CBSE Class 10 Notes | Maths Chapter 1 Real Numbers (Free PDF)

Title: CBSE Class 10 Notes | Real Numbers Chapter 1 (Maths)
Meta Description: Complete CBSE Class 10 Maths Notes for Chapter 1 Real Numbers. NCERT-based with Euclid's Division Lemma, HCF, LCM, proofs and solved examples. Continue to Chapter 2 Polynomials included.
Next Chapter: Chapter 2 - Polynomials Notes

CBSE Class 10 Maths Notes – Chapter 1: Real Numbers

1. Introduction to Real Numbers

Welcome to your comprehensive CBSE Class 10 Notes for Mathematics Chapter 1 - Real Numbers. This foundational chapter introduces you to the number system and its properties, carrying approximately 6-8 marks in the CBSE Board Examination.

What are Real Numbers?

Real Numbers comprise all rational and irrational numbers. They can be represented on the number line and include:

• Natural Numbers (N): 1, 2, 3, 4, ...
• Whole Numbers (W): 0, 1, 2, 3, ...
• Integers (Z): ..., -2, -1, 0, 1, 2, ...
• Rational Numbers (Q): Numbers of form p/q where q ≠ 0
• Irrational Numbers: Numbers that cannot be expressed as p/q

These CBSE Class 10 Maths Notes are strictly aligned with the latest NCERT syllabus (2024-25) and designed to help you master every concept from Euclid's Division Lemma to decimal expansions.

2. Chapter Overview: CBSE Class 10 Maths Notes

Key Topics Covered:

• Euclid's Division Lemma – Foundation of division algorithm
• Euclid's Division Algorithm – Method to find HCF
• Fundamental Theorem of Arithmetic – Prime factorization uniqueness
• HCF and LCM – Using prime factorization method
• Irrational Numbers – Properties and proofs
• Decimal Expansion of Rational Numbers – Terminating vs non-terminating

🎯 Up Next: After completing Real Numbers, proceed to Chapter 2: Introduction to Polynomials for 4-6 marks weightage content.

3. Key Concepts (NCERT Aligned)

3.1 Euclid's Division Lemma

Statement:

Given 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

Conditions:

• a = Dividend (any positive integer)
• b = Divisor (positive integer, b ≠ 0)
• q = Quotient (unique integer)
• r = Remainder (unique integer, 0 ≤ r < b)

Key Points:

• If r = 0, then b is a divisor of a (b divides a exactly)
• The remainder is always less than the divisor (r < b)
• This lemma forms the basis of the Division Algorithm

3.2 Euclid's Division Algorithm

Definition: A technique to compute the Highest Common Factor (HCF) of two positive integers using Euclid's Division Lemma repeatedly.

Steps to find HCF using Euclid's Division Algorithm:

Step 1: Apply Euclid's Division Lemma to integers a and b (where a > b)
a = bq + r, where 0 ≤ r < b

Step 2: If r = 0, then b is the HCF of a and b.
If r ≠ 0, apply the lemma again to b and r.

Step 3: Continue the process until remainder becomes zero.
The divisor at this stage will be the required HCF.

Example: Find HCF of 420 and 130

Step 1: 420 = 130 × 3 + 30
Step 2: 130 = 30 × 4 + 10
Step 3: 30 = 10 × 3 + 0

HCF = 10 (last non-zero remainder)

3.3 Fundamental Theorem of Arithmetic

Statement:

Every composite number greater than 1 can be expressed as a product of prime numbers, and this factorization is unique apart from the order in which the prime factors occur.

Key Points:

• Also known as the Unique Factorization Theorem
• Prime factorization of a number is unique (except for the order of factors)
• Example: 420 = 2² × 3 × 5 × 7 (only one way to express as product of primes)

3.4 HCF and LCM using Prime Factorization

Method to find HCF:

1. Express each number as product of prime factors
2. Identify common prime factors
3. Take the lowest power of each common prime factor
4. Multiply these together to get HCF

Method to find LCM:

1. Express each number as product of prime factors
2. Take all prime factors (common and uncommon)
3. Take the highest power of each prime factor
4. Multiply these together to get LCM

Important Relationship:

HCF(a, b) × LCM(a, b) = a × b

(For any two positive integers a and b)

Example: Find HCF and LCM of 36 and 48

36 = 2² × 3²
48 = 2⁴ × 3¹

HCF = 2² × 3¹ = 4 × 3 = 12
LCM = 2⁴ × 3² = 16 × 9 = 144

Verification: 12 × 144 = 1728 = 36 × 48 ✓

3.5 Irrational Numbers

Definition: Numbers that cannot be expressed in the form p/q where p and q are integers and q ≠ 0. Their decimal expansions are non-terminating and non-repeating.

