Real Numbers — Class 10 Notes

CBSE 2027 — dates, rules and official papers →

Maths · Ch 1Class 10 · CBSE 2026–27Medium weightage

Real Numbers

What the board actually asks: the Fundamental Theorem of Arithmetic and its two workhorses: HCF & LCM by prime factorisation (with HCF × LCM = product of the two numbers), and proving numbers like √2, √3, √5 irrational by contradiction. In the rationalised 2026-27 syllabus this chapter is short and mark-dense — prime-factorisation HCF/LCM and one irrationality proof carry almost every mark.

NCERT VERBATIM sourced, word-for-wordAUTHORED LazyTopper teaching scaffolding

The Big Idea AUTHORED

Break every number into primes — the rest follows.

The Fundamental Theorem of Arithmetic says every composite number is a product of primes in exactly one way (order aside). That single fact is the master key to this whole chapter: it gives you HCF and LCM by comparing prime powers, it explains why HCF × LCM = product of two numbers, and it powers the proofs that $2$ , $3$ and $5$ are irrational.

Formulae

Fundamental Theorem of Arithmetic

composite

= p_{1}^{a_{1}} p_{2}^{a_{2}} \dots p_{k}^{a_{k}}

(unique, order aside)

Tip: Use a factor tree and write primes in ascending order, then collect powers.

HCF

product of the smallest power of each common prime

Tip: Smallest power of each COMMON prime.

LCM

product of the greatest power of every prime

Tip: Greatest power of EVERY prime that appears.

HCF × LCM (two numbers)

HCF (a, b) \times LCM (a, b) = a \times b

Tip: Given both numbers and their HCF: LCM = (a × b) ÷ HCF. Given the HCF, the LCM and one number: the other number = (HCF × LCM) ÷ that number. Write the formula line first.

Prime divides a square

p prime, p ∣ a^{2} \Rightarrow p ∣ a

Tip: State it in words in every irrationality proof: "if a prime divides a², it divides a" (Theorem 1.2).

Proof that √2 is Irrational

Statement and proof outline

2

is irrational. Proof by contradiction: assume

2 = \frac{a}{b}

with

a, b

co-prime. Squaring gives

2 b^{2} = a^{2}

, so 2 divides

a^{2}

, hence 2 divides

a

. Write

a = 2 c

; then

b^{2} = 2 c^{2}

, so 2 divides

b

too. Now

a

and

b

share the factor 2, which contradicts co-primality. So

2

is irrational. The same argument works for

3

and

5

. The full proof is in the Proof tab.

Rational ± Irrational is Irrational

The sum or difference of a rational and an irrational number is irrational. The product (or quotient) of a non-zero rational and an irrational number is irrational. To prove it: assume the result is rational, isolate the surd, and reach a contradiction.

HCF and LCM of Three Numbers

For three numbers,

HCF \times LCM

is NOT equal to the product of the numbers. Find the HCF and LCM by prime factorisation directly.

When Does a Number End in 0?

A number ends in 0 only if its prime factorisation contains both 2 and 5. So

4^{n}

and

6^{n}

never end in 0, but

(2 \times 5)^{n}

always does.

Shortcuts & memory tricks →

Definitions the board rewards NCERT VERBATIM

State these in NCERT’s exact words

Fundamental Theorem of Arithmetic

NCERT verbatim — memorise this exact wording· Ch 1,

“Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.”

What it means: Any composite number breaks into primes in exactly ONE way — only the order can differ. This uniqueness is what makes prime factorisation a reliable tool for HCF, LCM and irrationality proofs.

Prime divides a square

NCERT verbatim — memorise this exact wording· Ch 1,

“Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.”

What it means: If a prime divides

a^{2}

, it must already divide

a

itself. This is the engine of every irrationality proof — it lets you conclude

2 ∣ a

from

2 ∣ a^{2}

Irrational number

NCERT verbatim — memorise this exact wording· Ch 1,

“a number 's' is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.”

What it means: A number you simply cannot write as one integer over another (with non-zero denominator).

2

3

π

are irrational; every terminating or repeating decimal is NOT.

Key terms — NCERT verbatim (state word-for-word)

Prime factorisation is unique: The prime factorisation of a natural number is unique, except for the order of its factors. (Ch 1, )
HCF = smallest powers: Product of the smallest power of each common prime factor in the numbers. (Ch 1, )
LCM = greatest powers: Product of the greatest power of each prime factor, involved in the numbers. (Ch 1, )
HCF × LCM = product: for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. (Ch 1, )

Must-know concepts AUTHORED FRAMING

…and where they’re tested

Fundamental Theorem of Arithmetic

Every composite number splits into primes in exactly one way (order aside). Build a factor tree — keep dividing by the smallest prime — until only primes remain, then collect them as powers, e.g.

