CBSE Class 10 Maths Revision Notes Chapter 1 Real Numbers

CBSE Class 10 Maths Revision Notes Chapter 1 explain Real Numbers through prime factorisation, HCF, LCM and irrationality proofs. For CBSE 2026 preparation, Chapter 1 builds the number theory base used in arithmetic reasoning and proof questions.

Real Numbers in Class 10 starts from a simple question: how can a positive integer be broken into prime factors? NCERT answers this through the Fundamental Theorem of Arithmetic, which says that every composite number has a unique prime factorisation. 

This rule explains why HCF takes the smallest common prime powers and LCM takes the greatest powers present in the numbers. It also supports proof-based questions, such as proving √2 irrational through contradiction. 

In CBSE 2026 Maths preparation, Chapter 1 builds the exact number logic needed for factorisation, divisibility and irrationality questions.

Key Takeaways

• 32760 factorisation: NCERT writes 32760 as 2³ × 3² × 5 × 7 × 13 to show unique prime factorisation.
• Two-number relation: HCF(a, b) × LCM(a, b) = a × b applies only to two positive integers.
• 6ⁿ ending in zero: 6ⁿ cannot end with 0 because its prime factorisation has no factor 5.
• √2 contradiction: The proof of √2 irrational shows both a and b become divisible by 2.

CBSE Class 10 Maths Revision Notes Chapter 1 Structure 2026

Real Numbers Chapter 1 NCERT Scope for 2026

Chapter 1 in the 2026 NCERT Class 10 Mathematics textbook studies positive integers through divisibility and prime factors. It connects Class 9 number systems with Class 10 proof-based arithmetic.

The chapter covers the Fundamental Theorem of Arithmetic, HCF-LCM by prime factorisation and irrationality of numbers like √2, √3 and √5. These three areas form the working base of real numbers class 10 revision notes.

What are real numbers in Class 10 Maths?

Real numbers are all numbers that can be shown on the number line. They include natural numbers, whole numbers, integers, rational numbers and irrational numbers.

Examples:

1. 5 is a natural number and a real number.
2. -7 is an integer and a real number.
3. 3/4 is a rational number and a real number.
4. √2 is an irrational number and a real number.

A rational number can be written as p/q, where p and q are integers and q ≠ 0. An irrational number has no p/q form.

Fundamental Theorem of Arithmetic Class 10

The Fundamental Theorem of Arithmetic says every composite number can be expressed as a product of primes uniquely. The prime factors stay fixed when their order changes.

This theorem gives Class 10 Real Numbers its main calculation rule. It supports prime factorisation, HCF-LCM and irrationality proofs.

Statement of Fundamental Theorem of Arithmetic

Every composite number can be expressed as a product of primes. This factorisation is unique, apart from the order of prime factors.

Example:

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13

32760 = 2³ × 3² × 5 × 7 × 13

The prime factors are 2, 3, 5, 7 and 13. Their powers show how many times each prime factor appears.

Prime and composite numbers

A prime number has exactly two factors: 1 and itself. Examples include 2, 3, 5, 7, 11 and 13.

A composite number has more than two factors. Examples include 4, 6, 8, 9, 10 and 12.

The number 1 has only one factor. It is neither prime nor composite.

Prime factorisation by factor tree

Prime factorisation breaks a composite number into prime numbers. A factor tree continues until every branch ends in a prime.

Example: Express 140 as prime factors.

140 = 2 × 70
70 = 2 × 35
35 = 5 × 7

So,

140 = 2 × 2 × 5 × 7
140 = 2² × 5 × 7

The final answer is 2² × 5 × 7.

HCF and LCM Class 10 by Prime Factorisation

HCF takes the smallest powers of common prime factors. LCM takes the greatest powers of all prime factors present in the given numbers.

This method works because prime factorisation is unique. Once the prime factors are written correctly, the HCF and LCM follow fixed rules.

HCF by prime factorisation

HCF means Highest Common Factor. It is the greatest number that divides all given numbers exactly.

Example: Find the HCF of 96 and 404.

96 = 2⁵ × 3
404 = 2² × 101

Common prime factor = 2
Smallest power of 2 = 2²

HCF(96, 404) = 2²
HCF(96, 404) = 4

LCM by prime factorisation

LCM means Least Common Multiple. It is the smallest number divisible by every given number.

Example: Find the LCM of 6, 72 and 120.

6 = 2 × 3
72 = 2³ × 3²
120 = 2³ × 3 × 5

Greatest powers = 2³, 3² and 5

LCM(6, 72, 120) = 2³ × 3² × 5
LCM(6, 72, 120) = 8 × 9 × 5
LCM(6, 72, 120) = 360

HCF and LCM formula for two positive integers

For two positive integers a and b:

HCF(a, b) × LCM(a, b) = a × b

Example:

6 = 2 × 3
20 = 2² × 5

HCF(6, 20) = 2
LCM(6, 20) = 2² × 3 × 5
LCM(6, 20) = 60

HCF × LCM = 2 × 60
HCF × LCM = 120

Product of numbers = 6 × 20
Product of numbers = 120

So,

HCF(6, 20) × LCM(6, 20) = 6 × 20

Real Numbers Class 10 Notes on Irrational Numbers

An irrational number cannot be written as p/q, where p and q are integers and q ≠ 0. NCERT proves irrationality in Chapter 1 using proof by contradiction.

The proof begins by assuming the number is rational. The calculation then creates a contradiction with coprime numbers.

Prime divisibility result used in irrationality proofs

If p is a prime number and p divides a², then p divides a.

Example:

If 2 divides a², then 2 divides a.

This result is used in the proof of √2. It connects square numbers with prime factors.

Proof that √2 is irrational

√2 is irrational because the rational assumption makes both numerator and denominator divisible by 2.

