be cd

here’s the line cd and here’s all the lines parallel to cd so actually we only

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have three pencils in this four point geometry we only have three

all right so um there’s some examples hope that gets you a little bit more

familiar with what a pencil is um and so now um the question is how

many pencils are there well it turns out in an alphine play there are at least

three pencils and in four point geometry we have exactly three pencils

but let’s show that no matter how large your geometry gets you always have at

least three all right so proof well let’s remember our axiom a3 a3 says

we have three nonconlinear points and actually if we look at the proof of

theorem two we took the lines through those three points and we said these

three lines cannot be equal to each other because all otherwise they would

be all collinear to some line and and then and they’re non-co linear points we

actually show that they’re non-concurrent lines too in theorem two

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so by pa any line l parallel to this one must intersect these two points because

a b and c are collinear so let’s try to understand

uh that sentence right there so so let’s see here we got uh points uh

non-collinear points abc and um and we have these three lines we

have the line through a b and the line through ac in the line through bc

and so those are just lines there none of these lines are equal to each other

if any of these lines are equal to each other then we would have them to be

co-linear and they’re non-collinear so none of these lines are equal to each

other so if we take a line any line parallel to a b so if i take

any line parallel to a b it has to let me put l down here if we take any

line parallel to a b it has to pass through

uh it has to intersect or pass through ac so ac could come up here and and hit

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it somewhere intersect or or pass through or must uh must pass through ac and

bc right here so there must be some point right there

on both lines uh because a b and c are non-collinear so let’s suppose that

they don’t intersect right here so this is going to be a common theme so i want

to make sure that you that you can follow this argument here so l is

parallel to a b right so l is parallel to the line through a b now um

what if b c is also parallel to l so what if this happens can this happen also

i think i said right here they that there has to be a point on both they

have to intersect but actually why why can’t they be parallel

well we know l is parallel to a b can’t this also happen let’s see

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um so both of these lines go through b and b is not on l

so how many lines go through b that are parallel to l well there’s only one and

a b is it so if if both if if this is if bc is parallel to l

then bc would have to be equal to a b because there’s only one line that goes

through b that’s parallel to l so i cannot have both of these b’s on both of

these and they’re both parallel to l so i cannot have both of these but we have

this one right here because that’s how we got l

so we cannot um have a line parallel to a b and it’s also parallel to bc so this

cannot happen which means they’re non-parallel which means they have a

unique point in common which means we got this point here

so we can make the same argument not just for bc

but why can’t we have this one right here also holding

why not we already got this one holding that’s how we got l why can’t we also

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have this one holding well that would be two different lines

uh parallel to l and they both pass through a and a is not on l

so by the pa by by the euclidean parallel property or or by par play first axiom

so these two lines would have to be equal um

yeah so so these two lines uh would have to be equal if this holds these two

lines have to be equal and they cannot be equal because we have three

nonconlinear points so ac also has to intersect i’ll just put it up here so

you can see it uh so there also has to be a point up

here okay so that’s an important thing to be able to do we’ll do that a couple

times in this episode um so if you want two lines to intersect

and you don’t have a theorem that says they intersect

then what you can do is just suppose they don’t intersect and use pa to to

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get a contradiction right so that’s a common theme there is all right so here

we go l is the only uh is is only in the pencil corresponding to a b so in other

words if i take my three points here i have and i throw down any line parallel

to a b it this line right here is not parallel

to bc because bc has to intersect that we just showed and ac has to intersect

that we just showed so the only so l can only be parallel to a b so in other

words l is only in the pencil corresponding to a b so

now we can do similarly for the other cases if we throw down a line that’s

parallel to ac then that line would only be in the pencil for ac

or similarly if we throw down a line that’s parallel to bc then l would only

be in the pencil corresponding to bc so here’s how the argument works when i

picked on this line uh in the in the cases uh for the other

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cases for it for this line and for this line you could use pa to make sure that

