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hi everyone welcome back in this episode we’re going to prove that every point

is incident with at least two distinct lines in other words if i put down any

point i want then i know there’s at least two

distinct lines going through that point there so let’s do some math

so we’re going to begin by briefly recapping what the incident axioms are

so we’re going to have a point line and an incidence

are going to be the undefined terms action one says if we’re given two

distinct points we have a unique line passing through them

axiom two says for every line we have two points on it

and a three uh says that there exists three distinct points with the property

that no line passes through all of them so we’re using the word incident

sometimes i’ll use the word passes through

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instead of incident or i’ll say p lies on a line but those are all just

shortcuts for the word incident all right and we have three definitions

here uh points are called collinear if they all lie on a

the same line and concurrent uh lines um they all have a point in common

and in parallel means they have no point in common

all right and so let’s get started but you know

actually before we get started though i want to mention though that if you’re

not clear on all this right here uh make sure and check out the link

below for the full uh series uh in the first video in this episode in the first

episode in the series we uh i went through these axioms very

carefully very slowly and so i recommend watching that episode

also we proved each of these theorems so this is our

episode on theorem one episode on theorem two and so on and right now

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we’re on theorem six so we’ve already proved theorems one through five

in the series uh in previous episodes and so now it’s time for theorem six

every point is incident with at least two lines

and so let’s get started here here we go and i’m going to start by doing this in

a column format so i’ll make a list of statements

and a list of justifications and then after we’re done with that we’ll get on

with the uh paragraph proof all right so here we go um the first

statement i’m going to make is number one is

p is a point so i’m saying every point right so i’m going to name my point it’s

just an arbitrary point and p is a point it could be any point at all

but i want to pick a name for that point so that i can refer to it

but again it’s any point all right justification so that’s just hypothesis and

number two now i’m going to say that a b and c are

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distinct collinear points a b and c are distinct collinear points

and that’s by a3 so a and b are different b and c are

different a and c are different we have three points

and uh they’re non-collinear points sorry meant to say non-collinear points

a3 says they’re non uh non-clean your points there that’s important

all right step three so i know a and b are different so i

know there’s a unique line passing through them and b and c and a and c so

i’m going to write all that out real quick in my proof so there exist lines

there exist lines and i’m going to say a b and then i’m going to say bc

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and i’m going to say ac and so and this this one is going to be incident with

and if you don’t like the word incident here you could say a b passes through

uh a and b and line bc is incident with b and c and incident with a and c

so all three of these statements right here which i’m combining into one

statement is true by axiom a1 so there’s actually a one there

all right so we got these three lines and a and b are on this line and b and c

are on this line and a and c on this line and we can say all these because a

b and c are distinct points they’re all distinct they’re also non-clean year

we’ll get to that here in a second so step four is

i’m going to say what happens with a b line a b and b c

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i know there’s only one line going through a b and i know there’s only one

line going through b c but how do we know about line a b and b

c well i’m going to do an r a hypothesis here i’m going to say line a b is equal

to line ac and that’s going to be our a hypothesis here now

that can’t happen though because uh a b and c are non-collinear if these

three if these two points are equal to each other then i got a line going

through all three points in other words a b and c are collinear

and that’s just definition of collinear and that’s a contradiction

and that would be steps um two and five two says nonconlinear steps

that’s five it says collinear so steps two and five are a direct contradiction

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and so a b and ac i assumed they were equal and we got a

contradiction so they’re not equal so a b is not equal to ac

in other words there we have two distinct lines here a b and a c

now we’re trying to get two distinct lines through p

and we haven’t shown that yet and so right now i’m kind of ignoring p

and what i’m working on are the a b and c and i’m just getting together facts

about them i got these three points here and i got these three lines here

but because we’re trying we’re interested in lines over here so i’m

going to try to get some of these lines going through p and

make sure that they’re different lines so that’s kind of the idea here

now now that i showed that a b line a b and line bc are not the same line

what about you know the b c and ac or what about a b and a e c

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so let’s work on those cases as well so uh let’s go to um

step eight now if i can squeeze that in so step eight will be uh a b and b c

oh i skipped to a b and a c let’s do a b and b c now

so what if these are equal a b and b c and so that these two equal so we’re

going to kind of repeat these steps here four through seven here

um but instead of a b and a c now i’m going to use a b and b c so what if

there is equal to each other so there’s my r a hypothesis

and so now if i get these are equal if these two lines are equal then like i

did right here i’ll get a b and c or collinear so i’ll write that out right

over here i’ll say about right here nine step nine a b and c are collinear are

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linear so if these two lines are equal right so

a is on this line that’s why i wrote it up here a is on this line and b is on

this line and c is on this line and the lines are equal so that means c is also

on this line you know so they’re all on one line

so i’ll say definition of collinear definition of collinear

and for step 10 i’m going to have a contradiction just like i had here and

this will be steps 2 which says it’s not clean year and

step 9 which says it is colleen year all right and then step 11 here

um so i got a contradiction so my r a hypothesis has been refuted

so these are not equal so line a b is not equal to line bc r a a conclusion

and then i’m going to repeat this one more time because we have one more

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possibility so a b is line a b is not equal to line ac

line a b is not equal to line b c which is [Music]

uh yeah we did that right there and so now the last one to check is is line bc

equal to the line ac so number 12 here bc is equal to ac

and that’s my r a hypothesis again and 13.

