Short Answer TypeThe relation R ⊆ N x N is defined by by (a. b)∈ R if and only if 5 divides b – a.
This means that R is a relation on N defined by , if a. b ∈ N then (a, b) ∈ R if and only if 5 divides b – a.
Let a, b, c belongs to N. Then (i) a – a = 0 = 5 . 0.
5 divides a – a.
⇒ (a. a) ∈ R .
⇒ R is reflexive.
(ii) Let (a, b) ∈ R.
∴ divides a – b.
⇒ a – b = 5 n for some n ∈ N.
⇒ b – a = 5 (–n).
⇒ 5 divides b – a ⇒ (b, a) ∈ R.
∴ R is symmetric.
(iii) Let (a, b) and (b, c) ∈ R.
5 divides a – b and b – c both
∴ a – 6 = 5 n1 and b – n = 5 n2 for some n1 and n2 ∈ N ∴ (a – b) + (b – c) = 5 n1 + 5 n2⇒ a – c = 5 (n1 + n2)
⇒ 5 divides a – c ⇒ (a, c) ∈ R
∴ R is transitive relation in N.
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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