Relations and Functions (Class 12) — Practice

Question 1

Show that the relation R in the set ℤ of integers defined by R = {(a,b) : 2 divides (a−b)} is an equivalence relation.

Accepted Answer

REFLEXIVE: For any a ∈ ℤ, a−a = 0 = 2×0. Since 2 divides 0, (a,a) ∈ R. ✓ SYMMETRIC: If (a,b) ∈ R, then 2 | (a−b). So a−b = 2k for some k ∈ ℤ. Then b−a = −2k = 2(−k), and since −k ∈ ℤ, 2 | (b−a). Therefore (b,a) ∈ R. ✓ TRANSITIVE: If (a,b) ∈ R and (b,c) ∈ R, then 2 | (a−b) and 2 | (b−c). So a−b = 2m and b−c = 2n for m,n ∈ ℤ. Adding: a−c = (a−b)+(b−c) = 2m+2n = 2(m+n). Since m+n ∈ ℤ, 2 | (a−c). So (a,c) ∈ R. ✓ Since R is reflexive, symmetric, and transitive, R is an EQUIVALENCE RELATION.

Question 2

Let f: ℝ → ℝ be defined by f(x) = 2x + 3. Show that f is bijective and find f⁻¹.

Accepted Answer

ONE-ONE: Let f(x₁) = f(x₂). Then 2x₁ + 3 = 2x₂ + 3. Subtracting 3: 2x₁ = 2x₂. Dividing by 2: x₁ = x₂. Hence f is ONE-ONE (injective). ✓ ONTO: Let y ∈ ℝ. We need x ∈ ℝ such that f(x) = y. Solving: 2x + 3 = y ⟹ x = (y−3)/2. Since y ∈ ℝ, (y−3)/2 ∈ ℝ = domain. And f((y−3)/2) = 2·(y−3)/2 + 3 = (y−3) + 3 = y. ✓ Hence f is ONTO (surjective). Since f is both one-one and onto, f is BIJECTIVE. FINDING f⁻¹: Set y = f(x) = 2x + 3. Solving for x: x = (y−3)/2. Replacing y by x: f⁻¹(x) = (x−3)/2. VERIFICATION: f(f⁻¹(x)) = f((x−3)/2) = 2·(x−3)/2 + 3 = x−3+3 = x ✓

Question 3

Let f: A → B and g: B → C be two functions. Prove that if both f and g are onto, then g ∘ f: A → C is also onto. Is the converse true? Justify.

Accepted Answer

THEOREM: If f: A→B and g: B→C are both onto, then g∘f: A→C is onto. PROOF: Let z be any arbitrary element in C. Since g: B→C is onto, there exists some y ∈ B such that g(y) = z. ... [1] Since f: A→B is onto, for this y ∈ B, there exists some x ∈ A such that f(x) = y. ... [2] Now, (g∘f)(x) = g(f(x)) = g(y) [from 2] = z [from 1]. Therefore, for every z ∈ C, there exists x ∈ A such that (g∘f)(x) = z. Hence g∘f is ONTO. ✓ CONVERSE — IS IT TRUE?: 'If g∘f is onto, are both f and g onto?' PARTIAL CONVERSE: If g∘f is onto, then g must be onto (but f may NOT be). PROOF that g must be onto: Let z ∈ C. Since g∘f is onto, there exists x ∈ A with (g∘f)(x) = z, i.e., g(f(x)) = z. Let y = f(x) ∈ B. Then g(y) = z. So for every z ∈ C there exists y ∈ B with g(y) = z. Therefore g is onto. COUNTEREXAMPLE that f need not be onto: Let A = {1}, B = {2,3}, C = {4}. Define f: A→B by f(1) = 2 (NOT onto since 3 ∈ B has no preimage). Define g: B→C by g(2) = g(3) = 4 (onto). Then g∘f: A→C with (g∘f)(1) = g(f(1)) = g(2) = 4, so g∘f IS onto. But f is NOT onto. CONCLUSION: The converse is PARTIALLY true — if g∘f is onto, then g is onto. But f need not be onto.

Practice — Relations and Functions (Class 12)

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