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anonymous
 4 years ago
is anybody good at discrete math?
anonymous
 4 years ago
is anybody good at discrete math?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright my question is how do you figure out if anything is antisymmetric

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0like okay R1 (1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1In mathematics, a binary relation R on a set X is antisymmetric if, for all a and b in X if aRb and bRa then a = b,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah but how does that explain if this is antisymmetric or not

perl
 4 years ago
Best ResponseYou've already chosen the best response.1well first you have to state what the relation is, equality? less than, greater than? division modulo n?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1I would say this is not antisymmetric because you have (3,4) ( 4,1) but 3 does not equal 1

perl
 4 years ago
Best ResponseYou've already chosen the best response.1if you delete the point (3,4) then it is antisymmetric

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what about (1,2) and (2,1) and (1,1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait so basically its anti symmetric if no points equal each other like a is not equal 2 b?

perl
 4 years ago
Best ResponseYou've already chosen the best response.11R2 and 2R1 so 1 =1 is true

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh so in that situation a = b?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and if thats the case then it would be antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1well the order relation <= is antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so the definition of antisymmetry generalises what we find in the usual order relation <= . for instance if a<=b and b <= a , then a = b , where a,b are integers and <= means less than or equal (usual ordering). but notice you can have unusual relations, like subset. if A is a subset of B and B is a subset of A, then A = B

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dnt get that sorry this type of math is really confusing it took me 3 weeks to understand sets and now im trying to understand relations bim pretty much behind in class :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That make sense but how does that apply to that problem

perl
 4 years ago
Best ResponseYou've already chosen the best response.1well we try to find in math the most general or abstract definitions. we find definitions that strip the terms of any specific meaning, so we can generalise to many different types of relations.

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so we find many 'things' that behave in math just like the ordering <= where 2<= 3 . in advanced math we look for patterns

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah thats why im being confused because im so used to specific examples that this abstract way of thinking leaves me lost in the dark

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so how do you define <= . when you define it, you realise that other mathematical objects behave similiar to how 2 <= 3 . for instance {1,2} is a subset of {1,2,3}

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so you could say {1,2} < {1,2,3} , or you could say <= to indicate that it is not a proper subset

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so using our austere definition we have R is antisymmetric if whenever (a,b) is in R and (b,a) is in R, then it must be the case that a = b

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OHHHH IGET IT !!!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol thanks but hold on lol hold on

perl
 4 years ago
Best ResponseYou've already chosen the best response.1right, aRb is the same thing as saying (a,b) is in R. or equivalent

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait but if thats the case where (a,b) element of R and (b,a) an element of R and a = b then isnt that kinda like being symmetric

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because (1,2) is symmetric to (2,1)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1R is symmetric when if (a,b) is in R, then (b,a) is in R

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats the same thing i said tho

perl
 4 years ago
Best ResponseYou've already chosen the best response.1right, sorry i didnt see what you wrote

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah but using that.. isnt that the same as antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1ok so here are some examples one sec , thinking

perl
 4 years ago
Best ResponseYou've already chosen the best response.1R = { (1,1) (2,1) (2,1) } is antisymmetric. R= { (1,1) , (1,2) (2,1) } is symmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1A relation can be both symmetric and antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1R= { (1,1) ( 1,2) } is anti symetric

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a null set would be everything but i dnt get why the first one is antisymmetric

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol careful you are speaking to an open mind at the moment

perl
 4 years ago
Best ResponseYou've already chosen the best response.1R= { (1,1) ( 2,1) } is antisyemmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1The terms 'symmetric' and 'antisymmetric' apply to a binary relation R: R symmetric means: if aRb then bRa. R antisymmetric means: if aRb and bRa, then a=b.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay i get that now because a in the R = {(1,1), (2,1)} is basically 1 and b is 1 as well so a is 1 in the (1,1) and b is 1 in the (2,1) right?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1well it satisfies it because it doesnt violate it

perl
 4 years ago
Best ResponseYou've already chosen the best response.1it satisfies it trivially, since there is no occurence of both (a,b) and (b,a) in R R = {(1,1), (2,1)}

perl
 4 years ago
Best ResponseYou've already chosen the best response.1the antisymmetry is an 'AND' condition. if you have both (a,b) AND (b,a) , then a=b. the symmetry condition is ifthen condition

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm i guess sheesh why is this math so hard i swear who thinks of creating stuff like this

perl
 4 years ago
Best ResponseYou've already chosen the best response.1well they are both if then conditions, but the antisymmetry has a little extra in it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay so symmetry mean something must happen first then something else happen.. its like a program

perl
 4 years ago
Best ResponseYou've already chosen the best response.1R is symmetric means: if (a,b) is in R then (b,a) is in R. R antisymmetric means: if (a,b) and (b,a) is in R, then a=b.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay so what about this one (3,4) what properties you give this in my eyes i see no reflexive no symmetry no antisymmetry and no transitive

perl
 4 years ago
Best ResponseYou've already chosen the best response.1yes its like rules for a checking program , you could program it to check

