anonymous
  • anonymous
is anybody good at discrete math?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
perl
  • perl
me
anonymous
  • anonymous
Thanks!!
anonymous
  • anonymous
alright my question is how do you figure out if anything is antisymmetric

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perl
  • perl
thats not very specific
anonymous
  • anonymous
like okay R1 (1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)
perl
  • perl
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,
anonymous
  • anonymous
yeah but how does that explain if this is antisymmetric or not
perl
  • perl
well first you have to state what the relation is, equality? less than, greater than? division modulo n?
perl
  • perl
I would say this is not anti-symmetric because you have (3,4) ( 4,1) but 3 does not equal 1
perl
  • perl
if you delete the point (3,4) then it is anti-symmetric
anonymous
  • anonymous
what about (1,2) and (2,1) and (1,1)
anonymous
  • anonymous
Wait so basically its anti symmetric if no points equal each other like a is not equal 2 b?
perl
  • perl
1R2 and 2R1 so 1 =1 is true
anonymous
  • anonymous
ohh so in that situation a = b?
anonymous
  • anonymous
and if thats the case then it would be antisymmetric
perl
  • perl
well the order relation <= is anti-symmetric
perl
  • perl
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
anonymous
  • anonymous
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 :/
anonymous
  • anonymous
That make sense but how does that apply to that problem
perl
  • perl
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.
perl
  • perl
so we find many 'things' that behave in math just like the ordering <= where 2<= 3 . in advanced math we look for patterns
anonymous
  • anonymous
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
anonymous
  • anonymous
ohhh okay
perl
  • perl
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}
anonymous
  • anonymous
Yeah i get that
perl
  • perl
so you could say {1,2} < {1,2,3} , or you could say <= to indicate that it is not a proper subset
anonymous
  • anonymous
alright
perl
  • perl
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
anonymous
  • anonymous
OHHHH IGET IT !!!!!!!
anonymous
  • anonymous
lol thanks but hold on lol hold on
perl
  • perl
lol
perl
  • perl
right, aRb is the same thing as saying (a,b) is in R. or equivalent
anonymous
  • anonymous
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
anonymous
  • anonymous
because (1,2) is symmetric to (2,1)
perl
  • perl
symmetric is different
anonymous
  • anonymous
how so
perl
  • perl
R is symmetric when if (a,b) is in R, then (b,a) is in R
anonymous
  • anonymous
thats the same thing i said tho
perl
  • perl
antisymmetry is different
perl
  • perl
right, sorry i didnt see what you wrote
anonymous
  • anonymous
yeah but using that.. isnt that the same as antisymmetric
perl
  • perl
ok so here are some examples one sec , thinking
anonymous
  • anonymous
okay
perl
  • perl
R = { (1,1) (2,1) (2,1) } is antisymmetric. R= { (1,1) , (1,2) (2,1) } is symmetric
perl
  • perl
A relation can be both symmetric and antisymmetric
perl
  • perl
woops
perl
  • perl
R= { (1,1) ( 1,2) } is anti symetric
anonymous
  • anonymous
a null set would be everything but i dnt get why the first one is antisymmetric
anonymous
  • anonymous
a = 2 right?
anonymous
  • anonymous
and b = 1?
anonymous
  • anonymous
lol careful you are speaking to an open mind at the moment
perl
  • perl
R= { (1,1) ( 2,1) } is antisyemmetric
perl
  • perl
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.
anonymous
  • anonymous
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?
perl
  • perl
well it satisfies it because it doesnt violate it
perl
  • perl
it satisfies it trivially, since there is no occurence of both (a,b) and (b,a) in R R = {(1,1), (2,1)}
perl
  • perl
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
anonymous
  • anonymous
hmm i guess sheesh why is this math so hard i swear who thinks of creating stuff like this
perl
  • perl
well they are both if then conditions, but the antisymmetry has a little extra in it
anonymous
  • anonymous
okay so symmetry mean something must happen first then something else happen.. its like a program
perl
  • perl
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.
