<p>Let <em>N</em> denote the set of all natural numbers. Define two binary relations on <em>N</em> as \(R_1 = \{(x, y) \in N \times N : 2x + y = 10\}\) and \(R_2 = \{(x, y) \in N \times N : x + 2y = 10\}\). Then</p>
<p>Range of \(R_1\) is \{2, 4, 8\}.</p>
<p>Range of \(R_2\) is \{1, 2, 3, 4\}.</p>
<p>Both \(R_1\) and \(R_2\) are symmetric relations.</p>
<p>Both \(R_1\) and \(R_2\) are transitive relations.</p>
Step-by-Step Solution
Key Concept: A binary relation is a set of ordered pairs satisfying the given condition. To find R₁ and R₂, substitute natural number values systematically into each equation and collect only valid pairs where both coordinates are natural numbers.
<p><strong>Step 1: Find R₁ = {(x, y) ∈ N × N : 2x + y = 10}</strong></p><p>From 2x + y = 10, we get y = 10 - 2x</p><p>For y ∈ N, we need 10 - 2x ≥ 1, so 2x ≤ 9, giving x ≤ 4.5</p><p>Since x ∈ N: x ∈ {1, 2, 3, 4}</p><p>• x = 1: y = 8 → (1, 8)</p><p>• x = 2: y = 6 → (2, 6)</p><p>• x = 3: y = 4 → (3, 4)</p><p>• x = 4: y = 2 → (4, 2)</p><p><strong>Therefore: R₁ = {(1, 8), (2, 6), (3, 4), (4, 2)}</strong></p><p><strong>Step 2: Find R₂ = {(x, y) ∈ N × N : x + 2y = 10}</strong></p><p>From x + 2y = 10, we get x = 10 - 2y</p><p>For x ∈ N, we need 10 - 2y ≥ 1, so 2y ≤ 9, giving y ≤ 4.5</p><p>Since y ∈ N: y ∈ {1, 2, 3, 4}</p><p>• y = 1: x = 8 → (8, 1)</p><p>• y = 2: x = 6 → (6, 2)</p><p>• y = 3: x = 4 → (4, 3)</p><p>• y = 4: x = 2 → (2, 4)</p><p><strong>Therefore: R₂ = {(8, 1), (6, 2), (4, 3), (2, 4)}</strong></p><p><strong>Step 3: Identify key properties</strong></p><p>• |R₁| = 4 and |R₂| = 4</p><p>• R₁ ∩ R₂ = ∅ (no common elements)</p><p>• Neither relation is reflexive, symmetric, nor transitive</p><p>∴ Answer: B</p>
Correct Answer: B