Let a relation R be defined on the set of integers Z as R = (x, y) : x, y Z and x^2 y^2 . Then R is :

Step-by-step solution

For reflexivity: Let x Z . Since x^2 x^2 is always true, (x, x) R . Thus, R is reflexive. For symmetry: Let x = 2 and y = 1 . Here 2^2 1^2 , so (2, 1) R . However, 1^2 2^2 , so (1, 2) R . Thus, R is not symmetric. For transitivity: Let (x, y) R and (y, z) R . This means x^2 y^2 and y^2 z^2 . Combining these inequalities gives x^2 z^2 , which implies (x, z) R . Thus, R is transitive. Therefore, R is reflexive and transitive but not symmetric. Answer: reflexive and transitive but not symmetric