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so just that one point? R = { ( 3,4) } ?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1then it is not symmetric, but it is antisymmetric (trivially)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its in an example in my book i hate that they dnt explain stuff

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes thats what they said but it makes no sense

perl
 4 years ago
Best ResponseYou've already chosen the best response.1it is anti symetric why? do you see, because it is an if then condition

perl
 4 years ago
Best ResponseYou've already chosen the best response.1if (a,b) and (b,a) are in R, then a=b. but you dont have (a,b) and (b,a) in R

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea okay you just have (a,b)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1because 'ifthen' is true when the 'if' part is false

perl
 4 years ago
Best ResponseYou've already chosen the best response.1i think thats called trivial or vacuously true. we just say true though :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are you talking about the logic statement if p is false but q is true then the whole statement ends up being true?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1p> q is true when p is false.

perl
 4 years ago
Best ResponseYou've already chosen the best response.1or another way to read the definition... is ,

perl
 4 years ago
Best ResponseYou've already chosen the best response.1Program: if you can find two points in your relation (a,b) & (b,a) .... then it must be the case that a = b. else true

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and a specific example would be (3,4) so basically any single point except (1,1) or other any points similar to that are antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1Program: if you can find two points in your relation (a,b) & (b,a) .... then it must be the case that a = b. (if a does not equal B then your antisymmetry is false.) otherwise true (if you cant find any two points (a,b) (b,a) )

perl
 4 years ago
Best ResponseYou've already chosen the best response.1right , any relation R = { (a,b) } is antisymmetric , trivially , since there is only one point

perl
 4 years ago
Best ResponseYou've already chosen the best response.1since you need at least two points to even check the condition

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0true cool so one answer down for the rest of my life a single point is antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1well, you have to be more precise. the relation R= { (a,b)} is anti symmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1and also better say its a binary operation, and 'point' is an element of your relation aRb or (a,b) ordered pair

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah lol that .. R = { (a,b) }

perl
 4 years ago
Best ResponseYou've already chosen the best response.1now, how can we make your original relation antisymmetric. can we delete something

perl
 4 years ago
Best ResponseYou've already chosen the best response.1R= { (1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4) } now what are the least amount of points we can delete to make this antisymmetric its no fair if we delete all but one , lol

perl
 4 years ago
Best ResponseYou've already chosen the best response.1ask yourself, why does it fail antisymmetry (clue)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(4,40 and (1,1) and (2,2) and (3,4)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0everything left will make it anti symmetric

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the elements shall be (1,2), (2,1), (4,1), (4,4)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1right, i guess there is more than one solution, ok can you write out the new R

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so R = { (1,2), (2,1) , (4,1) , (4,4)}

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wow thanks your like a teacher right.. you have to be.. if not you should Seriously!!! lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0huh but wont you need (4,1) to be with (4,4)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1thanks. we have to be careful though

perl
 4 years ago
Best ResponseYou've already chosen the best response.1lets look at the definitoin again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh yeah i forgot its a single element once the the other antisymmetric elements are taken out of play.. right?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1R antisymmetric means: if (a,b) and (b,a) is in R, then a=b.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ohh and even so a = 1 and b= 1 ohh okay!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah so its the one!!

perl
 4 years ago
Best ResponseYou've already chosen the best response.1but you have (1,2) (2,1) so 2 = 1 , that is false

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so (4,4) can be out or in right?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1or (1,2) and (2,1) are in R , but 1 != 2, i use != for not equal

perl
 4 years ago
Best ResponseYou've already chosen the best response.1ok lets start from left to right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think i get it.. it might be misconstrued but i get it better than how this conversation began

perl
 4 years ago
Best ResponseYou've already chosen the best response.1(1,1) is fine . now (1,2) and (2,1) is a problem, so we have to delete one or the other

perl
 4 years ago
Best ResponseYou've already chosen the best response.1remember we only look at pairs of (a,b) , (b,a)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1we dont care about points that are not in the form (a,b) (b,a)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so thats the only problem i see , (1,2) (2,1)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so there are two possible solutions for problem outline above R = { (1,1), (2,1),(2,2),(3,4),(4,1),(4,4) } R= { (1,1),(1,2) ,(2,2),(3,4),(4,1),(4,4) }

perl
 4 years ago
Best ResponseYou've already chosen the best response.1thats two antisymmetric relations , which i deleted minimal number of terms

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0See this is where i mess up and im just realizing i do this i look at (1,2) and call that (a,b) then i look at (2,1) and call that (b,a) lol like i see it in reverse.. okay am i looking at it wrong or what

