Class 10 Maths Chapter 1 ‘Real Numbers Notes’: Complete NCERT Guide for CBSE Board Exams

Class 10 Maths Chapter 1: Real Numbers Notes for CBSE exams are available in this article. This easy-to-follow guide explains core ideas like Euclid’s division lemma, the Fundamental Theorem of Arithmetic, and the difference between rational and irrational numbers in plain language. You’ll find clear worked examples (including proofs that √2 and √3 are irrational), tips for spotting recurring and terminating decimals, and exam-smart tricks that make tricky concepts feel simple. Ideal for students and teachers, the notes match the NCERT textbook and include quick revision points, practice problems, and little reminders to help you stay calm and confident on exam day.

 
Chapter 1: Real Numbers Notes for Class 10 Maths

Real Numbers: 

R = Real Numbers:

All rational and irrational numbers together form the set of real numbers.

I = Integers:

Numbers like …, -3, -2, -1, 0, 1, 2, 3 … are called integers.

Q = Rational Numbers: 

A rational number can be defined as any number that can be expressed in the form p/q, where p and q are integers and q ≠ 0 . Example: 3/2, 0.333…Their decimal expansions are either terminating or non‑terminating but repeating.

Q' = Irrational Numbers:

Real numbers that cannot be expressed as p/q, are irrational. Their decimal expansions are non‑terminating and non‑repeating. Like √2, √3, √7…

N = Natural Numbers:

Counting numbers: 

N = { 1,2,3,…}

W = Whole Numbers:

Natural numbers together with zero: 

W={0,1,2,3,…}.

 

Euclid's Division Lemma: 

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

When you divide 17 by 5, you get:

17 = 5 × 3 + 2

Here, a = 17, b = 5, q = 3, r = 2. And indeed, 0 ≤ 2 < 5.

Euclid's Division Algorithm (Finding HCF):

It gives us a step-by-step method to find the HCF (Highest Common Factor) of two positive integers.

To find HCF of two numbers a and b (where a > b):

Step 1: Apply Euclid's Division Lemma to a and b: a = bq + r

Step 2: If r = 0, then HCF = b. Stop.

Step 3: If r ≠ 0, replace a with b and b with r. Repeat from Step 1.

Keep going until the remainder becomes zero. The divisor at that step is the HCF.

Example 1: Find HCF of 56 and 72

Step 1: 72 = 56 × 1 + 16 (r = 16 ≠ 0, continue)

Step 2: 56 = 16 × 3 + 8 (r = 8 ≠ 0, continue)

Step 3: 16 = 8 × 2 + 0 (r = 0 )

HCF (56, 72) = 8

The Fundamental Theorem of Arithmetic 

Statement: 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.

Example: 270 = 2 × 135

    = 2 × 3 × 45

    = 2 × 3 × 3 × 15

    = 2 × 3 × 3 × 3 × 5

    = 2 × 3³ × 5

Question: Check whether 15ⁿ can end with the digit zero for any natural number n.

Solution: If 15ⁿ ends in 0, it must be divisible by 10.

10 = 2 × 5, so for any number to end in 0, its prime factorisation must include both 2 and 5.

15ⁿ = (3 × 5)ⁿ = 3ⁿ × 5ⁿ

The prime factorisation of 15ⁿ contains only 3s and 5s, there is no factor of 2.

By the Fundamental Theorem of Arithmetic, 15ⁿ can never end with the digit 0 for any natural number n.

 

HCF and LCM Using Prime Factorisation

 Express both numbers as products of prime factors.

HCF = product of the smallest power of each common prime factor.

LCM = product of the greatest power of each prime factor in either number.

 
Example 1: Find HCF and LCM of 120 and 144

Step 1: Prime factorisation

120 = 2³ × 3 × 5

144 = 2⁴ × 3²

Step 2: HCF (smallest power of common primes: 2 and 3)

HCF = 2³ × 3¹ = 8 × 3 = 24

Step 3: LCM (greatest power of every prime that appears)

LCM = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720

 

HCF × LCM = Product of Two Numbers

For any two positive integers a and b:

HCF(a, b) × LCM(a, b) = a × b

Question: If HCF(a, b) = 4 and LCM(a, b) = 48, and a = 12, find b.

