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hi everyone welcome back i’m dave in this episode we’re going to talk about

um this theorem right here in incidence geometry there exists three distinct

lines that are non-concurrent so let’s do some math

so let’s recap the incident axioms so we’re going to let point line and

incidence be undefined terms by the way this is just a recap so if this is going

too quickly for you you’re not following along make sure and check out the full

series the previous episodes the link is below in the description

so um axiom one which i just call a1 so for every point p and every point

uh q not equal to p right so if i take just uh sorry just a point and

i take another point q then there’s exist a new unique line

through those uh two points there so that’s incident axiom one

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it’s in axiom two for every line l so now if i just put down a random line

any line you know just pick one there exist at least two points on that line

so i’ll just say p and q there’s just two distinct points

and then the third axiom is if uh so there exists three distinct points

uh with the property uh that no line is incident with all three of them so i

have a point p i have a point q and i have a point r

and i’m just randomly naming them the names are unimportant but i know

that there’s three of them and one line can’t go through all of them in other

words they’re non-collinear and so those are

that’s that’s one of the definitions that we have

three or more points are called collinear if there exists a line that

passes through all of them whereas incident with all of them

and then we have the dual concept would you switch points and lines three or

more lines are concurrent if there’s just a point incident with all of them

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in other words the lines are concurrent if um they all

if a point if there’s at least one point incident with all of them

okay and lines l and m are parallel if they’re distinct right so they’re not

the same line and no point is incident with both of them right so here’s a line

l here’s a line m and they’re parallel meaning

they’re distinct and there’s no point on either one of them

all right so there’s just a really uh brief recap

and so what’s been proven in our geometry so far in the last episode we

proved this theorem here if we have two distinct lines that are not parallel

then there’s a unique point in common so l and m

um are different lines there’s a you know a point not on both of them

every point on l is on point m and every point on m is on l then the lines would

be equal but there are distinct lines and they

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have a unique point in common only one point in common and so we proved that in

the previous episode there all right and so in today’s episode

we’re going to show that there exists three distinct lines that

are non-concurrent um and then in the upcoming episodes

we’re going to prove theorem three every

point is incident with at least one line

so in other words there’s no points that are just hanging out in no man’s land

um every point goes through every every point is on at least one line

and every line every line has at least one point that’s

not on it so if i just pick any line i know there’s some points that are not on

it so in other words all of our points aren’t aren’t on

a single line um and so number five every point has at least one line

not incident with it so if i just pick a random point i’m going to be able to

find a line that’s you know doesn’t pass through it

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um and then theorem uh six um every point is incident with at least two

distinct lines so these are upcoming uh episodes uh

recommend checking them out in the series um and so yeah let’s get started on uh

theorem two there’s just three distinct lines that are non-concurrent

so we’re going to begin here um by making statements and justifications

and i’m going to number the statements so we’re going to make this in the

column format and then at the end we’ll do this again in

the paragraph format where the paragraph format the idea behind it is just to

contain the major important steps so that you can go back and reproduce

this proof which is the formal proof uh on your own

um and also i just want to say before we get started on this this isn’t

necessarily the shortest proof that you could write

this proof is just to help beginners get started writing proofs um and it’s just

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for clarity purposes um you know my feeling is that any proof

that’s you know smaller than say a hundred lines

um isn’t a long proof and doesn’t need to necessarily be made any shorter

um unless it’s just completely obvious that it should be made shorter

but if a line is over that if it becomes too long then obviously you want to

fight hard to make it shorter and more digestible more readable

so anyways this proof is not necessarily going to be the shortest one

but it’s going to be try to be very clear

all right so we’re going to start off by naming these points

so by axiom three there exists three distinct uh nonconlinear points and i’m

just gonna call them a b and c now i’m not establishing any meaning to

the labeling so i just want to be clear just i know there’s three of them and so

i’ll name them a b and c and so by distinct here mean i know that

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a is not b and a is not c and b is not c

and c is not a or b all right and so and they’re non-linear okay so we have that

by axiom three and now by axiom one see because we know that a and b are not

the same points so i have two distinct points here

and so action one says there’s a line that goes through them

i guess i could uh keep a diagram going over here so i have three nonconlinear

points by the way this over here is not just uh is not for

uh actual part of the proof it’s just to help with the words it’s just a

visualization aid so there’s just a line through a and b so i’ll put a line here

and there exists a line um through b and c and i’m not really naming the lines

other than just using this notation here this is

um the line because there’s a unique line so

this right here is just the unique line that goes through the points b and c

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and then as you might imagine there’s a line that goes through a and c

