RD Sharma Solutions for Class 12 Maths Chapter 1 – Relations

Mathematics forms the backbone of logical reasoning and problem-solving in our academic journey. Among the various important topics, Relations is a foundational concept in Class 12 Maths. It establishes the framework for understanding how elements from one set are associated with elements of another set. From pairing data in real life to analyzing mathematical patterns, Relations in Class 12 Maths plays a significant role in both academic study and practical scenarios.

In the RD Sharma Class 12 Maths textbook, Chapter on Relations introduces students to deeper mathematical relationships that are crucial for understanding advanced topics like Functions, Sets, and Mapping. The chapter explains how elements are connected using ordered pairs, Cartesian product, and explores various types of relations including reflexive, symmetric, transitive, and equivalence relations.

To make these concepts simpler, RD Sharma Solutions for Class 12 Chapter Relations offers step-by-step explanations and solved examples for all exercises in the textbook. Each solution is prepared in a clear and easy-to-follow manner, helping students strengthen their fundamentals and perform well in both board exams and competitive entrance tests like JEE Main.

Download RD Sharma Class 12 Chapter Relations PDF Solutions

Students can download RD Sharma Class 12 Maths Chapter Relations Solutions PDF with all exercises, solved examples, and additional practice questions. These PDF notes serve as a perfect revision tool before exams and help boost your confidence in tackling relations-based questions in Class 12 board exams.

Access Comprehensive RD Sharma Solutions for Class 12 Relations

Whether you are learning the basics of relations in Class 12, solving advanced problems, or preparing for entrance exams, these RD Sharma Solutions for Relations offer you a complete guide with easy explanations and multiple solving approaches.

Start your preparation today and build a strong foundation in Class 12 Maths Relations concepts, ensuring top performance in your exams!

RD Sharma Class 12 Relations - Step by Step Solutions

Question 1: Define a relation. Give examples of different types of relations (void, universal, identity, inverse, etc.) on a set.

Solution:

Definition: A relation R from set A to set B is a subset of the Cartesian product A × B. In other words, R ⊆ A × B.

Types of Relations:

1. Void/Empty Relation: R = ∅ (no elements are related)
   Example: Let A = {1, 2, 3}. R = ∅ is a void relation.
2. Universal Relation: R = A × A (every element is related to every element)
   Example: Let A = {1, 2}. R = {(1,1), (1,2), (2,1), (2,2)} is universal.
3. Identity Relation: R = {(a,a) : a ∈ A}
   Example: Let A = {1, 2, 3}. R = {(1,1), (2,2), (3,3)} is identity.
4. Inverse Relation: If R is a relation from A to B, then R⁻¹ = {(b,a) : (a,b) ∈ R}
   Example: If R = {(1,2), (3,4)}, then R⁻¹ = {(2,1), (4,3)}
5. Reflexive Relation: (a,a) ∈ R for all a ∈ A
   Example: R = {(1,1), (2,2), (3,3), (1,2)} on A = {1,2,3}

Question 2: Let A = {1, 2, 3}. Write all possible relations from A to A. How many such relations are there?

Solution:

Step 1: Find A × A
A × A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
|A × A| = 9

Step 2: Count possible relations
A relation R is a subset of A × A.
Number of subsets of a set with n elements = 2ⁿ
Therefore, number of relations = 2⁹ = 512

Step 3: Examples of some relations:
R₁ = ∅ (empty relation)
R₂ = {(1,1)}
R₃ = {(1,1), (2,2), (3,3)} (identity relation)
R₄ = A × A (universal relation)
And 508 more relations...

Question 3: Let R = {(x, y) : x and y work at the same place}. Is R reflexive, symmetric, and transitive?

Solution:

Step 1: Check Reflexivity
For reflexivity, (x,x) ∈ R for all x.
Since every person works at the same place as themselves, (x,x) ∈ R.
Therefore, R is reflexive.

Step 2: Check Symmetry
For symmetry, if (x,y) ∈ R, then (y,x) ∈ R.
If x and y work at the same place, then y and x also work at the same place.
Therefore, R is symmetric.

Step 3: Check Transitivity
For transitivity, if (x,y) ∈ R and (y,z) ∈ R, then (x,z) ∈ R.
If x and y work at the same place, and y and z work at the same place, then x and z work at the same place.
Therefore, R is transitive.

Conclusion: R is reflexive, symmetric, and transitive, hence it is an equivalence relation.

Question 4: Let R = {(x, y) : x is father of y} on the set of people in a town. Determine if R is reflexive, symmetric, or transitive.

