Let R be a relation on N × N defined by
(a, b) R (c, d) ⇔ a + d = b + c for all (a, b), (c, d) ∈ N × N
Show that:
(i) (a, b) R (a, b) for all (a, b) ∈ N × N
(ii) (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N
(iii) (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N
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 $c+f=d+e$
$\Rightarrow a+d+c+f=b+c+d+e$
$\Rightarrow a+f=b+e$
$\Rightarrow(a, b) R(e, f)$
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