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(D) $1$. Reflexivity: For any $(a, b) \in N 	imes N$,we have $a + b = b + a$,which implies $(a, b)R(a, b)$. Thus,$R$ is reflexive.$2$. Symmetry: Let

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An equivalence relation

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(D) $1$. Reflexivity: For any $(a, b) \in N 	imes N$,we have $a + b = b + a$,which implies $(a, b)R(a, b)$. Thus,$R$ is reflexive.$2$. Symmetry: Let $(a, b)R(c, d)$. Then $a + d = b + c$. This can be rewritten as $c + b = d + a$,which implies $c + b = d + a$,so $(c, d)R(a, b)$. Thus,$R$ is symmetric.$3$. Transitivity: Let $(a, b)R(c, d)$ and $(c, d)R(e, f)$. Then $a + d = b + c$ and $c + f = d + e$. Adding these two equations,we get $a + d + c + f = b + c + d + e$. Canceling $c + d$ from both sides,we obtain $a + f = b + e$,which implies $(a, b)R(e, f)$. Thus,$R$ is transitive.Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

Let $R$ be a relation over the set $N \times N$ and it is defined by $(a, b)R(c, d) \iff a + d = b + c$. Then $R$ is

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