Step 1: Assume √2 is rational.

√2 = a/b, where a and b are coprime and b ≠ 0.

Step 2: Square both sides.

2 = a²/b²

a² = 2b²

Step 3: Since a² is divisible by 2, a is divisible by 2.

Let a = 2c.

Step 4: Substitute a = 2c.

(2c)² = 2b²

4c² = 2b²

b² = 2c²

Step 5: Since b² is divisible by 2, b is divisible by 2.

So, a and b have 2 as a common factor.

Step 6: This contradicts the fact that a and b are coprime.

Therefore, √2 is irrational.

Proof pattern for √3

√3 is irrational because the rational assumption makes both numerator and denominator divisible by 3.

Step 1: Assume √3 is rational.

√3 = a/b, where a and b are coprime and b ≠ 0.

Step 2: Square both sides.

3 = a²/b²

a² = 3b²

Step 3: Since a² is divisible by 3, a is divisible by 3.

Let a = 3c.

Step 4: Substitute a = 3c.

(3c)² = 3b²

9c² = 3b²

b² = 3c²

Step 5: Since b² is divisible by 3, b is divisible by 3.

So, a and b have 3 as a common factor.

Step 6: This contradicts the fact that a and b are coprime.

Therefore, √3 is irrational.

Proof pattern for √5

√5 is irrational because the rational assumption makes both numerator and denominator divisible by 5.

Step 1: Assume √5 is rational.

√5 = a/b, where a and b are coprime and b ≠ 0.

Step 2: Square both sides.

5 = a²/b²

a² = 5b²

Step 3: Since a² is divisible by 5, a is divisible by 5.

Let a = 5c.

Step 4: Substitute a = 5c.

(5c)² = 5b²

25c² = 5b²

b² = 5c²

Step 5: Since b² is divisible by 5, b is divisible by 5.

So, a and b have 5 as a common factor.

Step 6: This contradicts the fact that a and b are coprime.

Therefore, √5 is irrational.

Class 10 Real Numbers Formulas and Results

CBSE Class 10 Maths Chapter 1 uses a small set of formulas and theorem results. These formulas appear in HCF-LCM calculations, factor questions and proof steps.

HCF and LCM relation

HCF(a, b) × LCM(a, b) = a × b

Use this relation for two positive integers. It helps when HCF is given and LCM is required.

Example:

HCF(306, 657) = 9

LCM(306, 657) = (306 × 657)/9

306 × 657 = 201042

LCM = 201042/9

LCM = 22338

HCF from prime factors

HCF = Product of the smallest powers of common prime factors.

Example:

72 = 2³ × 3²
120 = 2³ × 3 × 5

Common prime factors = 2 and 3
Smallest powers = 2³ and 3

HCF(72, 120) = 2³ × 3
HCF(72, 120) = 24

LCM from prime factors

LCM = Product of the greatest powers of all prime factors involved.

Example:

72 = 2³ × 3²
120 = 2³ × 3 × 5

Greatest powers = 2³, 3² and 5

LCM(72, 120) = 2³ × 3² × 5
LCM(72, 120) = 360

NCERT-Style Real Numbers Questions for Quick Practice

NCERT Chapter 1 questions often use exact number properties. Factorisation, HCF-LCM verification and number-ending logic are frequent school-test patterns.

Q1. Express 3825 as a product of prime factors.

3825 = 3² × 5² × 17.

Steps:

3825 = 3 × 1275
1275 = 3 × 425
425 = 5 × 85
85 = 5 × 17

So,

3825 = 3 × 3 × 5 × 5 × 17
3825 = 3² × 5² × 17

Q2. Find the HCF and LCM of 26 and 91.

HCF = 13 and LCM = 182.

Steps:

26 = 2 × 13
91 = 7 × 13

HCF = 13

LCM = 2 × 7 × 13
LCM = 182

Verification:

HCF × LCM = 13 × 182
HCF × LCM = 2366

Product = 26 × 91
Product = 2366

Q3. Given HCF(306, 657) = 9, find LCM(306, 657).

LCM(306, 657) = 22338.

Steps:

HCF × LCM = Product of two numbers

9 × LCM = 306 × 657

LCM = (306 × 657)/9

LCM = 201042/9

LCM = 22338

Q4. Can 6ⁿ end with digit 0 for any natural number n?

6ⁿ can never end with digit 0 for any natural number n.

Reason:

A number ending in 0 is divisible by 10.

10 = 2 × 5

6ⁿ = (2 × 3)ⁿ

The prime factors of 6ⁿ are only 2 and 3.

The prime factor 5 is required for a factor 10.

Therefore, 6ⁿ cannot end with digit 0.

Important Terms in CBSE Class 10 Maths Chapter 1 Real Numbers

Chapter 1 uses exact mathematical terms in definitions and proofs. These meanings help in one-mark answers, MCQs and proof-based questions.

Natural number

Natural numbers are counting numbers.

Examples: 1, 2, 3, 4, 5

Prime number

A prime number has exactly two factors: 1 and itself.

Examples: 2, 3, 5, 7, 11

Composite number

A composite number has more than two factors.

Examples: 4, 6, 8, 9, 10

Coprime numbers

Two numbers are coprime if their HCF is 1.

Example: 8 and 15 are coprime.

Rational number

A rational number can be written as p/q, where p and q are integers and q ≠ 0.

Examples: 2/3, -5/7, 8

Irrational number

An irrational number cannot be written as p/q, where p and q are integers and q ≠ 0.

Examples: √2, √3, √5

Prime factorisation

Prime factorisation means writing a composite number as a product of prime numbers.

Example:

84 = 2² × 3 × 7

Useful Links for Class 10 Maths