A relation $R$ such that $R\cup R^{-1}$ is not an equivalence relation I have a homework assignment to find a Relation $R$ over $A = \{1,2,3\}$
where $R\cup {{R}^{-1}}$ is not an equivalence relation (transitive, reflexive and symmetrical).
$R$ must be Transitive and Reflexive (${{I}_{A}}\subseteq R$)
Any clue anyone? 
I just am unable to find an example
Thanks
 A: Given that you are assuming that $R$ is reflexive, the only thing that can fail for $R\cup R^{-1}$ to be an equivalence relation is transitivity: you should verify that since $I_A\subseteq R$, then $I_A\subseteq R\cup R^{-1}$; and that $R\cup R^{-1}$ is symmetric for every relation $R$. So the only possible pitfall lies in transitivity.
Now, you are assuming that $R$ itself is transitive. So, how can transitivity fail? Say you have $(a,b),(b,c)\in R\cup R^{-1}$; if $(a,b),(b,c)\in R$, then since we are assuming $R$ is transitive, then $(a,c)\in R\subseteq R\cup R^{-1}$. If $(a,b),(b,c)\in R^{-1}$, then $(c,b),(b,a)\in R$, and again by transitivity we conclude $(c,a)\in R$, hence $(a,c)\in R^{-1}\subseteq R\cup R^{-1}$.
So what's left? What happens if $(a,b)\in R$, and $(b,c)\in R^{-1}$, but we do not have $(a,b)\in R^{-1}$ nor $(b,c)\in R$? Can you construct such an example? What will happen then?
A: One sometimes helpful way to approach questions like this, if you can't find a counterexample, is simply to try and prove the opposite: that is, prove that $R\cup R^{-1}$ is an equivalence relation. Well, it's obviously reflexive; it's also obviously symmetric, because anything in $R$ will have its inverse in $R^{-1}$. So transitivity must be where it breaks down.
Suppose there are two relations a~b and b~c in $R\cup R^{-1}$ whose product a~c is not in $R\cup R^{-1}$. (You might also write these relations (a,b), etc.) It's easy to see that a, b and c must all be distinct elements, so in fact, we might as well choose 1~2, 2~3 and 1~3: we want the first two to be in $R\cup R^{-1}$, but not the third.
As $R$ is transitive, we can't have both 1~2 and 2~3 in $R$. Likewise they can't both be in $R^{-1}$ (otherwise, flip them over to $R$ and use transitivity there). So one of them must be in $R$ and one in $R^{-1}$. Taking $R$ and $R^{-1}$ to be minimal relations containing 1~2 and 2~3 respectively works as a counterexample.
