These NCERT Solutions Class 10 Maths Chapter 1 Real Numbers provide complete, step-by-step answers to every exercise question in the 2025-26 CBSE syllabus. Whether you are preparing for your Class 10 board exam or building a strong foundation in number theory, this guide covers Euclid’s Division Lemma, the Fundamental Theorem of Arithmetic, HCF and LCM, and proofs of irrationality — all explained with full working and CBSE marking-scheme alignment. For all chapters, visit our NCERT Solutions for Class 10 hub, or explore the complete NCERT Solutions library. The official textbook is also available at the NCERT official website. This page covers the updated 2025-26 syllabus and exam pattern.
- Euclid’s Division Lemma: for any two positive integers a and b, there exist unique integers q and r such that \( a = bq + r,\ 0 \leq r < b \).
- The Euclid’s Division Algorithm repeatedly applies the lemma to find HCF.
- Fundamental Theorem of Arithmetic: every composite number has a unique prime factorisation.
- \( \text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b \) for any two positive integers.
- \( \sqrt{2},\ \sqrt{3},\ \sqrt{5} \) are irrational — proved by contradiction.
- A rational number \( \frac{p}{q} \) has a terminating decimal if and only if the prime factorisation of q contains only 2s and 5s.
- Non-terminating repeating decimals are rational; non-terminating non-repeating decimals are irrational.
Table of Contents
NCERT Solutions Class 10 Maths Chapter 1 Real Numbers — Chapter Overview
Chapter 1 of the NCERT Class 10 Mathematics textbook (published by NCERT, 2025-26 edition) introduces students to the deeper properties of real numbers — a concept first encountered in earlier classes. The chapter extends the understanding of rational and irrational numbers through two powerful tools: Euclid’s Division Lemma and the Fundamental Theorem of Arithmetic. It also explores how the prime factorisation of a denominator determines whether a rational number has a terminating or non-terminating decimal expansion.
Why it matters for CBSE Board Exams: The Number Systems unit (Unit 1) carries approximately 6 marks in the CBSE Class 10 Maths board exam 2025-26. Questions from this chapter appear as 1-mark MCQs or assertion-reason questions, 2-mark short answers, and occasionally a 3-mark proof or HCF/LCM problem. Mastering this chapter also builds the algebraic foundation needed for polynomials, quadratic equations, and arithmetic progressions later in the syllabus.
Prerequisites: Students should be comfortable with prime numbers, divisibility rules, factors and multiples from Class 6–8, and basic properties of rational and irrational numbers from Class 9.
Key Concepts and Theorems
Euclid’s Division Lemma
What it is: For any two positive integers \( a \) and \( b \), there exist unique integers \( q \) (quotient) and \( r \) (remainder) such that:
\[ a = bq + r, \quad 0 \leq r < b \]
Real-world analogy: If you distribute 17 chocolates equally among 5 children, each child gets 3 (quotient) and 2 chocolates are left over (remainder). So \( 17 = 5 \times 3 + 2 \).
Where it connects: This lemma is the foundation of Euclid’s Division Algorithm used to compute HCF efficiently — a method attributed to the ancient Greek mathematician Euclid.
Common mistake: Students sometimes write \( r \leq b \) instead of \( r < b \). The remainder must always be strictly less than the divisor.
Euclid’s Division Algorithm for HCF
How it works: To find \( \text{HCF}(a, b) \) where \( a > b \), apply the lemma repeatedly:
- Write \( a = bq_1 + r_1 \). If \( r_1 = 0 \), HCF \( = b \).
- If \( r_1 \neq 0 \), write \( b = r_1 q_2 + r_2 \). If \( r_2 = 0 \), HCF \( = r_1 \).
- Continue until remainder \( = 0 \). The last non-zero remainder is the HCF.
Why does this work? Each step reduces the pair of numbers while preserving their HCF. Since remainders strictly decrease, the process always terminates.
Fundamental Theorem of Arithmetic
Statement: Every composite number can be expressed as a product of primes in exactly one way (apart from the order of factors). This unique representation is called the prime factorisation of the number.
Example: \( 360 = 2^3 \times 3^2 \times 5 \). No other combination of primes gives 360.
Why it matters: This theorem guarantees that HCF and LCM calculations via prime factorisation always yield a unique, correct answer. It is also the key tool used to prove that certain numbers like \( \sqrt{2} \) are irrational.
HCF and LCM Using Prime Factorisation
Once you have the prime factorisations of two numbers:
- HCF = product of common prime factors with their lowest powers.
- LCM = product of all prime factors with their highest powers.
