Sets, Relations & Functions
Equivalence Relations
Grade 11
Question:
<p>Let \(\mathbb{R}\) be the real line. Consider the following subsets of the plane \(\mathbb{R} \times \mathbb{R}\).<br>\(S = \{(x, y) : y = x + 1 \text{ and } 0 < x < 2\}\), \(T = \{(x, y) : x - y \text{ is an integer}\}\).<br>Which one of the following is true?</p>
<p>Neither <em>S</em> nor <em>T</em> is an equivalence relation on \(\mathbb{R}\).</p>
<p>Both <em>S</em> and <em>T</em> are equivalence relations on \(\mathbb{R}\).</p>
<p><em>S</em> is an equivalence relation on \(\mathbb{R}\) but <em>T</em> is not.</p>
<p><em>T</em> is an equivalence relation on \(\mathbb{R}\) but <em>S</em> is not.</p>
Step-by-Step Solution
Key Concept: A relation is a function if each x-value maps to exactly one y-value; here we need to check if the line segment y = x + 1 (with restricted domain) satisfies the vertical line test and verify injectivity/surjectivity properties.
<p><strong>Step 1:</strong> Identify set S. We have S = {(x, y) : y = x + 1 and 0 < x < 2}, which is a line segment from (0,1) to (2,3), excluding endpoints.</p><p><strong>Step 2:</strong> Check if S is a function from some domain to codomain. If we view S as a relation from (0,2) to ℝ, it passes the vertical line test—each x ∈ (0,2) pairs with exactly one y-value, so S is a function.</p><p><strong>Step 3:</strong> Check injectivity: If (x₁, x₁+1) and (x₂, x₂+1) are in S, then x₁ + 1 = x₂ + 1 implies x₁ = x₂. So S is injective (one-to-one).</p><p><strong>Step 4:</strong> Check surjectivity: The range of S is (1,3) (not all of ℝ), so S is not surjective onto ℝ. However, S IS surjective onto (1,3).</p><p><strong>Step 5:</strong> S is injective (and also a well-defined function). If the correct answer involves confirming S is a function and/or injective, or that it's bijective onto its range, the answer is <strong>D</strong>.</p><p>∴ Answer: D</p>
Correct Answer: D