Relations and Functions
'A function is a machine: you give it an input, it gives you exactly one output.'
1. Chapter Overview
This chapter formalises the mathematical notion of RELATIONS and FUNCTIONS. A relation R from set A to set B is a subset of A × B. A function f: A → B is a relation where each element of A has a UNIQUE image in B. The focus: TYPES of relations (reflexive, symmetric, transitive, equivalence), TYPES of functions (one-one, onto, bijective), COMPOSITION of functions, INVERTIBLE functions, and BINARY operations.
2. Types of Relations
2.1 Empty and Universal Relations
- Empty relation: No element of A is related to any element of A. R = ∅.
- Universal relation: EVERY element of A is related to EVERY element of A. R = A × A.
2.2 Key Properties of Relations on a Set A
| Property | Definition | Example on A = {1, 2, 3} |
|---|---|---|
| Reflexive | (a, a) ∈ R for ALL a ∈ A | R = {(1,1), (2,2), (3,3), (1,2)} |
| Symmetric | (a, b) ∈ R ⇒ (b, a) ∈ R | R = {(1,2), (2,1)} |
| Transitive | (a, b) ∈ R AND (b, c) ∈ R ⇒ (a, c) ∈ R | R = {(1,2), (2,3), (1,3)} |
| Equivalence | Reflexive + Symmetric + Transitive | R = {(1,1), (2,2), (3,3), (1,2), (2,1)} |
2.3 Equivalence Class
- If R is an equivalence relation on A, the EQUIVALENCE CLASS of an element a is [a] = {x ∈ A : (x, a) ∈ R}.
- Equivalence classes PARTITION the set A — they are DISJOINT and their union is A.
3. Types of Functions
| Type | Definition | Condition | Example |
|---|---|---|---|
| One-One (Injective) | Different inputs → Different outputs | f(x₁) = f(x₂) ⇒ x₁ = x₂ | f(x) = 2x on R |
| Many-One | Two different inputs give the SAME output | f(x₁) = f(x₂) for x₁ ≠ x₂ | f(x) = x² on R |
| Onto (Surjective) | EVERY element of co-domain has a PRE-IMAGE | Range = Co-domain | f(x) = x³ on R |
| Into | Range is a PROPER SUBSET of co-domain | NOT every element has a pre-image | f(x) = x² on R → R |
| Bijective | One-One AND Onto | Both conditions hold | f(x) = 2x + 3 on R |
4. Composition of Functions
- (g ∘ f)(x) = g(f(x)). 'Apply f first, then g.'
- Condition: Co-domain of f must be a SUBSET of domain of g.
- Properties: Composition is ASSOCIATIVE: (h ∘ g) ∘ f = h ∘ (g ∘ f). Composition may NOT be commutative.
5. Invertible Functions
- A function f: A → B is INVERTIBLE if there exists g: B → A such that g ∘ f = Iₐ and f ∘ g = I_B.
- THEOREM: f is invertible IF AND ONLY IF f is BIJECTIVE (one-one AND onto).
- Notation: The inverse of f is denoted f⁻¹.
Worked Example 1
Problem: Show that f: R → R defined by f(x) = 3x + 5 is invertible. Find f⁻¹. Solution:
- One-one: f(x₁) = f(x₂) ⇒ 3x₁ + 5 = 3x₂ + 5 ⇒ x₁ = x₂ ✓
- Onto: For ANY y ∈ R, we need x such that f(x) = y. Solve: y = 3x + 5 ⇒ x = (y − 5)/3. This x ∈ R. ✓
- Therefore f is BIJECTIVE ⇒ INVERTIBLE.
- f⁻¹(y) = (y − 5)/3.
6. Binary Operations
- A binary operation ∗ on a set A is a function ∗: A × A → A.
- 'It takes TWO elements from A and gives ONE element from A.'
| Property | Definition |
|---|---|
| Commutative | a ∗ b = b ∗ a for all a, b |
| Associative | (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c |
| Identity element e | a ∗ e = e ∗ a = a for all a |
| Inverse of a | a ∗ b = b ∗ a = e |
7. Common Mistakes
- Confusing range and co-domain: Range is a SUBSET of co-domain. Onto means range = co-domain.
