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me

Thanks!!

alright my question is how do you figure out if anything is antisymmetric

thats not very specific

like okay R1 (1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)

yeah but how does that explain if this is antisymmetric or not

I would say this is not anti-symmetric because you have
(3,4) ( 4,1) but 3 does not equal 1

if you delete the point (3,4) then it is anti-symmetric

what about (1,2) and (2,1) and (1,1)

Wait so basically its anti symmetric if no points equal each other like a is not equal 2 b?

1R2 and 2R1 so 1 =1 is true

ohh so in that situation a = b?

and if thats the case then it would be antisymmetric

well the order relation <= is anti-symmetric

That make sense but how does that apply to that problem

ohhh okay

Yeah
i get that

so you could say {1,2} < {1,2,3} , or you could say <= to indicate that it is not a proper subset

alright

OHHHH IGET IT !!!!!!!

lol thanks but hold on
lol hold on

lol

right, aRb is the same thing as saying (a,b) is in R. or equivalent

because (1,2) is symmetric to (2,1)

symmetric is different

how so

R is symmetric when if (a,b) is in R, then (b,a) is in R

thats the same thing i said tho

antisymmetry is different

right, sorry i didnt see what you wrote

yeah but using that.. isnt that the same as antisymmetric

ok so here are some examples one sec , thinking

okay

R = { (1,1) (2,1) (2,1) } is antisymmetric.
R= { (1,1) , (1,2) (2,1) } is symmetric

A relation can be both symmetric and antisymmetric

woops

R= { (1,1) ( 1,2) } is anti symetric

a null set would be everything but i dnt get why the first one is antisymmetric

a = 2 right?

and b = 1?

lol careful you are speaking to an open mind at the moment

R= { (1,1) ( 2,1) } is antisyemmetric

well it satisfies it because it doesnt violate it

hmm i guess sheesh why is this math so hard i swear who thinks of creating stuff like this

well they are both if then conditions, but the antisymmetry has a little extra in it

okay so symmetry mean something must happen first then something else happen.. its like a program

yes its like rules for a checking program , you could program it to check

okay

so just that one point?
R = { ( 3,4) } ?

yea

then it is not symmetric, but it is anti-symmetric (trivially)

its in an example in my book i hate that they dnt explain stuff

Yes thats what they said but it makes no sense

it is anti symetric why? do you see, because it is an if then condition

how its only (3,4)

if (a,b) and (b,a) are in R, then a=b. but you dont have (a,b) and (b,a) in R

yea okay you just have (a,b)

because 'if-then' is true when the 'if' part is false

i think thats called trivial or vacuously true. we just say true though :)

p-> q is true when p is false.

or another way to read the definition... is
,

or true otherwise

right , any relation
R = { (a,b) } is antisymmetric , trivially , since there is only one point

since you need at least two points to even check the condition

true cool so one answer down for the rest of my life a single point is antisymmetric

lol

well, you have to be more precise.
the relation R= { (a,b)} is anti symmetric

yeah lol that .. R = { (a,b) }

now, how can we make your original relation antisymmetric. can we delete something

okay i get you

lol okay

all try my best

ill*

ask yourself, why does it fail anti-symmetry (clue)

(4,40 and (1,1) and (2,2) and (3,4)

(4,4) i meant

everything left will make it anti symmetric

let's see

wait no keep (4,4)

so the elements shall be (1,2), (2,1), (4,1), (4,4)

right, i guess there is more than one solution, ok can you write out the new R

actually (4,4) is fine

so R = { (1,2), (2,1) , (4,1) , (4,4)}

wow thanks your like a teacher right.. you have to be.. if not you should Seriously!!! lol

huh but wont you need (4,1) to be with (4,4)

thanks.
we have to be careful though

lets look at the definitoin again

R anti-symmetric means: if (a,b) and (b,a) is in R, then a=b.

Ohh and even so a = 1 and b= 1 ohh okay!!