HCF × LCM = a × b

4 × 48 = 12 × b

192 = 12b

b = 16

Revisiting Irrational Numbers

An irrational number is a real number 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, √7, π, e, √11, ∛2, √2 + √3

Key theorem (used in proofs):

If a prime number p divides a², then p also divides a, where a is a positive integer.

Proof 1: Prove that √2 is Irrational

Method: Proof by contradiction.

Assume √2 is rational. Then it can be written as:

√2 = p/q, where p and q are integers, q ≠ 0, and p/q is in its lowest terms (i.e., HCF(p, q) = 1 and they share no common factors).

Step 1: Squaring both sides:

2 = p²/q²

⇒ p² = 2q²

This tells us p² is even (since it equals 2q², which is a multiple of 2).

Step 2: If p² is even, then p must be even.

(Reason: If p were odd, p² would also be odd. Since p² is even, p must be even.)

Let p = 2k for some integer k.

Step 3: Substituting p = 2k into p² = 2q²:

(2k)² = 2q²

4k² = 2q²

q² = 2k²

Therefore q² is even, so q is also even.

Step 4: But if both p and q are even, they share a common factor of 2.

This contradicts our assumption that HCF(p, q) = 1.

Conclusion: Our assumption was wrong. √2 is irrational.

 

Proof 2: Prove that √3 is Irrational

Assume √3 = p/q in lowest terms (HCF(p, q) = 1).

Step 1: Squaring: p² = 3q²

⇒ 3 divides p²

⇒ 3 divides p (If a prime number p divides a², then p also divides a, where a is a positive integer)

Let p = 3k.

Step 2: (3k)² = 3q²

9k² = 3q²

q² = 3k²

⇒ 3 divides q² ⇒ 3 divides q

Step 3: Both p and q are divisible by 3. This contradicts HCF(p, q) = 1.

Conclusion: √3 is irrational.

Example 1: Prove that 2 + √5 is irrational.

Assume 2 + √5 = r (rational).

⇒ √5 = r − 2

r − 2 is rational (difference of two rationals), but √5 is irrational which is a contradiction to the fact that r − 2 is rational .

Therefore, 2 + √5 is irrational. 

 

Example 4: Prove that 1/√2 is irrational.

Assume 1/√2 = p/q (rational, in lowest terms).

⇒ √2 = q/p

q/p is rational. But √2 is irrational which is a contradiction.

Therefore, 1/√2 is irrational.

Key Properties: Sum and Product of Rationals and Irrationals

Decimal Expansion of Real Numbers

A. Terminating Decimals 

Every rational number, when divided, gives a decimal that either terminates or repeats. A decimal expansion that ends after a finite number of digits.

A rational number p/q (in standard form) has a terminating decimal expansion if and only if the prime factorisation of q contains only powers of 2 and/or 5 i.e., q is of the form 2ᵐ × 5ⁿ, where m, n are non-negative integers.

If q has any prime factor other than 2 or 5, the decimal will be non-terminating and repeating.

Examples:

1/2 = 0.5 

3/8 = 0.375

B. Non-Terminating Repeating (Recurring) Decimals

When the denominator has prime factors other than 2 and 5, the decimal goes on forever but repeats a pattern. 

Examples:

 1/3 = 0.3̄ = 0.333...

1/7 = 0.142857142857... = 0.̄1̄4̄2̄8̄5̄7̄

All non-terminating repeating decimals are rational numbers. They can always be expressed as p/q.

C. Non-Terminating Non-Repeating Decimals → Irrational

If a decimal goes on forever and has no repeating pattern, the number is irrational.

√2 = 1.41421356237...

π = 3.14159265358…

 

Here's everything compressed for quick revision.

 

Important Theorems at a Glance:

 Theorem 1: If a prime p divides a², then p divides a (where a is a positive integer).

(Used in every irrationality proof)

Theorem 2 (Euclid's Lemma): a = bq + r, 0 ≤ r < b, for unique integers q and r.

Theorem 3 (Fundamental Theorem of Arithmetic): Every composite number has a unique prime factorisation.

Theorem 4: Let p/q be a rational in standard form. Its decimal expansion is terminating iff q = 2ᵐ × 5ⁿ.

Theorem 5: Let p/q be a rational in standard form. If q has prime factors other than 2 and 5, its decimal expansion is non-terminating repeating.

 

Click below to download your free Class 10 Maths Chapter 1 Real Numbers PDF Notes perfect for last-minute CBSE board exam revision.

Class 10 Maths Chapter 1 Real Numbers PDF Notes