anonymous
  • anonymous
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
perl
  • perl
yes its like rules for a checking program , you could program it to check
anonymous
  • anonymous
okay
perl
  • perl
so just that one point? R = { ( 3,4) } ?
anonymous
  • anonymous
yea
perl
  • perl
then it is not symmetric, but it is anti-symmetric (trivially)
anonymous
  • anonymous
its in an example in my book i hate that they dnt explain stuff
anonymous
  • anonymous
Yes thats what they said but it makes no sense
perl
  • perl
it is anti symetric why? do you see, because it is an if then condition
anonymous
  • anonymous
how its only (3,4)
perl
  • perl
if (a,b) and (b,a) are in R, then a=b. but you dont have (a,b) and (b,a) in R
anonymous
  • anonymous
yea okay you just have (a,b)
perl
  • perl
because 'if-then' is true when the 'if' part is false
perl
  • perl
i think thats called trivial or vacuously true. we just say true though :)
anonymous
  • anonymous
are you talking about the logic statement if p is false but q is true then the whole statement ends up being true?
perl
  • perl
p-> q is true when p is false.
perl
  • perl
or another way to read the definition... is ,
perl
  • perl
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
perl
  • perl
or true otherwise
anonymous
  • anonymous
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
perl
  • perl
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) )
perl
  • perl
right , any relation R = { (a,b) } is antisymmetric , trivially , since there is only one point
perl
  • perl
since you need at least two points to even check the condition
anonymous
  • anonymous
true cool so one answer down for the rest of my life a single point is antisymmetric
perl
  • perl
lol
perl
  • perl
well, you have to be more precise. the relation R= { (a,b)} is anti symmetric
perl
  • perl
and also better say its a binary operation, and 'point' is an element of your relation aRb or (a,b) ordered pair
anonymous
  • anonymous
yeah lol that .. R = { (a,b) }
perl
  • perl
now, how can we make your original relation antisymmetric. can we delete something
anonymous
  • anonymous
okay i get you
perl
  • perl
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
anonymous
  • anonymous
lol okay
anonymous
  • anonymous
all try my best
anonymous
  • anonymous
ill*
perl
  • perl
ask yourself, why does it fail anti-symmetry (clue)
anonymous
  • anonymous
(4,40 and (1,1) and (2,2) and (3,4)
anonymous
  • anonymous
(4,4) i meant
anonymous
  • anonymous
everything left will make it anti symmetric
perl
  • perl
let's see
anonymous
  • anonymous
wait no keep (4,4)
anonymous
  • anonymous
so the elements shall be (1,2), (2,1), (4,1), (4,4)
perl
  • perl
right, i guess there is more than one solution, ok can you write out the new R
perl
  • perl
actually (4,4) is fine
anonymous
  • anonymous
so R = { (1,2), (2,1) , (4,1) , (4,4)}
anonymous
  • anonymous
wow thanks your like a teacher right.. you have to be.. if not you should Seriously!!! lol
anonymous
  • anonymous
huh but wont you need (4,1) to be with (4,4)
perl
  • perl
thanks. we have to be careful though
perl
  • perl
lets look at the definitoin again
anonymous
  • anonymous
ohh yeah i forgot its a single element once the the other antisymmetric elements are taken out of play.. right?
perl
  • perl
R anti-symmetric means: if (a,b) and (b,a) is in R, then a=b.
anonymous
  • anonymous
Ohh and even so a = 1 and b= 1 ohh okay!!
anonymous
  • anonymous
Yeah so its the one!!
perl
  • perl
but you have (1,2) (2,1) so 2 = 1 , that is false
anonymous
  • anonymous
so (4,4) can be out or in right?