that line parallel to this line would only be parallel to this line and not

the other two lines all right so therefore there exists at

least three distinct pencils so there’s a pencil corresponding to this line and

the pencil corresponding to this line and the pencil corresponding to this

line and you know um that gives you at least three there

could be more right because we could have more than just these three

nonconlinear points but there’s at least three

so that’s what that theorem says there let me know in the comments below if you

have any questions to any of these steps i’d be happy to uh include another

episode where i clear any up any questions you might have or i would just

respond to your comments all right so now let’s get on to the

main uh part of this episode right here now we’re going to talk about the finite

part here we haven’t talked about the finite part yet

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so what happens if you have a finite alpha plane well it turns out

that you can find an integer greater than or equal to two

and we’re going to have five very nice properties holding

so one of the properties is if you have a finite plane

then there’s going to be a perfect square number of points so

you know if you have an alpha plane and you know the total number of points has

to be a perfect square so you could have 25 points in your out in your finite

alpha plane right you could have 25 points in it

and so i’m gonna have more than just part one i’m going to have i think five

parts all total so let’s just go through the proof of each one at a time so the

first one here is um okay so how do we get um n squared

points in our plane right so um remember we have a theorem that

says all the points on the line all lines have the same

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number of points on them so let’s use that so there exists at least one line

right so we know that that’s true because

we have three nonclinical points and so we got two points at least and so we can

use axiom a1 to get a line through those two points right so we know we have at

least one line and we know that so we’re in a finite alpha plane

and we know that all lines have the same number of points so

i’m just going to say we have n points on it and we don’t know anything about

this n right now we don’t know n squared so i’m just going to say i have a line

and here’s all the points on it so i’ll just draw it over here so here’s our

line l and here’s all the points on it a1 a2 and then we have a n

now remember this drawing is not part of the proof in fact i drew this line like

it’s continuous but we don’t know all we know are these points here a1 a2

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a in we have eight we have endpoints on that right so n could be five or ten

but we got some number of points and this is all the points on the line okay

so there exists a point p b not on the line so that’s uh by a

previous theorem that we proved i think it’s theorem four that we proved

uh in an earlier episode now what i want to do is consider the

line through a1 and b and there’s a line through a1 and b because

of axiom a1 says that given two distinct points we have a unique line

now how do we know we have two distinct points well a1 is on this line l

and b is not on the line now so they cannot be equal to each other

and i’m going to call this line here m1 okay

so we’ve got a line l with and all the points on it

and we have a point not on the line and we have a line through it and now i’m

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going to call that m one all right so now for each of these um

points right here a2 a3 a4 all the way to a n so i’m going to use an index say

a i so for i equals 2 through n i’m going to use the euclidean parallel

postulate or or pa which is which is the way we have it

abbreviated pa to define mi to be the unique line through a i but parallel to m1

so here will be let me move my b down here so this will be m2 it goes through a2

and is parallel to m1 and then i’ll do another one so this will be m3

it goes through a3 right there and it’s parallel to m1

and i keep doing that all the way into the last one where i get m m line m sub n

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the nth one so we have m1 m2 m3 all the way to m sub n

and all of these lines are parallel to m1

all of them and i know those lines exist by by pa

okay so they’re all parallel to m1 and i got m1 just by picking a point not

on it and going through a1 right so we can get m1 and now i can get all these

lines right here okay so good now each of these lines m1 through mn

has n points on it so that’s my previous theorem that we proved that all lines

have the same number of points so l line l has n points on it so m1 has

endpoints on it this has endpoints on it this has endpoints on it

and so they’re in and there’s n of these lines right because we have line one

line two all the way through a in right so we have n of these lines

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now here it says no pair of lines has a point in common so why not why can’t why

can’t we just come up here and say here i’ll draw it down here temporarily we

have line l we got a one we got a point not on it so we got a line m1

and we got all these other points and i’m going to draw lines parallel to m1

parallel to m1 so we have m2 and then we have all the way to mn

and they’re all parallel to m1 and so what can we say

well no pair of these lines have a point in common let’s see why that’s true

so what if i have a line coming through here let’s call it a sub i

and i have another line coming through here and they oh they intersect or they

have a point in common right there because i claim no pair of these lines

has a point in common why not these two right here why not have a point in

common let’s call it point p there’s a point p

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and it’s it’s it’s in common to these parallel lines now these lines are