if these two lines are equal a b and c are collinear and step 14 oops arcelinia

are collinear and that’s the definition of collinear

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i’ll just abbreviate that there step 14 here is contradiction

and this will be steps 2 and 13. 13 says they are collinear

step 2 says they’re not so contradiction and then i have my r a hypothesis here

that these are not equal so bc and a ac these two right here are not equal

and this will be the raa conclusion there

so we have refuted this hypothesis here all right so we’re good to go so far

so what we have is these three lines existing

and we just showed that they’re off uh all three distinct lines we have three

distinct lines there um and so a quick diagram of our proof so far

i’ll erase this real quick and so what we have going on here is

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we haven’t picked a p just any point p and we have points a b and c

and what we know about them is they’re non-collinear

and i drew them as separate from p but we don’t know actually one of them might

be p we’ll deal with that in a second but what we’ve shown so far

is for these three points here we got these lines here

and they’re distinct lines they’re not the same lines

and that’s what we’ve shown so far is we’ve got three distinct lines now we

need to get some lines going through p though right all right so here we go so

step 16 here is going to be exactly one must hold exactly one must hold

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and p p is is one of a b or c or not in other words p is equal to one of

these three points or it’s not equal to any of those three points right so this

is just the law of excluded middle now by the way uh when i was writing up

this and thinking about this uh yeah there’s a lot of different paths to

write a proof no one proof has to be exactly the same this is just the way i

chose to do to do this um so now i got case one p is equal to a i’ll just call

this one here right here case one um actually let me let me bump that out

because i have an or not here so p so k step 17 here is p is one of the three so

this will be case one so case one is it’s equal to one of them

i don’t know which one but it’s equal to one of them

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and now step eighteen is well what if it’s actually equal to the a so i’m

gonna call this here case one a so now we know definitively we’re we’re

in the definitive case that p is equal to a so the diagram i had to go a minute

ago with p here and then we have a b and a c

and we got three distinct lines going through them because we’re in the case

where p is actually equal to the a okay and so then we have this this line

here we’ve already shown it and we have this line here we’ve already shown it

and we have two distinct lines so that that’s what we’ll say here so lines

lines a b and a c are distinct and pass through p and pass through point p

so why is this well we’ve already shown these lines are distinct

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and p is equal to a so it passes through a right a is on this one a is on this

one a is equal to p right so this is just steps of 3 and 11.

i’ll just say previous steps there um previous steps

right we’ve already shown those lines are distinct

and we’re in the case p is equal to a and a is on both of them we’ve already

said all that all right so now we’re on the case where p is equal to b

and we’re going to call this case 1 b 21 so now p is equal to b now p is p is

right here and so this is a a over here now we have the lines

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uh you know a b and b c so the lines a b and b c are distinct and pass through

p pass through p so lines a b and b c are distinct and pass through p

and again this is previous steps here just kind of summarizing our

previous steps there all right and then the last case for case one will be p is

equal to c and so this will be case one c and then 23 step 23 says lines so

if if p is equal to c then we’re going to have a line uh ac and bc so lines ac

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and bc are distinct and pass through p pass through p

and the justification for that is the previous steps because we’ve

already said everything in that uh were in that word and we’re

very shown that they’re distinct and in this case right here when p is

equal to c they pass through c because we said that one passes

through c and we already said that one passes through c all right so in

case one we’re finished and in case one is three sub cases and in each of the

sub cases we came up with two distinct lines passing through p

so let’s go on now to uh case two so i’ll uh get rid of this now and now

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we’re on case 2. so case 2 starts off with um

so what line number are we on so we’re on line number 22 20 23

and that says p is not equal to a p is not equal to b and p is not equal to c

so we’re going to call this here case 2.

