Let a relation R be defined on the set of integers ℤ as R = \ (x,...

For reflexivity: Let $x \in \mathbb{Z}$. Since $x^2 \ge x^2$ is always true, $(x, x) \in R$. Thus, $R$ is reflexive.

For symmetry: Let $x = 2$ and $y = 1$. Here $2^2 \ge 1^2$, so $(2, 1) \in R$. However, $1^2 \not\ge 2^2$, so $(1, 2) \notin R$. Thus, $R$ is not symmetric.

For transitivity: Let $(x, y) \in R$ and $(y, z) \in R$. This means $x^2 \ge y^2$ and $y^2 \ge z^2$. Combining these inequalities gives $x^2 \ge z^2$, which implies $(x, z) \in R$. Thus, $R$ is transitive.

Therefore, $R$ is reflexive and transitive but not symmetric.

Answer: reflexive and transitive but not symmetric