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Relations And Functions

Question
CBSEENMA12032263
Wired Faculty App

If R1 and R2 are equivalence relations in a set A, show that R1 ∩ R2 is also an equivalence relation.

Solution

Suppose that R1 and R2 are two equivalence relations on a non-empty set X.
First we prove that R1 ∩ R2 in an equivalence relation on X.
(i) R2 ∩ R2 is reflexive :
Let a ∈ X arbitrarily.
Then (a, a) ∈ R1 and (a, a) ∈ R2 , since R1, R2 both being equivalence relations are reflexive.
So. (a, a) ∈ R1 ∩ R2
⇒ R1 ∩ R2 is reflexive.
(ii) R1 ∩ R2 is symmetric :
Let a, b ∈ X such that (a, b) ∈ R1 ∩ R2 ∴ (a, b) ∈ R1 and (a, b) ∈ R2 ⇒ (b, a) ∈ R1and (b, a) ∈ R2, since R1 and R2 being equivalence relations are also symmetric.
(b, a) ∈ R1∩ R2
(a, b) ∈ R1 ∩ R2 implies that (b, a) ∈ R1 ∩ R2.
∴ R1 ∩ R2 is a symmetric relation.
(iii) R1 ∩ R2 is transitive :
Let a, b, c ∈ X such that (a, b) ∈ R1 ∩ R2 and (b, c) ∈ R1 ∩ R2.
(a, b) ∈ R1 ∩ R2 ⇒ (a, b) ∈ R1 and (a, b) ∈ R2    ...(i)
(b, c) ∈ R1 ∪ R2 ⇒ (b, c) ∈ R1 and (b, c) ∈ R2    ...(ii)
(i) and (ii) ⇒ (a, b) and (b, c) ∈ R1
⇒ (a. c) ∈ R1, since R1 being an equivalence relation is also transitive.
Similarly, we can prove that (a, c) ∈ R2 ∴ (a, c) ∈ R1 ∩ R2 So, R1 ∩ R2 is transitive.
Thus R1 ∩ R2 is reflexive, symmetric and also transitive. Thus R1 ∩ R2 is an equivalence relation.

Some More Questions From Relations and Functions Chapter

If a matrix A has 12 elements, what arc the possible orders it can have 7 What if it has 7 elements ?

Let A be the set of all students of a boys school. Show that the relation R in A given by R = {(a, b) : a is sister of b} is the empty relation and R’ = {(a, b) : the difference between heights of a and b is less than 3 meters} is the universal relation.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Give an example of a relation which is

(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither reflexive nor transitive.

 Determine whether each of the following relations are reflexive, symmetric and transitive :

(i) Relation R in the set A = {1, 2, 3,....., 13, 14} defined as

R = {(x, y) : 3 x – y = 0}

(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x,y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x,y) : x – y is an integer}

(v) Relation R in the set A of human beings in a town at a particular time given by
(a)    R = {(x, y) : x and y work at the same place}
(b)    R = {(x,y) : x and y live in the same locality}
(c)    R = {(x, y) : x is exactly 7 cm taller than y}
(d)    R = {(x, y) : x is wife of y}
(e)    R = {(x,y) : x is father of y}

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.

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