parallel to m1 so this line right here ai is parallel

to m1 and this other line let’s say call it aj

this aj here is parallel to m1 also but the euclidean parallel postulate

says that if these two lines are parallel to m1

and they both pass through the same point

then they have to be equal to each other there exists a unique

line through p and parallel to m1 so the if these two lines would have to

be equal to each other but they cannot be equal to each other because these two

a’s right here are not equal to each other and this line right here if these two

lines were equal to each other they would intersect l at only one place

but they intersect l at two different places right so we cannot have this

point p here that that will just cause a contradiction so no pair of lines has a

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point in common that’s important because

what we have now here is we have n lines and each um line has n points on it

right so therefore the the number of points is at least n squared

now why did we say at least in square right so we have n here in here in here

so we have n of them and we have endpoints on each line right so

so now the question is how do we know that we counted all of the points right

so we certainly counted all the points on all these lines but how do we know

that there isn’t some point sitting over

here that’s not one of these lines right there it’s not on any of those lines

so it’s not on line m1 it’s not on line m2 how do we know there’s not some other

point in other words how do we know we don’t have n squared plus one

points in our geometry right so let’s see why that that why that cannot happen

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so if p is any point then p is either on m1 or not

so let’s look at that so let’s let’s look at our line here we got our line l and

we have our uh a1 right so how do we get m1 right so we have a1

and a point not on this line b and that’s how we got our m1

and how do we know that some other point p here

is on the m1 or not right so it’s either on m1 now if p is on m1 well then we’ve

already counted p as part of the n squared right because how did we get the

n squared is we took the endpoints on this line on m1 and then the endpoints

on all those lines so p is on one of those lines we’ve already counted p

so if p is on m1 we’re really done so let’s say pose p is not on m1 here p is

not on m1 right so what we can do then is we can use pa again

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the euclidean parallel postulate so there’s a line through k there’s a line

k sorry through p so i got a line here and this line right here is parallel to

m1 so i got a line that’s parallel to m1 and it goes through p

now what i want and now there’s nothing wrong with that um

you know because you know p is just not on m1 and pa says there’s a line going

through here but as you can see right here i kind of want them to intersect

don’t i and the reason why is because we’ve already named all the points on l

a 2 all the way to a n so we want them to intersect down here because if they

intersect then where they intersect has to be one

of these points up here and then if it intersects at one of these points then

that means this line k was one of those lines up here so let’s

see if we can get them to intersect right here how do we know we can get

them to intersect right here i want them to intersect

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all right so let’s suppose they don’t intersect right so k must intersect l

at one of the points a through through a n all right and then let’s see why

and so i’m going to give you this kind of argument again i’ve been talking

about here right so if uh l and k are parallel right if l

and k are parallel and remember that k is parallel to m1 so both of these happen

and and both of these lines pass through a1 right l passes through a1

and m1 passes through a1 and they’re both parallel to k

if all that happens then line l has to be equal to m1 right they’re both

parallel to k and they both pass through a1 and a1 is not on the line k

then we have to have that these two lines are equal but these lines are not

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equal why why is l not equal to m1 well uh b is on m1 and b is not on l right so

so these two lines cannot be equal which means the line l and the line k

cannot be parallel this cannot happen right here

we already know this happens we already know this happens and we know this

cannot happen right so that cannot happen all right so

therefore this k here has to intersect but remember all the points on l are

already over here so this intersection point has to be one of these

and since k is parallel to m1 k has to be one of these lines up here that we

have already counted so when we did this counting process up

here we actually had all the points on the geometry already

so we tried to pick a point that wasn’t part of one of those lines but we’re

pulled back in so in fact there can only be n squared number of points um

in a finite alpha plane so that’s a very nice property notice it

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relies very heavily on pa um so i didn’t say that explicitly right here but um

you know when i say me means l equals m1 it’s really by pa right here by pa

okay so now let’s go to uh number two here

so not only do we have n squared number of points in the whole geometry

but also every point is incident with exactly n plus one lines

so if i look at my geometry and i just write a point down i know

that there are exactly this many lines passing through p

right so if for example if n is 5 then i would have 25 total points

and any point that i picked i would know that six lines would pass through it

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all right so let’s see why that’s true all right so suppose that b is an

