Can a function from ℝ to ℝ be one-one but not onto?

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CBSE Class 12 Mathematics★ 6-8 marks · CBSE Class 12 Mathematics Chapter 1

Relations and Functions (Class 12)

A deeper dive into RELATIONS (reflexive, symmetric, transitive — equivalence relations) and FUNCTIONS (one-one, onto, bijective — composition, invertibility). CBSE Class 12 Mathematics Chapter 1.

⏱ 20 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Prove whether a given relation is reflexive, symmetric, and/or transitive; identify equivalence relations
2Prove whether a given function is one-one (injective), onto (surjective), or bijective
3State and prove the condition for invertibility: f is invertible if and only if f is bijective
4Find and verify the composition of two functions (f ∘ g)(x) = f(g(x))
5Identify binary operations and check for commutativity, associativity, identity element, and inverses

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Why this chapter matters

Relations and Functions is the conceptual gateway to Class 12 Mathematics. Proving whether a relation is an equivalence relation, and proving whether a function is one-one and onto, are structured proof questions that appear in the board exam every year — they reward systematic approach over calculation skill.

Relations and Functions

"A relation groups things. A function maps them. Together, they are the grammar of mathematical structure."

1. Chapter Overview

Building on Class 11, this chapter takes relations and functions to a DEEPER level: types of relations (reflexive, symmetric, transitive, equivalence, empty, universal), types of functions (one-one/injective, onto/surjective, bijective), composition of functions, and invertibility (a function has an inverse IF AND ONLY IF it is bijective).

2. Relations — Deeper Properties

A relation R on a set A is:

Property	Definition	Example
Reflexive	(a,a) ∈ R for ALL a ∈ A	'is equal to' — every element relates to ITSELF
Symmetric	(a,b) ∈ R ⇒ (b,a) ∈ R	'is married to' — if a is married to b, b is married to a
Transitive	(a,b) ∈ R AND (b,c) ∈ R ⇒ (a,c) ∈ R	'is greater than' — if a>b and b>c, then a>c

Equivalence Relation

A relation that is REFLEXIVE, SYMMETRIC, AND TRANSITIVE. Examples: 'equality'. 'having the same birthday as'. Equivalence relations PARTITION a set into disjoint classes.

3. Functions — Types and Invertibility

One-One (Injective)

f(x₁) = f(x₂) ⇒ x₁ = x₂. Different inputs → DIFFERENT outputs.

Onto (Surjective)

Range = Codomain. Every element in the codomain is 'hit' by SOME input.

Bijective = One-One AND Onto

THE most important type. A function is INVERTIBLE iff it is BIJECTIVE.

Composition of Functions

(f ∘ g)(x) = f(g(x)). Apply g FIRST, then f.

Inverse of a Function

f⁻¹ EXISTS iff f is BIJECTIVE
f(f⁻¹(x)) = x. f⁻¹(f(x)) = x.

4. Binary Operations

A binary operation * on a set A assigns to each ordered pair (a,b) ∈ A×A an element a*b ∈ A
Properties: Commutative (ab = ba). Associative (a*(bc) = (ab)c). Identity element e (ae = e*a = a). Inverse.

5. Exam Focus

Relation properties — reflexive, symmetric, transitive. Equivalence relation.
Function types — one-one, onto, bijective. Proof strategies.
Composition of functions. Invertibility: f invertible ⇔ f bijective.
Binary operations — commutative, associative, identity, inverse.

6. Key Concepts

Equivalence relation = Reflexive + Symmetric + Transitive
f is invertible ⇔ f is bijective
(f ∘ g)(x) = f(g(x))
f⁻¹(f(x)) = x. f(f⁻¹(x)) = x.

7. Conclusion

Relations and functions are the FOUNDATIONAL LANGUAGE of mathematics:

Equivalence relations are the mathematical concept of 'sameness' with respect to a property
Bijective functions have PERFECT reversibility — every input maps uniquely to every output
Composition chains functions together. Inverse UNDOES a function.

