Chapter 1 of 14

CBSE Class 10 Mathematics★ 6 marks · CBSE Class 10 board

Real Numbers

Euclid's division algorithm, fundamental theorem of arithmetic, rational and irrational numbers, and revisiting decimal expansions.

⏱ 12 min readMedium difficulty✓ Free · NCERT-aligned

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

1Apply Euclid's division algorithm to find the HCF of two positive integers
2State and use the Fundamental Theorem of Arithmetic to factor any composite number
3Prove the irrationality of numbers like √2, √3 and 5 − 2√3
4Decide — without long division — whether a rational p/q has a terminating decimal expansion
5Use the identity HCF(a,b) × LCM(a,b) = a × b to solve word problems

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Why this chapter matters

Real Numbers is the foundation for everything that follows in Class 10 maths — polynomials use prime factorisation, quadratic equations use irrational roots, and coordinate geometry assumes you can reason about HCF and decimal expansions. It's also the chapter that quietly turns up in JEE Main number-theory problems.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Number System (Class 9)

Rational vs irrational, real number line

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Factors and Multiples (Class 6)

Primes, composites, divisibility

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HCF and LCM by listing (Class 6)

Intuition before the algorithm

Real Numbers

The set of real numbers is the union of rational and irrational numbers. This chapter formalises everything you already know about numbers — and proves a few results that you will reuse in every later topic.

1. Euclid's Division Lemma

Statement. Given positive integers $a$ and $b$ , there exist unique integers $q$ and $r$ satisfying $a = b q + r$ , where $0 \leq r < b$ .

In plain English: when you divide $a$ by $b$ , you always get a quotient $q$ and a remainder $r$ that is strictly less than the divisor.

Example. Take $a = 455$ , $b = 42$ . $455 = 42 \times 10 + 35, 0 \leq 35 < 42.$ So $q = 10$ and $r = 35$ .

Euclid's division algorithm (HCF)

To find $g cd (a, b)$ where $a > b$ :

Apply the lemma: $a = b q_{1} + r_{1}$ .
If $r_{1} = 0$ , then $g cd (a, b) = b$ . Done.
Otherwise apply the lemma to $b$ and $r_{1}$ , and continue.

The last non-zero remainder is the HCF.

Worked example — HCF(135, 225):

Step	Division	Remainder
1	$225 = 135 \times 1 + 90$	90
2	$135 = 90 \times 1 + 45$	45
3	$90 = 45 \times 2 + 0$	0

So $g cd (135, 225) = 45$ .

2. The Fundamental Theorem of Arithmetic

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

This is the result behind every HCF/LCM calculation you'll do.

HCF and LCM via prime factorisation.

For two numbers $a, b$ :

$HCF (a, b) =$ product of the smallest power of each common prime.
$LCM (a, b) =$ product of the greatest power of each prime that appears.

Worked example — $a = 96$ , $b = 404$ .

$96 = 2^{5} \cdot 3, 404 = 2^{2} \cdot 101.$ $HCF = 2^{2} = 4, LCM = 2^{5} \cdot 3 \cdot 101 = 9696.$

Sanity check: $HCF \times LCM = 4 \times 9696 = 38, 784 = a \cdot b$ . ✓

3. Revisiting Irrational Numbers

A number is irrational if it cannot be expressed as $p / q$ for integers $p$ and $q$ with $q \neq = 0$ .

Classic proof: $2$ is irrational

Assume for contradiction that $2 = p / q$ in lowest terms. Then $p^{2} = 2 q^{2}$ , so $p^{2}$ is even, so $p$ is even — write $p = 2 k$ . Substituting: $4 k^{2} = 2 q^{2} \Rightarrow q^{2} = 2 k^{2}$ , so $q$ is also even. But then both $p$ and $q$ are divisible by 2, contradicting "lowest terms". Hence $2$ is irrational.

The same argument works for $p$ where $p$ is any prime.

4. Decimal Expansions of Rational Numbers

A rational number $\frac{p}{q}$ (in lowest terms) has a terminating decimal expansion if and only if the prime factorisation of $q$ is of the form $2^{n} 5^{m}$ .

Quick check.

Fraction	Denominator factored	Terminates?
3/8	$2^{3}$	Yes
7/40	$2^{3} \cdot 5$	Yes
11/30	$2 \cdot 3 \cdot 5$	No (3 in denom)

Key formulas to memorise

$g cd (a, b) \cdot lcm (a, b) = a \cdot b$ (for two numbers).
For three numbers $a, b, c$ : $a \cdot b \cdot c \neq = g cd \cdot lcm$ in general — work via primes.
$a = b q + r$ , $0 \leq r < b$ .

