Question

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

Answer 1

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) ∈ N × N
$$\begin{gathered} \because a+b=b+a\mathrm{for}\mathrm{all}a,b\in N \\ \therefore (a,b)R(a,b)\mathrm{for}\mathrm{all}a,b\in N \end{gathered}$$

(ii) (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N
$$\begin{gathered} (a,b)R(c,d)\Rightarrow a+d=b+c \\ \Rightarrow c+b=d+a \\ \Rightarrow (c,d)R(a,b) \end{gathered}$$

(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
$$\begin{gathered} (a,b)R(c,d)\mathrm{and}(c,d)R(e,f) \\ \Rightarrow a+d=b+c\mathrm{and}c+f=d+e \\ \Rightarrow a+d+c+f=b+c+d+e \\ \Rightarrow a+f=b+e \\ \Rightarrow (a,b)R(e,f) \end{gathered}$$