$$R = \{(1,1), (1,3), (1,5), (4,4),(4,6)\}, \quad A = \{1,2,3,4,5,6\}$$
Determine the smallest set of pairs of elements which needs to be added to $R$ so that the resulting relation $\Bbb{R}^*$ is an equivalance relation.
My atttempt:
In order for it to equivalent relation it needs to be reflective, symmetric and transitive.
This would mean that the lowest reflective pair is $(1,1)$ which is already in $\Bbb{R}$.
There seem to be no transitive pairs within $\Bbb{R}$ and no symmetric pairs either.
Question: I am wondering how would this solved as I assumed you would write out all the other pairs that are not in $\Bbb{R}$ which would make up $\Bbb{R}^*$.
Hint: Think about this in terms of equivalence classes, rather than an equivalence relation: what is the most fine-grained partition of $A$ into equivalence classes that $R$ "allows"?
For instance, letting $A$ have only one equivalence class would be allowed by $R$, but it's probably not the most fine-grained. On the other hand, letting each element be alone in its own equivalence class is the most fine-grained you can get, but it is not allowed by $R$. You need to find the middle gound here.
Once you have the equivalence classes, you can find the corresponding equivalence relation, and figure out which pairs are in there.