perl
  • perl
or (1,2) and (2,1) are in R , but 1 != 2, i use != for not equal
perl
  • perl
(4,4) can be in
perl
  • perl
ok lets start from left to right
anonymous
  • anonymous
i think i get it.. it might be misconstrued but i get it better than how this conversation began
anonymous
  • anonymous
okay
perl
  • perl
(1,1) is fine . now (1,2) and (2,1) is a problem, so we have to delete one or the other
perl
  • perl
remember we only look at pairs of (a,b) , (b,a)
perl
  • perl
we dont care about points that are not in the form (a,b) (b,a)
anonymous
  • anonymous
ohh okay
perl
  • perl
so thats the only problem i see , (1,2) (2,1)
perl
  • perl
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) }
perl
  • perl
thats two antisymmetric relations , which i deleted minimal number of terms
perl
  • perl
given the original R
anonymous
  • anonymous
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
anonymous
  • anonymous
ahh i see so an easy way to look at it is to see where a = b or b = a?
perl
  • perl
yes if (a,b) is (1,2) then (b,a) is (2,1) , correct
perl
  • perl
so there is your only potential 'problem'
perl
  • perl
because clearly 1 is not equal to 2
anonymous
  • anonymous
yeah true
anonymous
  • anonymous
okay i understand
anonymous
  • anonymous
okay ill right out a set
perl
  • perl
ok :)
anonymous
  • anonymous
R5 = { (1,1), (1,2) , (1,3) , ( 1,4) , ( 2,2) , ( 2,3) ,( 2,4) , (3,3) , (3,4) ,( 4,4) }
anonymous
  • anonymous
well set A = {1,2,3,4}
anonymous
  • anonymous
So that R 5 is a relation of the set
perl
  • perl
why do you call it R5, thats just a label , like R_5
anonymous
  • anonymous
ehh i just say that because i cant make a sub script
perl
  • perl
no problem, so you want to know if R5 is antisymmetric?
anonymous
  • anonymous
and okay This is Reflexive, but not symmetric, this is transitive, but not antisymmetric
perl
  • perl
it looks antisymmetric, i dont see any 'inverse pairs', like (a,b) (b,a) .
perl
  • perl
where the (a,b) are switched
anonymous
  • anonymous
well i thought it has to be for all values of a and b not just for some?
perl
  • perl
whatever is in your relation ,
anonymous
  • anonymous
so the antisymmetric part would be (3,4) and (4,4)
perl
  • perl
no because (3,4) and (4,3) are not in the relation
anonymous
  • anonymous
ohh i see your point sorry i mixed that up
perl
  • perl
you have to find a pair like (1,2) (2,1) when it fails antisymmetry ,
anonymous
  • anonymous
(1,2) and (2,2)
perl
  • perl
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)
perl
  • perl
nope, (1,2) and (2,2) are fine. (1,2) and (2,1) are a problem
anonymous
  • anonymous
yeah we cant find a pair so because of that its antisymmetric?
perl
  • perl
yes !!
perl
  • perl
so it 'trivially' satisfies the antisymmetric condition
anonymous
  • anonymous
ohh okay i get this abstract thing however where my other deductions correct?
anonymous
  • anonymous
were*
perl
  • perl
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
anonymous
  • anonymous
okay makes sense.. so what about the transitive property
perl
  • perl
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
perl
  • perl
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
anonymous
  • anonymous
SO we have (2,3) (3,4) and (2,4) is that what makes it transitive
perl
  • perl
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) }
perl
  • perl
yes you are right
anonymous
  • anonymous
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?
perl
  • perl
(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
  • perl
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
perl
  • perl
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
anonymous
  • anonymous
ohh okay so (1,2) (2,3) (1,3)
perl
  • perl
yes that works
anonymous
  • anonymous
(1,3) (3,4) (1,4) that makes 3
anonymous
  • anonymous
but thats it how will i know when i satisfied the quota to make it transitive or any other property
perl
  • perl
so basically we have to check (1,2) (2,3) --> (1,3) is in R (1,3) (3,4) ---> (1,4) is in R...
perl
  • perl
when there is nothing left to check
anonymous
  • anonymous
but their is some thing left to check (1,1) (2,2) (3,3) (4,4)
perl
  • perl
and we dont need to check (a,a)
anonymous
  • anonymous
and those are not transitive they are reflexive
anonymous
  • anonymous
ohh okay
anonymous
  • anonymous
make sense
perl
  • perl
let me check reflexive relation, i believe a relation is reflexive when..
