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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


Solution

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
 a+b = b+a for all a,b N(a,b) R (a,b) for all a,b N

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

(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
(a,b) R (c,d) and (c,d) R (e,f)a+d = b+c and c+f = d+ea+d+c+f = b+c+d+e a+f = b+e (a,b) R (e,f)

Mathematics
RD Sharma XI (2015)
Standard XI

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