Solution:

Step 1: Check Reflexivity
For reflexivity, (x,x) ∈ R for all x.
This means x is father of x, which is impossible.
Therefore, R is not reflexive.

Step 2: Check Symmetry
For symmetry, if (x,y) ∈ R, then (y,x) ∈ R.
If x is father of y, then y is father of x, which is impossible.
Therefore, R is not symmetric.

Step 3: Check Transitivity
For transitivity, if (x,y) ∈ R and (y,z) ∈ R, then (x,z) ∈ R.
If x is father of y and y is father of z, then x is grandfather of z (not father).
Therefore, R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

Question 5: Given R = {(a, b) : a - b is divisible by 3, a, b ∈ ℤ}, show that R is an equivalence relation.

Solution:

Step 1: Check Reflexivity
For any a ∈ ℤ, we need to check if (a,a) ∈ R.
a - a = 0, and 0 is divisible by 3 (0 = 3 × 0).
Therefore, R is reflexive.

Step 2: Check Symmetry
Let (a,b) ∈ R. Then a - b is divisible by 3.
So a - b = 3k for some integer k.
Then b - a = -(a - b) = -3k = 3(-k).
Since -k is an integer, b - a is divisible by 3.
Therefore, (b,a) ∈ R, and R is symmetric.

Step 3: Check Transitivity
Let (a,b) ∈ R and (b,c) ∈ R.
Then a - b = 3k₁ and b - c = 3k₂ for some integers k₁, k₂.
Adding: (a - b) + (b - c) = 3k₁ + 3k₂
a - c = 3(k₁ + k₂)
Since k₁ + k₂ is an integer, a - c is divisible by 3.
Therefore, (a,c) ∈ R, and R is transitive.

Conclusion: R is reflexive, symmetric, and transitive, hence it is an equivalence relation.

Question 6: Let R be a relation on ℕ defined by aRb if and only if a + b is even. Is R an equivalence relation? Justify.

Solution:

Step 1: Check Reflexivity
For any a ∈ ℕ, we need a + a = 2a to be even.
Since 2a is always even for any natural number a, (a,a) ∈ R.
Therefore, R is reflexive.

Step 2: Check Symmetry
If (a,b) ∈ R, then a + b is even.
Since addition is commutative, b + a = a + b is also even.
Therefore, (b,a) ∈ R, and R is symmetric.

Step 3: Check Transitivity
Let (a,b) ∈ R and (b,c) ∈ R.
Then a + b is even and b + c is even.
Case 1: If b is even, then a is even and c is even, so a + c is even.
Case 2: If b is odd, then a is odd and c is odd, so a + c is even.
In both cases, a + c is even, so (a,c) ∈ R.
Therefore, R is transitive.

Conclusion: R is reflexive, symmetric, and transitive, hence it is an equivalence relation.

Question 7: For A = {1, 2, 3, 4}, define R = {(x, y) : x ≤ y}. Is R reflexive, symmetric, or transitive?

Solution:

Step 1: Write out the relation
R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)}

Step 2: Check Reflexivity
For reflexivity, (a,a) ∈ R for all a ∈ A.
Since a ≤ a for all a, we have (1,1), (2,2), (3,3), (4,4) ∈ R.
Therefore, R is reflexive.

Step 3: Check Symmetry
For symmetry, if (a,b) ∈ R, then (b,a) ∈ R.
We have (1,2) ∈ R (since 1 ≤ 2), but (2,1) ∉ R (since 2 ≰ 1).
Therefore, R is not symmetric.

Step 4: Check Transitivity
For transitivity, if (a,b) ∈ R and (b,c) ∈ R, then (a,c) ∈ R.
If a ≤ b and b ≤ c, then a ≤ c (by transitivity of ≤).
Therefore, R is transitive.

Conclusion: R is reflexive and transitive, but not symmetric.

Question 8: Represent the relation R = {(a, b) : a < b, a, b ∈ {1, 2, 3}} using an arrow diagram.

Solution:

Step 1: Find all pairs (a,b) where a < b
For a = 1: b can be 2, 3 → (1,2), (1,3)
For a = 2: b can be 3 → (2,3)
For a = 3: no valid b → no pairs

Step 2: The relation R
R = {(1,2), (1,3), (2,3)}

Step 3: Arrow diagram representation
Set A = {1, 2, 3} → Set B = {1, 2, 3}

A B

1 ────────→ 2

│

└─────────→ 3

│

2 ─────────→ 3

 

3 (no arrows from 3)

 

Explanation: Each arrow represents an ordered pair in the relation. Arrow from 1 to 2 represents (1,2), etc.

