(i) Since (x,y) R (x,y), ∀ (x, y) ∈ A, as x y = y x.
R is reflexive.
(ii) Again (x,y)R (u,v)
⇒ x v = y u ⇒ u y = v x and so (u, v) R (x, y).
∴ R is symmetric.
(iii) Again (x, y) R (u, v) and (u, v) R (a, b)
∴ (x, y)R (a,b).
∴ R is transitive.
From (i), (ii), (iii), it follows that R is an equivalence relation.
Given a non-empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows :
For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X) ? Justify your answer.
P(X) = { A : A is a subset of X }
(i) Since A C A ∀ ∈ P(X)
∴ ARA ∀ A ∈ P(X)
∴ R is reflexive relation (ii) Let A, B, C ∈ P(X) such that ARB and BRC A ⊂ B and B ⊂ C ⇒ A ⊂ C
⇒ A R C
∴ ARB and BRC ⇒ ARC
∴ R is transitive relation.
(iii) Now if A ⊂ B, then B may not be a subset of A i.e. ARB ⇏ BRA
∴ R is not a symmetric relation.
From (i), (ii), (iii), it follows that R is not an equivalence relation on P(X).
Show 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.
A= {x ∈ Z : 0 ≤ x ≤ 12}
= {0, 1, 2, 3, 4, 5, 6, 7, 8,9, 10, 11, 12 } (i) R = {(a, b) : | a – b | is a multiple of 4|
As | a – a | = 0 is divisible by 4 ∴ (a, a) ∈ R ∀ a ∈ A.
∴ R is reflexive.
Next, let (a, b) ∈ R
⇒ | a – b | is divisible by 4
⇒ | – (b – a) | is divisible by 4 ⇒ | b – a | is divisible by 4 ⇒ (b, a) ∈ R
∴ R is symmetric.
Again. (a, b) ∈ R and (b, c) ∈ R
⇒ | a – A | is a multiple of 4 and | b – c | is a multiple of 4 ⇒ a – b is a multiple of 4 and b – c is a multiple of 4 ⇒ (a – b) + (b – c) is a multiple of 4 ⇒ a – c is a multiple of 4 ⇒ | a – c | is a multiple of 4 ⇒ (a, c) ∈ R ∴ R is transitive.
∴ R is an equivalence relation.
Set of elements which are related to | = {a ∈ A : (a, 1) ∈ R}
= {a ∈ A : |a – 1| is a multiple of 4]
= {1, 5, 9}
[ ∵ | 1 – 1 | = 0, | 5 – 1 | = 4 and I 9 – 1 | = 8 are multiples of 4] (ii) R = {(a, b) : a = b} ∴ a = a ∀ a ∈ A,
∴ R is reflexive.
Again, (a, b) ∈ R ⇒ a = b ⇒ b = a ⇒ (b,a) ∈ R ∴ R is symmetric.
Next. (a, b) ∈ R and (b, c) ∈ R
⇒ a = b and b = c
⇒ a = c ⇒ (a, c) ∈ R
∴ R is transitive.
∴ R is an equivalence relation.
Set of elements of A which are related to I = {a ∈ A : (a,1) ∈ R }
= {a ∈ A : a = 1} = {1}.
A is the set of points in a plane.
R = {(P. Q) : distance of the point P from the origin is same as the distance of the point
Q from the origin}
= {(P, Q) : | OP | = | OQ | where O is origin}
Since | OP | = | OP |, (P, P) ∈ R ∀ P ∈ A.
∴ R is reflexive.
Also (P. Q) ∈ R
⇒ | OP | = | OQ |
⇒ | OQ | = | OP |
⇒ (Q.P) ∈ R ⇒ R is symmetric.
Next let (P, Q) ∈ R and (Q, T) ∈ R ⇒ | OP | = | OQ | and | OQ | = | OT |
⇒ | OP | = | OT |
⇒ (P,T) ∈ R
∴ R is transitive.
∴ R is an equivalence relation.
Set of points related to P ≠ O
= {Q ∈ A : (Q,P) ∈ R} = {Q ∈ A : | OQ | = | OP |}
= {Q ∈ A :Q lies on a circle through P with centre O}.
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.
We are given that
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 } (i) R is not reflexive as (x ,x) ∉ R ∀ x ∈ N
R is not symmetric as 1 < 2 ⇒ (1, 2) ∈ R but 2 < 1 ⇒ (2, 1) ∉ R
R is transitive as x and so ( x, y ) ∈ R and ( y, z ) ∈ R ⇒ ( x, z) ∈ R.
(ii) S is not reflexive as (x, x) ∉ R ∀ x ∈ IN
S is symmetric as x + y = 10 ⇒ y + x =10 and so (x, y) ∈ R ⇒ (y, x ) ∈ R ∀ x, y ∈ N S is not transitive as 3 + 7 = 10 ⇒ (3, 7) ∈ R and 7 + 3 = 10 ⇒ (7, 3) ∈ R but 3 + 3 ≠ 10 ⇒ (3,3) ∉ R (iii) T is not reflexive as x = x ∀ x ∈ N ⇒ (x, x) ∈ R
Now 2 – 1 = 1 as (2, 1) ∈ R but 1 ≠ 2 or 1 – 2 ≠ 1 ⇒ (1,2) ∉ R T is not symmetric.
Again 4 – 3=1 ⇒ (4,3) ∈ R and 3 – 2 = 1 ⇒(3,2) ∈ R but 4 ≠ 2 or 4 – 2 ≠ 1 ⇒ (4, 2) ∉ R.
T is not transitive.
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