NDA Math Sets Relations and Functions

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.