- Key relationship: \( \text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b \)
For three or more numbers, the HCF × LCM relationship does NOT hold directly — a common exam trap.
Irrational Numbers and Proofs
An irrational number cannot be expressed as \( \frac{p}{q} \) where \( p, q \) are integers and \( q \neq 0 \). The NCERT syllabus requires students to prove irrationality using proof by contradiction combined with the Fundamental Theorem of Arithmetic.
Standard proof structure for \( \sqrt{2} \): Assume \( \sqrt{2} = \frac{p}{q} \) in lowest terms. Then \( p^2 = 2q^2 \), so 2 divides \( p^2 \), hence 2 divides \( p \). Write \( p = 2m \); then \( 4m^2 = 2q^2 \Rightarrow q^2 = 2m^2 \), so 2 divides \( q \). This contradicts \( p \) and \( q \) being co-prime. Hence \( \sqrt{2} \) is irrational. The same method applies to \( \sqrt{3}, \sqrt{5} \), and combinations like \( 3 + 2\sqrt{5} \).
Rational Numbers and Decimal Expansions
A rational number \( \frac{p}{q} \) (in lowest terms) has:
- A terminating decimal if and only if \( q = 2^m \times 5^n \) for non-negative integers \( m, n \).
- A non-terminating repeating decimal if \( q \) has prime factors other than 2 and 5.
Irrational numbers have non-terminating, non-repeating decimal expansions (e.g., \( \pi, \sqrt{2} \)).
Exercise Solutions — Step by Step
Exercise 1.1 Solutions
Question 1
Use Euclid’s division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Step 1: Since 225 > 135, apply Euclid’s lemma to 225 and 135:
\[ 225 = 135 \times 1 + 90 \]
Step 2: Remainder \( 90 \neq 0 \), so apply to 135 and 90:
\[ 135 = 90 \times 1 + 45 \]
Step 3: Remainder \( 45 \neq 0 \), so apply to 90 and 45:
\[ 90 = 45 \times 2 + 0 \]
Why does this work? The remainder becomes 0, so the last non-zero remainder is the HCF.
\( \therefore \) HCF(135, 225) = 45
Step 1: 38220 > 196. Apply lemma:
\[ 38220 = 196 \times 195 + 0 \]
Why does this work? The remainder is 0 at the very first step, meaning 196 divides 38220 exactly.
\( \therefore \) HCF(196, 38220) = 196
Step 1: 867 > 255. Apply lemma:
\[ 867 = 255 \times 3 + 102 \]
Step 2: Apply to 255 and 102:
\[ 255 = 102 \times 2 + 51 \]
Step 3: Apply to 102 and 51:
\[ 102 = 51 \times 2 + 0 \]
\( \therefore \) HCF(867, 255) = 51
Question 2
Show that any positive odd integer is of the form \( 6q + 1 \), \( 6q + 3 \), or \( 6q + 5 \), where \( q \) is some integer.
Step 1: Let \( a \) be any positive integer and \( b = 6 \). By Euclid’s Division Lemma:
\[ a = 6q + r, \quad 0 \leq r < 6 \]
So \( r \) can be 0, 1, 2, 3, 4, or 5, giving: \( a = 6q,\ 6q+1,\ 6q+2,\ 6q+3,\ 6q+4,\ 6q+5 \).
Step 2: Identify odd values. \( 6q = 2(3q) \) — even. \( 6q+2 = 2(3q+1) \) — even. \( 6q+4 = 2(3q+2) \) — even.
Step 3: The remaining forms \( 6q+1,\ 6q+3,\ 6q+5 \) cannot be expressed as \( 2k \) for any integer \( k \), so they are odd.
\( \therefore \) Any positive odd integer is of the form \( 6q+1,\ 6q+3, \) or \( 6q+5 \). Proved.
Exercise 1.2 Solutions
Question 1
Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
\( 140 = 2^2 \times 5 \times 7 \)
\( 156 = 2^2 \times 3 \times 13 \)
\( 3825 = 3^2 \times 5^2 \times 17 \)
\( 5005 = 5 \times 7 \times 11 \times 13 \)
\( 7429 = 17 \times 19 \times 23 \)
Question 2
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers: (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Step 1: Prime factorisations: \( 26 = 2 \times 13 \) and \( 91 = 7 \times 13 \).
Step 2: \( \text{HCF} = 13 \) (common factor with lowest power). \( \text{LCM} = 2 \times 7 \times 13 = 182 \).
