Area Between Curves Vertical vs Horizontal (Explained)

Video Series: Applications of Integration (Complete In-Depth Tutorials For Calculus)

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

00:00
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

00:01
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

00:02
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

00:03
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

00:04
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

00:05
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

00:07
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

00:09
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

00:14
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

00:18
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

00:20
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

00:21
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

00:22
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

00:23
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

00:24
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

00:25
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

00:26
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

00:27
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

00:28
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

00:29
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

00:30
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

00:31
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

00:32
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

00:33
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

00:34
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

00:35
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

00:36
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

00:37
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

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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