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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:
(ii) (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ 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
(ii) (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N
\[(a, b) R (c, d) \Rightarrow a + d = b + c \]
\[ \Rightarrow c + b = d + a \]
\[ \Rightarrow (c, d) R (a, b)\]
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