Practice (try before checking)

Find the HCF of 196 and 38220 using Euclid's algorithm.
Show that $5 - 23$ is irrational.
Without long division, decide whether $\frac{17}{6250}$ terminates.
The HCF of two numbers is 27 and their LCM is 162. If one number is 54, find the other.

Answers

196 (verify by primes: $196 = 2^{2} \cdot 7^{2}$ , $38220 = 2^{2} \cdot 3 \cdot 5 \cdot 7^{2} \cdot 13$ , common $= 2^{2} \cdot 7^{2} = 196$ ).
Suppose $5 - 23 = r$ is rational. Then $3 = (5 - r) /2$ , which would make $3$ rational. Contradiction.
$6250 = 2 \cdot 5^{5}$ . The form is $2^{n} 5^{m}$ , so the decimal terminates.
$a \cdot b = HCF \cdot LCM \Rightarrow 54 \cdot b = 27 \cdot 162 \Rightarrow b = 81$ .

Euclid's Division Lemma — More Applications

Example — Show that the square of any odd positive integer is of the form 8k+1: Any odd positive integer = 4q+1 or 4q+3 (Euclid's division lemma by 4). Case 1: (4q+1)² = 16q²+8q+1 = 8(2q²+q)+1 = 8k+1. Case 2: (4q+3)² = 16q²+24q+9 =13 16q²+24q+8+1 = 8(2q²+3q+1)+1 = 8k+1. In both cases, square = 8k+1 ✓.

Example — Show that n²−n is divisible by 2 for any positive integer n: Case 1 (n even): n=2q → n²−n = 4q²−2q = 2(2q²−q) — divisible by 2. Case 2 (n odd): n=2q+1 → n²−n = (4q²+4q+1)−(2q+1) = 4q²+2q = 2(2q²+q) — divisible by 2. In both cases, n²−n is even ✓.

Fundamental Theorem of Arithmetic — More Worked Examples

Example — Find LCM and HCF of 510 and 92 by prime factorisation: 510 = 2×3×5×17. 92 = 2²×23. HCF = product of SMALLEST powers of common primes = 2¹ = 2. LCM = product of GREATEST powers of ALL primes = 2²×3×5×17×23 = 4×3×5×17×23 = 23460. Check: HCF × LCM = 2 × 23460 = 46920. Product of numbers = 510 × 92 = 46920 ✓.

Example — Given that HCF(306, 657) = 9, find LCM: LCM = (306×657)/9 = 201,042/9 = 22,338.

Example — Check if 6ⁿ can end with digit 0 for any natural number n: For a number to end in 0, it must be divisible by 10 = 2×5. So the prime factorisation must contain BOTH 2 and 5. 6ⁿ = (2×3)ⁿ = 2ⁿ×3ⁿ. The factor 5 is ABSENT. Therefore, 6ⁿ CANNOT end with the digit 0.

Example — Find the smallest number which when divided by 12, 15, 18 leaves remainder 5: Required number = LCM(12,15,18) + 5. 12 = 2²×3. 15 = 3×5. 18 = 2×3². LCM = 2²×3²×5 = 180. Number = 180+5 = 185.

Rational Numbers and Decimal Expansions

Theorem: Let p/q be a rational number in simplest form (no common factors). If q = 2ⁿ×5ᵐ, the decimal expansion TERMINATES. If q has any prime factor OTHER than 2 or 5, the decimal expansion is NON-TERMINATING RECURRING.

Examples: 13/8 = 13/2³ → terminates (=1.625). 17/5 = 17/5¹ → terminates (=3.4). 1/7 → 7 is not 2 or 5 → non-terminating recurring (=0.142857...). 7/12 = 7/(2²×3) → 3 is a factor → non-terminating recurring.

Check without long division: 17/3125 = 17/(5⁵). Denominator has only 5 → terminates. 29/343 = 29/(7³). 7 is not 2 or 5 → non-terminating recurring.

Common Mistakes

Euclid's lemma vs Fundamental Theorem: Euclid's lemma is about the relation a = bq+r. The Fundamental Theorem is about UNIQUE prime factorisation. They are different.
HCF×LCM = a×b works ONLY for TWO numbers: For three numbers, this relationship does NOT hold.
Irrationality proofs: When proving √n is irrational, you MUST state "assume p and q are COPRIME (HCF = 1)."
Terminating decimal check: Simplify the fraction FIRST, then check the denominator's prime factors.

