Symmetric Relation on Set

Here we will discuss about the symmetric relation on set.

Let A be a set in which the relation R defined. Then R is said to be a symmetric relation, if (a, b) ∈ R ⇒ (b, a) ∈ R, that is, aRb ⇒ bRa for all (a, b) ∈ R.

Consider, for example, the set A of natural numbers. If a relation A be defined by “x + y = 5”, then this relation is symmetric in A, for

a + b = 5 ⇒ b + a = 5

But in the set A of natural numbers if the relation R be defined as ‘x is a divisor of y’, then the relation R is not symmetric as 3R9 does not imply 9R3; for, 3 divides 9 but 9 does not divide 3.

For a symmetric relation R, R\(^{-1}\) = R.


Solved example on symmetric relation on set:

1. A relation R is defined on the set Z by “a R b if a – b is divisible by 5” for a, b ∈ Z. Examine if R is a symmetric relation on Z.

Solution:

Let a, b ∈ Z and aRb hold. Then a – b is divisible by 5 and therefore b – a is divisible by 5.

Thus, aRb ⇒ bRa and therefore R is symmetric.


2. A relation R is defined on the set Z (set of all integers) by “aRb if and only if 2a + 3b is divisible by 5”, for all a, b ∈ Z. Examine if R is a symmetric relation on Z.

Solution:

Let a, b ∈ Z and aRb holds i.e., 2a + 3a = 5a, which is divisible by 5. Now, 2a + 3a = 5a – 2a + 5b – 3b = 5(a + b) – (2a + 3b) is also divisible by 5.

Therefore aRa holds for all a in Z i.e. R is reflexive.


3. Let R be a relation on Q, defined by R = {(a, b) : a, b ∈ Q and a – b ∈ Z}. Show that R is Symmetric relation.

Solution:

Given R = {(a, b) : a, b ∈ Q, and a – b ∈ Z}.

Let ab ∈ R ⇒ (a – b) ∈ Z, i.e. (a – b) is an integer.

               ⇒ -(a – b) is an integer

               ⇒ (b – a) is an integer

               ⇒ (b, a) ∈ R

Thus, (a, b) ∈ R ⇒ (b, a) ∈ R

Therefore, R is symmetric.


4. Let m be given fixed positive integer.

Let R = {(a, a) : a, b  ∈ Z and (a – b) is divisible by m}.

Show that R is symmetric relation.

Solution:

Given R = {(a, b) : a, b ∈ Z, and (a – b) is divisible by m}.

Let ab ∈ R . Then,

     ab ∈ R ⇒ (a – b) is divisible by m

               ⇒ -(a – b) is divisible by m

               ⇒ (b – a) is divisible by m

               ⇒ (b, a) ∈ R

Thus, (a, b) ∈ R ⇒ (b, a) ∈ R

Therefore, R is symmetric relation on set Z.

Set Theory

Sets

Representation of a Set

Types of Sets

Pairs of Sets

Subset

Practice Test on Sets and Subsets

Complement of a Set

Problems on Operation on Sets

Operations on Sets

Practice Test on Operations on Sets

Word Problems on Sets

Venn Diagrams

Venn Diagrams in Different Situations

Relationship in Sets using Venn Diagram

Examples on Venn Diagram

Practice Test on Venn Diagrams

Cardinal Properties of Sets



7th Grade Math Problems

8th Grade Math Practice

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