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}.