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 prove that R is equivalence relation

Question

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

Answer 1

We have $(a, b) R(a, b)$ for all $(a, b) \in N \times N$

Since $a+b=b+a$. Hence, $R$ is reflexive.

$R$ is symmetric for all $(a, b),(c, d) \in N \times N \quad$ we have $(a, b) R(c, d)$

$\Rightarrow \quad a+d=b+c $

$\Rightarrow \quad c+b=d+a \quad \Rightarrow \quad(c, d) R(a, b) \text {. }$

$(a, b) R(c, d) \text { and }(c, d) R(e, f)$

$a+d=b+c \text { and } c+f=d+e \text {, }$

$\Rightarrow a+d+c+f=b+c+d+e \Rightarrow a+f=b+e$

$\Rightarrow(a, b) R(e, f) \Rightarrow R \text { is transitive }$

$\text { Thus, (a, b) R (c, d) and (c, d) R (e, f) } \quad \Rightarrow \quad(a, b) R(e, f) $