Question 1191317: Determine whether the relation R on the set of all real numbers is reflective, symmetric, and/or transitive, where open parentheses x comma y close parentheses element of R if and only if x y greater or equal than 0. Is the relation R an equivalence relation?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
|
|
|
