Let R be a relation over the set N × N | vedclass

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(d) We have $(a,b)R(a,b)$ for all $(a,b) \in N 	imes N$

Since $a + b = b + a$. Hence, $R$ is reflexive.

$R$ is symmetric for we have $(a,b)R(c,

Vedclass · Accepted Answer

An equivalence relation

Vedclass · Answer

(d) We have $(a,b)R(a,b)$ for all $(a,b) \in N 	imes N$

Since $a + b = b + a$. Hence, $R$ is reflexive.

$R$ is symmetric for we have $(a,b)R(c,d)$ ==> $a + d = b + c$

==> $d + a = c + b$ ==> $c + b = d + a \Rightarrow (c,d)R(e,f).$

Then by definition of $R$, we have

$a + d = b + c$ and $c + f = d + e$,

whence by addition, we get

$a + d + c + f = b + c + d + e$ or $a + f = b + e$

Hence,$(a,b)\,\,R\,(e,f)$

Thus, $(a, b)$ $R(c,d)$ and $(c,d)R(e,f) \Rightarrow (a,b)R\,(e,\,\,f)$.

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

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