Relations & Its Types | Mathematics (Maths) for JEE Main & Advanced PDF Download

Relations

Given any two non-empty sets A and B, a relation R from A to B is a subset of the Cartesian product A × B and is derived by describing a relationship between the first element (say x) and the other element (say y) of the ordered pairs in A & B.

Further, if (x, y) ∈ R, then we say that x is related to y and write this relation as x R y. Hence, R = {(x, y); x ∈ A, y ∈ B, x R y}.

Note: Ordered pairs means (x, y) and (y, x) are two different pairs.

Example: If R is a relation between two sets A = {1,2,3} and B = {1,4,9} defined as "square root of ". Here, 1R1, 2R4, 3R9
Then, R = {(1, 1), (2, 4), (3, 9)}.

Representation of a Relation

1. Roster form: In this form, a relation is represented by the set of all ordered pairs belonging to R. If R is a relation from set A = {1,2,3,4} to set B = {1,4,9,16,25} such that the second elements are square of the first elements. So, R can be written in roster form as 

R = {(1,1),(2,4),(3,9),(4,16)}

2. Set-builder form: In this form the relation R from set A to B is as

R = {(x,y): x∈A, y∈B; The rule which asociates x and y}

Example: R = {(1,1),(2,4),(3,9),(4,16)} can be written in set-builder form as

R = {(x,y): x∈A, y∈B; y=x²}, where A = {1,2,3,4}, B = {1,4,9,16}.

3. Visual Representation (Arrow Diagram)

In this form of representation, we draw arrows from first element to the second element for all ordered pairs belonging to R.

Example: Let R = {(1,5),(1,6),(2,5),(3,5)} from set A = {1,2,3} to set B = {5,6}

It can be represented by the following arrow diagram.

Domain and Range of a Relation

1. Domain of a Relation
Let R be a relation from A to B. The domain of relation R is the set of all those elements a ∈ A such that (a,b) ∈ R for some b ∈ B.

Thus domain of R = {a ∈ A: (a,b) ∈ R for some b ∈ B}

= set of first elements of all the ordered pairs belonging to R

2. Range of a Relation 

Let R be a relation from A to B. The range of R is the set of all those elements b ∈ B such that (a,b) ∈ R for some a ∈ A.

Thus range of R = {b ∈ B: (a,b) ∈ R for some a ∈ A}

= set of second elements of all the ordered pairs belonging to R.

3. Co-domain of a Relation

If R be a relation from A to B, then B is called the co-domain of relation R.

Clearly range of a relation ⊆ co-domain.

Total Number of Relations

For two non-empty finite sets A and B. If the number of elements in A is h i.e., n(A) = h & that of B is k i.e. n(B) = k, then the number of ordered pair in the Cartesian product will be n(A × B) = hk. The total number of relations is 2^hk.

Types of Relations

1. Empty Relation 
If no element of set X is related or mapped to any element of X, then the relation R in A is an empty relation, i.e, R = Φ. Think of an example of set A consisting of only 100 hens in a poultry farm. Is there any possibility of finding a relation R of getting any elephant in the farm? No! R is a void or empty relation since there are only 100 hens and no elephant.

2. Universal Relation 
A relation R in a set, say A is a universal relation if each element of A is related to every element of A, i.e., R = A × A. Also called Full relation. Suppose A is a set of all - natural numbers and B is a set of all whole numbers. The relation between A and B is universal as every element of A is in set B. Empty relation and Universal relation are sometimes called trivial relation.

3. Identity Relation 
In Identity relation, every element of set A is related to itself only. I = {(a, a), ∈ A}.
Example: If we throw two dice, we get 36 possible outcomes, (1, 1), (1, 2), … , (6, 6). If we define a relation as R: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, it is an identity relation.

4. Inverse Relation 
Let R be a relation from set A to set B i.e., R ⊆ A × B. The 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}. Considering the case of throwing of two dice if R = {(1, 2), (2, 3)}, R^-1 = {(2, 1), (3, 2)}. Here, the domain of R is the range of R^-1 and vice-versa.

5. Reflexive Relation
If every element of set A maps to itself, the relation is Reflexive Relation. For every a ∈ A, (a, a) ∈ R.
Example: Let A = {1, 2, 3}
Then R₁ = {(1, 1), (2, 2), (3, 3)}
R₂ = {(1, 1), (2, 2), (3, 3), (1, 2)}
R₃ = {(1, 1), (2, 2), (1, 2)}
Clearly ,R₁, R₂ both are reflexive relation, but R₃ is not reflexive relation because (3, 3) ∉ R₃

Note: Every identity relation is reflexive relation,but every reflexive relation is not identity relation.

6. Symmetric Relation 
A relation R on a set A is said to be symmetric if (a, b) ∈ R then (b, a) ∈ R, a & b ∈ A.
Example: Let A = {1, 2, 3}
Then R₁ = {(1, 1), (2, 2), (3, 3)}
R₂ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R₃ = {(1, 1), (2, 2), (1, 2)}
Clearly, R₁, R₂ both are symmetric relation, but R₃ is not symmetric relation because (1, 2) ∈ R₃ , but (2, 1) ∉ R₃

Note: A relation is called symmetric if R = R^-1.

7. Transitive Relation 
A relation in a set A is transitive if, (a, b) ∈ R, (b, c) ∈ R, then (a, c) ∈ R, a, b, c ∈ A
Example: Let A = {1, 2, 3}
Then R₁ = {(1, 1), (2, 2), (3, 3)}
R₂ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R₃ = {(1, 1), (2, 2), (1, 2), (2, 3)}
Clearly ,R₁, R₂ both are transitive relation, but R₃ is not transitive relation because(1, 2) & (2, 3) ∈ R₃, but (1, 3) ∉ R₃

Note:
Every empty relation  defined on a non empty set is always symmetric and transitive but not reflexive.
Every universal set defined on non empty set is always reflexive, symmetric and transitive.

