Every Line Has at Least One Point Not Incident With It

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

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

00:00
hi everyone welcome back in this episode
we’re going to prove that every line has at least one point not incident with it
let’s do some math so we’re going to begin by recapping the incident axioms
so we’re going to let point line and incidence be undefined terms
so uh the first axiom was or is every point p and for every point q not
equal to p there exists a unique line through p and q so if we pick any p and
q not equal to each other then there exists a unique line that
passes through them so there’s axiom one a one
and for the next one here we have a2 for every line
there exists at least two distinct points on it so if i draw any line say l
i know that i can find somewhere on it two distinct points

00:01
which i’ll name p and q again sometimes i use capital letters
and i often use lowercase letters for lines so no matter what your line is l m
whatever line you have there’s at least two distinct points on
it that’s what a2 says a3 says that there exists three distinct
non-collinear points so no uh so sorry non-concurrent points
in other words no line is incident with all three of them so somewhere along the
way which i’ll just name them a b and c i cannot put these points all in one
line there’s no line incident with all of them and that has the meaning of
non-concurrent um no line is incident with all three of
them oh sorry that is a non-collinear okay so
just to make sure we’re all good on that three or more points are called
collinear if there exists a line uh with all of them on it right and so a3 says

00:02
there exists three distinct non-collinear points we could rewrite it like that
um and then the uh second definition here is non-concurrent
three lines uh three or more lines um if there exists a point incident with all
of them so here’s example of concurrent lines so
there’s just a point incident with all of them and so in fact we can put as
many as we want so we can say all these lines are
concurrent because they all pass through uh the same point there and then
our third definition is l m are parallel and that means uh that they’re distinct
lines and have no point on both of them so these two lines are not parallel
simply because there’s a point that is on both of them so
what we’ve proven so far is the following
we’ve proven in our previous episodes by the way just very briefly here if this

00:03
was too quickly if this was too brisk then uh check out the series link below
uh the first episode of the series i go through these much slower
uh and i give much more detail in in the other episodes in the series so far
we’ve proven uh theorem one we’ve proven theorem two here and we’ve proven
theorem three and so today we’re going to prove
theorem four now when we’re proving theorem four
we can use any of these three axioms if we want in our proof and we can sorry we
can use any of these three theorems and we can also use any of the three axioms
that we want in our proof and so you know just make sure keep that
in mind is you know what’s already been proven
so that you can know what you can prove that is based upon that those theorems
all right and so in the next coming episodes we’re going to show that every
point has at least one line not incident with it
so it’s a little bit different than today today’s theorem and then and we’re

00:04
going to prove theorem six every point is incident with at least two distinct
lines uh you know we started proving something
like that in theorem three here every point is instant with at least one line
and now theorem six we’re going to go with at least two lines here so let’s
get to work on theorem four here actually this is
uh i don’t want to give anyone the impression that this is all you can
prove because that’s certainly not the case
it’s just that with these six here what i wanted to do is the goal of the series
is to help someone get started with these axioms and how to write rigorous
proofs so uh with d6 here we’re going to continue our uh right first write the
proof in column format where we lay out the detail for every single line and
then we kind of give a summary paragraph proof
and so let’s do that for theorem for right now
all right so every line has at least one point not incident with it now as

00:05
usual i’m going to have a diagram running over here to help us guide us
and approve but keep in mind that this is over here
is just for intuition it’s not part of the proof you cannot say anything in the
proof say oh look at the figure okay so you know the diagram over here is
actually very simple because we’re just going to say here every line right so i
need to start off with the line right so i’m just going to say i have a line l
and what i need to show is that there’s at least one point
uh not incident with it there there is at least one point so somewhere out here
i need to find a point so i’ll just draw
it up here and i’ll call that point p we need to show p exists
we don’t know that it exists and you might say well can’t you just
look at all this room up here surely there’s something up there no right we
have to prove it with our axioms and our theorems
so i’m going to write down a list of statements and a list of reasons why
they’re true and i’ll number everything so that we

