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Question

Let R be a relation on the set N of natural numbers defined by nRm

⇔ n is a factor of m (i.e. n(m). Then R is


A

Reflexive and symmetric

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B

Transitive and symmetric

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C

Equivalence

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D

Reflexive, transitive but not symmetric

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Solution

The correct option is D

Reflexive, transitive but not symmetric


Since n | n for all n in N,

therefore R is reflexive.

Since 2 | 6 but 6 | 2, therefore R is not symmetric.

Let n R m and m R p n|m and m|p n|p nRp

So. R is transitive.


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