Examples: √2, √3, √5, π, e, 0.1010010001...

Theorem: If p is a prime number, then √p is irrational.

Proof that √5 is irrational (Contradiction Method):

Step 1: Assume √5 is rational.

Then √5 = ab where a, b are co-prime integers (HCF = 1) and b ≠ 0.

Step 2: Squaring both sides:

5 = a²b²

⇒ a² = 5b² ... (i)

Step 3: From (i), a² is divisible by 5.

⇒ a is divisible by 5 (since 5 is prime)

Let a = 5c for some integer c.

Step 4: Substituting in (i):

(5c)² = 5b²

25c² = 5b²

5c² = b²

Step 5: From above, b² is divisible by 5.

⇒ b is divisible by 5.

Step 6: From steps 3 and 5, both a and b are divisible by 5.

This contradicts our assumption that a and b are co-prime.

Conclusion: Our assumption is wrong. Therefore, √5 is irrational.

3.6 Decimal Expansion of Rational Numbers

Theorem: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form pq where p and q are co-prime, and the prime factorization of q is of the form 2ⁿ × 5^m, where n and m are non-negative integers.

Condition for Terminating Decimal:

A rational number pq (in lowest terms) has a terminating decimal expansion if and only if the prime factorization of q is of the form 2ⁿ × 5^m.

If q has any prime factor other than 2 or 5, the decimal expansion is non-terminating repeating.

Examples:

Number	Prime Factorization of Denominator	Type of Decimal	Decimal Value
38	8 = 2³	Terminating	0.375
720	20 = 2² × 5	Terminating	0.35
13	3 = 3¹	Non-terminating repeating	0.333...
56	6 = 2 × 3	Non-terminating repeating	0.8333...

4. Important Formulas and Results

Essential Formulas for CBSE Class 10 Maths Notes

1. Euclid's Division Lemma:

a = bq + r, where 0 ≤ r < b

Use when: Finding relationship between dividend, divisor, quotient and remainder

2. HCF and LCM Relationship:

HCF(a, b) × LCM(a, b) = a × b

Use when: Finding LCM when HCF is known, or vice versa

3. Terminating Decimal Condition:

If pq is in lowest terms, decimal terminates iff q = 2ⁿ × 5^m

Use when: Determining if a fraction will have terminating or repeating decimal

4. Prime Factorization Steps:

• Divide by smallest prime (2) until odd
• Then divide by 3, 5, 7, ... successively
• Continue until quotient becomes 1

5. Solved Examples (CBSE Pattern)

Example 1: Euclid's Division Algorithm (Easy)
Question: Find the HCF of 867 and 255 using Euclid's Division Algorithm.

Solution:
Step 1: 867 = 255 × 3 + 102
Step 2: 255 = 102 × 2 + 51
Step 3: 102 = 51 × 2 + 0

Since remainder is 0, the divisor at this step is the HCF.
Therefore, HCF(867, 255) = 51

Example 2: HCF and LCM using Prime Factorization
Question: Find the HCF and LCM of 72, 120, and 168.

Solution:
Prime Factorization:
72 = 2³ × 3²
120 = 2³ × 3 × 5
168 = 2³ × 3 × 7

HCF: Take lowest power of common primes
HCF = 2³ × 3 = 8 × 3 = 24

LCM: Take highest power of all primes
LCM = 2³ × 3² × 5 × 7 = 8 × 9 × 5 × 7 = 2520

Example 3: Decimal Expansion Identification
Question: Without performing long division, determine whether the following rational numbers have terminating or non-terminating repeating decimal expansions:
(i) 13125 (ii) 178 (iii) 775

Solution:

(i) 13125:
125 = 5³ = 2⁰ × 5³
Since denominator is of form 2ⁿ × 5^m, decimal expansion is terminating.

(ii) 178:
8 = 2³ = 2³ × 5⁰
Since denominator is of form 2ⁿ × 5^m, decimal expansion is terminating.

(iii) 775:
75 = 3 × 25 = 3 × 5²
Since denominator contains prime factor 3 (other than 2 and 5), decimal expansion is non-terminating repeating.

Example 4: Irrational Proof Question
Question: Prove that 3 + 2√5 is irrational.

Solution:
Step 1: Assume 3 + 2√5 is rational.

Step 2: Then 3 + 2√5 = ab where a, b are co-prime integers, b ≠ 0.

Step 3: Rearranging:
2√5 = ab - 3 = a - 3bb
√5 = a - 3b2b

Step 4: Since a and b are integers, a - 3b2b is rational.
This implies √5 is rational.

