IIT JEE Main Maths -Unit 1- Relations: types and properties- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 1- Relations: types and properties – Study Notes – New syllabus
IIT JEE Main Maths -Unit 1- Relations: types and properties – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Relation
- Domain, Co-domain, and Range
- Representation of Relation
- Types of Relations
- Number of Relations
- Equivalence Class
- Partition of a Set
- Composition of Relations
Cartesian Product of Sets
If \( A \) and \( B \) are two non-empty sets, then the Cartesian product of \( A \) and \( B \), denoted by \( A \times B \), is the set of all ordered pairs \( (a, b) \) such that \( a \in A \) and \( b \in B \).
\( A \times B = \{(a,b) \mid a \in A, \, b \in B\} \)
The number of elements in \( A \times B \) is given by:
\( n(A \times B) = n(A) \times n(B) \)
Note: The order of elements in each pair matters. In general, \( A \times B \ne B \times A \).
Example
Let \( A = \{1, 2\} \) and \( B = \{x, y, z\} \). Find \( A \times B \) and \( B \times A \).
▶️ Answer / Explanation
Step 1: Form all possible ordered pairs \( (a,b) \) where \( a \in A \) and \( b \in B \).
\( A \times B = \{(1,x), (1,y), (1,z), (2,x), (2,y), (2,z)\} \)
Step 2: Reverse the order for \( B \times A \).
\( B \times A = \{(x,1), (x,2), (y,1), (y,2), (z,1), (z,2)\} \)
Conclusion: \( A \times B \ne B \times A \).
Relation
A relation from set \( A \) to set \( B \) is any subset of the Cartesian product \( A \times B \).
\( R \subseteq A \times B \)
If \( (a, b) \in R \), we say that \( a \) is related to \( b \) under the relation \( R \), and we write it as \( a \, R \, b \).
Example: If \( A = \{1, 2, 3\} \) and \( B = \{1, 4, 9\} \), a relation \( R \) can be defined by “\( b = a^2 \)”. Then, \( R = \{(1,1), (2,4), (3,9)\} \).
Thus, a relation describes how elements of one set are associated with elements of another set.
Example
Let \( A = \{2, 4, 6\} \) and \( B = \{1, 2, 3, 4, 5, 6\} \). Define a relation \( R \) from \( A \) to \( B \) as “\( a \) divides \( b \)”. Write \( R \).
▶️ Answer / Explanation
Step 1: For each \( a \in A \), find all \( b \in B \) such that \( b \) is divisible by \( a \).
- For \( a = 2 \): \( b = 2, 4, 6 \)
- For \( a = 4 \): \( b = 4 \)
- For \( a = 6 \): \( b = 6 \)
Step 2: Write the set of ordered pairs.
\( R = \{(2,2), (2,4), (2,6), (4,4), (6,6)\} \)
Step 3: Verify definition.
Each pair \( (a,b) \) satisfies \( a \) divides \( b \).
Conclusion: \( R \subseteq A \times B \) is a valid relation defined by “divides”.
Domain, Co-domain, and Range
For a relation \( R \) from set \( A \) to set \( B \), i.e., \( R \subseteq A \times B \):
- Domain: The set of all first elements of the ordered pairs in \( R \). \( \text{Domain}(R) = \{a \mid (a,b) \in R\} \)
- Co-domain: The set \( B \), i.e., the second set from which elements of ordered pairs are taken.
- Range: The set of all second elements of the ordered pairs in \( R \). \( \text{Range}(R) = \{b \mid (a,b) \in R\} \)
Note: Range \( \subseteq \) Co-domain
Example
For the relation \( R = \{(1,1), (2,4), (3,9)\} \) from \( A = \{1, 2, 3\} \) to \( B = \{1, 4, 9, 16\} \), find the domain, co-domain, and range.
▶️ Answer / Explanation
Step 1: Identify sets \( A \) and \( B \).
\( A = \{1, 2, 3\} \), \( B = \{1, 4, 9, 16\} \)
Step 2: Find domain and range.
Domain = \( \{1, 2, 3\} \)
Range = \( \{1, 4, 9\} \)
Step 3: Co-domain = \( \{1, 4, 9, 16\} \)
Step 4: Verify that Range \( \subseteq \) Co-domain.