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh i see so an easy way to look at it is to see where a = b or b = a?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1yes if (a,b) is (1,2) then (b,a) is (2,1) , correct

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so there is your only potential 'problem'

perl
 4 years ago
Best ResponseYou've already chosen the best response.1because clearly 1 is not equal to 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay ill right out a set

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0R5 = { (1,1), (1,2) , (1,3) , ( 1,4) , ( 2,2) , ( 2,3) ,( 2,4) , (3,3) , (3,4) ,( 4,4) }

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well set A = {1,2,3,4}

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So that R 5 is a relation of the set

perl
 4 years ago
Best ResponseYou've already chosen the best response.1why do you call it R5, thats just a label , like R_5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ehh i just say that because i cant make a sub script

perl
 4 years ago
Best ResponseYou've already chosen the best response.1no problem, so you want to know if R5 is antisymmetric?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and okay This is Reflexive, but not symmetric, this is transitive, but not antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1it looks antisymmetric, i dont see any 'inverse pairs', like (a,b) (b,a) .

perl
 4 years ago
Best ResponseYou've already chosen the best response.1where the (a,b) are switched

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well i thought it has to be for all values of a and b not just for some?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1whatever is in your relation ,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the antisymmetric part would be (3,4) and (4,4)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1no because (3,4) and (4,3) are not in the relation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh i see your point sorry i mixed that up

perl
 4 years ago
Best ResponseYou've already chosen the best response.1you have to find a pair like (1,2) (2,1) when it fails antisymmetry ,

perl
 4 years ago
Best ResponseYou've already chosen the best response.1dont want to call it 'anti antisymmetry'. lets just say it is 'not' antisymmetric if we can find a pair (a,b) (b,a)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1nope, (1,2) and (2,2) are fine. (1,2) and (2,1) are a problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah we cant find a pair so because of that its antisymmetric?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so it 'trivially' satisfies the antisymmetric condition

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh okay i get this abstract thing however where my other deductions correct?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1I found an equivalent condition. might be easier to read R is antisymmetric if whenever (a,b) is in R with a ≠ b then (b,a) must not be in R

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay makes sense.. so what about the transitive property

perl
 4 years ago
Best ResponseYou've already chosen the best response.1So if you can find at least two points (a,b) (b,a) with a not equal to b, then it is not antisymmetric relation. Otherwise it is antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1transitive, if aRb and bRc , then aRc if (a,b) is in R and (b,c) is in R, then it must be that (a,c) is in R

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0SO we have (2,3) (3,4) and (2,4) is that what makes it transitive

perl
 4 years ago
Best ResponseYou've already chosen the best response.1you have to check all possibilities R5 = { (1,1), (1,2) , (1,3) , ( 1,4) , ( 2,2) , ( 2,3) ,( 2,4) , (3,3) , (3,4) ,( 4,4) }

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh and i have a question as long as i find one thing inside the relation does that make the whole relation transitive or symmetric or whatever?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1(2,3) (3,4) are in R, so it must be that (2,4) in R. (and it is) , similarly we check other candidates

perl
 4 years ago
Best ResponseYou've already chosen the best response.1if you find one example? no you have to check to make sure that if (a,b) (b,c) are in R then (a,c) is in R. for any a,b,c

perl
 4 years ago
Best ResponseYou've already chosen the best response.1but you can do shortcuts. if you have (a,b) and (b,b) then transitive says (a,b) must be in it, and clearly that is already there

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh okay so (1,2) (2,3) (1,3)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(1,3) (3,4) (1,4) that makes 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but thats it how will i know when i satisfied the quota to make it transitive or any other property

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so basically we have to check (1,2) (2,3) > (1,3) is in R (1,3) (3,4) > (1,4) is in R...

perl
 4 years ago
Best ResponseYou've already chosen the best response.1when there is nothing left to check

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but their is some thing left to check (1,1) (2,2) (3,3) (4,4)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1and we dont need to check (a,a)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and those are not transitive they are reflexive

perl
 4 years ago
Best ResponseYou've already chosen the best response.1let me check reflexive relation, i believe a relation is reflexive when..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0great so i now understand transitive and antisymmetric and oddly enough the symmetric property thanks to you ^_^

perl
 4 years ago
Best ResponseYou've already chosen the best response.1In mathematics, a reflexive relation is a binary relation on a set for which every element is related to itself

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so we need (1,1 ) (2,2) (3,3) (4,4)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1if you are going to use elements 1,2,3,4 in your R

perl
 4 years ago
Best ResponseYou've already chosen the best response.1if we dont use 4 at all in our R, then we dont need (4,4)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1for instance, im going to delete terms

perl
 4 years ago
Best ResponseYou've already chosen the best response.1R = { (1,1), (1,2) , (1,3) , ( 2,2) , ( 2,3) , (3,3) }

perl
 4 years ago
Best ResponseYou've already chosen the best response.1this is antisymmetric, transitive, reflexive

