The relation R defined in the set of natural | vedclass

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(D) $1$. Reflexivity: For any $a \in N$,$a$ divides $a$ (since $a = a 	imes 1$). Thus,$aRa$ holds for all $a \in N$. So,$R$ is reflexive.$2$. Symmetr

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) $1$. Reflexivity: For any $a \in N$,$a$ divides $a$ (since $a = a 	imes 1$). Thus,$aRa$ holds for all $a \in N$. So,$R$ is reflexive.$2$. Symmetry: Consider $a=1$ and $b=2$. Here,$1|2$ is true,so $1R2$ holds. However,$2|1$ is false,so $2R1$ does not hold. Thus,$R$ is not symmetric.$3$. Transitivity: Let $aRb$ and $bRc$. This means $a|b$ and $b|c$. Since $a|b$,there exists $k_1 \in N$ such that $b = a 	imes k_1$. Since $b|c$,there exists $k_2 \in N$ such that $c = b 	imes k_2$. Substituting $b$,we get $c = (a 	imes k_1) 	imes k_2 = a 	imes (k_1 	imes k_2)$. Since $k_1 	imes k_2 \in N$,$a|c$ holds,so $aRc$ holds. Thus,$R$ is transitive.Conclusion: The relation is reflexive and transitive but not symmetric. Therefore,the correct option is $(d)$.

The relation $R$ defined in the set of natural numbers $N$ as $aRb \iff b$ is divisible by $a$ is:

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