Short Answer TypeR = {(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.
Let R be the relation defined on the set of natural numbers N as R = {(x, y) : x ∈ N, y ∈ N, 2 x + y = 41 }
Find the domain and range of this relation R. Also verify whether R is (i) reflexive (ii) symmetric (iii) transitive.
The following three relations are defined on the set of natural numbers :
R = {(x, y) : x < y, x ∈ N, y ∈ N}
S = { (x,y) : x + y = 10, x ∈ N, y ∈ N}
T = { (x, y) : x = y or x – y = 1, x ∈ N, y ∈ N } Explain clearly which of the above relations are (i) Reflexive (ii) Symmetric (iii) Transitive.
Long Answer TypeShow that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12 }, given by
(i) R = {(a, b) : | a – b | is a multiple of 4 }
(ii) R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.
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