CBSE Exam Focus

Topic	Marks
Euclid's Division Lemma	2-3
Fundamental Theorem of Arithmetic (HCF/LCM)	3-4
Proof of irrationality	4-5
Decimal expansion type (without actual division)	1-2

Quick Self-Test

Use Euclid's lemma to find HCF of 196 and 38220. (Answer: 196=2²×7². 38220=2²×3×5×7²×13. HCF=196.)
Show that 3√2 is irrational. (Answer: Assume 3√2 = p/q → √2 = p/(3q) → rational. But √2 is irrational. Contradiction.)
Without actual division, state whether 77/210 terminates. (Answer: 77/210 = 11/30 = 11/(2×3×5). Factor 3 is present → non-terminating.)
Prove that √5 is irrational. (Answer: Assume √5=p/q (coprime). 5q²=p² → 5 divides p → p=5k. 5q²=25k² → q²=5k² → 5 divides q. Contradiction — both divisible by 5.)
Find HCF and LCM of 12, 15, 21. (Answer: 12=2²×3, 15=3×5, 21=3×7. HCF=3. LCM=2²×3×5×7=420.)

Once Real Numbers is solid, Polynomials builds on the prime factorisation idea but applies it to algebraic expressions — zeroes, the division algorithm and the relationship between coefficients and roots.

Key formulas & results

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

Euclid's Division Lemma

a = bq + r,  0 ≤ r < b

q and r are unique once you fix a and b.

HCF × LCM identity (two numbers)

HCF(a, b) × LCM(a, b) = a × b

Only true for TWO numbers.

Fundamental Theorem of Arithmetic

n = p₁^a · p₂^b · p₃^c · …

Every composite number has a unique prime factorisation.

Terminating decimal condition

p/q terminates  ⇔  q = 2^n · 5^m (in lowest terms)

If q has any other prime factor → non-terminating recurring.

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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

✗ Writing Euclid's lemma without the constraint 0 ≤ r < b

✓ The constraint is what guarantees q and r are unique. Always state it — examiners deduct marks otherwise.

WATCH OUT

✗ Assuming any number containing √ is irrational

✓ Only √n where n is not a perfect square is irrational. √4 = 2 and √(9/16) = 3/4 are rational.

WATCH OUT

✗ Using HCF × LCM = a × b for three numbers

✓ The identity is only for two numbers. For three or more, compute HCF and LCM via prime factorisation.

WATCH OUT

✗ Stopping the proof of √2 irrational at 'p² is even, so p is even'

✓ You must then show q is also even, contradicting the 'p/q in lowest terms' assumption. Without that step the proof is incomplete.

WATCH OUT

✗ Saying 23/40 is non-terminating because 23 is prime

✓ It's the DENOMINATOR's prime factorisation that matters. 40 = 2³ · 5, which is 2ⁿ·5ᵐ, so 23/40 terminates.

NCERT exercises (with solutions)

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

Ex 1.1

Exercise 1.1

Euclid's division algorithm — finding HCF, showing certain forms

Questions

Ex 1.2

Exercise 1.2

Fundamental Theorem of Arithmetic — HCF and LCM via primes, word problems

Questions

Ex 1.3

Exercise 1.3

Proving irrationality — √2, √3, 3 + 2√5 style problems

Questions

Ex 1.4

Exercise 1.4

Decimal expansions — without long division, identify terminating vs non-terminating

Questions

Practice problems

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

Q1MEDIUM

Find the HCF of 96 and 404 using Euclid's division algorithm. Show all steps.

Q2MEDIUM

If HCF(72, 120) = 24, find LCM(72, 120) using the HCF × LCM identity.

Q3MEDIUM

Prove that √2 is irrational.

Q4MEDIUM

Prove that 5 − 2√3 is irrational, given that √3 is irrational.

Q5MEDIUM

Without performing long division, determine whether 17/6250 has a terminating or non-terminating decimal expansion.

Q6MEDIUM

Two tankers contain 850 L and 680 L of petrol. Find the maximum capacity of a container that can measure both quantities exactly (without any leftover).