Verification: \( \text{HCF} \times \text{LCM} = 13 \times 182 = 2366 \) and \( 26 \times 91 = 2366 \). ✓
HCF = 13, LCM = 182
Step 1: \( 510 = 2 \times 3 \times 5 \times 17 \) and \( 92 = 2^2 \times 23 \).
Step 2: \( \text{HCF} = 2 \). \( \text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460 \).
Verification: \( 2 \times 23460 = 46920 \) and \( 510 \times 92 = 46920 \). ✓
HCF = 2, LCM = 23460
Step 1: \( 336 = 2^4 \times 3 \times 7 \) and \( 54 = 2 \times 3^3 \).
Step 2: \( \text{HCF} = 2^1 \times 3^1 = 6 \). \( \text{LCM} = 2^4 \times 3^3 \times 7 = 3024 \).
Verification: \( 6 \times 3024 = 18144 \) and \( 336 \times 54 = 18144 \). ✓
HCF = 6, LCM = 3024
Exercise 1.3 Solutions
Question 1
Prove that \( \sqrt{5} \) is irrational.
Step 1 (Assumption): Assume \( \sqrt{5} \) is rational. Then \( \sqrt{5} = \frac{p}{q} \) where \( p \) and \( q \) are co-prime integers and \( q \neq 0 \).
Step 2: Squaring both sides: \( 5q^2 = p^2 \). So 5 divides \( p^2 \).
Step 3: By the Fundamental Theorem of Arithmetic, if a prime \( p \) divides \( a^2 \), it divides \( a \). So 5 divides \( p \). Write \( p = 5m \).
Step 4: Substituting: \( 5q^2 = 25m^2 \Rightarrow q^2 = 5m^2 \). So 5 divides \( q^2 \), hence 5 divides \( q \).
Step 5 (Contradiction): Both \( p \) and \( q \) are divisible by 5, contradicting the assumption that they are co-prime.
\( \therefore \) Our assumption is false. \( \sqrt{5} \) is irrational. Proved.
Question 2
Prove that \( 3 + 2\sqrt{5} \) is irrational.
Step 1: Assume \( 3 + 2\sqrt{5} \) is rational. Then \( 3 + 2\sqrt{5} = \frac{p}{q} \) for some co-prime integers \( p, q \) with \( q \neq 0 \).
Step 2: Rearranging:
\[ 2\sqrt{5} = \frac{p}{q} – 3 = \frac{p – 3q}{q} \]
\[ \sqrt{5} = \frac{p – 3q}{2q} \]
Step 3: Since \( p \) and \( q \) are integers, \( \frac{p – 3q}{2q} \) is rational. This means \( \sqrt{5} \) is rational — a contradiction (proved in Q1).
\( \therefore \) \( 3 + 2\sqrt{5} \) is irrational. Proved.
Question 3
Prove that the following are irrationals: (i) \( \frac{1}{\sqrt{2}} \) (ii) \( 7\sqrt{5} \) (iii) \( 6 + \sqrt{2} \)
Assume \( \frac{1}{\sqrt{2}} \) is rational, say \( \frac{p}{q} \). Then \( \sqrt{2} = \frac{q}{p} \), which is rational — contradicting the known irrationality of \( \sqrt{2} \).
\( \therefore \) \( \frac{1}{\sqrt{2}} \) is irrational.
Assume \( 7\sqrt{5} = \frac{p}{q} \). Then \( \sqrt{5} = \frac{p}{7q} \), which is rational — contradicting the irrationality of \( \sqrt{5} \).
\( \therefore \) \( 7\sqrt{5} \) is irrational.
Assume \( 6 + \sqrt{2} = \frac{p}{q} \). Then \( \sqrt{2} = \frac{p}{q} – 6 = \frac{p – 6q}{q} \), which is rational — contradicting the irrationality of \( \sqrt{2} \).
\( \therefore \) \( 6 + \sqrt{2} \) is irrational.
Formula Reference Table
| Formula Name | Formula | Variables Defined |
|---|---|---|
| Euclid’s Division Lemma | \( a = bq + r \) | \( a \) = dividend, \( b \) = divisor, \( q \) = quotient, \( r \) = remainder \( (0 \leq r < b) \) |
| HCF × LCM Relationship | \( \text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b \) | Valid for exactly two positive integers \( a \) and \( b \) |
| HCF via Prime Factorisation | Product of common primes with lowest powers | Apply after full prime factorisation of both numbers |
| LCM via Prime Factorisation | Product of all primes with highest powers | Apply after full prime factorisation of both numbers |
| Terminating Decimal Condition | \( q = 2^m \times 5^n \) | \( \frac{p}{q} \) in lowest terms; \( m, n \geq 0 \) integers |
Solved Examples Beyond NCERT
Example 1 — Easy
Find the HCF and LCM of 12, 15, and 21 using prime factorisation.
