# Let R be a relation on N × N defined by

Question:

Let R be a relation on N × N defined by

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

Show that:

(i) (ab) R (ab) for all (ab) ∈ N × N

(ii) (ab) R (cd) ⇒ (cd) R (ab) for all (ab), (cd) ∈ N × N

(iii) (ab) R (cd) and (cd) R (ef) ⇒ (ab) R (ef) for all (ab), (cd), (ef) ∈ N × N

Solution:

We are given ,

(ab) R (cd) ⇔ a + d = b + c for all (ab), (cd) ∈ 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 $c+f=d+e$

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

$\Rightarrow a+f=b+e$

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