Let R be a relation on the set N of natural numbers defined by nRm
⇔ n is a factor of m (i.e. n(m). Then R is
Reflexive, transitive but not symmetric
Since n | n for all n in N,
therefore R is reflexive.
Since 2 | 6 but 6 | 2, therefore R is not symmetric.
Let n R m and m R p ⇒ n|m and m|p ⇒ n|p ⇒ nRp
So. R is transitive.