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Q.Let RR be a relation defined over N\mathbb{N}, where N\mathbb{N} is the set of natural numbers, defined as "mRnmRn if and only if mm is a multiple of nn, m,n∈Nm, n \in \mathbb{N}." Find whether RR is reflexive, symmetric and transitive or not.

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The relation “mm is a multiple of nn” is reflexive and transitive but not symmetric — it is a partial order on N\mathbb{N}.


The core idea: “mm is a multiple of nn” means m=knm = kn for some natural number kk. This is a divisibility relation in reverse — usually we say “nn divides mm”. Here, mRnmRn means n∣mn \mid m.

We check the three properties one by one.


  1. Reflexive — Is every natural number a multiple of itself?

    For any m∈Nm \in \mathbb{N}, we have m=1⋅mm = 1 \cdot m, so mm is a multiple of mm.

    Hence mRmmRm holds for all mm.

    RR is reflexive.

  2. Symmetric — If mm is a multiple of nn, does it follow that nn is a multiple of mm?

    Take m=4m = 4, n=2n = 2. Then 44 is a multiple of 22 (since 4=2⋅24 = 2 \cdot 2), so 4R24R2 holds.

    But 22 is not a multiple of 44 (no natural kk gives 2=k⋅42 = k \cdot 4).

    So 2R42R4 is false.

    One counterexample is enough: RR is not symmetric.

    Watch out

    A common mistake: thinking “multiple of” is symmetric because multiplication is commutative. But the relation is directional — mm being a multiple of nn means m≥nm \ge n (except when n=0n=0, but here n∈Nn \in \mathbb{N} and typically N\mathbb{N} starts at 1). So symmetry would require m=nm=n always, which is false.

  3. Transitive — If mm is a multiple of nn, and nn is a multiple of pp, is mm necessarily a multiple of pp? …

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