Real Numbers

Learn Euclid's division lemma, HCF, LCM, and the fundamental theorem of arithmetic.

Chapter Notes

1. Euclid's Division Lemma and Algorithm

  • For any two positive integers a and b, there exist unique whole numbers q and r such that a = bq + r, where 0 ≤ r < b.
  • Euclid's division algorithm finds the HCF: divide the larger number by the smaller, then divide the divisor by the remainder, and repeat until the remainder is 0. The last non-zero divisor is the HCF.
  • Example: HCF(455, 42) → 455 = 42×10 + 35, 42 = 35×1 + 7, 35 = 7×5 + 0, so HCF = 7.
  • Useful in proving results such as 'every positive even integer is of the form 2q and every odd integer is of the form 2q + 1'.
a = bq + r, 0 ≤ r < b

Watch it happen — Euclid's Division Algorithm

Controls

455
42

Live result

Steps
455 = 42 × 10 + 35 | 42 = 35 × 1 + 7 | 35 = 7 × 5 + 0
HCF
7
Euclid's algorithm — repeated divisionSteps: 455 = 42 × 10 + 35 | 42 = 35 × 1 + 7 | 35 = 7 × 5 + 0
a = 45565 groups of 7b = 426 groups of 7HCF 7 — remainder becomes the new divisor
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The remainder must satisfy 0 ≤ r < b; when the remainder becomes 0, the divisor is the HCF.

Board-style practice questions

Q1.Use Euclid's division algorithm to find the HCF of 135 and 225.

2 marks

2. 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 factors.
  • Prime factorisation is used to find HCF and LCM: HCF = product of the smallest power of each common prime factor; LCM = product of the greatest power of each prime factor involved.
  • For two numbers only, HCF(a, b) × LCM(a, b) = a × b. This relation is not valid for three or more numbers.
  • Typical board question: find the least number that is divisible by given numbers, or the largest number dividing given numbers leaving remainders.
HCF(a, b) × LCM(a, b) = a × b

Watch it happen — HCF & LCM Machine

Controls

12
18

Live result

HCF
6
LCM
36
HCF × LCM
216
a × b
216
HCF & LCM as common blocksHCF: 6
a = 122 groups of 6b = 183 groups of 6common block = HCF 6
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For any two positive integers, HCF × LCM = product of the numbers.

Board-style practice questions

Q1.Find the HCF and LCM of 96 and 404 by prime factorisation and verify HCF × LCM = product of the numbers.

3 marks

3. Prime Factorisation in Practice

  • Write each number as a product of primes using a factor tree or repeated division.
  • Example: 96 = 2⁵ × 3 and 404 = 2² × 101 → HCF = 2² = 4, LCM = 2⁵ × 3 × 101 = 9696.
  • Check: 4 × 9696 = 38784 = 96 × 404, confirming the HCF × LCM property.

Watch it happen — Prime Factorisation Tree

Controls

360

Live result

Prime factorisation
2^3 × 3^2 × 5
Number of prime factors
3
Type
Composite
Splitting the number into equal prime groupsPrime factorisation: 2^3 × 3^2 × 5
a = 360180 groups of 2b = 360180 groups of 2Composite
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Fundamental Theorem of Arithmetic: every composite number has a unique prime factorisation.

Board-style practice questions

Q1.Three bells toll at intervals of 9, 12 and 15 minutes. If they toll together at 8 a.m., when will they next toll together?

3 marks

4. Irrational Numbers and their Proofs

  • A number that cannot be written as p/q with integers p, q (q ≠ 0) is irrational — for example √2, √3, √5, π.
  • Theorem used: if a prime p divides a², then p divides a.
  • Proof of √2 being irrational (contradiction): assume √2 = a/b in lowest terms → 2b² = a² → 2 divides a → a = 2c → 2b² = 4c² → b² = 2c² → 2 divides b. So a and b have 2 as a common factor, contradicting 'lowest terms'. Hence √2 is irrational.
  • Results: the sum or difference of a rational and an irrational number is irrational; the product of a non-zero rational with an irrational is irrational.
√2 is irrational (proof by contradiction)

Watch it happen — Rational or Irrational?

Controls

2

Live result

√n
1.41421356
Nature
Irrational
Reason
Non-terminating, non-recurring decimal
Splitting the number into equal prime groups√n: 1.41421356
a = 21 groups of 2b = 21 groups of 2Irrational
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√p is irrational for every prime p — proved by contradiction in the NCERT text.

Board-style practice questions

Q1.Prove that √5 is irrational.

3 marks

5. Decimal Expansions of Rational Numbers

  • Let x = p/q be a rational number in lowest terms. If q is of the form 2ⁿ5ᵐ, the decimal expansion terminates; otherwise it is non-terminating repeating.
  • Example: 13/3125 = 13/5⁵ terminates; 7/80 = 7/(2⁴ × 5) terminates; 17/6 does not, since 6 = 2 × 3.
  • Every terminating or non-terminating repeating decimal represents a rational number; a non-terminating non-repeating decimal is irrational.
p/q terminates ⇔ q = 2ⁿ5ᵐ

Watch it happen — Terminating or Recurring?

Controls

13
125

Live result

Simplified fraction
13/125
Decimal
0.104
Expansion
Terminating
p/q — does the denominator have only 2s and 5s?Simplified fraction: 13/125
a = 137 groups of 2b = 12563 groups of 20.104
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p/q terminates only when q (in lowest terms) is of the form 2ᵐ × 5ⁿ.

Board-style practice questions

Q1.Without long division, state whether 13/3125 has a terminating or non-terminating decimal expansion.

2 marks