all right there we go sorry i got to get that focused all right so

p is not any of them p is not equal to a

it’s not going to be it’s not equal to c

that’s case 2. so case 1 was p is one of them and we went through all the sub

cases all right now case two so now for case two now these are

different points here a is not p p is not a not b right so we can use

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axiom one again so here we go there exist lines and i’m going to say ap and bp

and cp and this one is going to be incident with a and p and incident with bmp

and this one is incident with c and p and this is of course by axiom one

so i use all those one three separate statements use axiom one

i put them all together in one step there so there we go

so we got these three lines ap bp and cp it has all these points on it

a ap is on ap and so on all right so there we go so now what i want i want

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two lines through p and it looks like we have three but actually it

doesn’t necessarily hold that they’re all different lines for example

if you have a b and c as we’ve been writing before our p who knows where it’s at

p actually might be like right here it might be on that line uh we’re under

the case where p is not a b and c so p is not here not here not here but that

doesn’t mean p isn’t on that line um but but even if p is on this line

we would still have two distinct lines going through p um it would be the line

you know bp or ap or ab they would all be equal to each other because they’re

clean here and the line uh cp so i would have two distinct lines going through p

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um but you know we don’t know that we have three you know just because

we write three down here that doesn’t mean any of them are the same or not all

right so long story short let’s assume that they’re all one line

so i’m going to do an r a hypothesis that all three lines are the same

actually we just need to get rid of this case that they’re all the same line

if they’re not all the same line then at

least there’s two lines here there could

be more they all three might be distinct but um if we get rid of the case that

they’re all the same line and that’s false then we’ll have at least two lines

among these three so let’s go here with our a hypothesis

what’s so bad about them all being the same

so a would be on this line and b would be on this line and c would be on this

line and so we would have them that they’re collinear so are collinear

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a b and c are collinear and that’s just by definition of collinear

and then now step 27 well that’s a contradiction because steps two

uh steps two and twenty six are negations of each other this one

says there are collinear step 2 says they’re not go linear

all right and then so step 8 28 is the raa conclusion we need to negate this

now how would we negate this so this is really an and here

it’s line ap is equal to line bp and line bp is equal to line cp and

line ap is equal to line cp and so to negate that

you know we’re just going to have that one of them is

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it’s going to be an or so ap is not equal to bp and then bp is not equal to cp

or ap is not equal to cp we don’t know which one of these three holds

but at least we get two points uh sorry at least we get two lines going through

p and so that’s the r a conclusion and so you know that’s basically the proof

right there either these two are not equal or these

two are not equal or these two are not equal but p is on all of them

so it doesn’t matter which ones are not equal you’re going to get at least two

it’s not the case one whole so they can’t be all all these cannot be one

line right all right so there’s at least two lines there could be more okay

so there are the steps to this theorem six here and now let’s look at

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a write-up of the uh proof here so i’m going to

get rid of this here for us and let’s look at a proof now so the first thing

is i’m going to say p is a point and i’m not going to say anything else about p

um it’s just a point i’m not saying if it’s equal to anything else or

anything else i’m just starting off with a point that’s important there

and then now i’m going to try to get some more points axiom three says we

have three distinct non-collinear points

and and like i said a minute ago i don’t know if p’s any of them are not

so it could be or could be not we have to look at all those cases all right so

since that since a b and c are uh non-collinear uh i put clean you

there so it should be non-clean you’re here non-linear here

that needs to be fixed there so since a b and c are

non-collinear it follows that line a b line ac and line bc are three distinct

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lines now we went through those proofs line by line case by case

and so i’m just kind of summarizing that here each case was very short in itself

and so i’m just summarizing that so uh you know for example it follows these

are three distinct lines so the argument was these two are not the same line if

these two are the same line then a b and c are collinear

but they’re non-collinear axiom a3 says these three lines these three points are

non-collinear and so we made that argument for each pair of lines right there so

there’s the next sentence so now we’re going to suppose

that p is one of these points so we went through the case p is equal

to a p is equal to b p is equal to c we went through each of those cases and

they all ended up exactly the same so i’m going to say without loss of

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generality that p is equal to the a and in that case by axiom one it follows

that these right here are two distinct lines uh that are incident with p so

by axiom one um you know we have um you know p is equal to a

so we know that they’re incident with with p and we know we have

uh those two lines right there so when when

when p is equal to a we’re done we got the two lines that go through p they go

through p because uh p is equal to a in this case you know p is equal to a

um and we have two distinct lines as we’ve already proven up here

all right and so now what if we have four distinct points so by axiom one

there exists line ap incident with a and p and b p incident with b and p and c p

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incident with c and p so we got axiom one giving us those three lines

we don’t know if those three lines are all different

or some of them are the same or in fact the worst case would be that they’re all

equal to each other so let’s look at the

worst case what if all these three lines are equal to each other

so then that would put a b and c all on the same line that would make them

collinear but we know that they’re non-collinear

so hence these cannot be the same line so at least two of these are different

from each other so we have at least two lines passing through p and there we go

there’s the proof and hey you want to do some more math

together let’s do let’s do some right here i’ll see you in that episode