Step 5: But this contradicts the fact that √5 is irrational.

Conclusion: Our assumption is wrong. Therefore, 3 + 2√5 is irrational.

Example 5: Board-Level Mixed Question
Question: The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other number.

Solution:
Given:
HCF = 23
LCM = 1449
First number (a) = 161
Let second number = b

Using the formula:
HCF × LCM = a × b
23 × 1449 = 161 × b
33327 = 161 × b
b = 33327161 = 207

Therefore, the other number is 207.

6. Smart Tricks and Memory Aids

Quick HCF Trick

For two consecutive integers, HCF is always 1.

For two consecutive even integers, HCF is always 2.

For two consecutive odd integers, HCF is always 1.

Decimal Expansion Quick-Check Trick

Look at the denominator (in lowest terms):

• If it ends in 0, 2, 4, 5, 6, 8 → Check if only 2 and 5 are prime factors
• If it ends in 1, 3, 7, 9 → Likely non-terminating (check for other factors)
• Shortcut: If denominator = 2ⁿ × 5^m, decimal terminates

Proof Writing Presentation Trick

Always structure irrational proofs in 6 steps:

1. Assume number is rational (= p/q)
2. Manipulate to isolate irrational part
3. Show the other side is rational
4. Conclude irrational part is rational
5. State the contradiction
6. Conclude original assumption is false

Time-Saving Board Strategy

• Always write the statement of theorem before solving
• For HCF by Euclid's algorithm, show each step clearly
• In prime factorization, use division method for speed
• For terminating decimal questions, don't actually divide - just check denominator
• Box your final answer for visibility

7. Visual Learning: Important Diagrams

Diagram 1: Euclid's Division Algorithm Flowchart

Key Steps in Flowchart:

• Start with two numbers a and b (a > b)
• Apply division: a = bq + r
• Check if r = 0
• If yes, b is HCF
• If no, replace a with b, b with r, and repeat

Diagram 2: Prime Factorization Tree

How to use Factor Tree:

1. Write the number at top
2. Break into any two factors
3. Continue breaking composite factors
4. Stop when all factors are prime
5. Multiply all prime factors at the end

Diagram 3: Decimal Expansion Classification

Diagram 4: HCF and LCM by Prime Factorization

8. Most Important Board Questions

1 Mark Questions (Very Short Answer)

Q1. State Euclid's Division Lemma.
Ans: For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

Q2. What is the HCF of two consecutive integers?
Ans: 1

Q3. Write the condition for a rational number to have terminating decimal expansion.
Ans: In the rational number p/q (in lowest terms), the prime factorization of q must be of the form 2ⁿ × 5^m, where n and m are non-negative integers.

Q4. If HCF(26, 169) = 13, find LCM(26, 169).
Ans: HCF × LCM = 26 × 169
13 × LCM = 4394
LCM = 338

Q5. Is 7 × 11 × 13 + 11 a composite number? Justify.
Ans: Yes, because 7 × 11 × 13 + 11 = 11(7 × 13 + 1) = 11 × 92, which has factors other than 1 and itself.

2-3 Mark Questions (Short Answer)

Q1. Find the HCF of 4052 and 12576 using Euclid's Division Algorithm.
Ans:
12576 = 4052 × 3 + 420
4052 = 420 × 9 + 272
420 = 272 × 1 + 148
272 = 148 × 1 + 124
148 = 124 × 1 + 24
124 = 24 × 5 + 4
24 = 4 × 6 + 0
HCF = 4

Q2. Find the LCM and HCF of 510 and 92. Verify that LCM × HCF = Product of the two numbers.
Ans:
510 = 2 × 3 × 5 × 17
92 = 2² × 23
HCF = 2
LCM = 2² × 3 × 5 × 17 × 23 = 23460

Verification:
LCM × HCF = 23460 × 2 = 46920
Product = 510 × 92 = 46920
LHS = RHS (Verified)

Q3. Check whether 6ⁿ can end with the digit 0 for any natural number n.
Ans:
If 6ⁿ ends with 0, it must be divisible by 5.
6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
Prime factorization of 6ⁿ contains only 2 and 3.
It does not contain 5 as a factor.
Therefore, 6ⁿ cannot end with digit 0 for any natural number n.

4-5 Mark Questions (Long Answer/Proofs)

Q1. Prove that √3 is irrational.
Ans:
Step 1: Assume √3 is rational.
Then √3 = ab where a, b are co-prime integers, b ≠ 0.

Step 2: Squaring: 3 = a²b²
⇒ a² = 3b² ... (i)

Step 3: From (i), a² is divisible by 3 ⇒ a is divisible by 3.
Let a = 3c.