Conclusion: Domain represents inputs, Co-domain represents all possible outputs, and Range represents actual outputs related through \( R \).
Representation of Relation
A relation from a set \( A \) to a set \( B \) can be represented in different forms for better understanding. The most common methods are:
Roster (Tabular) Form:
In this method, the relation is represented by listing all the ordered pairs explicitly.
Example: If \( A = \{1, 2, 3\} \) and \( B = \{2, 4, 6\} \), and \( R = \{(1,2), (2,4), (3,6)\} \), this is the roster form of relation \( R \).
Set-builder Form:
In this form, a common property is described that all ordered pairs satisfy.
Example: \( R = \{(a,b) \mid b = 2a,\, a \in A,\, b \in B\} \)
Arrow (Pictorial) Diagram:
In this representation, elements of set \( A \) and \( B \) are written in two ovals. Arrows are drawn from each element of \( A \) to its related elements in \( B \).
Example: For \( R = \{(1,a), (1,d), (2,b), (3,c)\} \):
- An arrow from 1 → a
- An arrow from 1 → d
- An arrow from 2 → b
- An arrow from 3 → c represent the relation pictorially.
Example
Let \( A = \{1, 2, 3\} \) and \( B = \{1, 2, 3\} \). Define a relation \( R \) from \( A \) to \( B \) as “\( a \) is less than \( b \)”. Represent \( R \) in roster and matrix form.
▶️ Answer / Explanation
Step 1: List all ordered pairs where \( a < b \).
\( R = \{(1,2), (1,3), (2,3)\} \)
Step 2: Write in matrix form.
Since \( A = B = \{1,2,3\} \), both have 3 elements.
| A\B | 1 | 2 | 3 |
|---|---|---|---|
| 1 | 0 | 1 | 1 |
| 2 | 0 | 0 | 1 |
| 3 | 0 | 0 | 0 |
Step 3: Interpretation:
‘1’ in position \( (i,j) \) shows that \( a_i < b_j \). For example, \( (1,2) \) and \( (2,3) \) have 1s indicating true relations.
Conclusion: The relation \( R = \{(1,2), (1,3), (2,3)\} \) is represented by the above 3×3 matrix.
Types of Relations
A relation on a set \( A \) is a subset of \( A \times A \). Depending on the properties it satisfies, a relation can be classified as Reflexive, Identity, Symmetric, Transitive, Antisymmetric, Equivalence, or Inverse relation.
Reflexive Relation
A relation \( R \) on a set \( A \) is said to be reflexive if every element of \( A \) is related to itself.
That is, \( (a,a) \in R \) for all \( a \in A \).
Mathematically: \( \forall a \in A,\, (a,a) \in R \).
Example
Let \( A = \{1,2,3\} \) and \( R = \{(1,1), (2,2), (3,3)\} \).Describe the type of Relation.
▶️ Explanation
Each element is related to itself, so \( R \) is reflexive on \( A \).
Identity Relation
The identity relation on a set \( A \) is the relation that contains only pairs where both elements are the same.
It is denoted by \( I_A = \{(a,a) \mid a \in A\} \).
Every identity relation is reflexive.
Example
If \( A = \{x, y, z\} \), then \( I_A = \{(x,x), (y,y), (z,z)\} \).Describe the type of Relation.
▶️ Explanation
This relation includes only pairs of the form \( (a,a) \), hence it’s the identity relation on \( A \).
Symmetric Relation
A relation \( R \) on \( A \) is symmetric if for every \( (a,b) \in R \), we also have \( (b,a) \in R \).
Mathematically: \( \forall a,b \in A,\ (a,b) \in R \Rightarrow (b,a) \in R \).
Example
Let \( A = \{1,2,3\} \) and \( R = \{(1,2), (2,1), (2,3), (3,2)\} \).Describe the type of Relation.
▶️ Explanation
For every pair \( (a,b) \), the pair \( (b,a) \) is also present. Hence, \( R \) is symmetric.
Transitive Relation
A relation \( R \) on \( A \) is transitive if whenever \( (a,b) \in R \) and \( (b,c) \in R \), then \( (a,c) \in R \).
Mathematically: \( \forall a,b,c \in A,\ (a,b) \in R \text{ and } (b,c) \in R \Rightarrow (a,c) \in R \).
