[deprosun]

Let R and S be two equivalence relations on the same set A. Define a new relation U such that U(x,y)⇔ [R(x,y) V S(x,y)]. Is U necessarily an equivalence relation? Either prove that it is or give an example where it's not.

Will checking this U(x,y)⇔ [R(x,y) V S(x,y)] with a truth table help?
How should I think for this problem? Does the relations R and S give the set A two partitions?
I need a jump start on this.

[mojo666]

I think it would be better to just trace through the logic. I would break it down into parts.

By definition U(x,y)->[R(x,y) V S(x,y)] ^ [R(x,y) V S(x,y)]->U(x,y)

Because of "U(x,y)->[R(x,y) V S(x,y)] ", if (x,y) is in U then (x,y) must be in either R or S (or both). Since both R and S are equivalence relations, then what other combinations must be in that relation? What is the implication of the presence of those combinations (a,b ) when accounting for the reverse statement "[R(a,b ) ^ S(a,b )]->U(a,b )"? How does that implication affect the status of U's equivalence?

This post has been edited by mojo666: 01 October 2013 - 09:39 AM

[macosxnerd101]

A biconditional is always an equivalence relation. "A if and only if B" can also be read as "A is equivalent to B" or "A is defined to be B." As always, start with the definitions.

[deprosun]

I really apologize for making you guys wait.
I haven't been able to solve this problem out.
I have put this to aside in order to "freshen" my mind and
I am working on some "easy and straight-forward" problems.
I will be come back with an sensible attempt on this question.


PS: There are so many I's, I apologize for that too. />

This post has been edited by deprosun: 03 October 2013 - 06:39 PM