Concept Topic GuideCBSE & ICSEClass 10MathematicsReal NumbersVerified Faculty Notes & Solutions
Real Numbers – Class 10 Maths Chapter 1 NCERT Notes & Formulas
Class 10 Real Numbers notes: Euclid's division lemma, HCF, LCM, Fundamental Theorem of Arithmetic, irrational proofs for CBSE & ICSE boards.
8 students readUpdated 4 September 2026
Chapter 1: Real Numbers
Number System Hierarchy: Natural Numbers ⊂ Whole Numbers ⊂ Integers ⊂ Rational Numbers ⊂ Real Numbers. Real numbers include both rational (expressible as p/q) and irrational numbers (non-terminating, non-repeating decimals like √2, π).
1.1 Euclid's Division Lemma
For any two positive integers a and b, there exist unique integers q (quotient) and r (remainder) such that:
a = b × q + r, where 0 ≤ r < b
Euclid's Division Algorithm (to find HCF)
To find the HCF of two positive integers, say 455 and 42:
- Divide the larger number by the smaller: 455 = 42 × 10 + 35
- Now divide the divisor (42) by the remainder (35): 42 = 35 × 1 + 7
- Continue: 35 = 7 × 5 + 0
- When remainder = 0, the divisor at that step is the HCF. HCF(455, 42) = 7
1.2 The Fundamental Theorem of Arithmetic
Statement: Every composite number can be expressed as a product of prime numbers, and this factorisation is unique (apart from the order of the primes).
Example: 1260 = 2² × 3² × 5 × 7 — This is the only way to express 1260 as a product of primes.
Example: 1260 = 2² × 3² × 5 × 7 — This is the only way to express 1260 as a product of primes.
Using Prime Factorisation to Find HCF and LCM
Example: Find HCF and LCM of 12, 15, and 21.
- 12 = 2² × 3
- 15 = 3 × 5
- 21 = 3 × 7
HCF = Product of the smallest power of each common prime factor = 3
LCM = Product of the greatest power of each prime factor = 2² × 3 × 5 × 7 = 420
Key Relationship: For any two positive integers a and b:
HCF(a, b) × LCM(a, b) = a × b
HCF(a, b) × LCM(a, b) = a × b
1.3 Revisiting Irrational Numbers
Proving √2 is Irrational (Proof by Contradiction)
- Assume √2 is rational, i.e., √2 = p/q where p, q are co-prime integers (HCF = 1).
- Squaring: 2 = p²/q², so p² = 2q².
- This means p² is even, so p must be even. Let p = 2m.
- Then (2m)² = 2q² → 4m² = 2q² → q² = 2m².
- So q² is also even, meaning q is even.
- But if both p and q are even, they have a common factor 2 — contradicting our assumption that they are co-prime.
- Therefore, √2 is irrational. ∎
1.4 Revisiting Rational Numbers and Their Decimal Expansions
Theorem: Let x = p/q be a rational number where p and q are co-prime.
• If q = 2ⁿ × 5ᵐ (only factors 2 and 5), then x has a terminating decimal expansion.
• If q has prime factors other than 2 or 5, then x has a non-terminating repeating decimal expansion.
Examples: 7/8 = 7/(2³) = 0.875 (terminating); 1/3 = 0.333... (non-terminating repeating)
• If q = 2ⁿ × 5ᵐ (only factors 2 and 5), then x has a terminating decimal expansion.
• If q has prime factors other than 2 or 5, then x has a non-terminating repeating decimal expansion.
Examples: 7/8 = 7/(2³) = 0.875 (terminating); 1/3 = 0.333... (non-terminating repeating)
Indexed Topics & Examination Keywords
#real numbers class 10#euclid division lemma#HCF LCM#fundamental theorem arithmetic#irrational numbers proof#NCERT maths#CBSE#ICSE#prime factorisation
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