Reflexivity and symmetry on set {1,2,3} I'm trying to think up of a relation on the set ${1,2,3}$ that is reflexive, symmtetric and NOT transitive. Does anyone know of one off the top of their because I cannot think for the life of me whether there is such a relation or not?
 A: Suppose every element is related to every other element, except that 1 is not related to 3 (or vice versa)
i.e. $\{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)\}$
It is reflexive if every element is related to itself).
It is symmetric, if $(x,y)$ then $(y,x)$
But it is not transitive as $(1,2)$ and $(2,3)$ but not $(1,3)$
If the relation is reflexive and symmetric and transitive, then it is an "equivalence relation" and we can partition the set such that every element in a class is related to every other element in the class (including itself).  If you can find a relation that does not allow you to partition based on that relation then you know that one of properties has not been met.
A: In addition to Doug's answer, here is a concrete example you can think of.
For $x$, $y$ $\in \{1,2,3\}$, let us say $x$ is related to $y$ iff $|x-y| \leq 1$. 
You can see it is reflexive since $|x-x| = 0 \leq 1$
You can see it is symmetric since $|x-y| = |y-x|$
You can come up with an example of it not being transitive. 1 is related to 2 and 2 is related to 3 but 1 is not related to 3 because the absolute difference is 2.
A: To be reflexive the relation must contain at least $\{(1,1),(2,2),(3,3)\}$.  
However, just that alone is also symmetric and transitive.   So we need to add elements which maintain symmetry but break transitivity.
To be symmetric, for any pair which we add, then we must also add its mirror.    Then we need to only consider wether to add any from the "upper triangle" elements $\{(1,2),(1,3),(2,3)\}$ (and their mirrors).
If we add only one from these, then the result will a transitive relation.   If we add all three, the result will also be a transitive relation.
So the answer, if any, lies in adding exactly two* from $(1,2),(1,3),(2,3)$ in a way that breaks transitivity. But which two? Hmmm....
(*and their mirrors)
