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Official PYQ · 2015

GATE CS 2015 Set 2 Q16 — Sets, Relations and Functions

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Let RR be the relation on the set of positive integers such that aRbaRb and only if aa and bb are distinct and let have a common divisor other than 1.1. Which one of the following statements about RR is true?

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Answer: D

Let the relation RR be defined on the set of positive integers Z+\mathbb{Z}^+ as:

aRb  ⟺  a≠b and gcd⁡(a,b)>1aRb \iff a \neq b \text{ and } \gcd(a, b) > 1

We analyze the properties of RR as follows:

  1. Reflexivity: For any a∈Z+a \in \mathbb{Z}^+, the condition a≠aa \neq a is always false. Thus, aRaaRa does not hold for any element.
    Therefore, RR is not reflexive.
  2. Symmetry: Suppose aRbaRb holds. This means a≠ba \neq b and gcd⁡(a,b)>1\gcd(a, b) > 1. Since inequality and greatest common divisor are symmetric operations: b≠aandgcd⁡(b,a)=gcd⁡(a,b)>1b \neq a \quad \text{and} \quad \gcd(b, a) = \gcd(a, b) > 1 This implies bRabRa holds.
    Therefore, RR is symmetric.
  3. Transitivity: Suppose aRbaRb and bRcbRc hold. For RR to be transitive, aRcaRc must also hold. Let us test this with a counterexample:
    • Let a=2a = 2, b=6b = 6, and c=3c = 3.
    • 2R62R6 is true because 2≠62 \neq 6 and gcd⁡(2,6)=2>1\gcd(2, 6) = 2 > 1.
    • 6R36R3 is true because 6≠36 \neq 3 and gcd⁡(6,3)=3>1\gcd(6, 3) = 3 > 1.
    • However, 2R32R3 is false because gcd⁡(2,3)=1\gcd(2, 3) = 1.
    Therefore, RR is not transitive.

Hence, RR is symmetric but not reflexive and not transitive, which corresponds to Option D.

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