Relations and Functions

Answer

Step 1: For reflexive relation, check if (a,a) ∈ R for all a ∈ A. Step 2: Check each element: - (1,1) ∈ R ✓ - (2,2) ∈ R ✓ - (3,3) ∈ R ✓ Step 3: Since all pairs (a,a) are present, R is reflexive. Ans

Answer

Step 1: For symmetric relation, if (a,b) ∈ R, then (b,a) ∈ R. Step 2: Check each pair: - (1,2) ∈ R and (2,1) ∈ R ✓ - (2,3) ∈ R and (3,2) ∈ R ✓ Step 3: For every (a,b), the corresponding (b,a) exists.

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Step 1: For transitive relation, if (a,b) ∈ R and (b,c) ∈ R, then (a,c) ∈ R. Step 2: Check all combinations: - (1,2) and (2,3) → need (1,3) → (1,3) ∈ R ✓ - (1,2) and (2,4) → need (1,4) → (1,4) ∈ R ✓ -

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Step 1: Check Reflexive: a ≡ a (mod 3) ✓ (3 divides a-a = 0) Step 2: Check Symmetric: If a ≡ b (mod 3), then 3|(a-b) ∴ 3|(b-a) ∴ b ≡ a (mod 3) ✓ Step 3: Check Transitive: If a ≡ b (mod 3) and b ≡ c (m

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Step 1: Two numbers are related if their difference is even. Step 2: This happens when both are odd or both are even. Step 3: Partition A into equivalence classes: - [1] = {1, 3, 5} (all odd numbers)

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Step 1: For one-one function, if f(x₁) = f(x₂), then x₁ = x₂. Step 2: Assume f(x₁) = f(x₂) 2x₁ + 3 = 2x₂ + 3 Step 3: Solve: 2x₁ = 2x₂ x₁ = x₂ Step 4: Since f(x₁) = f(x₂) implies x₁ = x₂, the function

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Step 1: For onto function, range = codomain. Step 2: Codomain = R (all real numbers) Step 3: Range = {y ∈ R: y = x² for some x ∈ R} Since x² ≥ 0 for all real x, range = [0, ∞) Step 4: Negative numbers

Answer

Step 1: (g∘f)(x) = g(f(x)) Step 2: First find f(x) = x + 2 Step 3: Substitute f(x) into g: g(f(x)) = g(x + 2) Step 4: Apply g to (x + 2): g(x + 2) = 3(x + 2) - 1 Step 5: Simplify: = 3x + 6 - 1 = 3x +