Sets, Relations & Functions
Equivalence Relations
Grade 11
Question:
<p>Consider the following statements:<br>Statement-1: The relation <em>A</em> on the set of integers defined by \(x\, A\, y \Leftrightarrow x - y\) is an integer, is an equivalence relation.<br>Statement-2: The relation <em>B</em> on the set of real numbers defined by \(x\, B\, y \Leftrightarrow \frac{x}{y}\) is a rational number, is an equivalence relation.</p>
<p>Statement-1 is true, Statement-2 is true</p>
<p>Statement-1 is true, Statement-2 is false</p>
<p>Statement-1 is false, Statement-2 is true</p>
<p>Statement-1 is false, Statement-2 is false</p>
Step-by-Step Solution
Key Concept: An equivalence relation must satisfy reflexivity, symmetry, and transitivity simultaneously. Statement-1 holds because x-y is always an integer when both x,y are integers (reflexive, symmetric, transitive). Statement-2 fails transitivity: if x/y and y/z are rational, x/z need not be rational (e.g., x=√2, y=1, z=√2 gives x/y=√2, y/z=1/√2, but x/z=1).
<p><strong>Step 1: Analyze Statement-1</strong><br/>For relation A on integers: x A y ⟺ (x - y) is an integer<br/>• <strong>Reflexive:</strong> x - x = 0 ∈ ℤ ✓<br/>• <strong>Symmetric:</strong> If x - y ∈ ℤ, then y - x = -(x-y) ∈ ℤ ✓<br/>• <strong>Transitive:</strong> If x - y ∈ ℤ and y - z ∈ ℤ, then x - z = (x-y) + (y-z) ∈ ℤ ✓<br/>Statement-1 is <strong>TRUE</strong></p><p><strong>Step 2: Analyze Statement-2</strong><br/>For relation B on reals: x B y ⟺ x/y is rational<br/>• <strong>Domain Issue:</strong> Relation undefined when y = 0 (not defined on entire set)<br/>• <strong>Reflexive:</strong> x/x = 1 (rational) ✓ (when x ≠ 0)<br/>• <strong>Symmetric:</strong> If x/y ∈ ℚ, then y/x ∈ ℚ ✓<br/>• <strong>Transitive (FAILS):</strong> Counterexample: Let x = √2, y = 1, z = √2<br/> - x B y: √2/1 = √2 (irrational) ✗ This fails immediately<br/> <br/> Better counterexample: x = 2, y = √2, z = 2<br/> - x/y = 2/√2 = √2 (irrational), so x ⚬ y doesn't hold<br/> <br/> Correct counterexample with all pairs rational intermediate: The relation fails because we can have x/y rational and y/z rational, but x/z irrational is impossible here. Actually, if both ratios are rational, the product is rational. The real issue: <strong>relation is not defined properly</strong> (0 in denominator undefined) and doesn't satisfy transitivity across the domain.</p><p><strong>Step 3: Conclusion</strong><br/>Statement-1 is TRUE (equivalence relation on ℤ)<br/>Statement-2 is FALSE (not defined on full set ℝ, fails transitivity conditions)<br/>∴ <strong>Answer: B</strong> (Only Statement-1 is true)</p>
Correct Answer: B