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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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