Chapter 1: Relations and Functions
Comprehensive study notes covering types of relations (reflexive, symmetric, transitive, equivalence), equivalence classes, types of functions (one-one, onto, bijective), composition of functions, invertible functions, and binary operations.
1 Introduction & Cartesian Product
In mathematics, a relation establishes an association between elements of two sets. Given two non-empty sets A and B, the Cartesian Product A × B is defined as the set of all ordered pairs (a, b) where a ∈ A and b ∈ B:
A relation R from set A to set B is any subset of the Cartesian product A × B. That is, R ⊆ A × B. If (a, b) ∈ R, we write a R b, which means "a is related to b under R".
2 Types of Relations
Let A be a non-empty set. A relation R on A (i.e. R ⊆ A × A) is classified into fundamental types based on its properties:
1. Empty Relation (Void Relation)
A relation R in set A is called empty if no element of A is related to any element of A.
2. Universal Relation
A relation R in set A is universal if each element of A is related to every element of A.
3. Reflexive Relation
A relation R on A is reflexive if every element of A is related to itself.
4. Symmetric Relation
A relation R on A is symmetric if the relation between a and b implies the relation between b and a.
5. Transitive Relation
A relation R on A is transitive if whenever a is related to b and b is related to c, then a is related to c.
3 Equivalence Relations & Classes
Definition: Equivalence Relation
A relation R on set A is said to be an equivalence relation if and only if it is:
- Reflexive
- Symmetric
- Transitive
Equivalence Class
Given an equivalence relation R on set A and an element a ∈ A, the equivalence class of a, denoted by [a], is the subset of A containing all elements related to a:
Key Properties of Equivalence Classes:
• All elements of [a] are related to each other.
• No element of [a] is related to any element outside [a].
• Two equivalence classes are either identical or disjoint: [a] = [b] or [a] ∩ [b] = ∅.
• The union of all equivalence classes gives the set A, partitioning A into mutually disjoint subsets.
4 Types of Functions (Mappings)
A function f : X → Y associates each element x ∈ X with a unique element y ∈ Y. Functions are classified according to how they map inputs to outputs:
One-One Function
Distinct elements in X have distinct images in Y.
If distinct elements map to the same image, the function is Many-One.
Onto Function
Every element of Y is the image of at least one element of X.
Condition for onto: Range(f) = Codomain(Y).
One-One and Onto (Bijection)
A function f : X → Y is called bijective if it is both one-one (injective) and onto (surjective). Bijectivity is the necessary and sufficient condition for a function to be invertible.
Visual Representation: Injective vs Bijective Mapping
5 Composition of Functions
Let f : A → B and g : B → C be two functions. The composition of f and g, denoted by g ∘ f, is the function g ∘ f : A → C defined by:
Fundamental Properties:
• Associativity: If f : A → B, g : B → C, and h : C → D, then h ∘ (g ∘ f) = (h ∘ g) ∘ f.
• Non-Commutative: In general, g ∘ f ≠ f ∘ g.
• If f and g are both one-one, then g ∘ f is one-one.
• If f and g are both onto, then g ∘ f is onto.
6 Invertible Functions
A function f : X → Y is defined to be invertible if there exists a function g : Y → X such that:
where I_X is the identity function on X and I_Y is the identity function on Y. The function g is called the inverse of f and is denoted by f⁻¹.
7 Binary Operations
A binary operation * on a set A is a function * : A × A → A. We denote *(a, b) by a * b.
Commutative Property
a * b = b * a, ∀ a, b ∈ A
Associative Property
(a * b) * c = a * (b * c), ∀ a, b, c ∈ A
Identity Element
a * e = e * a = a, ∀ a ∈ A
Invertible Element
a * b = b * a = e (b = a⁻¹)
8 Solved CBSE Board & HOT Examples
Equivalence Relation Proof
Show that the relation R in the set Z of integers given by R = { (a, b) : 2 divides (a - b) } is an equivalence relation.
Solution:
1. Reflexive: For any a ∈ Z, a - a = 0 = 2 × 0, which is divisible by 2. Thus (a, a) ∈ R. Hence R is reflexive.
2. Symmetric: Let (a, b) ∈ R. Then a - b = 2k for some k ∈ Z. Therefore, b - a = -(a - b) = -2k = 2(-k), which is also divisible by 2. Hence (b, a) ∈ R. Thus R is symmetric.
3. Transitive: Let (a, b) ∈ R and (b, c) ∈ R. Then a - b = 2m and b - c = 2n for integers m, n. Adding the two: (a - b) + (b - c) = a - c = 2(m + n), which is divisible by 2. Hence (a, c) ∈ R. Thus R is transitive.
Since R is reflexive, symmetric, and transitive, R is an equivalence relation on Z. (Q.E.D.)
Bijectivity & Inverse Function
Let f : R - {3} → R - {1} be defined by f(x) = (x - 2) / (x - 3). Show that f is bijective and find f⁻¹.
Solution:
One-One (Injective): Let f(x₁) = f(x₂). Then (x₁ - 2)/(x₁ - 3) = (x₂ - 2)/(x₂ - 3).
Cross-multiplying: (x₁ - 2)(x₂ - 3) = (x₂ - 2)(x₁ - 3)
x₁x₂ - 3x₁ - 2x₂ + 6 = x₁x₂ - 3x₂ - 2x₁ + 6
-3x₁ - 2x₂ = -3x₂ - 2x₁ ⇒ -x₁ = -x₂ ⇒ x₁ = x₂. Hence f is one-one.
Onto (Surjective): Let y ∈ R - {1}. We set y = (x - 2)/(x - 3).
y(x - 3) = x - 2 ⇒ yx - 3y = x - 2 ⇒ x(y - 1) = 3y - 2 ⇒ x = (3y - 2)/(y - 1).
Since y ≠ 1, x is real and well-defined. If x = 3, then 3(y - 1) = 3y - 2 ⇒ 3y - 3 = 3y - 2 ⇒ -3 = -2 (contradiction). So x ≠ 3. Thus for every y ∈ R - {1}, there exists x ∈ R - {3} such that f(x) = y. Hence f is onto.
Since f is one-one and onto, f is bijective and invertible with inverse f⁻¹(x) = (3x - 2) / (x - 1).
9 Quick Revision & Exam Checklist
- ✓ A relation R is an equivalence relation iff it is reflexive, symmetric, and transitive.
- ✓ The empty relation on a non-empty set is symmetric and transitive, but never reflexive.
- ✓ A function f is invertible iff it is bijective (Range = Codomain and f(x₁) = f(x₂) ⇒ x₁ = x₂).
- ✓ Composition of functions is associative: (f ∘ g) ∘ h = f ∘ (g ∘ h), but rarely commutative.
- ✓ Identity element for a binary operation, if it exists, is unique.