5-minute revision

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

•Euclid's lemma: a = bq + r, 0 ≤ r < b — q and r are unique.
•Euclid's algorithm: repeat the lemma; last non-zero remainder is the HCF.
•Fundamental Theorem of Arithmetic: every composite has a unique prime factorisation.
•HCF = product of smallest powers of common primes. LCM = product of greatest powers of all primes.
•HCF(a,b) × LCM(a,b) = a × b — only for TWO numbers.
•√p is irrational for any prime p — proof by contradiction (p² even ⇒ p even ⇒ q even ⇒ contradiction).
•Sum, difference and rational multiple of an irrational with a non-zero rational is irrational.
•p/q (lowest terms) terminates ⟺ q has only 2s and 5s as prime factors. Otherwise non-terminating recurring.

CBSE marks blueprint

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

Typical chapter weightage: Real Numbers sits in Unit I (Number Systems), worth 6 marks total.

Question type	Marks each	Typical count	What it tests
MCQ (1 mark)	1	1	Terminating vs non-terminating identification, or HCF/LCM MCQ
Short Answer (2 marks)	2	1	Euclid's algorithm steps, or HCF × LCM identity application
Long Answer (3 marks)	3	1	Irrationality proof OR word problem using HCF/LCM

Prep strategy

Irrationality proofs are the most reliably tested 3-mark questions. Memorise the STRUCTURE: assume rational in lowest terms → square/simplify → show both numerator and denominator share a factor → contradicts 'lowest terms.' The logic is always the same.
Euclid's algorithm: always write EACH STEP as a = bq + r. Do not skip steps or guess. Board examiners check each line.
Terminating decimal: check ONLY the denominator's prime factors (after reducing the fraction). If denominator = 2ⁿ × 5ᵐ only → terminates. Any other prime factor → non-terminating recurring.
HCF and LCM via prime factorisation: HCF = SMALLEST powers of COMMON primes. LCM = LARGEST powers of ALL primes. This is the reverse of each other — remember both.

Where this shows up in the real world

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

RSA encryption — prime factorisation is hard to reverse

Music and HCF/LCM — rhythmic cycles

Gear ratios and engineering

Exam strategy

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

Irrationality proofs: the grader checks for THREE things — (1) the 'assume rational in lowest terms' set-up, (2) the contradiction step (showing both are even/share a factor), (3) the final 'contradiction' statement. Missing any one drops you 1 mark.
Euclid's algorithm: write each step as a complete equation (404 = 96 × 4 + 20), not just the remainder. Examiners follow your steps — if a step is missing you lose process marks.
Terminating decimal questions are free marks: factorise the denominator, check for 2s and 5s only. Common trap: students check the numerator (irrelevant — only denominator matters). Another trap: not reducing the fraction first (23/46 = 1/2 = 0.5; denominator 2 is 2¹·5⁰, terminates).
HCF and LCM word problems: read carefully whether the question asks for HCF (largest that divides both — 'measuring' problems, 'tiling' problems) or LCM (smallest that both divide — 'meeting again' problems, 'cycling' problems). The word 'maximum' usually signals HCF; 'minimum' usually signals LCM.

Going beyond the textbook

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

Research the Euclidean Algorithm's efficiency: Euclid's algorithm is one of the oldest algorithms in mathematics (~300 BCE) and also one of the most efficient. The Lamé theorem (1844) proved that the algorithm takes at most 5 times the number of digits in the smaller number as steps. For a 100-digit number, this means at most 500 steps — vastly more efficient than trial division. How does this compare to the fastest known factorisation algorithm? This comparison is at the heart of modern cryptography.
Explore Cantor's proof that the real numbers are uncountable (1874): using diagonalisation, Cantor showed that you cannot list all real numbers even in an infinite list. Since the rationals ARE countable, the irrationals must form an 'uncountably infinite' set — so in a precise sense, irrational numbers are 'almost all' of the real line, and rational numbers are the exceptions (measure zero). This makes the chapter's irrationality proofs even more interesting: you're proving membership in the 'typical' set.
Investigate the Collatz Conjecture — one of the most famous unsolved problems in mathematics: take any positive integer; if even, divide by 2; if odd, multiply by 3 and add 1. Repeat. Does the sequence always reach 1? This has been verified for numbers up to 2⁶⁸, but no proof exists. The connection to this chapter: the problem involves fundamental properties of integers under division — the same number-theory mindset you're developing here.
Research Waring's Problem (1770): every positive integer can be expressed as the sum of at most 4 perfect squares (e.g., 7 = 4 + 1 + 1 + 1). Can every integer be expressed as the sum of at most k cubes, k fourth powers, etc.? This connects the chapter's study of primes and integer properties to the deep study of additive number theory.

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