Consider the following relations: R1(a,b) if (a+b) is even over the set of…

2001

Consider the following relations:

R1(a,b) if (a+b) is even over the set of integers
R2(a,b) if (a+b) is odd over the set of integers
R3(a,b) if a.b > 0 over the set of non-zero rational numbers
R4(a,b) if |a - b| <= 2 over the set of natural numbers

Which of the following statements is correct?

  1. A.

    R1 and R2 are equivalence relations, R3 and R4 are not

  2. B.

    R1 and R3 are equivalence relations, R2 and R4 are not

  3. C.

    R1 and R4 are equivalence relations, R2 and R3 are not

  4. D.

    R1, R2, R3 and R4 are all equivalence relations

Attempted by 35 students.

Show answer & explanation

Correct answer: B

R1(a, b): (a + b) is even over integers (Z)

  • Reflexive: a + a = 2a. Since 2a is always divisible by 2, it is always even. Thus, (a, a) ∈. (True)

  • Symmetric: If a + b is even, then b + a is also clearly even because addition is commutative. Thus, (b, a) ∈ (True)

  • Transitive: If a + b is even and b + c is even, then their sum (a + b) + (b + c) = a + 2b + c must also be even. Because 2b is inherently even, subtracting it tells us that a + c must be even. Thus, (a, c) ∈ (True)

  • Result: R_1 is an equivalence relation. (Conceptually, it partitions integers into two equivalence classes: all evens and all odds).

R2(a, b): (a + b) is odd over integers (Z)

  • Reflexive: For any integer a, a + a = 2a, which is always even, never odd. Therefore, (a, a) ∉ R2.

  • Result: R2 is NOT an equivalence relation because it completely fails reflexivity

R3(a, b): a . b > 0 over non-zero rational numbers (Q*)

  • Reflexive: For any non-zero rational number a, squaring it always yields a positive result: a . a = a2 > 0. Thus, (a, a) ∈. (True)

  • Symmetric: If a . b > 0, then b . a > 0 because multiplication is commutative. Thus, (b, a) ∈. (True)

  • Transitive: If a . b > 0 and b . a > 0, then multiplying them gives (a . b) . (b . c) > 0 => a . b2 . c > 0 . Since b2 is strictly positive, we can divide by it to find that a .c > 0. Thus, (a, c) ∈ R3. (True)

  • Result: R3 is an equivalence relation. (Conceptually, it partitions non-zero rationals into two blocks: all positive numbers and all negative numbers).

R4(a, b): |a - b| <= 2 over natural numbers (N)

  • Reflexive: |a - a| = 0 <=. (True)

  • Symmetric: If |a - b| <= 2, then |b - a| <= 2. (True)

  • Transitive: Let's test this with a simple counterexample:

    Let a = 1, b = 3, c = 5.

    • |1 - 3| = 2 <= 2 ⟹ (1, 3) ∈ R4

    • $|3 - 5| <= 2 2 ⟹ (3, 5) ∈ R4

    • However, |1 - 5| = 4, which is not <= 2. Thus, (1, 5) ∉ R4.

  • Result: R4 is NOT an equivalence relation because it fails transitivity.

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