Every Point Has at Least One Line Not Incident With It

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

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

hi everyone welcome back in this episode we’re going to prove
that every point has at least one line not incident with it let’s do some math
so first we’re going to begin by recapping what the incident axioms are
so we’re going to let point line and incident be undefined terms
and we have three axioms here so the first axiom says that
for every point in p and q right so they’re not equal so we have two
distinct points so there exists a unique line that passes through them and
axiom two says for every line l i can find two points on them
has at least two points on them okay and then axiom three says
there exists three non-collinear points and so we define what non-collinear
means three or more points are non-clean year

if there exists a line incident with all three of them
now i’m going through these brisk briskly because we’ve already covered these
axioms and these definitions in a previous in the previous episodes
so if you haven’t seen the series link yet the playlist link
it’s below in the description and you can catch out catch up on all the
episodes that we’ve done so far so we have also concurrent three or more
lines are concurrent if we can find a point that’s on all of them
and lines are parallel if they have no points in common
all right and so let’s get started um you know but first let’s check out what
we’ve proven so far right so we’ve proven theorem one and two and three and four
and in today’s episode we’re going to prove that every point has at least one
line not incident so if i put down a point here i know that i can find some line

that doesn’t pass through it and that’s we’re going to prove in
theorem five here today now the previous episodes you know we talked about all
the incident axioms and we had an episode for each one of these theorems
and we’re going to go through this theorem five here just like we did the
other four first we’re going to prove it in column
format and then we’re going to prove it in paragraph format
so theorem six will be the last one here and we’ll do that in the next episode
so let’s get started with our proof here for every point and i’m just going to
name it a there exists at least one line not passing through a
now i can give this a just just an arbitrary name
as long as we don’t specify anything about it other than it’s a point
okay and so um i’m going to make a list of statements and justifications for
each statement and i’m going to um you know right number each step out
here as we go so the first step here is i’m going to say a is a point

a is a point and i usually follow the convention that
i use capital letters for points and lowercase for lines so a is a point
and this is going to be our hypothesis right because we need every point so
let’s just say we have a point and now step two is i’m going to say
every line passes through every line passes through a
every line passes through a now if you don’t like to use the words passes
through then you can use the word incident suppose a
um is incident with every lie every line passes through a so
what we’re going to say is that’s an r a hypothesis we have this point here

that every line passes through it and that seems you know crazy but it’s
going to be our hypothesis and the reason why is because you know
we we want to show that there’s there’s at least one line not passing through it
so we want the negation of the statement right here to be true
so let’s what’s so wrong with it what’s so wrong with it if we just say every
line passes through a so let’s put a diagram over here now
remember the diagram is not part of the proof it’s just help help us guide guide
us through it so what we have so far is a is a point and no matter
you know no matter how many lines or wherever our line is somewhere along the
way it has to go through a every line goes through a that’s our hypothesis
now i’m going to use axiom a3 so there exists three distinct points there exists

three non-collinear points and i’m just going to name them e d and f
now i didn’t want to name one of them a and use like a b and c
because i’m already using a from my point uh that i’m assuming exists right
here and that every line passes through it and axiom a3 doesn’t say anything
about this line a all axiom a3 says is that i have three points i don’t know
how they’re related to a or not i just know that i have three non-clinical
points so it’s important to not just say a b and c
and give it the same name because we really don’t know
all right but we do know is that there are three non-linear points and i just
gave them random name e d and f all right and so now what i can say is
i’m going to use axiom 1. i’m going to get a line through e d
so there exists a line there exists a line
and i’m going to denote it by e d it’s unique so it doesn’t matter what

notation we use for it incident with e and d incident with e and e
and this is axiom a1 here and i’m going to on step 5 here uh do
the same thing again i have a dnf so there exists a line there exists a line ef
and that’s incident with enf incident with enf
or you can say the line passes through e and f and that’s also by axiom one here
okay and so now what i want to do is i want to make a step six up here
and um so we didn’t have room right here so actually let’s uh
let’s just follow along with it uh written up already a is a point

