For example, reflexive: We must have aRa for all possible values of a. R2 is obviously not reflexive, for example, because 5R1.

Transitive is probably the one that will take the most work. For transitive we must have that if aRb and bRc then aRc for all a, b, c. Again, R2 is not transitive because 1R2 and 2R3 but we don't have that 1R3.

I'm not sure what to say about R1 as far as transitivity. I suppose it is transitive because it fits the theorem, but the only possible case in R1 is of the type: if aRa and aRa then aRa, which is rather trivial.

-Dan