Answer 1

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

(iii) (a, b) R (c, d) \text{ and }  (c, d) R (e, f) ⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N

\[(a, b) R (c, d) \text{ and } (c, d) R (e, f)\]
\[ \Rightarrow a + d = b + c \text{ and } c + f = d + e\]
\[ \Rightarrow a + d + c + f = b + c + d + e \]
\[ \Rightarrow a + f = b + e \]
\[ \Rightarrow (a, b) R (e, f)\]