Short Answer TypeLet 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.
2 x + y = 41 ⇒ y = 41 – 2 x
x = 1 y = 41 – 2 (1) = 41 – 2 = 39 x = 2 ⇒ y = 41 – 2 (2) = 41 – 4 = 37 x = 3 ⇒ y = 41 – 2 (3) = 41 – 6 = 35 x = 4 ⇒ y = 41 – 2 (4) = 41 – 8 = 33
x =19 ⇒ y = 41 – 2 (19) = 41 – 38 = 3 x = 20 ⇒ y = 41 – 2 (20) = 41 – 40 = 1 x = 21 ⇒ y = 41 – 2 (21) = 41 – 42 = –1 ∉ N
∴ R = {(1,39), (2, 37), (3, 35), (4, 33)., (20, 1)}
domain of R = {1,2,3,4,........., 20}
and range of R = {1, 3, 5, 7,......,39}
(i) Now 1∉ N but (1, 1) ∉ R
(ii) (1,39), ∈ R but (39, 1) ∴ R ∴ R is not symmetric (iii) (20,1), (1,30) ∈ R but (20. 39) ∉ R ∴ R is not 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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