Let R be a relation on N x N defined by (a, b) R (c, d) a + d = b + c for all (a, b), (c, d) N x N. Show that: i. (a, b) R (a, b) for all (a, b) N x N ii. (a, b) R (c, d) (c, d) R (a, b) for all (a, b), (c, d) N x 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

Question

Philoid · Accepted Answer

Given,

(a, b) R (c, d) a + d = b + c for all (a, b), (c, d) N x N

i. (a, b) R (a, b)

⇒ a + b = b + a for all (a, b) N x N

∴ (a, b) R (a, b) for all (a, b) N x N

ii. (a, b) R (c, d)

⇒ a + d = b + c ⇒ c + b = d + a

⇒ (c, d) R (a, b) for all (c, d), (a, b) N x N

iii. (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 + c + d + e – c – d

⇒ a + f = b + e

⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) N × N