Let $R=\{(3,3),(6,6),(9,9),(12,12)(6,12),(3,9),(3,12),(3,6)\}$ be relation on the set $A=\{3,6,9,12\}$. Then the relation R is
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Let $R=\{(3,3),(6,6),(9,9),(12,12)(6,12),(3,9),(3,12),(3,6)\}$ be relation on the set $A=\{3,6,9,12\}$. Then the relation $R$ is

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SOLUTION: $(a, a) \in R$ for each $a \in A$

$\therefore \quad$ reflexive

$(6,12) \in R$ but $(12,6) \notin R$

$\therefore \quad$ not symmetric

$(a, b) \in R$ and $(b, c) \in R \quad \Rightarrow \quad(a, c) \in R \Rightarrow$ transitive

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