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(D) $1$. Reflexivity: For any $a \in N$,$2a + 3a = 5a$,which is a multiple of $5$. Thus,$a R a$ holds for all $a \in N$. So,$R$ is reflexive.$2$. Symm

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an equivalence relation

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(D) $1$. Reflexivity: For any $a \in N$,$2a + 3a = 5a$,which is a multiple of $5$. Thus,$a R a$ holds for all $a \in N$. So,$R$ is reflexive.$2$. Symmetry: Let $a R b$,then $2a + 3b = 5k$ for some integer $k$. We want to check if $b R a$,i.e.,if $2b + 3a$ is a multiple of $5$.Note that $(2a + 3b) + (2b + 3a) = 5a + 5b = 5(a + b)$.Since $2a + 3b = 5k$,we have $2b + 3a = 5(a + b) - 5k = 5(a + b - k)$.Since $a, b, k$ are integers,$5(a + b - k)$ is a multiple of $5$. Thus,$b R a$ holds. So,$R$ is symmetric.$3$. Transitivity: Let $a R b$ and $b R c$. Then $2a + 3b = 5k_1$ and $2b + 3c = 5k_2$ for some integers $k_1, k_2$.We want to check if $a R c$,i.e.,if $2a + 3c$ is a multiple of $5$.From $2a + 3b = 5k_1$,we have $2a = 5k_1 - 3b$.From $2b + 3c = 5k_2$,we have $3c = 5k_2 - 2b$.Adding these,$2a + 3c = 5k_1 + 5k_2 - 5b = 5(k_1 + k_2 - b)$.Since $k_1, k_2, b$ are integers,$2a + 3c$ is a multiple of $5$. Thus,$a R c$ holds. So,$R$ is transitive.Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

Let $R$ be a relation defined on $N$ such that $a R b$ if $2a + 3b$ is a multiple of $5$,where $a, b \in N$. Then $R$ is

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