00:06
can refer to each of the steps if we need to so step one here is
you know how would be how would be a what would be a good approach
to starting this proof how how would we start this proof
so what i’m going to do is i’m going to say you know
why why does there have to be a point out there here’s my thinking here why
does there have to be a point out there i mean what if what’s so wrong with all
the points being on one line so we have an axiom that says uh or we have
some actions and theorems that says that can’t happen
but let’s just start off with that is our our a hypothesis suppose l is a line
every line has at least one point on it right
so let’s start off here l is a line incident with every point
l is a line incident with every point what’s so bad about that

00:07
why can’t that happen why does there have to be a point off the line
right so l is a line incident with every point notice we
didn’t use any other words like uh oh you know there has to be a
point you know off the line or anything like that right so i’m just making an r
a hypothesis and that’s something that we don’t want
to be true we want so we want to come up
with a contradiction and then we’ll come over here and be able to negate this
what’s the negation the negation is you know there isn’t a line with every
point on it in other words if there isn’t a line with every point on it that
means there exists um a line uh no that means that every line will
have uh at least one point not on it all right and so here we go
we’re going to use axiom a3 so i’m going to say a b and c are distinct

00:08
non-collinear points and this is the axiom a3 so i’ll just put over here a3
so a3 says we have three distinct non conlinear points and i’m
just going to put them somewhere because we don’t know where they’re at we don’t
know the relationship between these three points and line out we just don’t
know because a3 doesn’t say anything about a line
it just says there’s three distinct non-linear points that exist
uh step one here is l is a line and every point has to go through that line
so actually um by step one we know that actually all three of these points have
to go through this line here because this is our hypothesis right here so
here we’re going to so we’re going to say a b and c are incident with line l
and our justification for that is just step one

00:09
l is a line that’s incident with every point a b and c are points so a b and c
are incident with line l and now what we know is that if you have
a line that goes through all three points then that means a b and c are
collinear so a b and c are collinear and this is by definition of collinear
so i’ll just say definition of co-linear and now step five
we have a clear contradiction here because we have that a b and c are
non-clean year and we have a b and c arc are collinear
right and so steps two and four give us a contradiction so i’m going to put
contradiction and i’m going to say the steps that have
to be negations of each other they are they are collinear and the negation is
they are not linear so steps two and four

00:10
all right and so what we’ve done is say there isn’t a line with every point on
it there is no line with every point on it right so every line has to have at
least one point not on it so there exists a point not incident with l
not incident with l all right there we go
there’s just a point not incident with l and that’s exactly what we’re trying to
prove so i’ll just squish that over here there exist a point uh not incident
with line l just so that you can see that there
there’s the step six the last step and the justification for this last step here

00:11
is the r r a a conclusion we negated the r a hypothesis right here in step one
the negation is uh so the negation of a for all is there exists all right so
every point we have there exist a point all right so there we go there’s the uh
column proof for theorem four there and now let’s see if we can write a
paragraph proof right here so let’s write this in paragraph form here
and we want to uh summarize the proof for somebody and have put all the main
points of logic so that someone could easily come and put in here
um all the steps so here’s what i’m going to do i’m going to say suppose so
a major thing of our proof that we just did the column proof was that it was a
proof by contradiction right and so that’s what i’m going to put in the
proof is i’m going to say suppose that suppose there exists a line say l

00:12
that is incident with every point that is incident with every point
um so by a3 uh let’s put it over here by a3 there exists three distinct and
non-non-collinear points there exists three distinct non-clean your points and
let’s call them say a b and c there exists three distinct
non-conlinear points column a b c and then i’ll just say contrary to hypothesis
so this could be one way of writing up the proof
is suppose we exist that that it can’t happen come up with a contradiction

00:13
right so we have a line that’s instant with every point so they’re collinear
but a3 says they’re non-collinear there exists three distinct uh non-collinear
points by a3 there is just three distinct non-collinear points
um yeah and so that contradiction right there
uh is is contrary uh you know leads to a
contradiction right and so you know this could be one way of writing up your
proof here all right so you know just to see it
typed up or written up suppose there exists a line say l that just goes
through every point every single point right so by a3 there exists three
distinct non-linear points and that’s contrary to this hypothesis
right here that we just said if l goes through every three all points
it goes through these three but a3 says these three are non-collinear so that
contradiction says that there cannot there there does not exist a line that

00:14
is instant with every point so in other words every line has at least one point
not on it all right so that’s it uh for this episode if you
like the proofs we’re gonna we’re gonna do another one right here check it out
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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