Step 4: Substituting: (3c)² = 3b²
9c² = 3b²
3c² = b²

Step 5: b² is divisible by 3 ⇒ b is divisible by 3.

Step 6: Both a and b are divisible by 3, contradicting that they are co-prime.

Conclusion: √3 is irrational.

Q2. Prove that 5 - √3 is irrational.
Ans:
Step 1: Assume 5 - √3 is rational = ab.

Step 2: Then √3 = 5 - ab = 5b - ab.

Step 3: Since a, b are integers, RHS is rational.
This implies √3 is rational.

Step 4: This contradicts the fact that √3 is irrational.

Conclusion: 5 - √3 is irrational.

Q3. Three bells toll at intervals of 9, 12, and 15 minutes respectively. If they start tolling together, after how many minutes will they next toll together?
Ans:
Find LCM of 9, 12, and 15.

9 = 3²
12 = 2² × 3
15 = 3 × 5

LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes (or 3 hours)

Case-Study Based Question (CBSE Pattern)

Read the following and answer the questions:

A mathematics teacher asked students to find the HCF of two numbers 1260 and 1386. Student A used the prime factorization method, while Student B used Euclid's Division Algorithm.

(a) Find the HCF using both methods.

Method 1 - Prime Factorization:
1260 = 2² × 3² × 5 × 7
1386 = 2 × 3² × 7 × 11
HCF = 2 × 3² × 7 = 126

Method 2 - Euclid's Algorithm:
1386 = 1260 × 1 + 126
1260 = 126 × 10 + 0
HCF = 126

(b) If the teacher wants to arrange 1260 books and 1386 pens into identical stacks, what is the maximum number of items in each stack?

Maximum items in each stack = HCF(1260, 1386) = 126
Number of book stacks = 1260 ÷ 126 = 10
Number of pen stacks = 1386 ÷ 126 = 11

(c) Find the LCM of 1260 and 1386.

Using HCF × LCM = Product
126 × LCM = 1260 × 1386
LCM = 1260 × 1386126 = 10 × 1386 = 13860

9. Common Mistakes to Avoid

Mistake 1: Writing Wrong Remainder Condition

• Wrong: Writing 0 < r ≤ b or r > b
• Correct: Always write 0 ≤ r < b (remainder is less than divisor, can be zero)

Mistake 2: Incorrect Prime Factorization

• Not dividing completely until all factors are prime
• Forgetting that 1 is not a prime number
• Tip: Always verify by multiplying factors to get original number

Mistake 3: Incomplete Proof Steps

• Not stating the assumption clearly
• Forgetting to mention that a and b are co-prime
• Missing the contradiction statement
• Remember: Every proof must end with a clear conclusion

Mistake 4: Not Writing Theorem Statement Properly

• Forgetting to mention "unique integers q and r"
• Missing the condition "0 ≤ r < b"
• Always quote the complete statement with all conditions

Mistake 5: Decimal Expansion Errors

• Not reducing fraction to lowest terms before checking denominator
• Forgetting that denominator must contain only 2 and 5 for terminating decimals
• Example: 720 terminates (20 = 2² × 5), but 730 doesn't (30 = 2 × 3 × 5)

10. Practice Section

MCQs (Multiple Choice Questions)

Q1. The decimal expansion of 178 will terminate after how many places?
(a) 1 (b) 2 (c) 3 (d) 4
Ans: (c) 3
Explanation: 8 = 2³, so it terminates after 3 decimal places. (17/8 = 2.125)

Q2. If HCF(a, b) = 12 and LCM(a, b) = 180, and a = 36, then b = ?
(a) 60 (b) 54 (c) 45 (d) 90
Ans: (a) 60
Explanation: 12 × 180 = 36 × b ⇒ b = 2160/36 = 60

Q3. Which of the following is not an irrational number?
(a) 2 + √3 (b) √4 + √5 (c) √2 × √8 (d) √7
Ans: (c) √2 × √8
Explanation: √2 × √8 = √16 = 4, which is rational.

Q4. The HCF of two consecutive odd numbers is:
(a) 1 (b) 2 (c) 0 (d) undefined
Ans: (a) 1

Assertion-Reason Questions

Q1. Assertion (A): The HCF of two numbers is 5 and their LCM is 150. If one number is 25, then the other number is 30.
Reason (R): For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Ans: (a) Both true and R explains A
Verification: 5 × 150 = 25 × 30 ⇒ 750 = 750 ✓

Q2. Assertion (A): 13/3125 is a terminating decimal.
Reason (R): If q = 2ⁿ × 5^m where n, m are non-negative integers, then p/q is a terminating decimal.
Ans: (a) Both true and R explains A
Explanation: 3125 = 5⁵, so it satisfies the condition.