32760 = 2^{3} \times 3^{2} \times 5 \times 7 \times 13

FigureFactor tree · 32760 = 2³ × 3² × 5 × 7 × 13

Figure asset pending extraction — the sourced original is Figure.

Split off the smallest prime at each step; the green leaves (2, 2, 2, 3, 3, 5, 7, 13) are the prime factors — collected as powers this is the Fundamental Theorem of Arithmetic in action.

1-2 mark 'express as a product of primes' / composite-number reasoning

HCF & LCM by prime factorisation

Prime-factorise each number. HCF = product of the smallest power of each COMMON prime; LCM = product of the greatest power of EVERY prime that appears. Line the powers up in a column and read them off.

2-3 mark 'find the HCF and LCM by prime factorisation'

HCF × LCM = product of the two numbers

For any two positive integers,

HCF (a, b) \times LCM (a, b) = a \times b

. So once you have the HCF you can get the LCM without re-factorising:

LCM = \frac{a \times b}{HCF}

. ⚑ Warning: this holds for TWO numbers only, never three.

1-2 mark 'given HCF, find LCM' (e.g. HCF(306,657)=9 → find LCM)

Proving irrationality by contradiction

To prove

2

irrational: assume it is rational,

2 = a / b

with

a, b

coprime. Square and rearrange to get

2 ∣ a^{2}

, so (by Theorem 1.2)

2 ∣ a

; substitute back to force

2 ∣ b

too. Now

a

and

b

share the factor

2

— contradicting 'coprime'. Hence

2

is irrational. The same steps prove it for

3

and

5

3-mark 'prove that √2 / √3 / √5 is irrational'

'Can it end in 0/5?' — uniqueness in action

A number ends in 0 only if its prime factorisation contains BOTH 2 and 5. Since

4^{n} = 2^{2 n}

has no factor 5, it can never end in 0. The uniqueness half of the theorem is what lets you rule this out for every

n

at once.

1-2 mark competency: 'check whether 6ⁿ (or 4ⁿ) can end with the digit 0'

Worked examples — real solved problems NCERT VERBATIM

Solved exactly as the source does them

Every problem is a real sourced item, quoted verbatim with its page; the step-by-step decoding and mark-logic are LazyTopper-authored coaching.

Ch 1, Example 2,

“Find the LCM and HCF of 6 and 20 by the prime factorisation method.”

Prime-factorise both numbers.

6 = 2^{1} \times 3^{1}

and

20 = 2 \times 2 \times 5 = 2^{2} \times 5^{1}

.Concept: prime factorisation

1 mark

HCF = smallest power of each COMMON prime.The only common prime is 2, smallest power

2^{1}

. So

HCF (6, 20) = 2

.Concept: smallest common powers

½ mark

LCM = greatest power of EVERY prime.

LCM (6, 20) = 2^{2} \times 3 \times 5 = 60

.Concept: greatest powers

½ mark

Mark-scheme logic (authored): 2-mark shape: 1 for the two prime factorisations, ½ for the HCF (smallest common power) and ½ for the LCM (greatest powers of all primes).

Ch 1, Example 3,

“Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.”

Ch 1, Example 4,

“Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.”

Ch 1, Example 5,

“Prove that √3 is irrational.”

Ch 1, Example 6,

“Show that 5 – √3 is irrational.”

Ch 1, Example 1,

“Consider the numbers 4ⁿ, where n is a natural number. Check whether there is any value of n for which 4ⁿ ends with the digit zero.”

Problem-solving strategies AUTHORED

'Find HCF and LCM'?Prime-factorise, stack the powers in a column: HCF = smallest common power, LCM = greatest power of every prime.

'Given HCF, find LCM' (two numbers)?Use

LCM = \frac{a \times b}{HCF}

— don't re-factorise. (Two numbers ONLY.)

'Prove √3 is irrational' (same steps for √2, √5, √7)?Assume

3 = a / b

coprime → square →

3 ∣ a^{2} \Rightarrow 3 ∣ a

(Theorem 1.2) → sub back →

3 ∣ b

→ contradiction.

'Show rational ± irrational is irrational'?Assume rational, isolate the surd = ratio of integers → 'rational surd' contradicts the known irrational.

'Can it end in 0 / prove composite'?Prime-factorise: ends in 0 needs BOTH 2 and 5; a sum with a common factor is composite.

Common mistakes that cost marks

Where students lose easy marks

Claiming HCF × LCM = product for THREE numbers. It holds for two numbers only.

Forgetting 'coprime' in an irrationality proof. The whole contradiction rests on assuming

a, b

share no factor — skip it and the proof collapses.