8. Equivalence Relation 
A relation is said to be equivalence if and only if it is Reflexive, Symmetric, and Transitive.
Example: If we throw two dices A & B and note down all the possible outcome.

Define a relation R= {(a, b): a ∈ A, b ∈ B}, we find that {(1, 1), (2, 2), …, (6, 6) ∈ R} (reflexive). If {(a, b) = (1, 2) ∈ R} then, {(b, a) = (2, 1) ∈ R} (symmetry). If {(a, b) = (1, 2) ∈ R} and {(b, c) = (2, 3) ∈ R} then {(a, c) = (1, 3) ∈ R} (transitive)

Solved Examples

Q.1. If A is a set of all triangles and the relation R is defined by “is congruent to” prove that R is an equivalence relation.

Ans.

(i) R is reflexives as every triangle is congruent to itself.

(ii) R is symmetric: if a triangle x is congruent to another triangle y, then the triangle y is congruent to the triangle x.

(iii) If ‘a triangle x is congruent to a triangle y and y is congruent to a third triangle z, then x is also congruent to z.

Hence the relation R is transitive. (i), (ii) and (iii) ⇒ that the relation R is an equivalence relation.

Q.2. Three friends A, B, and C live near each other at a distance of 5 km from one another. We define a relation R between the distances of their houses. Is R an equivalence relation?

Ans. 

• For an equivalence Relation, R must be reflexive, symmetric and transitive.
• R is not reflexive as A cannot be 5 km away to itself.
• The relation, R is symmetric as the distance between A & B is 5 km which is the same as the distance between B & A.
• R is transitive as the distance between A & B is 5 km, the distance between B & C is 5 km and the distance between A & C is also 5 km. Therefore, this relation is not equivalent.

Equivalence class

An equivalence class is subset of a given set in which each element is related with each other. Moreover equivalence class can be found if the given relation on given set is an equivalence relation. 

Example: Let A = {1, 3, 5, 9, 11, 18} and a relation on A given by
R = {(a, b) | a - b is divisible by 4 } find all equivalence classes.
Solution. Clearly R ={(1, 1), (3, 3), (5, 5), (9, 9), (11, 11), (18, 18), (1, 5)(5, 1), (1, 9), (3, 11)(11, 3), (5, 9), (9, 5)}
Now A₁ = {1, 5, 9}, A₂ = {3, 11}, A₃= {18}
Hence A₁, A₂, A₃ are called equivalence classes of a set A on the given relation R
Moreover we can represent these classes in a very specific way as below:
A₁ = [1] or [5] or [9]
A₂ = [3] or [11]
A₃ = [18]
Clearly, A₁, A₂ , A₃  are disjoint and A₁∪ A₂ ∪ A₃ = A  and A₁ ∩ A₂ ∩ A₃= ∅

Note:

• All elements of A_i are related to each other, for all i.
• No element of A_i is related to any element of A_j, i≠ j. 
• ∪ A_j = X and A_i ∩A_j = ∅, i ≠ j.

The subsets A_i are called equivalence classes and are called partitions or subdivisions of set A and these are mutually disjoint to one another.

Short Answer Type Questions

Q.1. Show that the number of equivalence relation in the set {1, 2, 3} containing (1, 2) and (2, 1) is two.
Ans. A = {1, 2, 3}
The maximum possible relation (i.e. universal relation) is
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}
The smallest equivalence relation  R₁ containing (1 , 2) and (2 , 1) is
 R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
we are left with four pairs (from universal relation) i.e. (2, 3), (3, 2), (1, 3) and (3, 1)
If we add (2, 3) to R₁, then for symmetric by we must add (3, 2) and now for transitivity we are forced to add (1, 3) and (3, 1)
Thus the only relation bigger than R₁ is universal relation i.e R
∴ The no. of equivalence relations containing (1, 2) and (2, 1) is two.

Q.2. If 𝑅 = {(𝑥, 𝑦) ∶ 𝑥² + 𝑦² ≤ 4 ; 𝑥, 𝑦 ∈ 𝑧} is a relation on 𝑧. Write the domain of R.
Ans. R = {(0, 1), (0, -1), (0, 2), (0, -2), (1, 1), (1, -1), (-1, 0), (-1, 1), (-1, -1), (2, 0), (-2, 0)}
∴ Domain of R = {0, 1 , -1, 2, -2}
(i.e the first domain of each ordered pairs)

Q.3. Let R = {(𝑥, 𝑦) : ∣𝑥² − 𝑦²∣ < 1} be a relation on set A = {1, 2, 3, 4, 5}. Write R as a set of ordered pairs.
Ans. A = {1, 2, 3, 4, 5}
for ∣𝑥² − 𝑦²∣ < 1 : 𝑥 should be equal to 𝑦
∴ R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}

Q.4. R is a relation in Z defined as (𝑎, 𝑏) ∈ 𝑅 ⇔ 𝑎² + 𝑏² = 25. Find the range.
Ans. We have, 𝑎² + 𝑏²  = 25 and 𝑎 , 𝑏 ∈ 𝑧
∴ R = {(0, 5), (0, -5), (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5, 0), (-5, 0)}
∴ Range = {-5, 5, 4, -4, 4, 3, -3, 0}
(i.e. second elements of each order pairs)