'Mathematics is the art of giving the same name to different things.' — Henri Poincaré. Relations and functions give mathematical structure to that art.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Relation Properties

REFLEXIVE: (a,a) ∈ R for all a ∈ A. SYMMETRIC: (a,b) ∈ R ⟹ (b,a) ∈ R. TRANSITIVE: (a,b) ∈ R and (b,c) ∈ R ⟹ (a,c) ∈ R. EQUIVALENCE RELATION = Reflexive + Symmetric + Transitive. EMPTY RELATION: no pair (a,b) ∈ R. Empty relation is symmetric and transitive but NOT reflexive. UNIVERSAL RELATION: all pairs (a,b) ∈ R. Universal relation is an equivalence relation.

Exam trap: EMPTY relation — it is symmetric and transitive (vacuously true, since there are no pairs to violate the conditions) but NOT reflexive (requires (a,a) for all a, which is missing). Students often say empty relation is an equivalence relation — WRONG.

Function Types

ONE-ONE (INJECTIVE): f(x₁) = f(x₂) ⟹ x₁ = x₂. Proof strategy: assume f(x₁) = f(x₂), deduce x₁ = x₂. ONTO (SURJECTIVE): For every y in the codomain, there exists x in the domain such that f(x) = y. Proof strategy: let y be arbitrary in codomain; find x = f⁻¹(y) in domain. BIJECTIVE: Both one-one AND onto. INVERTIBLE ⟺ BIJECTIVE.

Standard exam proof structure for one-one: 'Let f(x₁) = f(x₂). Then [work] ... = ..., therefore x₁ = x₂. Hence f is one-one.' For onto: 'Let y ∈ codomain. We need x such that f(x) = y. Solving: x = [expression]. Verify x is in domain. Hence f is onto.'

Composition of Functions

(f ∘ g)(x) = f(g(x)). Apply g FIRST, then f. (g ∘ f)(x) = g(f(x)). Generally: f ∘ g ≠ g ∘ f (NOT commutative). Associative: f ∘ (g ∘ h) = (f ∘ g) ∘ h. If f and g are bijections, then f ∘ g is also a bijection. (f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹.

Order matters in composition: (f ∘ g)(x) = f(g(x)) means g acts on x first. Memory aid: read right to left, like functions in parentheses. The inverse of a composition REVERSES order: (f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹.

Binary Operations

A binary operation * on set A: A × A → A. CLOSED if a*b ∈ A for all a,b ∈ A. COMMUTATIVE: a*b = b*a. ASSOCIATIVE: a*(b*c) = (a*b)*c. IDENTITY ELEMENT e: a*e = e*a = a for all a ∈ A. INVERSE of a: b such that a*b = b*a = e. (b = a⁻¹)

Standard MCQ pattern: 'Is operation * defined as a*b = a + b − ab on ℝ commutative?' Check: b*a = b + a − ba = a + b − ab = a*b. YES, commutative. Then check: identity (a*e = a ⟹ e = 0). Inverse: a*b = 0 ⟹ b = a/(a−1). Work through systematically.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Saying an empty relation is an equivalence relation

✓ An equivalence relation must be REFLEXIVE, which requires (a,a) ∈ R for ALL a ∈ A. The EMPTY relation has NO pairs — so (a,a) is missing for every a. Therefore: empty relation is NOT reflexive, and hence NOT an equivalence relation. It IS symmetric and transitive (vacuously, since there are no pairs to violate these conditions).

WATCH OUT

✗ Confusing domain with codomain in onto proofs

✓ ONTO means range = CODOMAIN (not domain). 'Range' is the set of actual values that f takes. 'Codomain' is the TARGET set specified in the function's definition. f: ℝ → ℝ defined by f(x) = x² is NOT onto because negative numbers are in the codomain (ℝ) but not in the range. But f: ℝ → [0,∞) defined by f(x) = x² IS onto because now the codomain matches the range.