Yeah so its the one!!

but you have (1,2) (2,1) so 2 = 1 , that is false

so (4,4) can be out or in right?

or (1,2) and (2,1) are in R , but 1 != 2, i use != for not equal

(4,4) can be in

ok lets start from left to right

i think i get it.. it might be misconstrued but i get it better than how this conversation began

okay

(1,1) is fine . now (1,2) and (2,1) is a problem, so we have to delete one or the other

remember we only look at pairs of (a,b) , (b,a)

we dont care about points that are not in the form (a,b) (b,a)

ohh okay

so thats the only problem i see , (1,2) (2,1)

thats two antisymmetric relations , which i deleted minimal number of terms

given the original R

ahh i see so an easy way to look at it is to see where a = b or b = a?

yes if (a,b) is (1,2) then (b,a) is (2,1) , correct

so there is your only potential 'problem'

because clearly 1 is not equal to 2

yeah true

okay i understand

okay ill right out a set

ok :)

R5 = { (1,1), (1,2) , (1,3) , ( 1,4) , ( 2,2) , ( 2,3) ,( 2,4) , (3,3) , (3,4) ,( 4,4) }

well set A = {1,2,3,4}

So that R 5 is a relation of the set

why do you call it R5, thats just a label , like R_5

ehh i just say that because i cant make a sub script

no problem, so you want to know if R5 is antisymmetric?

and okay This is Reflexive, but not symmetric, this is transitive, but not antisymmetric

it looks antisymmetric, i dont see any 'inverse pairs', like (a,b) (b,a) .

where the (a,b) are switched

well i thought it has to be for all values of a and b not just for some?

whatever is in your relation ,

so the antisymmetric part would be (3,4) and (4,4)

no because (3,4) and (4,3) are not in the relation

ohh i see your point sorry i mixed that up

you have to find a pair like (1,2) (2,1) when it fails antisymmetry
,

(1,2) and (2,2)

nope, (1,2) and (2,2) are fine. (1,2) and (2,1) are a problem

yeah we cant find a pair
so because of that its antisymmetric?

yes !!

so it 'trivially' satisfies the antisymmetric condition

ohh okay i get this abstract thing however where my other deductions correct?

were*

okay makes sense.. so what about the transitive property

SO we have (2,3) (3,4) and (2,4) is that what makes it transitive

yes you are right

ohh okay so (1,2) (2,3) (1,3)

yes that works

(1,3) (3,4) (1,4) that makes 3

but thats it how will i know when i satisfied the quota to make it transitive or any other property

so basically we have to check
(1,2) (2,3) --> (1,3) is in R
(1,3) (3,4) ---> (1,4) is in R...

when there is nothing left to check

but their is some thing left to check (1,1) (2,2) (3,3) (4,4)

and we dont need to check (a,a)

and those are not transitive they are reflexive

ohh okay

make sense

let me check reflexive relation, i believe a relation is reflexive when..

so we need (1,1 ) (2,2) (3,3) (4,4)

if you are going to use elements 1,2,3,4 in your R

if we dont use 4 at all in our R, then we dont need (4,4)

ohkay

makes sense

for instance, im going to delete terms

alright

R = { (1,1), (1,2) , (1,3) , ( 2,2) , ( 2,3) , (3,3) }

this is antisymmetric, transitive, reflexive

okay shall i prove it

sure , you can check exhaustively

that just means you exhaust the possibilities

(claps hands) time to work..

dammit i always mess that one up .. lol

heheh

ill learn over dam lol

ill learn over time i meant to say

if both (a,b) and (b,a) are in R, then it is not antisymmetric , (where a and b are distinct)

ohh okay but if (b,a) is in R then a must equal to b for to be antisymmetric

yes, which brings you back to the old definition

great this makes sense now

many paths that lead to Rome

by the way what is irreflexive?

so R = { (1,1) } is not irreflexive
R = { (1,2) } is irreflexive

ohh that easy!!

okay i understood that off rip

yeah now that i understand them it does take me a bit longer to figure out the transitive ones

ahh yes i was just reading about that

there are also 'order' relations,. partially orders , and total orders

yeah i have no idea about those havent read up to that part yet

Yeah just about to bad those i might have to get into tomorrow its like 4 am right nw

well now you a grasp of this :)

Yup!! :)

thanks

It is often easier to use 'equivalent' definitions. also for programming :)

Hmm how so?

computer programmers are not as concerned with proving things . computer scientists, maybe. :)

I am
a programmer well tryna be

lol well thanks for the help i must go to bed thanks ^_^

gnite :)

yea hthe upside down A means for all right

right

R(a,b) means (a,b) is in R

I like this part
R is antisymmetric
.. if R(a,b) with a ≠ b, then R(b,a) must not hold.

i got that from wikipedia. thats the easiest way to look at it