Jumping from

p ∣ a^{2}

straight to

a = p c

without citing Theorem 1.2. The step

p ∣ a^{2} \Rightarrow p ∣ a

is exactly what earns the mark.

Swapping the HCF/LCM rules — HCF takes the SMALLEST powers of COMMON primes; LCM takes the GREATEST powers of ALL primes.

Formulae

Fundamental Theorem of Arithmetic

composite

= p_{1}^{a_{1}} p_{2}^{a_{2}} \dots p_{k}^{a_{k}}

(unique, order aside)

Tip: Use a factor tree and write primes in ascending order, then collect powers.

HCF

product of the smallest power of each common prime

Tip: Smallest power of each COMMON prime.

LCM

product of the greatest power of every prime

Tip: Greatest power of EVERY prime that appears.

HCF × LCM (two numbers)

HCF (a, b) \times LCM (a, b) = a \times b

Tip: Given both numbers and their HCF: LCM = (a × b) ÷ HCF. Given the HCF, the LCM and one number: the other number = (HCF × LCM) ÷ that number. Write the formula line first.

Prime divides a square

p prime, p ∣ a^{2} \Rightarrow p ∣ a

Tip: State it in words in every irrationality proof: "if a prime divides a², it divides a" (Theorem 1.2).

Proof that √2 is Irrational

Statement and proof outline

2

is irrational. Proof by contradiction: assume

2 = \frac{a}{b}

with

a, b

co-prime. Squaring gives

2 b^{2} = a^{2}

, so 2 divides

a^{2}

, hence 2 divides

a

. Write

a = 2 c

; then

b^{2} = 2 c^{2}

, so 2 divides

b

too. Now

a

and

b

share the factor 2, which contradicts co-primality. So

2

is irrational. The same argument works for

3

and

5

. The full proof is in the Proof tab.

Rational ± Irrational is Irrational

HCF and LCM of Three Numbers

For three numbers,

HCF \times LCM

is NOT equal to the product of the numbers. Find the HCF and LCM by prime factorisation directly.

When Does a Number End in 0?

A number ends in 0 only if its prime factorisation contains both 2 and 5. So

4^{n}

and

6^{n}

never end in 0, but

(2 \times 5)^{n}

always does.

Shortcuts & Tricks

If HCF(a,b) = 1 (co-prime) then LCM = a × b.

Safe in written answers

Two numbers in ratio m:n (m,n co-prime) are mk and nk: HCF = k, LCM = mnk, so HCF = LCM ÷ (mn).

MCQ / check only: don't write this as working

A number ends in 0 only if both 2 and 5 divide it; test 5ⁿ, 4ⁿ, 6ⁿ at a glance.

MCQ / check only: don't write this as working

HCF divides LCM: HCF is a factor of both numbers and LCM is a multiple of both, so HCF ≤ smaller number and LCM ≥ larger number. Use to drop wrong options.

MCQ / check only: don't write this as working

Never use HCF × LCM = product for three numbers.

Safe in written answers

Memory Tricks

HCF = Chhota power, Common prime. LCM = Bada power, Sab prime.

HCF: smallest power, common primes only. LCM: greatest power, every prime. C=Chhota (smallest power), C=Common (common primes only); B=Bada (greatest power), S=Sab (all primes).

Mummy Se Dahi Vala Nashta

Proof order. M = Maan lo √2 = a/b, a,b co-prime · S = Square it (2b² = a²) · D = Divides: 2 divides a² so 2 divides a; write a = 2c, then b² = 2c², so 2 divides b · V = Virodh (a and b share factor 2, contradiction) · N = Nishkarsh (so √2 is irrational).

Do ki jodi mein HCF × LCM = a × b; teen ki bheed mein nahi.

Product rule holds for two numbers only. Do = two numbers (rule holds); Teen = three (fails).

Mindmap

- - Every composite = product of primes
  - Unique, apart from order
  - Factor tree → powers of primes
- - HCF = smallest power of common primes
  - LCM = greatest power of all primes
  - Stack the powers in a column
- - LCM = a×b / HCF
  - Two numbers ONLY
  - 'Given HCF find LCM'
- - √2, √3, √5 by contradiction
  - p | a² ⇒ p | a (Theorem 1.2)
  - assume coprime → force shared factor
  - rational ± irrational = irrational
- - HCF×LCM=product for 3 numbers (wrong)
  - forgetting 'coprime'
  - skipping the Theorem-1.2 citation
  - swapping smallest/greatest powers

√2 is irrational — the proof by contradiction

This is NCERT's Theorem 1.3 (p.6-7). It is the model every irrationality proof in this chapter follows, and rests entirely on Theorem 1.2 ( $p ∣ a^{2} \Rightarrow p ∣ a$ ), which itself comes from the Fundamental Theorem of Arithmetic. Learn this one and √3, √5 are the same argument with a different prime.

