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