Sets, Relations & Functions Question - None

Question

Let $\mathbb{R}$ be the set of real numbers.
Statement-1: $A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : y - x \text{ is an integer}\}$ is an equivalence relation on $\mathbb{R}$.
Statement-2: $B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x = \alpha y \text{ for some rational number } \alpha\}$ is an equivalence relation on $\mathbb{R}$.

Accepted Answer

Step 1: Analyze Relation A: {(x,y) : y - x ∈ ℤ}

• Reflexive: y - x = 0 ∈ ℤ for all x ∈ ℝ. ✓

• Symmetric: If y - x ∈ ℤ, then x - y = -(y-x) ∈ ℤ. ✓

• Transitive: If y - x ∈ ℤ and z - y ∈ ℤ, then z - x = (z-y) + (y-x) ∈ ℤ. ✓

Statement-1 is TRUE.

Step 2: Analyze Relation B: {(x,y) : x = αy for some rational α}

• Reflexive: x = 1·x (1 is rational). ✓

• Symmetric: If x = αy (α rational), need y = βx for rational β.

Then y = x/α. For this to be rational times x, we need 1/α rational.

But 1/α is rational only if α ≠ 0. When α = 0: x = 0, but y can be any real number. If y ≠ 0, then y ≠ β·0 for any β. So (0, π) ∈ B but (π, 0) ∉ B. ✗

Statement-2 is FALSE.

Conclusion: Statement-1 TRUE, Statement-2 FALSE.

∴ Answer: A

Question:

Step-by-Step Solution

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