# 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

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

## Recent Articles

1. ### Method of H.C.F. |Highest Common Factor|Factorization &Division Method

Apr 13, 24 05:12 PM

We will discuss here about the method of h.c.f. (highest common factor). The highest common factor or HCF of two or more numbers is the greatest number which divides exactly the given numbers. Let us…

2. ### Factors | Understand the Factors of the Product | Concept of Factors

Apr 13, 24 03:29 PM

Factors of a number are discussed here so that students can understand the factors of the product. What are factors? (i) If a dividend, when divided by a divisor, is divided completely

3. ### Methods of Prime Factorization | Division Method | Factor Tree Method

Apr 13, 24 01:27 PM

In prime factorization, we factorise the numbers into prime numbers, called prime factors. There are two methods of prime factorization: 1. Division Method 2. Factor Tree Method

4. ### Divisibility Rules | Divisibility Test|Divisibility Rules From 2 to 18

Apr 13, 24 12:41 PM

To find out factors of larger numbers quickly, we perform divisibility test. There are certain rules to check divisibility of numbers. Divisibility tests of a given number by any of the number 2, 3, 4…