<p>The relation R = {(a, b) : gcd(a, b) = 1, 2a ̸= b, a, b ∈Z} is:
11
JEE Main 2019–2024 | Relations
Complete Solutions Booklet</p>
Transitive but not reflexive
Symmetric but not transitive
Reflexive but not symmetric
Neither symmetric nor transitive
Step-by-Step Solution
Key Concept: gcd(a, a) = |a| = 1 only for a = \pm1 — not reflexive. The condition 2a ̸= b breaks symmetry:
(2, 1) \in R but (1, 2) /\in R. gcd is never transitive in general.
<p><strong>Step 1</strong>: Reflexivity fails: (a, a) \in R requires gcd(a, a) = |a| = 1, which holds only for a = \pm1. For a = 2:</p><br>gcd(2, 2) = 2 ̸= 1. NOT reflexive.<p><strong>Step 2</strong>: Symmetry fails: Check (2, 1): gcd(2, 1) = 1 ✓, 2(2) = 4 ̸= 1 ✓\Rightarrow (2, 1) \in R.</p><br>Check (1, 2): gcd(1, 2) = 1 ✓, but 2(1) = 2 = b ✗(condition 2a ̸= b violated) \Rightarrow (1, 2) /\in R.<br>R is NOT symmetric.<p><strong>Step 3</strong>: Transitivity fails: Take a = 2, b = 3, c = 4.</p><br>• gcd(2, 3) = 1 ✓; 2(2) = 4 ̸= 3 ✓\Rightarrow (2, 3) \in R.<br>• gcd(3, 4) = 1 ✓; 2(3) = 6 ̸= 4 ✓\Rightarrow (3, 4) \in R.<br>• gcd(2, 4) = 2 ̸= 1 ✗\Rightarrow (2, 4) /\in R.<br>NOT transitive.
Correct Answer: 4