Question 9: If R is a relation on set A = {a, b, c} defined as R = {(a, a), (b, b), (c, c), (a, b), (b, a)}, check if R is symmetric and/or transitive.

Solution:

Step 1: Check Symmetry
For symmetry, if (x,y) ∈ R, then (y,x) ∈ R.
Check all pairs:
(a,a) ∈ R and (a,a) ∈ R ✓
(b,b) ∈ R and (b,b) ∈ R ✓
(c,c) ∈ R and (c,c) ∈ R ✓
(a,b) ∈ R and (b,a) ∈ R ✓
(b,a) ∈ R and (a,b) ∈ R ✓

Therefore, R is symmetric.

Step 2: Check Transitivity
For transitivity, if (x,y) ∈ R and (y,z) ∈ R, then (x,z) ∈ R.
Check all combinations:
(a,a) ∈ R and (a,a) ∈ R → (a,a) ∈ R ✓
(a,a) ∈ R and (a,b) ∈ R → (a,b) ∈ R ✓
(a,b) ∈ R and (b,a) ∈ R → (a,a) ∈ R ✓
(a,b) ∈ R and (b,b) ∈ R → (a,b) ∈ R ✓
(b,a) ∈ R and (a,a) ∈ R → (b,a) ∈ R ✓
(b,a) ∈ R and (a,b) ∈ R → (b,b) ∈ R ✓
(b,b) ∈ R and (b,a) ∈ R → (b,a) ∈ R ✓
(b,b) ∈ R and (b,b) ∈ R → (b,b) ∈ R ✓
(c,c) ∈ R and (c,c) ∈ R → (c,c) ∈ R ✓

Therefore, R is transitive.

Conclusion: R is both symmetric and transitive.

Question 10: Let R be a relation on the set of polygons, where P₁RP₂ if P₁ and P₂ have the same number of sides. Prove R is an equivalence relation and describe the equivalence classes.

Solution:

Step 1: Check Reflexivity
For any polygon P, P has the same number of sides as itself.
Therefore, (P,P) ∈ R for all polygons P.
R is reflexive.

Step 2: Check Symmetry
If P₁RP₂, then P₁ and P₂ have the same number of sides.
This means P₂ and P₁ also have the same number of sides.
Therefore, P₂RP₁.
R is symmetric.

Step 3: Check Transitivity
If P₁RP₂ and P₂RP₃, then:
P₁ and P₂ have the same number of sides, and
P₂ and P₃ have the same number of sides.
Therefore, P₁ and P₃ have the same number of sides.
Hence, P₁RP₃.
R is transitive.

Step 4: Describe Equivalence Classes
The equivalence classes are:
[Triangle] = {all triangles} (3 sides)
[Quadrilateral] = {all quadrilaterals} (4 sides)
[Pentagon] = {all pentagons} (5 sides)
[Hexagon] = {all hexagons} (6 sides)
And so on...

Conclusion: R is an equivalence relation that partitions polygons into classes based on the number of sides.

Question 11: Given a relation R on ℤ defined by aRb if a - b is divisible by 5, list all elements of the equivalence class containing 2.

Solution:

Step 1: Understand the equivalence class
The equivalence class [2] = {x ∈ ℤ : xR2}
This means [2] = {x ∈ ℤ : x - 2 is divisible by 5}

Step 2: Find the pattern
x - 2 is divisible by 5 means x - 2 = 5k for some integer k.
Therefore, x = 2 + 5k

Step 3: List all elements
For k = ..., -2, -1, 0, 1, 2, ...
k = -2: x = 2 + 5(-2) = -8
k = -1: x = 2 + 5(-1) = -3
k = 0: x = 2 + 5(0) = 2
k = 1: x = 2 + 5(1) = 7
k = 2: x = 2 + 5(2) = 12
...

Step 4: Write the equivalence class
[2] = {..., -13, -8, -3, 2, 7, 12, 17, 22, ...}
[2] = {2 + 5k : k ∈ ℤ}

Conclusion: The equivalence class containing 2 consists of all integers that leave remainder 2 when divided by 5.

Question 12: Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.

Solution:

Statement: False. A relation can be symmetric and transitive without being reflexive.

Counterexample:
Let A = {1, 2, 3} and R = {(1,1), (2,2)}.

Step 1: Check Symmetry
For (1,1) ∈ R: (1,1) ∈ R ✓
For (2,2) ∈ R: (2,2) ∈ R ✓
R is symmetric.