arbitrary point i just pick the random point and i want to show there’s n plus

one lines through it all right so well we know there’s a line

not uh containing b that’s my previous theorem that we proved one of the

theorems one through eleven and so we have this line l

and it has all these points on it and these are all the points on the line l

so that’s what i mean by l is equal to this you know set here all right so a 1

and a n and all the points are right here on it and b is not on it

and we started with b b was just a random point so we have a random point

we know we have a line and we got these number of points on

it right so um by a1 and theorem one so the lines going through b

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and ai right so i’m going to call this li so here’s l1

because it goes through a1 and then here’s l2 and l3 and ln so there’s ln

so i got lines one through lines ln and they’re they’re basically going

through the a1 a2 a3 all the way to a n so this will give us n lines already

and why aren’t they any of them equal to each other how do we know we actually

have n lines well none of these points are equal to each other and

if these two lines were equal to each other for example l1 it was equal to l2

well that would put those two lines equal to each other

and so then we would have one line and another line and they’re only going to

intersect in one point but we got two points where they intersect so that

can’t happen those two lines cannot be the same line

and you can make the argument for any pair of of lines here if any pair of

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lines are equal then the ai’s would have to be equal but the ai’s are all

different right that’s the number of points on the line all right so

these lines right here um are all different lines we have n of them by a1

and theorem 1. um and now we’re going to get one more line

so how do we know we get one more line well here’s a point b not on line l

and by pa there’s a line through b that’s parallel to it

so this is different line than all of these n lines here

and the reason why is because these n lines here just lines l one l through l

n they all intersect l but here’s one given to us by pa

that says it’s parallel to l so it cannot be any of those other lines so we

have n lines and now we have another line

so now we’re going to have all together m plus one lines um

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so any line through b intersects l and then is therefore one of the li

lines or is parallel to l and is therefore ln right so we’ve

constructed m plus one lines through b and this sentence right here says that

has to be all the lines through b let’s just make sure we understand this

last line here so i can draw n lines through b just by parameterizing

with a line a1 through a n and i get get these in lines through b

here line one through b line n through b and then i get one more line through b

parallel all right so that’s how i get m plus one

lines how do we know we don’t have more lines though is the question so now if i

throw down another line any line through b all right any line through b either

intersects line l or is parallel to l but there’s only one line parallel to l

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that goes through b so it has to be that line right there

and any other line that intersects l has to be one of those

because that’s all of the lines that are passing through l and b

all right so that gives us exactly n plus one lines

all right so now let’s go on to number three um and we talked about pencils

before so let’s uh understand uh number three here

every pencil contains exactly in lines all right so for this one right here

suppose that we draw a line uh l is any line

and c one is a point on l so let’s put a c one here

and c two is a point not on l so i’ll put c2 here

i’m going to draw the line through them and now i’m going to write all the

points out on this line right here the line going through c1 and c2 i can say

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here’s all the points on it c1 c2 and c3 and dotted to cn now remember this is

just a diagram it’s just helping us with the words

i’m in no way saying that the points are ordered on the line this way

in fact you know the diagram doesn’t even make sense because i’m drawing the

line like it’s continuous like there’s a whole bunch of points on here but in

fact there’s only these points on here so the diagram is just as an

illustration don’t think that um you know c c3 has to be over here hey why

not c3 over here you know this is just a guide it’s just a guide we do know

that there is a line through c1 and c2 and we do know that there is a finite

number of points on this line and here’s all of them

all right so for each of these i’s here you know 2 through n

let l i be the line through c i and parallel to l

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right so this is a l right here this is the line l and

so i’m going to be drawing parallel lines again parallel line parallel line

they’re all parallel to l imagine and they go through these ci’s here right so

this gives us how many uh lines in our pencil so here’s c1 this line l’s in our

pencil and all these lines are parallel to it so that gives us in lines

in our pencil we need inlines in our pencil right here so

uh the there’s n minus one from right here but then there’s also one right here

and these are all different because these points are distinct

all these c c i’s are are distinct so in other words if you try to make them

uh intersect and to then the c’s will have to end up being

the same so we’ve kind of ran through that argument in the previous episode uh

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sorry in the previous uh part part two i

