If \(\mathrm{S}\) is defined on \(\mathrm{R}\) by \((\mathrm{x}, \mathrm{y}) \in \mathrm{S} \Leftrightarrow \mathrm{xy} \geq 0\). Then \(\mathrm{S}\) is \(\ldots \ldots\) (a) an equivalence relation (b) reflexive only (c) symmetric only (d) transitive only

A relation is reflexive if and only if (x, x) ∈ S for all x in R. Consider the given condition xy ≥ 0. If we replace y with x and evaluate the expression, we get: x * x ≥ 0 Since x is a real number and squaring any real number will always result in a value greater than or equal to 0, this condition is true. Therefore, the relation is reflexive.

A relation is symmetric if and only if (x, y) ∈ S implies (y, x) ∈ S for all pairs (x, y) in R. We will use the given condition xy ≥ 0. If (x, y) ∈ S then xy ≥ 0. We need to test if the condition still holds if we swap the positions of x and y: y * x ≥ 0. As multiplication is commutative (xy = yx), the condition still holds. Therefore, the relation is symmetric.

A relation is transitive if and only if (x, y) ∈ S and (y, z) ∈ S imply (x, z) ∈ S for all x, y, z in R. Using the given condition xy ≥ 0, we have: 1. (x, y) ∈ S: xy ≥ 0 2. (y, z) ∈ S: yz ≥ 0 We now need to test if these two conditions imply that (x, z) ∈ S: xz ≥ 0. Let's consider the possible signs of x, y, and z: - If x, y, and z have the same sign, then xz will also have the same sign and xz ≥ 0. In this case, the relation is transitive. - If any two of x, y, and z have different signs, then xy or yz (or both) will be negative, which means (x, y) or (y, z), or both, are not in the relation S, and the transitive condition is not applicable. Thus, the relation is transitive in this case as well. As the relation is found to be transitive in all cases, we can conclude that the relation S is indeed transitive. After testing for reflexivity, symmetry, and transitivity, we find that the relation S possesses all three properties. Therefore, the correct answer is: (a) an equivalence relation