also by axiom one so again you know you could probably put

all that in one statement uh and it’s just appealing to axiom one

but again i just you know want to make it you know nice clear and slow and easy

proof so now what i’m going to do is i’m going to make an r a hypothesis

and i’m going to say that a and b these two lines are actually equal to each

other so we know these three lines exist but when i say three lines i don’t

necessarily mean distinct so for example i might have here one

one and two and i wrote down three numbers but i only wrote down two distinct

numbers right so you know we want to be extra extra clear and very precise

so i want to say okay i have this line i have this line have

this line and i don’t know if they’re going to be equal to each other

using the notation that i wrote down it looks like they’re different but keep in

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mind that that’s just notation so i want to say hey what’s so wrong

with them being equal what if they’re what if they are equal to each other so

that’s my my ra hypothesis which i want to come up with a contradiction now

so uh you know because i have a and b and c if if these lines are equal what

that means is that every point on this line is also on this line and every

point on this line is also on this line so step five right here tells us that

we have three collinear points now because a and b are on this line by step two

and b and c are on this line by step three

and step five says well they’re all on the same line so they’re all collinear

but that’s a contradiction because step one clearly says that they’re

non-collinear and step six says they are

collinear and so you know if you look at

the definition with uh clean cleaning or non-clean here they’re the you know

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negation so um we have a contradiction there so in fact now we can conclude

that these are two different lines here now why did i pick on a and b uh sorry

why did i pick on this line through a and b and this line through b and c

because actually i you know could have done another one well we’re going to do

the other ones too what happens if the line through a and b

is equal to the line through a through a and c

well we’re going to get the same case again we’re going to get then that all

three lines are collinear and then we’re going to get that that’s

a contradiction because we know these three points are non-collinear

so now i know that these two lines are different

so and when you when we look at the diagram over here it’s it’s obvious our

intuition is oh yeah the lines are different right

but we’re actually going through the process what happens if ac is equal to

the line bc well then they’re all collinear they’ll

pass through the same line this line right here

and so then we get a contradiction again

so by looking at all three possibilities there

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we know that we have three different lines here

so these are three distinct lines so there’s a good example of something

that you know by just looking at diagrams

you might have jumped to the conclusion from um you know step you know one

um you know a b and c are three distinct and uncle linear points and you might

have jumped to step 17 but if you wanted to see

how to actually fill in those steps well there’s some there’s some ideas on how

to do that okay so the next step is so there exists

a point incident with all three lines um and this is going to be our a

hypothesis this is exactly what we don’t

want to happen we don’t want there to be any point that’s on all three lines

because that’s what non-concurrent means so we’re assuming that it’s not

non-concurrent and that’s what this statement right

here means it’s not non-concurrent and we’re going to get a contradiction which

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means it will be non-concurrent all right so i got this point x by assumption

all right so one and one uh must hold and i’m just gonna pick on the a we

could you could pick on the b or the c but i’m just gonna pick on the a so

because i have this point x and i have these other three points and i’m

wondering if x is any one of these three points or if x is not equal to the a

so that’s the law of excluded middle so this point x here

which is is incident with all three lines it’s somewhere on all three lines

and the first thing i’m going to do is say okay what if it’s a in other words

what if a is on all three lines here that’s case one

well then if a is all incident with all three lines then in fact they’re

collinear and we already saw that that can’t happen

so in fact that’s a contradiction by steps 1 and 22 they’re not collinear

now so i know that x is not equal to a so that’s case 2.

so this point here that’s on all three lines it cannot be a

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all right so here we go um lines a b and a c are not parallel

and you can you know kind of get the into intuition from the diagram over

here the reason why they’re not parallels of course because point a is

on both of them um and and what we’ve uh sorry that should say

theorem one here by theorem one what we proved in the uh previous episode is

that if two lines are not parallel they have a unique point in common

and that’s this point here a is on both of them we can see that from right here

in the previous steps that a was on this one and a was on this one

um and then now because we’re under the assumption that a is on all three lines

it’s definitely on these two lines here also

and so here’s this point a that’s on these two non-parallel lines and x is on

these two non-parallel line so the uniqueness of theorem one says that in

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fact those two points have to be the same but we’ve already shown that x is not