Example
Let \( A = \{1,2,3\} \) and \( R = \{(1,2), (2,3), (1,3)\} \).Describe the type of Relation.
▶️ Explanation
Since \( (1,2) \) and \( (2,3) \) imply \( (1,3) \) is in \( R \), the relation is transitive.
Antisymmetric Relation
A relation \( R \) on \( A \) is antisymmetric if for all \( a,b \in A \), \( (a,b) \in R \) and \( (b,a) \in R \) together imply \( a = b \).
Mathematically: \( (a,b) \in R \text{ and } (b,a) \in R \Rightarrow a = b \).
Example
Let \( A = \{1,2,3\} \) and \( R = \{(1,1), (2,2), (3,3), (1,2)\} \).Describe the type of Relation.
▶️ Explanation
\( (1,2) \in R \), but \( (2,1) \notin R \). Only equal pairs like \( (1,1), (2,2), (3,3) \) have both directions, so \( R \) is antisymmetric.
Equivalence Relation
A relation \( R \) on \( A \) is an equivalence relation if it is:
- Reflexive
- Symmetric
- Transitive
Equivalence relations partition a set into disjoint subsets called equivalence classes.
Example
Let \( A = \mathbb{Z} \) and define \( R = \{(a,b) \mid a – b \text{ is divisible by } 5\} \). Describe the type of Relation.
▶️ Explanation
Reflexive: \( a – a = 0 \) is divisible by 5.
Symmetric: If \( a – b \) divisible by 5, then \( b – a \) is also divisible by 5.
Transitive: If \( a – b \) and \( b – c \) divisible by 5, then \( a – c \) is divisible by 5.
Hence, \( R \) is an equivalence relation.
Inverse Relation
If \( R \) is a relation from set \( A \) to set \( B \), then the inverse relation \( R^{-1} \) is the relation from \( B \) to \( A \) defined as:
\( R^{-1} = \{(b,a) \mid (a,b) \in R\} \).
That is, all ordered pairs in \( R \) are reversed in \( R^{-1} \).
Example
Let \( R = \{(1,2), (2,3), (3,4)\} \).Find its Inverse Relation.
▶️ Explanation
Then \( R^{-1} = \{(2,1), (3,2), (4,3)\} \). The direction of relation is reversed.
Number of Relations
This concept helps in counting the total number of possible relations that can be formed between elements of sets. It is one of the most frequently asked topics in JEE Main.
Relation on a Set
If a set \( A \) has \( n \) elements, then the number of ordered pairs in \( A \times A \) is:
\( n^2 \)
Each ordered pair can either be included in a relation or not included.
So, total number of relations on \( A \) is:
\( 2^{n^2} \)
Relation from One Set to Another
If \( A \) has \( m \) elements and \( B \) has \( n \) elements, then:
Number of ordered pairs in \( A \times B = mn \)
Each pair has 2 choices → include or exclude
So, total number of relations from \( A \to B \):
\( 2^{mn} \)
Important Notes:
- A relation is any subset of the Cartesian product.
- Total subsets = total relations.
- Use power of 2 because of inclusion/exclusion of each pair.
Example
If \( A = \{1,2,3\} \), find the number of relations on \( A \).
▶️ Answer / Explanation
Step 1: Number of elements \( n(A) = 3 \)
Step 2: Number of ordered pairs in \( A \times A = 3^2 = 9 \)
Step 3: Number of relations \( = 2^9 = 512 \)
Conclusion: There are 512 relations on set \( A \).
Example
If \( A = \{1,2\} \) and \( B = \{a,b,c\} \), find the number of relations from \( A \to B \).
▶️ Answer / Explanation
Step 1: \( n(A) = 2, \quad n(B) = 3 \)
Step 2: Number of ordered pairs in \( A \times B = 2 \times 3 = 6 \)
Step 3: Number of relations \( = 2^6 = 64 \)
Conclusion: There are 64 relations from \( A \to B \).
Equivalence Class
An equivalence class is defined for an equivalence relation. It groups together all elements that are related to a given element.
If \( R \) is an equivalence relation on a set \( A \), then the equivalence class of an element \( a \in A \) is:
\( [a] = \{x \in A \mid (x,a) \in R\} \)
Meaning:
The equivalence class \( [a] \) contains all elements of \( A \) that are related to \( a \).
