Real Numbers Class 10 Notes & Mind Maps | Chapter 1 Maths (CBSE)

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What Is Class 10 Chapter 1 Real Numbers About?

Real Numbers is the first chapter of Class 10 CBSE Mathematics and one of the highest-scoring topics in the board exam. It builds directly on your Class 9 number system knowledge and introduces two powerful tools — Euclid's Division Algorithm and the Fundamental Theorem of Arithmetic — that you will use throughout secondary and senior secondary maths.

The chapter carries approximately 6 marks in the CBSE Class 10 board paper (Unit 1: Number Systems). Questions typically appear as 1-mark MCQs, 2-mark short answers, and occasionally a 3-mark problem on HCF/LCM. Students who understand the why behind each theorem — not just the formula — consistently score full marks here.

These short notes and mind maps cover every definition, theorem, and result from the NCERT textbook. Work through each section, download the mind map PDF, and use the solved examples to test yourself before your exam.

For detailed step-by-step solutions to every NCERT exercise question, visit the NCERT Solutions for Class 10 Maths page.

Key Topics at a Glance — Chapter 1 Weightage Table

Euclid's Division Lemma and Algorithm — How Does It Work?

Euclid's Division Lemma states: For any two positive integers a and b, there exist unique integers q (quotient) and r (remainder) such that:

a = bq + r, where 0 ≤ r < b

This single statement is the foundation for Euclid's Division Algorithm, which gives you a systematic method to find the HCF of any two positive integers without listing all factors.

Steps of Euclid's Division Algorithm

1. Given two positive integers a and b (where a > b), write: a = bq₁ + r₁
2. If r₁ = 0, then HCF(a, b) = b. Stop.
3. If r₁ ≠ 0, apply the lemma again: b = r₁q₂ + r₂
4. Continue until the remainder becomes 0. The last non-zero remainder is the HCF.

Worked Example

Find HCF(270, 192):

• 270 = 192 × 1 + 78
• 192 = 78 × 2 + 36
• 78 = 36 × 2 + 6
• 36 = 6 × 6 + 0

∴ HCF(270, 192) = 6

Important Points to Remember

• Euclid's algorithm works only for positive integers.
• The algorithm always terminates because the remainder decreases at every step.
• This method is especially efficient for large numbers where listing all factors is impractical.

Board Exam Tip: CBSE frequently asks questions involving numbers of the form 6q, 6q+1, etc. Always begin with Euclid's Division Lemma and list all possible remainders to earn full step marks.

Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic (FTA) is one of the most important results in number theory and forms the foundation of several Class 10 board exam questions.

It states:

Every composite number can be expressed as a product of prime numbers, and this factorisation is unique apart from the order in which the prime factors occur.

Two Important Parts of the Theorem

Existence

Every composite number can be written as a product of prime numbers.

Uniqueness

The prime factorisation of a number is unique except for the order of factors.

Why Is This Theorem Important for CBSE Exams?

The Fundamental Theorem of Arithmetic is used directly in several high-frequency board exam questions.

Key Formula Linking HCF and LCM

HCF(a, b) × LCM(a, b) = a × b

This formula applies only to two positive integers. Many students mistakenly apply it directly to three numbers, which is not valid.

Board Exam Alert: Whenever you calculate HCF and LCM in a question, verify your answer using this formula. This verification step often earns an additional mark.

HCF and LCM Using Prime Factorisation

Prime factorisation is one of the easiest and most reliable methods to calculate HCF and LCM. It is directly based on the Fundamental Theorem of Arithmetic.

How to Find HCF by Prime Factorisation

Steps:

1. Write the prime factorisation of each number.
2. Identify the common prime factors.
3. Select the lowest power of each common prime factor.
4. Multiply them to obtain the HCF.

Worked Example: HCF(12, 18)

• 12 = 2² × 3
• 18 = 2 × 3²

Common prime factors: 2 and 3

HCF = 2¹ × 3¹ = 6

How to Find LCM by Prime Factorisation

Steps:

1. Write the prime factorisation of each number.
2. Select all prime factors appearing in any factorisation.
3. Choose the highest power of each prime factor.
4. Multiply them to obtain the LCM.

Worked Example: LCM(12, 18)

• 12 = 2² × 3
• 18 = 2 × 3²

Highest powers:

LCM = 2² × 3² = 36

Verification Using the Formula

HCF × LCM = 6 × 36

= 216

= 12 × 18 ✓

Quick Comparison: HCF vs LCM

Exam Tip: Students often lose marks by forgetting the verification step:

HCF × LCM = Product of the Numbers

Always write this line after solving HCF and LCM questions.

Rational and Irrational Numbers — What Is the Key Difference?

One of the most important concepts in Chapter 1 is understanding the difference between rational and irrational numbers. Questions based on number classification frequently appear in CBSE board exams and MCQs.

What Is a Rational Number?

A rational number is any number that can be written in the form:

p/q, where p and q are integers and q ≠ 0

Examples

• 3/4
• −7/2
• 0 = 0/1
• 5 = 5/1
• 0.75 = 3/4

What Is an Irrational Number?

An irrational number cannot be expressed in the form p/q.

Decimal expansion is non-terminating and non-repeating.

Examples

• √2
• √3
• √5
• π
• 0.1011011101111...

Quick Comparison: Rational vs Irrational Numbers

How to Prove a Number Is Irrational?

Proofs involving irrational numbers are among the most frequently asked long-answer questions in CBSE Class 10 board examinations.

The standard method uses contradiction and the Fundamental Theorem of Arithmetic.

Proof That √2 Is Irrational

1. Assume √2 = p/q, where p and q are co-prime integers and q ≠ 0.
2. Squaring both sides:
   2 = p²/q²
3. Therefore:
   p² = 2q²
4. Since p² is divisible by 2, p must also be divisible by 2.
5. Let p = 2m.
6. Substituting:
   (2m)² = 2q²
   4m² = 2q²
   q² = 2m²
7. Therefore, q is also divisible by 2.
8. Both p and q are divisible by 2, contradicting the assumption that they are co-prime.
9. Therefore, our assumption is false.

∴ √2 is irrational.

Board Exam Tip: The same proof structure can be used for √3, √5, √7 and other non-perfect square roots. Practise this proof until you can reproduce it without looking at notes.

Decimal Expansions of Rational Numbers

The decimal expansion of a rational number depends entirely on the prime factorisation of its denominator when written in lowest form.

When Is a Decimal Expansion Terminating?

A rational number p/q (in simplest form) has a terminating decimal expansion if and only if:

q = 2ⁿ × 5ᵐ

where n and m are non-negative integers.

Examples

Key Rule to Memorise

If the denominator (in lowest terms) contains any prime factor other than 2 or 5, the decimal expansion is non-terminating and repeating.

This rule allows you to answer most decimal expansion MCQs within seconds.

Chapter 1 Quick Revision Sheet

• Euclid's Division Lemma: a = bq + r
• HCF is obtained using Euclid's Division Algorithm.
• Every composite number has a unique prime factorisation.
• HCF = Product of common prime factors with the lowest powers.
• LCM = Product of all prime factors with the highest powers.
• HCF × LCM = Product of the two numbers.
• Rational numbers can be written as p/q.
• Irrational numbers cannot be expressed as p/q.
• √2, √3, √5 and π are irrational.
• Terminating decimal ⇒ denominator contains only 2 and/or 5.
• Any other prime factor ⇒ non-terminating repeating decimal.

For step-by-step worked solutions on this exact chapter, the NCERT solutions for class 10 Maths page is the natural next stop after this revision.

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