In the set N,a relation R is defined as aRb L | vedclass

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(A) $1$. Reflexivity: For any $a \in N$,$a$ is divisible by $a$. Thus,$aRa$ holds for all $a \in N$. So,$R$ is reflexive.$2$. Symmetry: Let $aRb$. Thi

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Reflexive but not symmetric

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(A) $1$. Reflexivity: For any $a \in N$,$a$ is divisible by $a$. Thus,$aRa$ holds for all $a \in N$. So,$R$ is reflexive.$2$. Symmetry: Let $aRb$. This means $b$ is divisible by $a$. For example,$1R2$ holds because $2$ is divisible by $1$,but $2R1$ does not hold because $1$ is not divisible by $2$. Thus,$R$ is not symmetric.$3$. Transitivity: Let $aRb$ and $bRc$. This means $b = ka$ and $c = mb$ for some integers $k, m \in N$. Then $c = m(ka) = (mk)a$. Since $mk$ is an integer,$c$ is divisible by $a$. Thus,$aRc$ holds. So,$R$ is transitive.Conclusion: $R$ is reflexive and transitive,but not symmetric.

In the set $N$,a relation $R$ is defined as $aRb \Leftrightarrow b$ is divisible by $a$. Then $R$ is:

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