Let R be a relation over the set N × N and it is defined by (a, b) R (c, d) ⇒ a + d = b + c. Then R is
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We have (a, b)R (a, b) for all (a, b) ∈ N × N
Since a + b = b + a. Hence, R is reflexive.
R is symmetric for all (a, b), (c, d) ∈ N × N we have (a, b) R (c, d)
⇒ a + d = b + c
⇒ c + b = d + a ⇒ (c, d) R (a, b).
(a, b)R (c, d) and (c, d)R (e, f)
a + d = b + c and c + f = d + e,
⇒ a + d + c + f = b + c + d + e ⇒ a + f = b + e
⇒ (a, b) R (e, f) ⇒ R is transitive
Thus, (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f)
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