Assume, to the contrary, that √2 is rational.

2 = \frac{r}{s}, s \neq = 0

Proof by contradiction: we suppose the negation and drive it to an impossibility.

Reduce to lowest terms, so the numerator and denominator are coprime.

2 = \frac{a}{b}, a, b coprime

Divide out any common factor to get a and b with no common factor other than 1.

Multiply across and square.

2 b^{2} = a^{2} \Rightarrow 2 ∣ a^{2}

From b√2 = a, squaring gives 2b² = a².

So 2 divides a.

2 ∣ a \Rightarrow a = 2 c

By Theorem 1.2, since the prime 2 divides a², it divides a. Write a = 2c.

Substitute back — now 2 divides b as well.

b^{2} = 2 c^{2} \Rightarrow 2 ∣ b

2b² = 4c² gives b² = 2c², so 2 | b² and (again by Theorem 1.2) 2 | b.

Contradiction: a and b share the factor 2, yet we chose them coprime.

∴ 2 is irrational.

The contradiction arose from assuming √2 rational — so that assumption is false.

3 competency questions from the CBSE board pattern, with the full step-marked solution for each — the marks are shown per step, as the official scheme awards them.

1 markSection A

Shilpi wants to organize a party. She has 36 kiwis and 60 oranges at home and decided to distribute them equally among all. She decides to add 42 apples also. In this case, how many maximum guests can she invite?

(A) 6
(B) 12
(C) 120
(D) 180

Step-marked solution

[1 mark] Maximum guests = HCF(36, 60, 42) = 6. Answer: A.

1 markSection A

Shweta wants to organize a party. She has 336 guavas and 54 oranges at home and decided to distribute them equally among all. How many maximum guests can she invite?

(A) 6
(B) 9
(C) 56
(D) 3024

Step-marked solution

[1 mark] 336 = 56 × 6 and 54 = 9 × 6, so HCF(336, 54) = 6 = maximum guests. Answer: A.

1 markSection A

There are five bells placed at different swings in a park, which toll at intervals of 2, 3, 5, 6, and 10 minutes, respectively. They all toll together when the park is open for visitors at 10:00 AM. How many more times do they all toll together till the park is closed at 8:00 PM?

(A) 10
(B) 20
(C) 30
(D) 60

Step-marked solution

[1 mark] LCM(2, 3, 5, 6, 10) = 30, so the bells toll together every 30 minutes. From 10:00 AM to 8:00 PM is 10 hours = 600 minutes; 600 ÷ 30 = 20 more times. Answer: B.

Practise Real Numbers competency questions

Read NCERT Chapter 1 on LazyTopper (jemh101)

Official NCERT chapter PDF (jemh101) — ncert.nic.in

Sources: NCERT 2026-27 (tap to view)

Source ledger PROVENANCE

Source edition: NCERT Reprint 2026-27, Class 10 Mathematics, Ch 1 (jemh101.pdf, printed pages 1-9). One row per sourced item — every verbatim, example and NCERT figure traces to its page.

Item	Type	Source
Fundamental Theorem of Arithmetic	verbatim	NCERT 2026-27 Ch1 (Theorem 1.1)
Prime divides a square	verbatim	NCERT 2026-27 Ch1 (Theorem 1.2)
Irrational number	verbatim	NCERT 2026-27 Ch1
Prime factorisation is unique	verbatim	NCERT 2026-27 Ch1
HCF = smallest powers	verbatim	NCERT 2026-27 Ch1
LCM = greatest powers	verbatim	NCERT 2026-27 Ch1
HCF × LCM = product	verbatim	NCERT 2026-27 Ch1
HCF & LCM by prime factorisation · NCERT Example 2	example	NCERT 2026-27 Ch1 Example 2
HCF then LCM via HCF×LCM=product · NCERT Example 3	example	NCERT 2026-27 Ch1 Example 3 -5
HCF & LCM of three numbers · NCERT Example 4	example	NCERT 2026-27 Ch1 Example 4
Prove irrationality by contradiction · NCERT Example 5	example	NCERT 2026-27 Ch1 Example 5
Rational ± irrational is irrational · NCERT Example 6	example	NCERT 2026-27 Ch1 Example 6
Uniqueness / 'ends with 0' reasoning · NCERT Example 1	example	NCERT 2026-27 Ch1 Example 1

Real Numbers · spec-rendered note · NCERT Reprint 2026-27, Class 10 Mathematics, Ch 1 (jemh101.pdf, printed pages 1-9)

Practise writing answers the way examiners mark them → Guide