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay shall i prove it

perl
 4 years ago
Best ResponseYou've already chosen the best response.1sure , you can check exhaustively

perl
 4 years ago
Best ResponseYou've already chosen the best response.1that just means you exhaust the possibilities

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(claps hands) time to work..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its antisymmetric because (2,3) and (3,3) and its reflexive because (1,1) (2,2) (3,3) and its transitive because (1,2) (2,3) (1,3)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1it is antisymmetric because there are no pairs (a,b) and (b,a) in R . so it satisfies antisymmetry trivially

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dammit i always mess that one up .. lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ill learn over dam lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ill learn over time i meant to say

perl
 4 years ago
Best ResponseYou've already chosen the best response.1try the new equivalent definition, whenever (a,b) is in R then (b,a) cannot be in R. if it is in R , then it is not antisymmetric relation

perl
 4 years ago
Best ResponseYou've already chosen the best response.1if both (a,b) and (b,a) are in R, then it is not antisymmetric , (where a and b are distinct)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh okay but if (b,a) is in R then a must equal to b for to be antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so in other words, you are looking for counterexamples to antisymmetry. if you cannot find any counterexamples (a,b) and (b,a) , then it is true antisymmetry

perl
 4 years ago
Best ResponseYou've already chosen the best response.1yes, which brings you back to the old definition

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0great this makes sense now

perl
 4 years ago
Best ResponseYou've already chosen the best response.1many paths that lead to Rome

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0by the way what is irreflexive?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1An irreflexive, or antireflexive, relation is the opposite of a reflexive relation. It is a binary relation on a set where no element is related to itself. An example is the "greater than" relation (x>y).

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so R = { (1,1) } is not irreflexive R = { (1,2) } is irreflexive

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay i understood that off rip

perl
 4 years ago
Best ResponseYou've already chosen the best response.1actually, reflexive, irreflexive, and antisymmetry are easy to check. the one that takes a bit of work is transitive

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah now that i understand them it does take me a bit longer to figure out the transitive ones

perl
 4 years ago
Best ResponseYou've already chosen the best response.1now, we define an 'equivalence relation' as a relation which is reflexive, symmetric, and transitive.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh yes i was just reading about that

perl
 4 years ago
Best ResponseYou've already chosen the best response.1there are also 'order' relations,. partially orders , and total orders

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i have no idea about those havent read up to that part yet

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so now we see the point of all this work . we can define equivalence relations, and order relations using the definitions we used earlier (transitive, reflexive, symmetric, antisymmetric, etc)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah just about to bad those i might have to get into tomorrow its like 4 am right nw

perl
 4 years ago
Best ResponseYou've already chosen the best response.1well now you a grasp of this :)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1yeah the book should be clear about the 'trivial' satisfying . that a relation can trivially satisfy antisymmetry (since there are no pairs (a,b) (b,a) ).

perl
 4 years ago
Best ResponseYou've already chosen the best response.1this gets into the logic thing, p>q is true in the case that p cannot be satisfied. for instance , the empty set is a subset of any set.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it does but i understand best when some one is pointing me in the right direction especially when i have questions which the book doesnt ever want to answer

perl
 4 years ago
Best ResponseYou've already chosen the best response.1its like, the definition says for all a,b in R if (a,b ) & (b,a) is in R then a=b. well, what if there are no (a,b), (b,a) in R. then it is taken to be true vacuously or trivially/

perl
 4 years ago
Best ResponseYou've already chosen the best response.1but it is easier to just look for (a,b) (b,a) where a not equal to b. If you find it, it is not antisymmetric, if you cannot find it then it is antisymmetric

perl
 4 years ago
Best ResponseYou've already chosen the best response.1It is often easier to use 'equivalent' definitions. also for programming :)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1computer programmers have different needs than mathematicians . mathematicians want precise rigor , to be able to prove things. but at a cost , it might be harder to understand what the mathematician is talking about .

perl
 4 years ago
Best ResponseYou've already chosen the best response.1computer programmers are not as concerned with proving things . computer scientists, maybe. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am a programmer well tryna be

perl
 4 years ago
Best ResponseYou've already chosen the best response.1here is the way mathematicians define antisymmetry , http://en.wikipedia.org/wiki/Antisymmetric_relation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol well thanks for the help i must go to bed thanks ^_^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea hthe upside down A means for all right

perl
 4 years ago
Best ResponseYou've already chosen the best response.1R(a,b) means (a,b) is in R

perl
 4 years ago
Best ResponseYou've already chosen the best response.1I like this part R is antisymmetric .. if R(a,b) with a ≠ b, then R(b,a) must not hold.

perl
 4 years ago
Best ResponseYou've already chosen the best response.1i got that from wikipedia. thats the easiest way to look at it
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