HOTS (Higher Order Thinking Skills)

Q1. If the HCF of 65 and 117 is expressible in the form 65m - 117, find the value of m.
Ans:
Using Euclid's algorithm:
117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0
HCF = 13

Given: 65m - 117 = 13
65m = 130
m = 2

Q2. Prove that one of every three consecutive positive integers is divisible by 3.
Ans:
Let the three consecutive integers be n, n+1, n+2.
By Euclid's Division Lemma, n = 3q + r where 0 ≤ r < 3.

Case 1: If r = 0, then n = 3q (divisible by 3)
Case 2: If r = 1, then n+2 = 3q + 3 = 3(q+1) (divisible by 3)
Case 3: If r = 2, then n+1 = 3q + 3 = 3(q+1) (divisible by 3)

In all cases, one of the three consecutive integers is divisible by 3.

11. Frequently Asked Questions (FAQ)

Q: What is Euclid's Division Lemma?

Euclid's Division Lemma states that 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 lemma is the foundation of the division algorithm and is used to find the HCF of two numbers.

Q: How to identify terminating decimals?

To identify if a rational number p/q has a terminating decimal expansion:

1. First, reduce the fraction to its lowest terms
2. Find the prime factorization of the denominator q
3. If q = 2ⁿ × 5^m (where n and m are whole numbers), the decimal terminates
4. If q has any prime factor other than 2 or 5, the decimal is non-terminating repeating

Q: Are these CBSE Class 10 Maths Notes based on NCERT?

Yes, these CBSE Class 10 Maths Notes are 100% aligned with the latest NCERT syllabus (2024-25). All theorems, proofs, and solved examples follow NCERT guidelines strictly. These notes cover every topic from Chapter 1 Real Numbers including Euclid's Division Lemma, Fundamental Theorem of Arithmetic, HCF, LCM, irrational numbers, and decimal expansions.

Q: What is the relationship between HCF and LCM?

For any two positive integers a and b:

HCF(a, b) × LCM(a, b) = a × b

This relationship is very useful when you know one of HCF or LCM and need to find the other. It is also helpful in verifying your calculations.

Q: How many marks does Chapter 1 carry in CBSE Board Exam?

Chapter 1 Real Numbers carries approximately 6-8 marks in the CBSE Class 10 Mathematics Board Examination. This typically includes 1-mark MCQs, 2-mark short answer questions, and 3-4 mark long answer questions including proofs.

12. What's Next: Introduction to Polynomials (Chapter 2 Preview)

🎓 Chapter 2: Introduction to Polynomials

After mastering Real Numbers, the next logical step in your CBSE Class 10 Maths journey is Chapter 2 - Introduction to Polynomials. This chapter builds upon your algebraic foundation and carries 4-6 marks in board exams.

Key Topics You'll Learn:

• Definition & Types: Linear, Quadratic, and Cubic polynomials
• Zeroes of Polynomial: Finding roots and their geometric meaning
• Critical Relationship: Zeroes and Coefficients connection (α + β = -b/a, αβ = c/a)
• Graphical Interpretation: Understanding parabolas and x-intercepts
• Division Algorithm: Dividing polynomials and finding factors

🔗 Connection with Chapter 1:

Just as you learned about factors in Real Numbers (HCF/LCM), Polynomials extends this concept to algebraic expressions. The factorization techniques you mastered in Chapter 1 will help you factorize polynomials like x² - 5x + 6 into (x-2)(x-3).

📖 Read Complete Chapter 2 Notes →

13. Conclusion

Mastering Real Numbers is crucial for building a strong foundation in mathematics for your CBSE Class 10 Board Examination. This chapter introduces fundamental concepts that are essential for advanced mathematics in higher classes.

Key Takeaways from CBSE Class 10 Maths Notes:

• Master Euclid's Division Lemma and its application in finding HCF
• Understand the Fundamental Theorem of Arithmetic for prime factorization
• Practice the relationship HCF × LCM = Product of numbers
• Learn the condition for terminating decimal expansions
• Master the contradiction method for proving irrationality
• Solve all NCERT examples and exercises thoroughly

Revision Strategy:

1. Revise these CBSE Class 10 Notes regularly
2. Practice Euclid's Division Algorithm with different numbers
3. Memorize the proofs of irrationality step-by-step
4. Solve previous year board questions (at least 5 years)
5. Attempt sample papers under exam conditions
6. Focus on presentation in proofs and algorithms