Let R = (x, y): x, y N . and .x^2-4 x y+3 y^2=0 , where N is the set of all natural numbers. Then the relation R is :

Step-by-step solution

R = (x, y): x, y N . and .x^2-4 x y+3 y^2=0 Now, x^2-4 x y+3 y^2=0 (x-y)(x-3 y)=0 x=y or x=3 y 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.