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NCERT Exemplar · Q22

Q.Let us define a relation RR in R\mathbb{R} as aRbaRb if a≥ba \geq b. Then RR is
(A) an equivalence relation
(B) reflexive, transitive but not symmetric
(C) symmetric, transitive but not reflexive
(D) neither transitive nor reflexive but symmetric

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The relation aRbaRb defined by a≥ba \geq b is reflexive (every number is ≥\geq itself) and transitive (if a≥ba \geq b and b≥cb \geq c, then a≥ca \geq c), but it is not symmetric (e.g., 3≥23 \geq 2 does not imply 2≥32 \geq 3). So the correct classification is reflexive, transitive but not symmetric.

The core of this problem is checking three properties — reflexivity, symmetry, and transitivity — against the definition aRb  ⟺  a≥baRb \iff a \geq b. Each property tests a different logical condition, and the key is to apply them to real numbers without overcomplicating.

Let’s go step by step.

  1. Reflexivity: A relation RR on a set is reflexive if every element is related to itself. For aRbaRb to hold when a=ba = b, we need a≥aa \geq a. Since any real number is equal to itself, a≥aa \geq a is always true. So RR is reflexive.

  2. Symmetry: A relation is symmetric if whenever aRbaRb holds, then bRabRa must also hold. Here, aRbaRb means a≥ba \geq b. Does a≥ba \geq b always imply b≥ab \geq a? Only if a=ba = b. For a counterexample, take a=5a = 5 and b=3b = 3: 5≥35 \geq 3 is true, but 3≥53 \geq 5 is false. So symmetry fails.

Watch out

A common mistake is to think that because a≥ba \geq b and b≥ab \geq a can both be true (when a=ba = b), the relation is symmetric. But symmetry requires the implication to hold for all pairs — one counterexample is enough to break it. …

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