Question 1191317
Let's analyze the relation R defined on the set of real numbers, where (x, y) ∈ R if and only if xy ≥ 0.

* **Reflexive:** A relation is reflexive if for all x, (x, x) ∈ R.  In our case, this means x*x = x² ≥ 0. Since the square of any real number is non-negative, this condition is always true. Therefore, R *is* reflexive.

* **Symmetric:** A relation is symmetric if for all x and y, if (x, y) ∈ R, then (y, x) ∈ R.  If xy ≥ 0, then yx ≥ 0 (since multiplication is commutative). So, if (x, y) ∈ R, then (y, x) ∈ R. Therefore, R *is* symmetric.

* **Transitive:** A relation is transitive if for all x, y, and z, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R. Let's consider a counterexample:

    * Let x = -1, y = 0, and z = 1.
    * x*y = (-1)*0 = 0 ≥ 0, so (x, y) ∈ R.
    * y*z = 0*1 = 0 ≥ 0, so (y, z) ∈ R.
    * x*z = (-1)*1 = -1 < 0, so (x, z) ∉ R.

Since we found a case where (x, y) and (y, z) are in R, but (x, z) is not, R is *not* transitive.

* **Equivalence Relation:** A relation is an equivalence relation if it is reflexive, symmetric, *and* transitive. Since R is not transitive, R is *not* an equivalence relation.

**In summary:**

* R is reflexive.
* R is symmetric.
* R is *not* transitive.
* R is *not* an equivalence relation.