Equivalence Relations Analysis

Question

A relation R is defined on a set $ A:(a, b, c) as R: ((a, c), (b, b))$ then the relation R is ?

A) transitive and antisymmetric B)transitive and symmetric C)symmetric and antisymmetric D)reflexive and transitive

I couldn't relate this question with any choices. To me it is not symmetric because does not include (c,a) and not transitive because of (a,c) does not included in R as (c,?).

score 1 · Answer 1 · answered Jun 06 '20 at 05:12

1

You're right that it's not symmetric.

But it IS transitive. Because there is no pair of the form $(c,?)$, there is nothing to check for transitivity.

So what you've said rules out choices B and C. But now you just have to decide if it's reflexive or antisymmetric to decide if it's A or D.

answered Jun 06 '20 at 05:12

Nick

5,618

I said A and accordint to the answers it seems wrong. Then I will go with D. Thanks for checking. – wave Jun 06 '20 at 05:14
It cannot be D. It is not reflexive since there is no (a,a). I think A is correct. – Nick Jun 06 '20 at 05:16
Thanks, I will discuss with the instructor too. – wave Jun 06 '20 at 06:09

score 0 · Answer 2 · answered Jun 06 '20 at 06:38

To be transitive then, because of the $(a,c)$ for every $(c,?)$ it must include $(a,?)$. But there is no need for there to be any $(c,?)$ in the first place. If there is no $(c,?)$ there's nothing to check.

It is transitive. For $(a,c)$ there aren't any $(c,?)$ so there aren't any thing to check. For $(b,b)$ then only $(b,?)$ there is is $(b,b)$ and we have $(b,b)$ so it passes.

Its not symmetric as There is an $(a,c)$ so to be symmetric there would need to be an $(c,a)$ and there isn't.

To be antisymmetric for every $(x,y); x\ne y$ we can not have $(y,x)$. The only $(x,y); x\ne y$ we have is $(a,c)$ and we don't have $(c,a)$. So it is antisymmetric.

To be reflexive we need $(a,a), (b,b)$ and $(c,c)$ and we don't have those all so it's not reflexive

Equivalence Relations Analysis

2 Answers2