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.
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.
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.
The 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.
L is the set of all lines in XY plane.
R = {(L1, L2) : L1 is parallel to L2}
Since every line l ∈ L is parallel to itself,
∴ (l,l) ∴ R ∀ l ∈ L
∴ R is reflexive.
Let (L1, L2) ∈ R ∴ L1 || L2 ⇒ L2 || L1
⇒ (L2, L1) ∈ R.
∴ R is symmetric.
Next, let (L1 L2) ∈ R and (L2, L3) ∈ R ∴ L1 || L2 and L2 || L3
∴ L1 || L3 (L1 , L3) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
Let P be the set of all lines related to the line y = 2 x + 4.
∴ P = {l : l is a line related to the line y = 2 x + 4}
= {l : l is a line parallel to the line y = 2 x + 4}
= { l : l is a line with equation y = 2 x + c, where c is an arbitrary constant }
A is the set of all polygons
R = {(P1, P2) : P1 and P2 have same number of sides }
Since P and P have the same number of sides
∴ (P.P) ∈ R ∀ P ∈ A.
∴ R is reflexive.
Let (P1, P2) ∴ R
⇒ P1 and P2 have the same number of sides ⇒ P2 and P1 have the same number of sides ⇒ (P2, P1) ∈ R
∴ (P1, P2) ∈ R ⇒ (P2, P1) ∈ R ∴ R is symmetric.
Let (P1, P2) ∈ R and (P2, P3) ∈ R.
⇒ P1 and P2 have the same number of sides and P2 and P3 have same number of sides
⇒ P1 and P3 have the same number of sides
⇒ (P1, P3) ∈ R
∴ (P1, P2), (P2, P3) ∈ R ∈ (P1, P3) ∈ R ∴ R is transitive.
∴ R is an equivalence relation.
Now T is a triangle.
Let P be any element of A.
Now P ∈ A is related to T iff P and T have the same number of sides P is a triangle
required set is the set of all triangles in A.
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