Prove that the relation R in …
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Gaurav Seth 4 years, 1 month ago
R = {(a, b) : 2 divides a – b}
where R is in the set Z of integers.
(i) a – a = 0 = 2 .0
∴ 2 divides a – a ⇒ (a, a) ∈ R ⇒ R is reflexive.
(ii) Let (a, a) ∈ R ∴ 2 divides a – b ⇒ a – b = 2 n for some n ∈ Z ⇒ b – a = 2 (–n)
⇒ 2 divides b – a ⇒ (b. a) ∈ R
(a, ft) G R ⇒ (b, a) ∈ R ∴ R is symmetric.
(iii) Let (a, b) and (b, c) ∈ R
2 divides a – b and b – c both ∴ a – b = 2 n1 and b – c = 2 n2 for some n1, n2 ∈ Z ∴ (a – b) + (b – c)= 2 n1 + 2 n2 ⇒ a – c = 2 (n1 + n2 )
⇒ 2 divides a – c
⇒ (a, c) ∈ R
∴ (a,b), (b,c) ∈ R ⇒ (a, c) ∈ R
∴ R is transitive
From (i), (ii), (iii) it follows that R is an equivalence relation.
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