WATCH OUT

✗ Writing (f ∘ g)(x) = g(f(x))

✓ (f ∘ g)(x) = f(g(x)), NOT g(f(x)). The RIGHT-MOST function in the notation is applied FIRST. Think: '∘' is read as 'after': 'f after g' = apply g first, then f. (g ∘ f)(x) = g(f(x)) — g after f.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 1.1 and 1.2

Exercise 1.1 and 1.2

Proving relation properties (reflexive, symmetric, transitive, equivalence); proving one-one and onto; composition of functions; binary operations

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· equivalence-relation

Show that the relation R in the set ℤ of integers defined by R = {(a,b) : 2 divides (a−b)} is an equivalence relation.

▶ Show solution

REFLEXIVE: For any a ∈ ℤ, a−a = 0 = 2×0. Since 2 divides 0, (a,a) ∈ R. ✓ SYMMETRIC: If (a,b) ∈ R, then 2 | (a−b). So a−b = 2k for some k ∈ ℤ. Then b−a = −2k = 2(−k), and since −k ∈ ℤ, 2 | (b−a). Therefore (b,a) ∈ R. ✓ TRANSITIVE: If (a,b) ∈ R and (b,c) ∈ R, then 2 | (a−b) and 2 | (b−c). So a−b = 2m and b−c = 2n for m,n ∈ ℤ. Adding: a−c = (a−b)+(b−c) = 2m+2n = 2(m+n). Since m+n ∈ ℤ, 2 | (a−c). So (a,c) ∈ R. ✓ Since R is reflexive, symmetric, and transitive, R is an EQUIVALENCE RELATION.

Q2MEDIUM· bijection-proof

Let f: ℝ → ℝ be defined by f(x) = 2x + 3. Show that f is bijective and find f⁻¹.

▶ Show solution

ONE-ONE: Let f(x₁) = f(x₂). Then 2x₁ + 3 = 2x₂ + 3. Subtracting 3: 2x₁ = 2x₂. Dividing by 2: x₁ = x₂. Hence f is ONE-ONE (injective). ✓ ONTO: Let y ∈ ℝ. We need x ∈ ℝ such that f(x) = y. Solving: 2x + 3 = y ⟹ x = (y−3)/2. Since y ∈ ℝ, (y−3)/2 ∈ ℝ = domain. And f((y−3)/2) = 2·(y−3)/2 + 3 = (y−3) + 3 = y. ✓ Hence f is ONTO (surjective). Since f is both one-one and onto, f is BIJECTIVE. FINDING f⁻¹: Set y = f(x) = 2x + 3. Solving for x: x = (y−3)/2. Replacing y by x: f⁻¹(x) = (x−3)/2. VERIFICATION: f(f⁻¹(x)) = f((x−3)/2) = 2·(x−3)/2 + 3 = x−3+3 = x ✓

Q3HARD· long-answer

Let f: A → B and g: B → C be two functions. Prove that if both f and g are onto, then g ∘ f: A → C is also onto. Is the converse true? Justify.

