Question 1191688
To prove that the relation R on S = ℝ (real numbers), where (a, b) ∈ R if and only if a - b = 0, is an equivalence relation, we need to show that R is reflexive, symmetric, and transitive.

**1. Reflexivity:**

We need to show that for all a ∈ S, (a, a) ∈ R.

If a ∈ S, then a - a = 0.  Since a - a = 0, (a, a) ∈ R by the definition of R.  Therefore, R is reflexive.

**2. Symmetry:**

We need to show that for all a, b ∈ S, if (a, b) ∈ R, then (b, a) ∈ R.

Suppose (a, b) ∈ R.  This means that a - b = 0.
If a - b = 0, then we can multiply both sides by -1 to get -(a - b) = -0, which simplifies to b - a = 0.
Since b - a = 0, (b, a) ∈ R by the definition of R.  Therefore, R is symmetric.

**3. Transitivity:**

We need to show that for all a, b, c ∈ S, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.

Suppose (a, b) ∈ R and (b, c) ∈ R.
This means that a - b = 0 and b - c = 0.
From a - b = 0, we have a = b.
From b - c = 0, we have b = c.
Substituting b = c into a = b, we get a = c.
If a = c, then a - c = 0.
Since a - c = 0, (a, c) ∈ R by the definition of R.  Therefore, R is transitive.

**Conclusion:**

Since R is reflexive, symmetric, and transitive, R is an equivalence relation on S = ℝ.  In simpler terms, this relation defines equality:  (a, b) ∈ R if and only if a = b.