Let R be a relation on mathbb{Z} times mathbb | vedclass

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(A) $1$. Reflexivity: For any $(a, b) \in \mathbb{Z} 	imes \mathbb{Z}$,we have $ab - ba = 0$,which is divisible by $5$. Thus,$(a, b) R (a, b)$ holds.

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Reflexive and symmetric but not transitive

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(A) $1$. Reflexivity: For any $(a, b) \in \mathbb{Z} 	imes \mathbb{Z}$,we have $ab - ba = 0$,which is divisible by $5$. Thus,$(a, b) R (a, b)$ holds. So,$R$ is reflexive.$2$. Symmetry: Let $(a, b) R (c, d)$. Then $ad - bc = 5k$ for some integer $k$. This implies $bc - ad = 5(-k)$,which is also divisible by $5$. Thus,$(c, d) R (a, b)$ holds. So,$R$ is symmetric.$3$. Transitivity: Consider $(3, 1) R (10, 5)$ because $3(5) - 1(10) = 15 - 10 = 5$,which is divisible by $5$. Also,$(10, 5) R (1, 1)$ because $10(1) - 5(1) = 5$,which is divisible by $5$. However,for $(3, 1)$ and $(1, 1)$,we have $3(1) - 1(1) = 2$,which is not divisible by $5$. Thus,$(3, 1)$ is not related to $(1, 1)$. Therefore,$R$ is not transitive.

Let $R$ be a relation on $\mathbb{Z} \times \mathbb{Z}$ defined by $(a, b) R (c, d)$ if and only if $ad - bc$ is divisible by $5$. Then $R$ is

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