think it was but anyways um all these uh lines have to be um distinct

so they’re all they’re all different because these points are different all

right so let’s draw just any line parallel to l and let’s call it k

so i’ll draw a k say right here so here’s k it’s a line parallel to l

and i drew it so that it passed through this line right here c1 c2

but how do you know that it intersects right there right so that’s another

argument that we have to make like we’ve been making

so k must intersect this line right so and that’s because l is the only line

containing c1 which is parallel to k so we’ve kind of made uh

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that kind of argument before all right uh so since uh all right so

therefore k in this line right here are incident at some c c i

and that’s how we know that the way that we constructed these n lines in this

pencil and then any other line that’s also parallel to l

by the parallel by by pa you know has to be one of those lines

and so that’s how we determine that there’s only n lines in a pencil first

we establish that there are at least n lines and then we use pa to establish

that there cannot be any more that’s basically the overall picture

it’s a very common theme we we’ve done that uh two times before and so

you know you can use that strategy to also prove number three here

so there are exactly um are the only lines parallel to l all

right and so if you include l then you get uh n of them

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all right so now for number four there are exactly n squared plus n lines

all right so number four here so suppose that b is any point and

i’ll try to put it down here try to get the idea of cross of how the

uh how the proof works so b is any point and we know that any point is incident

with n plus one lines that’s part two right there so i got m plus one lines

going through here i’ll just draw three but you get the idea we got m plus one

lines going through here and what we can say is that each of these

lines is in some pencil um consisting of n lines right so this

line right here that goes through b that line and all the lines parallel to

it is one pencil and so this line right here is in a pencil

this line right here that goes through b this line

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and all the lines parallel to it and so let’s actually just number the lines

let’s call this line here l1 here’s l2 and then here’s l3

and i’ll just draw three but imagine n of them so this line here l1 and all the

lines parallel to it that’s one pencil and here’s l2 passing through b and all

the lines parallel to it that’s another pencil

and there’s n plus one line so we’re going to have m plus one pencils

um so the lines through b together with the lines parallel through them

account for n plus one times n right so we’re going to have n plus 1

lines and for each of these lines we’re going to have n lines that are parallel

to it so for example l1 here there’s going to be remember each pencil

contains inlines so l1 and all the lines parallel to it that’s in lines and l2

and all the lines parallel to it that’s another n lines and so we’re going to

00:38

have n plus 1 times n distinct lines and so for any line m so now

now now that we’ve established that we have at least this many of them how do

we know that’s all of them so now i’m going to take another line

and let’s draw it up here now so here’s our line uh here’s our point b

and we got lines going through it say l1 l2 you know ln plus one and

this is going to generate for us n plus one lines

and then if i include all the lines parallel to all of these lines i get n

times n plus one distinct lines now what we’re going to do here is for

any line m right so now i’m just going to throw down another line here

00:39

just any line how do we know right because we’re trying to count all the

lines we know we have at least this many so how do we know we can’t have more

so let’s just throw down a line now either b is on that line or

not right so let’s say b is on the line so that means the line goes through b

but we’ve already counted all the lines through b so if m goes through b

we’ve already counted this line right here there’s no more lines to to add

so if the line doesn’t go through b that means that

it’s going to it’s going to have by the euclidean parallel postulate it’s

parallel to um you know or there’s a line all right so we’ve got

a line and we have a point not on the line and so there’s going to be a line

going through b that’s parallel to m and that’s by pa

00:40

so or there’s a line through containing b and parallel to m

and that’s by p a so i should say by p a right here

but that case has already also been counted because we’ve already counted

all those parallel lines when we did our counting process right there so

therefore there are exactly n times n plus one lines right there

okay so number five i think there’s a fifth one here yeah

all right so there’s exactly this many pencils

so we know um every pencil how many is in every pencil

but how many actual pencils do we have so let’s see number five here number five

is gonna go quickly because we’ve already proved one through four here

so since each of the uh this number of lines right so that’s part four

each line is in one and only one pencil and this has something to do with the

fact that we have an equivalent relation

00:41

remember the pencils are the equivalence classes so you can only be in one

equivalence class at a time so each line and there’s this many lines

is in one and only one pencil and each pencil contains in lines

and so there has to be n plus one pencils

and so that really relies upon the fact there that you kind of understand

uh equivalence classes right there all right that’s it let’s do some more

geometry let’s talk about projective planes i’ll see you in that episode