every line passes through a that’s our raa hypothesis i just erased it
um there exists three non-continual points ed and f i use axiom three
now through these two points right here that are not the same point
right two distinct points i get a line through them axiom a1 axiom one
same thing with these two points here d and f
now you might say why didn’t i go and do e and f well i will but i’ll do that
later when i’m going to use it so right now i’m just going to
concentrate on these two lines e d and d f now what i’m going to say is that
law of excluded middle this is the logic rule
and it says that one and only one must hold either a is this equal to this
point d or it’s not equal to point d so remember
we don’t have any relationship between a and e d and f axiom a3
doesn’t know anything about our the ra hypothesis that we’re making

axiom three just is just a point blank statement there’s three points they’re
non-linear they have to exist somewhere and this point and this point here a is
in our our hypothesis so i’m trying to make a connection between steps two and
three i’m saying here a and i’m just picking on d i could have picked on e or
f but i’m just going to pick on d i’m going to say a is either d or not d
one of the two has to hold all right and so now i’m going to go
into two different cases so i’ll use this is one case and this is another case
so case one what if a is not d so if a is not d then what we have here is that
a and d are incident with this line here e d and it’s also incident with d f
so why is this true so by the previous steps right so first
off d is incident with e d as we said in a previous step

and d is incident with df and remember step two says
a is incident with every line so it’s going to be incident with these lines
right here for sure so what this says is that that this line here e d and d f
actually have to be the same line so why
is that well we have two distinct points a and d
and they’re both and they’re and they’re on these two lines but remember axiom
one says there’s only one line that goes through two distinct points so if a and
d are on both of these lines and the points are different from each
other then action one applies we have two distinct points and there’s a unique
line that passes through them so what that tells us is that
f which is on this line right here has to be on this line right here now
and so that means we they’re all three collinear
that’s the definition of collinear they are they’re all on the same line which

is the line ed and so that’s a contradiction because uh
step three said that e d and f are non-collinear so
you know we have a contradiction a very clear contradiction between steps three
and eleven so we have a contradiction which means we get to negate um
or which means we’re finished with case one now case one cannot happen if i
assume that a is not d i get a contradiction
so now it’s uh let’s work in case two now so case two is a is equal to d
now remember um the previous step said that every line passes through a
that was our our a hypothesis well if a is d
now we know that every line that that d passes through every line so every line
is incident with uh point d steps two and three
so what mean uh so now what i can say is there exists a line through e and f by

axiom one because remember e and f are different points they’re you know
two distinct points so i have a line passing through e and d e f
and d because this incident with every line is on this line also
right d is equal to a a passes through every line
every line passes through a so d is on this line now
well again that means that e d and f are collinear
by definition of collinear because i have a line that they all pass through
and so that contradicts steps 3 and 17 right 3 and 17 are a contradiction
because step 3 says they’re not collinear and step 17 says they are collinear
so this step here cannot happen either so what we’ve done is
what we’ve done is in our r a hypothesis we broke it up into two cases
both cases gave us a contradiction one of the cases has to hold a is either

equal to d or not equal to d so since both cases lead to a
contradiction the r a hypothesis has been refuted and so that means the
negation of the ra hypothesis has to be true so there has to exist a line not
incident with a and that’s our r a a conclusion right there
all right so very good so let’s write up a proof now
in a paragraph form so what i’m going to do is i’m going to assume for a
contradiction so i’m letting everyone know right up front in the proof that
it’s going to be done by contradiction that i have a point that every line is
incident with and what we’re going to do now is get a contradiction to that
so by a3 there exists three distinct non-collinear points say e d and f and
i’m going to go through the cases right so we’re going to use axiom a1 and we’re
going to get these three lines here and then we’re going to say what happens

if a is equal to d right so then all three lines are
incident uh all three points are incident with this line
and if a is not equal to d then these three lines are actually
equal to each other and remember remember axiom eight one gives us a
unique line so that means that these signs have to be equal to each other
hence we get these three points are incident with this line
so in either case this case or this case the points are collinear which is a
contradiction because they’re non-collinear so this contradiction says that the
negation of this right here has to hold so there exists a line that is not
incident with a in other words every line it’s not true
that every line passes through a so there’s some line that does not pass
through a and there we go there’s our uh theorem
five all proven for us let’s check out this uh episode right here and we’ll get
on with the next theorem see you there

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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