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 Options: A . &nbsp Reflexive and symmetric B . &nbsp Transitive and symmetric C . &nbsp Equivalence D . &nbsp Reflexive, transitive but not symmetric

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

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

Options:

A . Reflexive and symmetric

B . Transitive and symmetric

C . Equivalence

D . Reflexive, transitive but not symmetric

Answer: Option D
:
D

Since n | n for all n $i n$ N,

therefore R is reflexive.

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

Let n R m and m R p $\Rightarrow$ n|m and m|p $\Rightarrow$ n|p $\Rightarrow$ nRp

So. R is transitive.

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