nRm ⇔ n is a factor of m. ⇒ m is divisible by n. Reflexivity We know that n is divisible by n∀n∊N (n,n)∊R∀n∊N R is reflexive. Symmetric n,m∈N Let n=2,m=6 m is divisible by n but n is not divisible by m. Hence R is not symmetric. Transitivity Let (n,m)∊R and (m,p)∊R then (n,m)∊R and (m,p)∊R⇒(,=n,p)∊R or lf m is divisible by n and p is divisible by m. Hence p is divisible by n. (n,p)∊R∀n,p∊N R is transitive relation on N. Hence R is reflexive, transitive but not symmetric.