is anybody good at discrete math?

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is anybody good at discrete math?

Mathematics
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me
Thanks!!
alright my question is how do you figure out if anything is antisymmetric

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thats not very specific
like okay R1 (1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)
In 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,
yeah but how does that explain if this is antisymmetric or not
well first you have to state what the relation is, equality? less than, greater than? division modulo n?
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
so the definition of anti-symmetry 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
i 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 :/
That make sense but how does that apply to that problem
well 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.
so we find many 'things' that behave in math just like the ordering <= where 2<= 3 . in advanced math we look for patterns
Yeah thats why im being confused because im so used to specific examples that this abstract way of thinking leaves me lost in the dark
ohhh okay
so 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}
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
so 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
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
wait 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
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
The terms 'symmetric' and 'anti-symmetric' apply to a binary relation R: R symmetric means: if aRb then bRa. R anti-symmetric means: if aRb and bRa, then a=b.
okay 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?
well it satisfies it because it doesnt violate it
it satisfies it trivially, since there is no occurence of both (a,b) and (b,a) in R R = {(1,1), (2,1)}
the antisymmetry is an 'AND' condition. if you have both (a,b) AND (b,a) , then a=b. the symmetry condition is if-then condition
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
R is symmetric means: if (a,b) is in R then (b,a) is in R. R anti-symmetric means: if (a,b) and (b,a) is in R, then a=b.
okay 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
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 :)
are you talking about the logic statement if p is false but q is true then the whole statement ends up being true?
p-> q is true when p is false.
or another way to read the definition... is ,
Program: if you can find two points in your relation (a,b) & (b,a) .... then it must be the case that a = b. else true
or true otherwise
and a specific example would be (3,4) so basically any single point except (1,1) or other any points similar to that are antisymmetric
Program: 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 anti-symmetry is false.) otherwise true (if you cant find any two points (a,b) (b,a) )
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
and also better say its a binary operation, and 'point' is an element of your relation aRb or (a,b) ordered pair
yeah lol that .. R = { (a,b) }
now, how can we make your original relation antisymmetric. can we delete something
okay i get you
R= { (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 anti-symmetric its no fair if we delete all but one , lol
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
ohh yeah i forgot its a single element once the the other antisymmetric elements are taken out of play.. right?
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)
so 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) }
thats two antisymmetric relations , which i deleted minimal number of terms
given the original R
See 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
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)
dont want to call it 'anti anti-symmetry'. lets just say it is 'not' antisymmetric if we can find a pair (a,b) (b,a)
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*
I 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
okay makes sense.. so what about the transitive property
So 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
transitive, 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
SO we have (2,3) (3,4) and (2,4) is that what makes it transitive
you 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) }
yes you are right
ohh 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?
(2,3) (3,4) are in R, so it must be that (2,4) in R. (and it is) , similarly we check other candidates
if 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
but 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
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..
great so i now understand transitive and antisymmetric and oddly enough the symmetric property thanks to you ^_^
In mathematics, a reflexive relation is a binary relation on a set for which every element is related to itself
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..
its 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)
it is antisymmetric because there are no pairs (a,b) and (b,a) in R . so it satisfies antisymmetry trivially
dammit i always mess that one up .. lol
heheh
ill learn over dam lol
ill learn over time i meant to say
try 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
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
so 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
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?
An irreflexive, or anti-reflexive, 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).
so R = { (1,1) } is not irreflexive R = { (1,2) } is irreflexive
ohh that easy!!
okay i understood that off rip
actually, reflexive, irreflexive, and antisymmetry are easy to check. the one that takes a bit of work is transitive
yeah now that i understand them it does take me a bit longer to figure out the transitive ones
now, we define an 'equivalence relation' as a relation which is reflexive, symmetric, and transitive.
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
so 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)
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
yeah 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) ).
this 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.
it 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
its 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/
but 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
It is often easier to use 'equivalent' definitions. also for programming :)
Hmm how so?
computer 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 .
computer programmers are not as concerned with proving things . computer scientists, maybe. :)
I am a programmer well tryna be
here is the way mathematicians define antisymmetry , http://en.wikipedia.org/wiki/Antisymmetric_relation
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

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