equal to a we’ve already tried to do that and we got a contradiction so

there does not exist a point in other words this um

hypothesis the r a hypothesis that we had in step [Music] 18

um cannot um you know cannot happen so there’s not a point incident with all

three lines right so that’s the negation of step 18 which was our our

which was the raa hypothesis all right very good so these lines are

not concurrent and that’s why the definition of non-concurrent and that’s

our conclusion is that we have three non-concurrent lines so um

you know just to make sure i didn’t lose anybody when we got to the x here

um we’re assuming here that step 18 is there exists a point somewhere in all

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three lines and my argument was was if x is a well we get a contradiction

if x is not a well then you know because a is on both of these

two lines and so is a then in fact x has to be a and then

but that leads us back to the same contradiction that we had before

so x you know it can’t exist um and so they’re uh non-concurrent

all right and so now let’s look at a paragraph proof

um something you know because this took us 29 lines now again it could be

written it could have been written up shorter that’s not really the point the

point is though is what were the big main ideas of the proof

certainly using axiom a3 was an important part of the way that i

organized this proof and then the uh you know there’s three points here and then

another uh big point of this proof was we got these lines through these two

points here through these three points and then another part was we use theorem

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one also that was theorem 1 what used right here so

let me see here if i can actually go and uh theorem one here

all right so let’s see if i can fix that um

error right there real quick there we go all right so the point is is that those

are the three important steps that you know the

proof should have here so let’s go through a paragraph proof here now so

this is how you might write it up um and in in my opinion um the proof that i’m

going to write for you is to give you the main ideas so that

you can go back on your own without much difficulty at all

and write the line-by-line proof so i’m going to say hey i’m going to use

axiom a3 i got three justine to not clean your points so that’s important

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because you know if you have say a hundred axioms which we don’t or if you

have a hundred theorems to sort through and realize you know what theorems do

you need you know that could be very important part of your proof is by

saying what am i using all right and then by axiom a1 and i

applied it three times we got this we got these lines here here i’m not saying

three distinct lines because we don’t know that right now at this point right

here so if any one of these lines is equal to the other

then all three points are collinear so this sentence right here kind of

summarizes several of the steps that we’ve laid out line by line by line in

the column proof and because you know not everyone wants

to read the 29 lines because some people are able to prove this on their own and

all they need are these ideas hence these three lines are distinct

so you know that just summarizes the steps that we did all right um

hence these three lines are distinct and then what did we have after they were

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distinct so we have here uh the remainder of the proof here

where we’re going to go here and say step 18 was so

hence all three lines are distinct so now we have three distinct lines

and now we can say we don’t want to end the proof yet so

scratch that right there the end of the proof here what we’re going to say is

that assume for a contradiction so this is another way of saying raa

hypothesis and conclusion assume for a contradiction there exists a point

and i’m going to call it x there exists a point x incident with all three lines

all right and so the three lines the only three lines i mentioned are these

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three right here so i’ll just say with all three lines here

all right and so then we can say if x equals a then a is incident

then if x equals a then a b and c are collinear are collinear and this would be

collinear and then i will say um are collinear contrary to hypothesis

in other words that’s a dead end because you know we’re assuming that

where they’re not linear if x is not equal to a then um [Music] lines a b

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and uh and then then distinct lines then distinct lines a b and a c um

do not have a unique point in common do not have a unique point in common

contrary to theorem one so if x is not equal to a right so we got a b and and

and uh ac so we have these two lines here and if x is not equal to a um

but it’s on all three lines so you know then then we’ll have uh we won’t have a

unique point in common because x is not equal to a

so um you know we’ve already proved here one is true right so um

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therefore or hence so hence um x does not exist and so what that says is that

so therefore the these lines here therefore a b a c and bc

are non-concurrent just by definition all right and so there’s the uh all the

main ideas and that proof right there and and you know your paragraph proof

can be shorter than that um it just kind of you know needs to

summarize all the steps in the proof uh the more drafts you write of it the

the quicker it will get and you’ll get the hang of it keep writing drafts don’t

settle on your first draft of your proof and as we work through this uh series

our proofs will get shorter and shorter as you become more capable improving

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them on your own all right so that’s it for this episode

right here uh check out the uh next theorem and the next episode right here

we’re gonna prove um some more interesting theorems in our

incidence geometry and i’ll see you then