Sets, Relations & Functions
Types of Functions
Grade 11
Question:
<p>A function \(f\) from the set of natural numbers to integers defined by<br>\[f(n) = \begin{cases} \dfrac{n-1}{2}, & \text{when } n \text{ is odd} \\ -\dfrac{n}{2}, & \text{when } n \text{ is even} \end{cases}\]<br>is</p>
<p>one–one but not onto</p>
<p>onto but not one–one</p>
<p>one–one and onto both</p>
<p>neither one–one nor onto</p>
Step-by-Step Solution
Key Concept: Verify injectivity by checking if different inputs map to different outputs, and verify surjectivity by checking if every integer in the codomain has a preimage in the domain (natural numbers).
<p><strong>Step 1: Check Injectivity</strong></p><p>Assume f(n₁) = f(n₂). We need to verify n₁ = n₂.</p><p><u>Case 1:</u> Both n₁, n₂ odd: (n₁-1)/2 = (n₂-1)/2 ⟹ n₁ = n₂ ✓</p><p><u>Case 2:</u> Both n₁, n₂ even: -n₁/2 = -n₂/2 ⟹ n₁ = n₂ ✓</p><p><u>Case 3:</u> n₁ odd, n₂ even: (n₁-1)/2 = -n₂/2. Since LHS ≥ 0 and RHS ≤ 0, both must equal 0, giving n₁=1, n₂=0. But 0∉ℕ, contradiction ✓</p><p>∴ <strong>f is injective</strong></p><p><strong>Step 2: Check Surjectivity</strong></p><p>For any integer m, find n∈ℕ such that f(n) = m:</p><p><u>If m ≥ 0:</u> Use n = 2m+1 (odd). Then f(2m+1) = (2m+1-1)/2 = m ✓</p><p><u>If m < 0:</u> Use n = -2m (even). Then f(-2m) = -(-2m)/2 = m ✓</p><p>∴ <strong>f is surjective</strong></p><p><strong>Step 3: Conclusion</strong></p><p>Since f is both injective and surjective, f is bijective.</p><p>∴ Answer: C</p>
Correct Answer: C