• # question_answer Let R be a relation on the set N of natural numbers defined by nRm $\Leftrightarrow$ n is a factor of m (i.e., n|m). Then R is A) Reflexive and symmetricB) Transitive and symmetricC) EquivalenceD) Reflexive, transitive but not symmetric

Since n | n for all $n\in N$, therefore R is reflexive. Since 2 | 6 but $6\not{|}2$, therefore R is not symmetric. Let n R m and m R p Þ n|m and m|p Þ n|p Þ nRp. So, R is transitive.