Sets, Relations & Functions
General
Grade None
Question:
<p>Let A = {2, 3, 6, 8, 9, 11}, B = {1, 4, 5, 10, 15}. R on A × B: (a, b) R (c, d) iff 3ad −7bc is even.
Then R is:</p>
Reflexive but not symmetric
Transitive but not symmetric
Reflexive and symmetric but not transitive
An equivalence relation
Step-by-Step Solution
Key Concept: 3ad -7bc is even \Leftrightarrow 3ad \equiv7bc (mod 2). Since 3 \equiv1 and 7 \equiv1 (mod 2): this reduces to ad \equivbc
(mod 2).
<p><strong>Step 1</strong>: Simplify: 3ad -7bc even \Leftrightarrow ad \equivbc (mod 2), since odd \times anything has the same parity as anything.</p><p><strong>Step 2</strong>: Reflexive ✓: (a, b) R (a, b): 3ab -7ab = -4ab, always even (divisible by 4).</p><p><strong>Step 3</strong>: Symmetric ✓: If ad \equivbc (mod 2), then cb \equivda (mod 2) since cb = bc and da = ad.</p><p><strong>Step 4</strong>: Transitivity FAILS with specific elements from the given sets. Take:</p><br>• (a, b) = (2, 1), (c, d) = (6, 4): 3(2)(4) -7(1)(6) = 24 -42 = -18 — even ✓\Rightarrow (2, 1) R (6, 4).<br>• (c, d) = (6, 4), (e, f ) = (3, 1): 3(6)(1) -7(4)(3) = 18 -84 = -66 — even ✓\Rightarrow (6, 4) R (3, 1).<br>• (a, b) = (2, 1), (e, f ) = (3, 1): 3(2)(1) -7(1)(3) = 6 -21 = -15 — odd ✗\Rightarrow (2, 1) ̸ R (3, 1).<br>Transitivity fails.
Correct Answer: 3