Important Properties:
- Every element belongs to exactly one equivalence class.
- Two equivalence classes are either equal or disjoint.
- The union of all equivalence classes gives the original set.
Understanding Idea:
Equivalence relation groups elements into “families” where all members are related to each other.
Example
Let \( A = \{1,2,3,4,5,6\} \) and define relation \( R \) as:
\( aRb \) if \( a \equiv b \pmod{2} \). Find equivalence classes.
▶️ Answer / Explanation
Step 1: Understand relation
Numbers are related if they have same remainder when divided by 2.
Step 2: Group elements
- Even numbers: \( \{2,4,6\} \)
- Odd numbers: \( \{1,3,5\} \)
Step 3: Write equivalence classes
\( [1] = \{1,3,5\} \)
\( [2] = \{2,4,6\} \)
Conclusion: Set is partitioned into two equivalence classes.
Example
Let \( A = \{1,2,3\} \) and \( R = \{(1,1),(2,2),(3,3)\} \). Find equivalence classes.
▶️ Answer / Explanation
Each element is related only to itself.
\( [1] = \{1\}, \quad [2] = \{2\}, \quad [3] = \{3\} \)
Conclusion: Each element forms its own equivalence class.
Partition of a Set
A partition of a set \( A \) is a division of the set into non-empty, disjoint subsets such that every element of \( A \) belongs to exactly one subset.
Conditions for Partition:
- Each subset is non-empty.
- Subsets are mutually disjoint (no common elements).
- The union of all subsets equals the original set.
Connection with Equivalence Relation:
Every equivalence relation on a set defines a partition of that set into equivalence classes.
Important Idea:
Partition divides a set into “groups” such that no element is left out and no element appears in more than one group.
Example
Let \( A = \{1,2,3,4,5,6\} \). Check whether the collection \( \{\{1,2\}, \{3,4\}, \{5,6\}\} \) is a partition of \( A \).
▶️ Answer / Explanation
Step 1: Check non-empty subsets → all subsets have elements ✔
Step 2: Check disjointness → no element is repeated ✔
Step 3: Check union:
\( \{1,2\} \cup \{3,4\} \cup \{5,6\} = \{1,2,3,4,5,6\} \)
Conclusion: It is a partition of \( A \).
Example
Check whether \( \{\{1,2\}, \{2,3\}, \{4,5,6\}\} \) is a partition of \( A = \{1,2,3,4,5,6\} \).
▶️ Answer / Explanation
Element 2 appears in both \( \{1,2\} \) and \( \{2,3\} \).
So subsets are not disjoint.
Conclusion: Not a partition.
Composition of Relations
The composition of relations combines two relations to form a new relation.
If \( R \subseteq A \times B \) and \( S \subseteq B \times C \), then the composition of \( R \) and \( S \), denoted by \( S \circ R \), is defined as:
\( S \circ R = \{(a,c) \mid (a,b) \in R \text{ and } (b,c) \in S\} \)
Meaning:
If \( a \) is related to \( b \) by \( R \), and \( b \) is related to \( c \) by \( S \), then \( a \) is related to \( c \) by \( S \circ R \).
Important Notes:
- Order matters: \( S \circ R \ne R \circ S \) in general.
- The middle set must match (relation chain must be valid).
- Used in chaining relations.
Example
Let \( A = \{1,2,3\} \), \( B = \{a,b\} \), \( C = \{x,y\} \)
\( R = \{(1,a),(2,b)\} \), \( S = \{(a,x),(b,y)\} \)
Find \( S \circ R \).
▶️ Answer / Explanation
Step 1: Match pairs
- \( (1,a) \in R \) and \( (a,x) \in S \Rightarrow (1,x) \)
- \( (2,b) \in R \) and \( (b,y) \in S \Rightarrow (2,y) \)
Step 2: Write result
\( S \circ R = \{(1,x),(2,y)\} \)
Conclusion: Relations are composed through matching middle element.
Example
Let \( R = \{(1,2),(2,3)\} \). Find \( R \circ R \).
▶️ Answer / Explanation
Step 1: Look for chaining
\( (1,2) \) and \( (2,3) \Rightarrow (1,3) \)
Step 2: Final relation
\( R \circ R = \{(1,3)\} \)