Relations & Functions — Class 12 Notes

RSetu.link — clear, example-driven notes with diagrams and practice problems

Introduction

This chapter covers Relations and Functions in a compact, revision-friendly format: precise definitions, important theorems with proofs, formulas, diagrams, solved examples, a summary table and practice questions so the whole chapter sticks after one strong study session.

1. Relation — Definition & Types

1.1 Definition

Given sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b). A relation R from A to B is any subset R ⊆ A × B. If (a, b) ∈ R we write aRb.

1.2 Properties of relations

Theorem (Relation properties): A relation on a set A may have properties such as reflexive, symmetric and transitive.

Definitions:

  • Reflexive: ∀a ∈ A, (a,a) ∈ R.
  • Symmetric: If (a,b) ∈ R then (b,a) ∈ R.
  • Transitive: If (a,b) ∈ R and (b,c) ∈ R then (a,c) ∈ R.

1.3 Special relations

  • Empty relation, universal relation, identity relation
  • Equivalence relation: A relation that is reflexive, symmetric and transitive. Equivalence relations partition the set into equivalence classes.

Mapping diagram (visual)

Use a mapping diagram for clarity: draw two columns (Domain and Codomain), draw arrows to show ordered pairs. Example images/diagrams help memory a lot — add mapping diagrams as PNG/SVG files here:

Mapping diagram example (replace with your image)

2. Function — Definition, Types & Properties

2.1 Definition

A function f from A to B (written f: A → B) is a relation that assigns to each element a ∈ A exactly one element b ∈ B. That is, for every a ∈ A there exists a unique b such that (a, b) ∈ f.

2.2 Types of functions

  • Injective (one-to-one): distinct inputs map to distinct outputs.
  • Surjective (onto): range = codomain; every element of codomain has a preimage.
  • Bijective: both injective and surjective (invertible).
  • Other examples: constant function, identity function, many-to-one.

2.3 Composition and inverse

If f: A → B and g: B → C then the composition g ∘ f: A → C is defined by (g ∘ f)(a) = g(f(a)). If f is bijective then inverse f⁻¹: B → A exists with f⁻¹(f(a)) = a.

2.4 Domain, Codomain, Range (Image)

Domain — set of inputs. Codomain — set where outputs live. Range/Image — actual outputs attained by the function (subset of codomain).

3. Important Theorems & Formulas

Theorem 1: If f and g are injective, then g ∘ f is injective.
Proof: Suppose (g ∘ f)(x1) = (g ∘ f)(x2). Then g(f(x1)) = g(f(x2)). Since g is injective, f(x1) = f(x2). Since f is injective, x1 = x2. Hence composition is injective.
Theorem 2: If f and g are surjective, then g ∘ f is surjective.
Proof: For c in C, since g is surjective there exists b in B with g(b)=c. Since f is surjective, there exists a in A with f(a)=b. Then (g ∘ f)(a)=c. Hence surjective.

Counting relations & functions

If |A| = m and |B| = n, total possible relations from A to B = 2^(m·n). Number of possible functions from A to B = n^m (each of the m domain elements choose one of n codomain elements).

4. Solved Examples

Example 1: A = {1,2,3}, B = {a,b}. Relation R = {(1,a),(2,a),(3,b)}. Is R a function? Domain & Range?

Solution: Domain = {1,2,3}. Each element of domain maps to exactly one element in B, so R is a function. Range = {a,b}.

Example 2: f: ℤ → ℤ defined by f(x)=x². Is f injective or surjective?

Solution: Not injective (f(2)=f(−2)). Not surjective (no negative integer is a square). So f is many-to-one and not onto.

Add more examples: composition examples, inverse, mapping diagrams, equivalence classes, partitions.

5. Summary & Practice

5.1 Summary Table

TopicKey Points
RelationSubset of A×B; properties: reflexive, symmetric, transitive
FunctionEach domain element maps to exactly one codomain element; domain, range, codomain
Injective / Surjective / BijectiveOne-to-one, onto, both (invertible)
Composition(g ∘ f)(x) = g(f(x)); composition preserves injectivity and surjectivity under conditions
CountingRelations: 2^(m·n). Functions: n^m.

5.2 Practice Questions

  1. Define relation and function with a mapping diagram example.
  2. Prove: composition of two injective functions is injective.
  3. Let f: {1,2,3} → {a,b,c}. How many relations and how many functions exist?
  4. Give an example of an equivalence relation on set {1,2,3} and list equivalence classes.
  5. If f(x)=2x+1 and g(x)=x², compute (g ∘ f)(x) and (f ∘ g)(x).

FAQ

Q: Can every relation be called a function?

A: No — a function requires exactly one output per input; relations may allow multiple outputs for a single input.

Q: What makes a function bijective?

A: If it is both injective (no two inputs share the same output) and surjective (every codomain value is reached), then it’s bijective.

Q: How do equivalence relations help?

A: They partition the original set into disjoint equivalence classes; each class collects elements equivalent under the relation.