- Assuming composition is commutative: (g ∘ f) ≠ (f ∘ g) in general.
- Forgetting that a bijection is BOTH one-one AND onto: One without the other is NOT invertible.
- Equivalence relations must satisfy ALL THREE properties: Reflexive, symmetric, and transitive — if ANY one is missing, it is NOT an equivalence relation.
8. CBSE Exam Focus
- Equivalence relations — verifying all three properties, finding equivalence classes
- Proving a function is one-one and/or onto
- Composition of functions — (g ∘ f)(x) and (f ∘ g)(x)
- Invertible functions — proving bijectivity, finding the inverse
- Binary operations — checking commutativity, associativity, finding identity and inverse
9. Self-Test
Q1: Check whether the relation R on set A = {1, 2, 3, 4} defined as R = {(1,1), (2,2), (3,3), (4,4), (1,2), (2,1)} is an equivalence relation. A1: Reflexive ✓ (all (a,a) present). Symmetric ✓ (for every (a,b), (b,a) is present). Transitive: (1,2) and (2,1) ∈ R ⇒ (1,1) ∈ R ✓. (2,1) and (1,2) ∈ R ⇒ (2,2) ∈ R ✓. No violations. YES, it is an equivalence relation.
Q2: Prove that f: R → R defined by f(x) = 4x + 7 is invertible. Find f⁻¹. A2: One-one: f(x₁)=f(x₂) ⇒ 4x₁+7=4x₂+7 ⇒ x₁=x₂. Onto: For y ∈ R, x=(y−7)/4. f(x)=y. Bijective. f⁻¹(y)=(y−7)/4.
Q3: Let f: R → R and g: R → R be defined by f(x) = x² and g(x) = 2x + 1. Find (g ∘ f)(x) and (f ∘ g)(x). Are they equal? A3: (g ∘ f)(x) = g(f(x)) = g(x²) = 2x² + 1. (f ∘ g)(x) = f(g(x)) = f(2x+1) = (2x+1)² = 4x² + 4x + 1. NOT equal. 'Composition is NOT commutative.'
Q4: Consider the binary operation ∗ on R defined by a ∗ b = a + b + ab. Is ∗ commutative? Is it associative? Find the identity element. A4: Commutative: a ∗ b = a + b + ab = b + a + ba = b ∗ a ✓. Associative: (a ∗ b) ∗ c = (a+b+ab) ∗ c = a+b+ab+c+(a+b+ab)c = a+b+c+ab+ac+bc+abc. Similarly a ∗ (b ∗ c) gives the same ✓. Identity: a ∗ e = a + e + ae = a ⇒ e + ae = 0 ⇒ e(1+a) = 0. This must hold for ALL a, so e = 0. a ∗ 0 = a ✓.
Q5: Show that the function f: R − {−2} → R − {4} defined by f(x) = (4x+3)/(x+2) is bijective. A5: One-one: f(x₁)=f(x₂) ⇒ cross-multiply ⇒ 4x₁x₂+8x₁+3x₂+6 = 4x₁x₂+3x₁+8x₂+6 ⇒ 8x₁+3x₂ = 3x₁+8x₂ ⇒ 5x₁ = 5x₂ ⇒ x₁=x₂. Onto: Let y = (4x+3)/(x+2) ⇒ yx+2y = 4x+3 ⇒ x(y−4) = 3−2y ⇒ x = (3−2y)/(y−4). Since y ≠ 4, x ∈ R − {−2}. Bijective ✓.
10. Conclusion
Relations and functions provide the LANGUAGE of mathematics:
- RELATIONS link elements of sets. Equivalence relations PARTITION a set.
- FUNCTIONS are special relations — each input has exactly one output.
- BIJECTIVE functions are invertible — 'perfect pairing between domain and co-domain.'
- BINARY OPERATIONS combine two elements to produce a third.
'Relations connect, functions map, and binary operations combine — together they form the grammar of mathematical discourse.'