Step 2: Check Transitivity
The only pairs to check are:
(1,1) ∈ R and (1,1) ∈ R → (1,1) ∈ R ✓
(2,2) ∈ R and (2,2) ∈ R → (2,2) ∈ R ✓
R is transitive.

Step 3: Check Reflexivity
For reflexivity, we need (a,a) ∈ R for all a ∈ A.
We have (1,1) ∈ R and (2,2) ∈ R, but (3,3) ∉ R.
R is not reflexive.

Conclusion: The statement is false. R is symmetric and transitive but not reflexive.

Question 13: Find the total number of reflexive relations on a set with n elements.

Solution:

Step 1: Understand reflexive relations
A relation R on set A is reflexive if (a,a) ∈ R for all a ∈ A.
For a set with n elements, there are n diagonal elements that must be in R.

Step 2: Count total pairs
For a set with n elements, A × A has n² ordered pairs.
The diagonal elements are: (a₁,a₁), (a₂,a₂), ..., (aₙ,aₙ)
Number of diagonal elements = n

Step 3: Count choices
For a reflexive relation:
- All n diagonal elements must be included (no choice)
- The remaining (n² - n) elements can be either included or not included
- Each of the (n² - n) elements has 2 choices: include or exclude

Step 4: Apply multiplication principle
Total number of reflexive relations = 2^(n² - n)

Verification:
For n = 2: 2^(4-2) = 2² = 4
For n = 3: 2^(9-3) = 2⁶ = 64

Answer: 2^(n² - n) reflexive relations on a set with n elements.

Question 14: If R is a relation on A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}, is R symmetric? Is it transitive?

Solution:

Step 1: Check Symmetry
For symmetry, if (a,b) ∈ R, then (b,a) ∈ R.
Check all pairs:
(1,1) ∈ R and (1,1) ∈ R ✓
(2,2) ∈ R and (2,2) ∈ R ✓
(3,3) ∈ R and (3,3) ∈ R ✓
(1,2) ∈ R and (2,1) ∈ R ✓
(2,1) ∈ R and (1,2) ∈ R ✓

Therefore, R is symmetric.

Step 2: Check Transitivity
For transitivity, if (a,b) ∈ R and (b,c) ∈ R, then (a,c) ∈ R.
Check all combinations:
(1,1) ∈ R and (1,1) ∈ R → (1,1) ∈ R ✓
(1,1) ∈ R and (1,2) ∈ R → (1,2) ∈ R ✓
(1,2) ∈ R and (2,1) ∈ R → (1,1) ∈ R ✓
(1,2) ∈ R and (2,2) ∈ R → (1,2) ∈ R ✓
(2,1) ∈ R and (1,1) ∈ R → (2,1) ∈ R ✓
(2,1) ∈ R and (1,2) ∈ R → (2,2) ∈ R ✓
(2,2) ∈ R and (2,1) ∈ R → (2,1) ∈ R ✓
(2,2) ∈ R and (2,2) ∈ R → (2,2) ∈ R ✓
(3,3) ∈ R and (3,3) ∈ R → (3,3) ∈ R ✓

Therefore, R is transitive.

Conclusion: R is both symmetric and transitive.

Question 15: Let R be the relation on ℝ defined by xRy if and only if |x - y| ≤ 2. Determine if (3, 5) ∈ R and whether R is reflexive, symmetric, and/or transitive.

Solution:

Step 1: Check if (3, 5) ∈ R
We need to check if |3 - 5| ≤ 2
|3 - 5| = |-2| = 2 ≤ 2 ✓
Therefore, (3, 5) ∈ R.

Step 2: Check Reflexivity
For any x ∈ ℝ, we need (x,x) ∈ R
|x - x| = |0| = 0 ≤ 2 ✓
Therefore, R is reflexive.

Step 3: Check Symmetry
If (x,y) ∈ R, then |x - y| ≤ 2
Since |x - y| = |y - x|, we have |y - x| ≤ 2
Therefore, (y,x) ∈ R
Therefore, R is symmetric.

Step 4: Check Transitivity
Let's find a counterexample:
Take x = 1, y = 3, z = 5
Check (1,3) ∈ R: |1 - 3| = 2 ≤ 2 ✓
Check (3,5) ∈ R: |3 - 5| = 2 ≤ 2 ✓
Check (1,5) ∈ R: |1 - 5| = 4 > 2 ✗
Since (1,3) ∈ R and (3,5) ∈ R but (1,5) ∉ R, transitivity fails.
Therefore, R is not transitive.

Conclusion:
• (3, 5) ∈ R
• R is reflexive and symmetric but not transitive