Relations and Functions
1. Introduction
Relations and functions form the backbone of higher mathematics. This chapter covers the fundamental concepts of relations, types of functions, composition, invertibility, and binary operations as per the ISC Class 12 syllabus.
2. Relations
2.1 Definition
A relation R from a set A to a set B is a subset of A × B. If (a, b) ∈ R, we write a R b, read as 'a is related to b'.
2.2 Types of Relations on a Set A
Let A be a non-empty set. A relation R on A is:
Reflexive: if (a, a) ∈ R for every a ∈ A. Example: R = {(a, b) : a ≤ b} on real numbers. Since a ≤ a, it is reflexive.
Symmetric: if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A. Example: R = {(a, b) : a is parallel to b} on set of lines.
Transitive: if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A. Example: R = {(a, b) : a < b} on real numbers.
Equivalence Relation: A relation that is reflexive, symmetric, and transitive simultaneously. Example: Equality relation on any set. R = {(a, b) : a - b is divisible by m} (congruence modulo m).
2.3 Equivalence Classes
For an equivalence relation R on A, the equivalence class of a ∈ A is [a] = {x ∈ A : (a, x) ∈ R}. Equivalence classes partition the set A.
2.4 Common Mistake
'Students often confuse symmetry with anti-symmetry. A relation like "≤" is not symmetric — if a ≤ b, it does NOT imply b ≤ a (unless a = b).'
3. Functions
3.1 Definition
A function f : A → B is a relation where each element of A has a unique image in B. Domain = A, Codomain = B, Range = {f(a) : a ∈ A}.
3.2 Types of Functions
One-one (Injective): Different elements have different images. f(a₁) = f(a₂) ⇒ a₁ = a₂. Test: Horizontal line test. If a horizontal line cuts the graph at most once, the function is injective.
Onto (Surjective): Every element of B has at least one pre-image in A. Range = Codomain. Test: The range must equal the codomain.
Bijective: Both one-one and onto. Invertible.
3.3 Composition of Functions
If f : A → B and g : B → C, then gof : A → C is defined as (gof)(x) = g(f(x)) for all x ∈ A.
Composition is associative: (hog)of = ho(gof). Composition is generally NOT commutative.
3.4 Invertible Functions
A function f : A → B is invertible iff it is bijective. The inverse f^{-1} : B → A satisfies f^{-1}(y) = x whenever f(x) = y.
Property: f^{-1}of = I_A and fof^{-1} = I_B.
4. Binary Operations
4.1 Definition
A binary operation ∗ on a non-empty set A is a function ∗ : A × A → A. For each (a, b) ∈ A × A, we denote the image as a ∗ b.
4.2 Properties
Commutative: a ∗ b = b ∗ a for all a, b ∈ A. Associative: (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ A. Identity element: e ∈ A such that a ∗ e = e ∗ a = a for all a ∈ A. Inverse: For a ∈ A, b ∈ A such that a ∗ b = b ∗ a = e.
Example: On real numbers, addition is commutative and associative with identity 0, and inverse of a is -a.
5. Worked Problems
Problem 1: Check if R = {(x, y) : x - y is an integer} is an equivalence relation on real numbers. Solution: (i) Reflexive: x - x = 0 ∈ Z. (ii) Symmetric: x - y ∈ Z ⇒ y - x = -(x - y) ∈ Z. (iii) Transitive: If x - y ∈ Z and y - z ∈ Z, then (x - y) + (y - z) = x - z ∈ Z. Hence R is an equivalence relation.
Problem 2: Determine if f(x) = 2x + 3, f : R → R is bijective. Solution: One-one: f(a) = f(b) ⇒ 2a + 3 = 2b + 3 ⇒ a = b. Onto: For any y ∈ R, y = 2x + 3 ⇒ x = (y - 3)/2 ∈ R. Hence bijective. f^{-1}(y) = (y - 3)/2.
Problem 3: Let f(x) = x + 2, g(x) = x². Find (gof)(x) and (fog)(x). Solution: (gof)(x) = g(f(x)) = g(x + 2) = (x + 2)². (fog)(x) = f(g(x)) = f(x²) = x² + 2. Note gof ≠ fog.
6. ISC Exam Focus
| Topic | Theory Marks | Practical/Application Marks |
|---|---|---|
| Types of relations | 2-3 | 1 |
| Types of functions | 3-4 | 1 |
| Composition and invertible | 4 | 2 |
| Binary operations | 3 | 1 |
7. Self-Test Questions
- Show that the relation R on Z defined by a R b iff a² + b² is even is an equivalence relation.
- Determine whether f(x) = x³ - 1 is bijective. Find its inverse if it exists.
- If f(x) = √(x - 1) and g(x) = x² + 1, find (gof)(x) and its domain.
- Check if ∗ defined on Q as a ∗ b = ab/2 is commutative and associative.
- Show that the function f : R → R defined by f(x) = e^x is injective but not surjective.
8. Key Points for Revision
- A relation that is reflexive, symmetric, and transitive is an equivalence relation.
- Equivalence classes partition the set into disjoint subsets.
- A function must be bijective to be invertible.
- Composition of functions is associative but not commutative.
- Binary operations require closure — the result must belong to the same set.
- Identity element is unique if it exists. Inverse of an element is unique.
- For a binary operation on a finite set, check closure using a Cayley table.
