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
⇔ n is a factor of m (i.e. n(m). Then R is
Answer: Option D
:
D
Since n | n for all n inN,
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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:
D
Since n | n for all n inN,
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.
Was this answer helpful ?
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