▶ Show solution

THEOREM: If f: A→B and g: B→C are both onto, then g∘f: A→C is onto. PROOF: Let z be any arbitrary element in C. Since g: B→C is onto, there exists some y ∈ B such that g(y) = z. ... [1] Since f: A→B is onto, for this y ∈ B, there exists some x ∈ A such that f(x) = y. ... [2] Now, (g∘f)(x) = g(f(x)) = g(y) [from 2] = z [from 1]. Therefore, for every z ∈ C, there exists x ∈ A such that (g∘f)(x) = z. Hence g∘f is ONTO. ✓ CONVERSE — IS IT TRUE?: 'If g∘f is onto, are both f and g onto?' PARTIAL CONVERSE: If g∘f is onto, then g must be onto (but f may NOT be). PROOF that g must be onto: Let z ∈ C. Since g∘f is onto, there exists x ∈ A with (g∘f)(x) = z, i.e., g(f(x)) = z. Let y = f(x) ∈ B. Then g(y) = z. So for every z ∈ C there exists y ∈ B with g(y) = z. Therefore g is onto. COUNTEREXAMPLE that f need not be onto: Let A = {1}, B = {2,3}, C = {4}. Define f: A→B by f(1) = 2 (NOT onto since 3 ∈ B has no preimage). Define g: B→C by g(2) = g(3) = 4 (onto). Then g∘f: A→C with (g∘f)(1) = g(f(1)) = g(2) = 4, so g∘f IS onto. But f is NOT onto. CONCLUSION: The converse is PARTIALLY true — if g∘f is onto, then g is onto. But f need not be onto.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Reflexive: (a,a) ∈ R for all a. Symmetric: (a,b) ⟹ (b,a). Transitive: (a,b) and (b,c) ⟹ (a,c)
•Equivalence = Reflexive + Symmetric + Transitive. Empty relation: NOT reflexive, so NOT equivalence.
•One-one: f(x₁)=f(x₂) ⟹ x₁=x₂. Onto: range = codomain. Bijective = one-one + onto.
•Invertible ⟺ Bijective. If f is bijective, f⁻¹ exists and is also bijective.
•(f∘g)(x) = f(g(x)). Apply g first. Not commutative. (f∘g)⁻¹ = g⁻¹∘f⁻¹.
•Binary operation *: commutative (a*b=b*a), associative (a*(b*c)=(a*b)*c), identity e (a*e=a), inverse b (a*b=e)

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks

Question type	Marks each	Typical count	What it tests
Proof (relations)	3-5	1	Show a relation is/is not an equivalence relation; prove reflexive/symmetric/transitive for a given relation
Proof (functions)	3-5	1	Prove f is one-one and onto; find f⁻¹; prove composition properties

Prep strategy

ALWAYS use the three-step proof structure for equivalence relation: (1) Reflexive → show (a,a) ∈ R. (2) Symmetric → assume (a,b), show (b,a). (3) Transitive → assume (a,b) and (b,c), show (a,c). Each step must be explicit.
For onto proofs: 'Let y be arbitrary in the codomain. We need to find x in the domain such that f(x) = y.' Solve for x, check it's in the domain, verify f(x) = y. This template earns full marks.
Know the composition order: (f∘g)(x) = f(g(x)). Practice with specific numbers to avoid confusion.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Cryptography and Bijective Functions

All encryption and decryption relies on bijective functions. Encryption maps plaintext to ciphertext (one-one: each message maps to a unique encrypted version). Decryption is the inverse function (possible only because the function is bijective). RSA encryption, AES, and all modern cryptography are built on the mathematics of bijective functions studied in this chapter.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Relations: ALWAYS check all three properties (R/S/T) systematically, even if you suspect it's not an equivalence relation. Partial marks are given for each property correctly proved.
Functions: for 'show f is bijective', prove one-one and onto SEPARATELY with clear headers. Then state 'since f is one-one and onto, f is bijective'.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Explore GROUP THEORY: a set with a binary operation satisfying closure, associativity, identity, and inverses is called a GROUP. The integers under addition form a group. This chapter's binary operations are the foundation of abstract algebra.
Investigate Cantor's diagonal argument — it uses the concept of bijection to prove that ℝ (real numbers) and ℕ (natural numbers) cannot be in bijection, meaning ℝ is a 'larger' infinite set. The notion of bijection as the definition of 'same size' for infinite sets is one of mathematics' most profound ideas.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Board (Mathematics)High

JEE Main (Sets, Relations, Functions)High

CUET (Mathematics)High

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

YES. f(x) = eˣ: ℝ → ℝ. One-one: if eˣ¹ = eˣ², then x₁ = x₂. But NOT onto: the value −1 is in the codomain ℝ but there is no real x with eˣ = −1 (exponential is always positive). Range = (0,∞) ≠ ℝ. So f is one-one but not onto.

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