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1.Relation and Function
medium
Let $R$ be a relation on the set $N$ of natural numbers defined by $nRm $$\Leftrightarrow$ $n$ is a factor of $m$ (i.e.,$ n|m$). Then $R$ is
A
Reflexive and symmetric
B
Transitive and symmetric
C
Equivalence
D
Reflexive, transitive but not symmetric
Solution
(d) 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.
Standard 12
Mathematics