Step 1: \( 12 = 2^2 \times 3 \), \( 15 = 3 \times 5 \), \( 21 = 3 \times 7 \).
Step 2: HCF = common factor with lowest power = \( 3^1 = 3 \).
Step 3: LCM = all primes with highest powers = \( 2^2 \times 3 \times 5 \times 7 = 420 \).
HCF = 3, LCM = 420
Example 2 — Medium
Check whether \( \frac{17}{8} \) and \( \frac{64}{455} \) have terminating decimal expansions.
For \( \frac{17}{8} \): \( 8 = 2^3 \). Only prime factor is 2. \( \therefore \) Terminating decimal. \( \frac{17}{8} = 2.125 \).
For \( \frac{64}{455} \): \( 455 = 5 \times 7 \times 13 \). Contains prime factor 7 and 13 (not just 2 and 5). \( \therefore \) Non-terminating repeating decimal.
\( \frac{17}{8} \) terminates; \( \frac{64}{455} \) does not terminate.
Example 3 — Hard
Prove that \( \sqrt{3} + \sqrt{7} \) is irrational. (NCERT Exemplar Class 10 Maths style)
Step 1: Assume \( \sqrt{3} + \sqrt{7} = r \) where \( r \) is rational.
Step 2: Squaring: \( 3 + 2\sqrt{21} + 7 = r^2 \Rightarrow 2\sqrt{21} = r^2 – 10 \Rightarrow \sqrt{21} = \frac{r^2 – 10}{2} \).
Step 3: The right side is rational (difference/sum of rationals). So \( \sqrt{21} \) would be rational.
Step 4: But \( 21 = 3 \times 7 \). By the Fundamental Theorem of Arithmetic, \( \sqrt{21} \) is irrational — a contradiction.
\( \therefore \) \( \sqrt{3} + \sqrt{7} \) is irrational. Proved.
Topic-Wise Important Questions for Board Exam
Euclid’s Division Lemma and Algorithm
1-mark questions:
- State Euclid’s Division Lemma.
- What is the HCF of two consecutive even numbers?
- Using Euclid’s algorithm, what is HCF(4, 16)?
3-mark questions:
- Use Euclid’s Division Algorithm to find HCF(455, 42).
- Show that every positive even integer is of the form \( 2q \) and every positive odd integer is of the form \( 2q + 1 \).
5-mark question:
- Show that the square of any positive integer is of the form \( 3m \) or \( 3m + 1 \) for some integer \( m \).
Fundamental Theorem of Arithmetic and Irrationality
1-mark questions:
- Write the prime factorisation of 2520.
- Is \( \sqrt{4} \) rational or irrational? Justify.
- If HCF(a, b) = 6 and LCM(a, b) = 120, find \( a \times b \).
3-mark questions:
- Prove that \( \sqrt{2} \) is irrational.
- Find LCM and HCF of 96 and 404 by prime factorisation and verify.
5-mark question:
- Prove that \( 5 – 3\sqrt{2} \) is irrational. Also state the theorem you used.
Common Mistakes Students Make
Mistake 1: Writing \( r \leq b \) in Euclid’s Lemma instead of \( r < b \).
Why it’s wrong: If \( r = b \), the quotient can be increased by 1 and \( r \) becomes 0, so the representation is not unique.
Correct approach: Always write \( 0 \leq r < b \) — the remainder is strictly less than the divisor.
Mistake 2: Applying HCF × LCM = a × b to three numbers.
Why it’s wrong: This relationship holds only for exactly two positive integers.
Correct approach: For three numbers, compute HCF and LCM separately using prime factorisation.
Mistake 3: Concluding \( \sqrt{4} \) is irrational because 4 is under a root.
Why it’s wrong: \( \sqrt{4} = 2 \), which is a rational number. Not all square roots are irrational.
Correct approach: Check whether the number under the root is a perfect square.
Mistake 4: Skipping the co-prime assumption in irrationality proofs.
Why it’s wrong: Without stating \( p \) and \( q \) are co-prime, the contradiction (both divisible by the prime) cannot be established.
Correct approach: Always begin with “Let \( \sqrt{p} = \frac{a}{b} \) where \( a \) and \( b \) are co-prime integers, \( b \neq 0 \).”
Mistake 5: Forgetting to verify HCF × LCM = product of numbers in Exercise 1.2.
Why it’s wrong: CBSE marking schemes award 1 mark specifically for the verification step.
