<p>(x1, y1) R (x2, y2) iff x1 ≤x2 or y1 ≤y2. Consider: (I) R is reflexive but not symmetric. (II)
R is transitive. Which statement is correct?</p>
Step-by-Step Solution
Key Concept: “OR” makes the relation very permissive — reflexivity holds trivially. Symmetry fails via a pair
that relates in one direction but not the other. Transitivity fails by mixing which coordinate carries
the relation at each step.
<p><strong>Step 1</strong>: Statement I — Reflexive ✓: x \leqx OR y \leqy — always true.</p><br>Not symmetric: (1, 1) R (2, 3): 1 \leq2 ✓. But (2, 3) R (1, 1): 2 \leq1? No. 3 \leq1? No. ✗<br>Statement I is TRUE.<p><strong>Step 2</strong>: Statement II — Transitivity fails: Let A = (3, 1), B = (2, 2), C = (2, 0).</p><br>A R B : 3 \leq2? No. 1 \leq2? Yes. ✓<br>B R C : 2 \leq2? Yes. ✓<br>A R C : 3 \leq2? No. 1 \leq0? No. 7<br>Statement II is FALSE.
Correct Answer: 2