Question 1191688: Given a set S = ℝ, prove that a relation R on S where (a , b) ∈ R and a - b = 0 is an equivalence relation
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! 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.
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