anonymous
  • anonymous
great so i now understand transitive and antisymmetric and oddly enough the symmetric property thanks to you ^_^
perl
  • perl
In mathematics, a reflexive relation is a binary relation on a set for which every element is related to itself
perl
  • perl
so we need (1,1 ) (2,2) (3,3) (4,4)
perl
  • perl
if you are going to use elements 1,2,3,4 in your R
perl
  • perl
if we dont use 4 at all in our R, then we dont need (4,4)
anonymous
  • anonymous
ohkay
anonymous
  • anonymous
makes sense
perl
  • perl
for instance, im going to delete terms
anonymous
  • anonymous
alright
perl
  • perl
R = { (1,1), (1,2) , (1,3) , ( 2,2) , ( 2,3) , (3,3) }
perl
  • perl
this is antisymmetric, transitive, reflexive
anonymous
  • anonymous
okay shall i prove it
perl
  • perl
sure , you can check exhaustively
perl
  • perl
that just means you exhaust the possibilities
anonymous
  • anonymous
(claps hands) time to work..
anonymous
  • anonymous
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)
perl
  • perl
it is antisymmetric because there are no pairs (a,b) and (b,a) in R . so it satisfies antisymmetry trivially
anonymous
  • anonymous
dammit i always mess that one up .. lol
perl
  • perl
heheh
anonymous
  • anonymous
ill learn over dam lol
anonymous
  • anonymous
ill learn over time i meant to say
perl
  • perl
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
perl
  • perl
if both (a,b) and (b,a) are in R, then it is not antisymmetric , (where a and b are distinct)
anonymous
  • anonymous
ohh okay but if (b,a) is in R then a must equal to b for to be antisymmetric
perl
  • perl
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
perl
  • perl
yes, which brings you back to the old definition
anonymous
  • anonymous
great this makes sense now
perl
  • perl
many paths that lead to Rome
anonymous
  • anonymous
by the way what is irreflexive?
perl
  • perl
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).
perl
  • perl
so R = { (1,1) } is not irreflexive R = { (1,2) } is irreflexive
anonymous
  • anonymous
ohh that easy!!
anonymous
  • anonymous
okay i understood that off rip
perl
  • perl
actually, reflexive, irreflexive, and antisymmetry are easy to check. the one that takes a bit of work is transitive
anonymous
  • anonymous
yeah now that i understand them it does take me a bit longer to figure out the transitive ones
perl
  • perl
now, we define an 'equivalence relation' as a relation which is reflexive, symmetric, and transitive.
anonymous
  • anonymous
ahh yes i was just reading about that
perl
  • perl
there are also 'order' relations,. partially orders , and total orders
anonymous
  • anonymous
yeah i have no idea about those havent read up to that part yet
perl
  • perl
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)
anonymous
  • anonymous
Yeah just about to bad those i might have to get into tomorrow its like 4 am right nw
perl
  • perl
well now you a grasp of this :)
anonymous
  • anonymous
Yup!! :)
anonymous
  • anonymous
thanks
perl
  • perl
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) ).
perl
  • perl
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.
anonymous
  • anonymous
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
perl
  • perl
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/
perl
  • perl
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
perl
  • perl
It is often easier to use 'equivalent' definitions. also for programming :)
anonymous
  • anonymous
Hmm how so?
perl
  • perl
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 .
perl
  • perl
computer programmers are not as concerned with proving things . computer scientists, maybe. :)
anonymous
  • anonymous
I am a programmer well tryna be
perl
  • perl
here is the way mathematicians define antisymmetry , http://en.wikipedia.org/wiki/Antisymmetric_relation
anonymous
  • anonymous
lol well thanks for the help i must go to bed thanks ^_^
perl
  • perl
gnite :)
anonymous
  • anonymous
yea hthe upside down A means for all right
perl
  • perl
right
perl
  • perl
R(a,b) means (a,b) is in R
perl
  • perl
I like this part R is antisymmetric .. if R(a,b) with a ≠ b, then R(b,a) must not hold.
perl
  • perl
i got that from wikipedia. thats the easiest way to look at it

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