R={(x,y):x,y∈N and x2−4xy+3y2=0} Now, x2−4xy+3y2=0 ⇒(x−y)(x−3y)=0 ∴x=y or x=3y ∴R={(1,1),(3,1),(2,2),(6,2),(3,3), (9,3),……} Since (1,1),(2,2),(3,3),…… are present in the relation, therefore R is reflexive. Since (3,1) is an element of R but (1,3) is not the element of R, therefore R is not symmetric Here (3,1)∈R and (1,1)∈R⇒(3,1)∈R (6,2)∈R and (2,2)∈R⇒(6,2)∈R For all such (a,b)∈R and (b,c)∈R ⇒(a,c)∈R Hence R is transitive.