Correct approach: Always include the verification as the final step in HCF/LCM problems.
Exam Tips for 2025-26
Based on the 2025-26 CBSE marking scheme, here is what to expect from Real Numbers in the board paper:
- Marks weightage: Unit 1 (Number Systems) carries 6 marks. Expect 1–2 questions from Chapter 1.
- Most frequent question types: MCQ on terminating/non-terminating decimals (1 mark), short proof of irrationality (2–3 marks), and HCF/LCM via prime factorisation with verification (3 marks).
- Assertion-Reason questions: The 2025-26 paper pattern includes A-R type questions. Practice statements like “All irrational numbers are real numbers” and “HCF of two prime numbers is always 1.”
- Proof questions: Write proofs in clear numbered steps. CBSE awards 1 mark per logical step — do not skip the contradiction statement.
- Decimal expansion: Always factorise the denominator first before declaring terminating or non-terminating — a 1-mark step examiners check.
Last-minute revision checklist:
- Memorise Euclid’s Division Lemma statement verbatim.
- Practise at least 3 irrationality proofs (\( \sqrt{2}, \sqrt{3}, 3+2\sqrt{5} \)).
- Know the terminating decimal condition: \( q = 2^m \times 5^n \).
- Remember: HCF × LCM = product (two numbers only).
- Review the NCERT Exemplar Class 10 Maths solutions for higher-order problems — they frequently appear in CBSE papers.
You can also check the CBSE Academic website for the latest 2025-26 sample papers and marking schemes.
Key Points to Remember
- Euclid’s Division Lemma guarantees a unique pair \( (q, r) \) for every pair \( (a, b) \).
- The Fundamental Theorem of Arithmetic guarantees unique prime factorisation for every composite number.
- HCF can be found by Euclid’s Algorithm (faster for large numbers) or prime factorisation.
- For two numbers: \( \text{HCF} \times \text{LCM} = a \times b \). This does NOT extend to three or more numbers.
- Irrationality proofs always use proof by contradiction + the property that if a prime divides \( n^2 \), it divides \( n \).
- A rational number \( \frac{p}{q} \) (lowest terms) terminates iff \( q \) has only 2 and 5 as prime factors.
- Every real number is either rational (terminating or repeating decimal) or irrational (non-terminating, non-repeating).
For more chapter-wise solutions, explore NCERT Solutions for Class 10 covering Maths, Science, Social Science, and English. You may also find our solutions for Chapter 2 Polynomials and Chapter 3 Pair of Linear Equations helpful as you progress through the Class 10 Maths syllabus.
Frequently Asked Questions — Real Numbers Class 10
How to find HCF using Euclid’s Division Algorithm in Class 10?
Apply Euclid’s Division Lemma: write the larger number as (divisor × quotient + remainder). Repeat with divisor and remainder until remainder = 0. The last non-zero remainder is the HCF. For example, to find HCF(135, 225): \( 225 = 135 \times 1 + 90 \), \( 135 = 90 \times 1 + 45 \), \( 90 = 45 \times 2 + 0 \). So HCF = 45.
How do you prove that root 2 is irrational in Class 10?
Assume \( \sqrt{2} = \frac{p}{q} \) where \( p, q \) are co-prime. Then \( p^2 = 2q^2 \), so 2 divides \( p \). Write \( p = 2m \); substituting gives \( q^2 = 2m^2 \), so 2 divides \( q \). This contradicts \( p, q \) being co-prime. Hence \( \sqrt{2} \) is irrational. This standard proof by contradiction is frequently tested in the CBSE board exam.
Where can I download NCERT maths book class 10 solutions PDF free?
You can access free NCERT maths book class 10 solutions on ncertbooks.net. The official NCERT textbook PDF is available at ncert.nic.in. Our solutions cover all exercises with step-by-step working aligned to the 2025-26 CBSE syllabus and exam pattern.
What is the Fundamental Theorem of Arithmetic and why is it important?
The Fundamental Theorem of Arithmetic states that every composite number has a unique prime factorisation. It is important because it guarantees that HCF and LCM calculations via prime factorisation always yield a unique answer, and it is the logical foundation for all irrationality proofs in this chapter.
How many marks does Chapter 1 Real Numbers carry in the CBSE Class 10 board exam?
In the CBSE Class 10 Maths board exam 2025-26, Unit 1 (Number Systems, which includes Real Numbers) carries approximately 6 marks. Questions typically appear as 1-mark MCQs or assertion-reason items, 2-mark short answers, and occasionally a 3-mark proof or HCF/LCM problem.