Let R be a relation on N × N defined by

We are given ,

(a, b) R (c, d) ⇔ a + d = b + c for all (a, b), (c, d) ∈ N × N

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

$\because a+b=b+a$ for all $a, b \in N$

$\therefore(a, b) R(a, b)$ for all $a, b \in N$

(ii) $(a, b) \mathrm{R}(c, d) \Rightarrow(c, d) \mathrm{R}(a, b)$ for all $(a, b),(c, d) \in \mathrm{N} \times \mathrm{N}$

$(a, b) R(c, d) \Rightarrow a+d=b+c$

$\Rightarrow c+b=d+a$

$\Rightarrow(c, d) R(a, b)$

(iii) $(a, b) \mathrm{R}(c, d)$ and $(c, d) \mathrm{R}(e, f) \Rightarrow(a, b) \mathrm{R}(e, f)$ for all $(a, b),(c, d),(e, f) \in \mathrm{N} \times \mathrm{N}$

$(a, b) R(c, d)$ and $(c, d) R(e, f)$

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

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

$\Rightarrow a+f=b+e$

$\Rightarrow(a, b) R(e, f)$