Asymmetric Relation: Domain, Range & Properties

Ans. Different types of Relations are:

• Empty Relation: If no element of any set X is mapped or related with any element of X, then the relation is considered to be an empty relation. 
• Universal Relation: A relation R in a set such as A is a universal relation if each element of A i.e. R = A*A. It is also known as full relation.
• Identity Relation: In this, every element of any set is related to itself only. I = {(a, a), ∈ A}. 
• Inverse Relation: Let R be a relation defined from set A to set B. A relation R-1 is said to be an inverse relation if R-1 from set B to A is denoted by R-1 = {(b, a):(a, b) ∈ R}.
• Reflexive Relation: If each element of a set X maps to itself, the relation is said to be a reflexive relation. 
• Transitive Relation: A relation in a given set A is transitive if, (a, b) ∈ R, (b, c) ∈ R, then (a, c) ∈ R for all a, b, c ∈ A. 
• Equivalence Relation: A relation is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive. 

Ans. Given, X = {3, 4} and {3, 4} ∈ R

Here, 3 is less than 4 but 4 is not less than 3, so

{3, 4} ∈ R => {4, 3} ∉ R

Therefore, it is proved that the relation on set X is symmetric. 

Ques. If Y = (7, 9) and relation R on a set Y is (7, 9), so prove that the relation is asymmetric. (2 Marks)

Ans. Given, Y = {7, 9} and {7, 9} ∈ R

Here, 7 is less than 9 but 9 is not less than 7, so

{7, 9} ∈ R => {9, 7} ∉ R

Therefore, it is proved that the relation on set Y is symmetric. 

Ans. A symmetric relation between two or more elements of a given set is such that if the first element is related to the second element, then the second element will also be related to the first element as defined in the relation. 

Ans. Assume that (x, y) and (y, x) are in a relation and then we have to show that x is equal to y. So, to prove x = y, we need to show that x is divisible by y and y is divisible by x to prove for the relation to be antisymmetric. 

Ans. Let A = {1, 2, 3}

A relation R on A is defined as R = {(1, 2), (2, 1)}

It is seen that (1, 1), (2, 2), (3, 3) ∈ R

So, R is not reflexive. 

Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric. 

Now, (1, 2) and (2, 1) ∈ R

Therefore, 

(1, 1) ∉ R

R is not transitive. 

Hence, R is symmetric but neither reflexive nor transitive. 

Ans. Let A = {1, 2, 3, 4, 5, 6}

A relation R defined on set A is:

R = {(a, b): b = a+1}

R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

We can find (a, a) ∉ R where a ∈ A

Essentially,

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R

So, R is not reflexive.

We can see that (1, 2) ∉ R but (2, 1) ∉ R

So, R is not symmetric.

Now, (1, 2), (2, 3) ∉ R

But, (1, 3) ∉ R

Hence, R is not transitive

Therefore, R is neither reflexive, nor symmetric, nor transitive. 

Ans. Set A is the set of all books in the library of a college. 

R = {x, y}: x and y have same number of pages

Now, R is reflexive since (x, x) ∈ R as x and x will have the same number of pages. 

Let (x, y) ∈ R => x and y will have the same number of pages.

=> y and x have the same number of pages.

=> (y, x) ∈ R

Hence, R is symmetric.

Now, let (x, y) ∈ R and (y, z) ∈ R

=> x and y will have the same number of pages and y and z will have the same number of pages. 

=> x and z will have the same number of pages.

=> (x, z) ∈ R

Hence, R is transitive.

Therefore, R is an equivalence relation. 

Ans. Initially, we will set the denominator to 0, then we will solve for x. 

3-x = 0

-x = -3 => x = 3

So, 3 will be excluded from the domain. 

The domain will be the set of real numbers x, where (x<3) and (x>3)

Now, we will find the range of y=(x+1)/(3-x)

Solving the above equation,

(3-x)y = x+1

3y-xy = x+1

3y-1 = x+xy

x(1+y) = 3y-1

x = (3y-1)/(1+y)

The final equation obtained will be a fraction but the fraction is undefined when its denominator is zero. 

So, 1+y ≠ 0 => y ≠ -1

Therefore, the range of a given function is set of all real numbers excluding -1.

Domain = (-∞, 3) ? (3, ∞)

Range = (-∞, -1) ? (-1, ∞)

Ans. X = {(a, b): a≤b}

Clearly, (a, a) ∈ R as a=a

So, R is reflexive

Now, (2, 4) ∈ R as (2 < 4)

But, (4, 2) ∉ R as 4 is greater than 2.

So, R is not symmetric

Now, let (a, b), (b, c) ∉ R

Then, a≤b and b≤c => a≤c

(a, c) ∈ R

So, R is transitive. 

Therefore, R is reflexive, transitive but not symmetric.