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Set relation: is it reflexive, symmetric or transitive?

Hi

First time doing sets and relations in discrete Maths.

If S={2,4,6} and R is a relation defined on S so that R={(2,2),(2,4),(4,2),(4,6)}

Is R reflexive, symmetric or transitive?

I don't know if a relation can be all 3 as there are elements for all three.

I'm ruling out reflexive since it is not true for all xRx but I'm not sure about the other two.

If I'm going with gut feeling alone, I'd say it's transitive based only on the fact that there isn't enough elements in R for me to confidently say it's symmetric.

Can a relation be symmetric and transitive? Also is it transitive?
Original post by 12yold
Hi

First time doing sets and relations in discrete Maths.

If S={2,4,6} and R is a relation defined on S so that R={(2,2),(2,4),(4,2),(4,6)}

Is R reflexive, symmetric or transitive?

I don't know if a relation can be all 3 as there are elements for all three.

I'm ruling out reflexive since it is not true for all xRx but I'm not sure about the other two.

If I'm going with gut feeling alone, I'd say it's transitive based only on the fact that there isn't enough elements in R for me to confidently say it's symmetric.

Can a relation be symmetric and transitive? Also is it transitive?


The three conditions are independent of each other and a relation can be any combination of the three or none of them. So, you need to check each condition separately.

Reflexivity.

As you've correctly deduced the relation is not reflexive, since (4,4), and (6,6) are not elements of the relations.

Symmetry.

The condtion of symmetry does not require every combination (a,b). It says IF (a,b) is present, then (b,a) must be present. Is this true for the given relation? You only need to find one instance of it not being true to show a relation isn't symmetric.

Transitivity.

Says IF (a,b) and (b,c) are present, then (a,c) must be present. Is this true? To show it's false all you need to do is find a pair (a,b) and (b,c) which are present, but (a,c) isn't.
(edited 4 years ago)
Reply 2
So would it be correct to state it is symmetric as we have (2,4) and (4,2) and no conflicting elements? Still a little confused about how to treat (4,6)

And that we can rule out transitive as it lacks (2,6) to complete xRy, yRz, and xRz?
Reply 3
Original post by 12yold
So would it be correct to state it is symmetric as we have (2,4) and (4,2) and no conflicting elements? Still a little confused about how to treat (4,6)

And that we can rule out transitive as it lacks (2,6) to complete xRy, yRz, and xRz?

What does "no conflicting elements" mean here? Symmetric would mean whenever (a,b) is in R then (b,a) is in R? Is this always the case in this example?
Reply 4
That's what I was confused about from ghostwalker. I wasn't sure what "The condtion of symmetry does not require every combination (a,b)." meant.

Ok, so seeing as (6,4) is not present it is not symmetric.

Which now brings me to a new question, can a relation be neither reflexive, symmetric or transitive? Seeing as there is not enough elements in R to show otherwise???
Original post by 12yold
That's what I was confused about from ghostwalker. I wasn't sure what "The condtion of symmetry does not require every combination (a,b)." meant.

Ok, so seeing as (6,4) is not present it is not symmetric.

Which now brings me to a new question, can a relation be neither reflexive, symmetric or transitive? Seeing as there is not enough elements in R to show otherwise???

A relation can fail all 3 conditions simultaneously. E.g. the relation on {1,2,3,} that merely says 1~2, 2~3.

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