# Math Help - Sets and Relations

1. ## Sets and Relations

Sorry not much of a maths wiz so I don't even know if I've posted this in the correct section. Apologies if not.

Just doing some revision for a first year maths exam and there is a question on sets and relations asking to determine if something is reflective, symmetric and transitive.

For example

S = {1,2,3} R = {(1,1),(1,2),(2,1),(2,2)}

Is symmetric and transitive but not reflective. I don't understand how this is worked out. Could some one please explain this to me?

2. Originally Posted by eddielee
Sorry not much of a maths wiz so I don't even know if I've posted this in the correct section. Apologies if not.

Just doing some revision for a first year maths exam and there is a question on sets and relations asking to determine if something is reflective, symmetric and transitive.

For example

S = {1,2,3} R = {(1,1),(1,2),(2,1),(2,2)}

Is symmetric and transitive but not reflective. I don't understand how this is worked out. Could some one please explain this to me?
I believe you meant reflexive not reflective.

A relation R on a set A is called reflexive if (a,a) $\epsilon$ R for every element a $\epsilon$ A.

R does not contain (3,3) so it is not reflexive.

A relation R on a set A is called symmetric if (b,a) $\epsilon$ R whenever (a,b) $\epsilon$ R, for all a,b $\epsilon$ A. Since (1,2) $\epsilon$ R and (2,1) $\epsilon$ R, R is symmetric.

I'll leave transitive to you.

3. Thank you so much.

Even though 3 isnt in R is still symmetric? And sorry to sound stupid but, I presume that this symbol means = or something along those lines but what is $
\epsilon
$

4. Sorry I understand the symmetric thing now. It just clicked. Out of the groups in R they all have to either be like (1,1) so the same number of if there like (1,2) there needs to be a (2,1).

5. Originally Posted by eddielee
Thank you so much.

Even though 3 isnt in R is still symmetric? And sorry to sound stupid but, I presume that this symbol means = or something along those lines but what is $
\epsilon
$

You're welcome. $\epsilon$ means is a member or element of.

6. Originally Posted by oldguynewstudent
You're welcome. $\epsilon$ means is a member or element of.
$$\in$$ gives the element symbol $\in$.

7. Right okay, So for example a $\in$ X means a is an eliminate of X.

With transitive if it was {(1,2),(3,4),(4,5)} it would be transitive and {(1,2),(3,4),(5,6)} wouldn't be?

8. Or do they have to go round in a loop? So it would have to be {(1,2),(3,4),(4,1)} to be transitive?

9. Originally Posted by eddielee
With transitive if it was {(1,2),(3,4),(4,5)} it would be transitive and {(1,2),(3,4),(5,6)} wouldn't be?
{(1, 2), (3, 4), (4, 5)} is NOT a transitive relation, since (3, 5) is not a member.

{(1, 2), (3, 4), (5, 6)} IS a transitive relation.

I suggest you review the definition of 'transitive relation'.

R = {(1, 2), (3, 4), (4, 5)} has in its field the "linking" member 4, but R LACKS the "linked" ordered pair (3, 5).

T = {(1, 2), (3, 4), (5, 6)} has no "linking" members in its field, so it can't fail to be transitive, since it can't miss having any "linked" ordered pair, since, as mentioned, there are no "linking" members in the field.

10. Originally Posted by eddielee
Or do they have to go round in a loop? So it would have to be {(1,2),(3,4),(4,1)} to be transitive?
No, "looping" is not required for transitivity. Also, the relation you just mentioned is not even transitive, since it lacks (3, 1); also it lacks (4, 2), and having (4, 2) would require also having (3, 2) .

You need to review the exact definition of 'transitive relation'.

11. Originally Posted by eddielee

S = {1,2,3} R = {(1,1),(1,2),(2,1),(2,2)}

Is symmetric and transitive but not reflective.
No, R is reflexive; however R is not reflexive on S. That is, R is reflexive on the field of R, but R is not reflexive on S, since R lacks (3, 3) as a member.