3
$\begingroup$

I'm studying properties of relations and there is one area that i'm kind of unsure about regarding the properties of asymmetry and anti-symmetry.

Suppose R = {(1,2),(3,4),(2,2)}

It would follow that R is:

Not reflexive, Not irreflexive, Not symmetric,

I would say it is not asymmetric and not antisymmetric also, but I get hung up on the (2,2) element.

Does (2,2), or any reflexive ordered pair, count as a = 2 and b = 2, such that a R b and b R a ?

$\endgroup$
  • 1
    $\begingroup$ It's transitive and antisymmetric. $\endgroup$ – Git Gud Feb 7 '15 at 17:02
  • $\begingroup$ So what you are saying is (2,2) means (a,a) and it does not mean (a,b) ? $\endgroup$ – Mike Feb 7 '15 at 17:05
1
$\begingroup$

Not reflexive Correct. Not every element is related to itself.

Not irreflexive Correct. Since we have $(2, 2)$. Irreflexivity requires that no element should be related to itself.

Not symmetric Correct. We have $(3, 4)$ but not $(4, 3)$.

Antisymmetric Correct, the only instance where $(a, b) \in R$ and $(b, a) \in R$ is $a = b$ (the pair$(2, 2))$

Not transitive ?

It is transitive because $(1, 2)$ and $(2, 2)$.

Not that this relation is defined on a set, which has a unique element 2, it seems to be the point of confusion from the comments.

$\endgroup$
  • $\begingroup$ hey axiom, thanks for the reply. Yes I think that is my point of confusion. Suppose that the relation is defined from set A = {1,2,3,4,5}. would (2,2) mean (a,b) or (a,a). $\endgroup$ – Mike Feb 7 '15 at 17:17
  • $\begingroup$ @Mike it would mean (a, a). $\endgroup$ – axiom Feb 7 '15 at 17:19
  • $\begingroup$ This would mean that the relation is Not antisymmetric then because there is no a = b it is a = a ? $\endgroup$ – Mike Feb 7 '15 at 17:22
  • $\begingroup$ @Mike The moment I add a pair $(a, b)$ and $(b, a)$, the relation ceases to be antisymmetric. Make sure you pay attention to the "for all" in the definition of antisymmetric. $\endgroup$ – axiom Feb 7 '15 at 17:24
  • $\begingroup$ Boom, that comment just made it clear for me haha. Thanks! $\endgroup$ – Mike Feb 7 '15 at 17:28
0
$\begingroup$

Taken any $a$, $b$ and $c$, we have:

$(a,b)$ and $(b,a)$ $\in R$ $\Rightarrow$ $a = b = 2$ because there are no other such couples in $R$. This shows $R$ to be anti-symmetric.

$(a,b)$ and $(b,c)$ $\in R$ $\Rightarrow$ $a = 1$, $b=c=2$ or $a=b=c=2$. In both cases: $(a,c) \in$ $R$. This is why $R$ is transitive.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.