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Class 10 Maths Real Numbers – NCERT Formulas & Solutions

 

Real Numbers — Class 10 | FUZY MATH ACADEMY (LMS + 24h AI Math Solver)

Real Numbers — Class 10

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Introduction

Real numbers include rational and irrational numbers and form the backbone of many algebra and number theory problems in Class 10. This post covers the key concepts, formulas, worked examples, and practice questions — including 10 NCERT-style problems with explained solutions you can toggle open. Read, practice, and use our AI solver whenever you get stuck.

Key ideas at a glance

  • Euclid's division algorithm: For integers a and b (a > b), there exist q (quotient) and r (remainder) such that $a = bq + r$ and $0 \le r < b$.
  • HCF by Euclid's algorithm: Repeated division until remainder is 0; last non-zero remainder is HCF (GCD).
  • Fundamental Theorem of Arithmetic: Every integer greater than 1 can be expressed uniquely as a product of primes (up to order).
  • Irrational numbers: Numbers that cannot be written as $\dfrac{p}{q}$ where p, q are integers and $q \ne 0$. Examples: $\sqrt{2}, \sqrt{3}, \sqrt{5},$ etc.
  • Decimal expansions: Rational numbers either terminate or repeat; irrational numbers have non-terminating, non-repeating decimals.

Important formulas and facts

  • Euclid division: $a = bq + r, \quad 0 \le r < b$
  • HCF and LCM relation: For integers $a,b$: $\text{HCF}(a,b) \times \text{LCM}(a,b) = |a \times b|$ (when numbers positive).
  • Prime factorization: Use $a = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}$ to find HCF and LCM:
    • HCF: product of common primes with minimum exponents.
    • LCM: product of primes with maximum exponents.
  • Decimal termination test: A rational number $\dfrac{p}{q}$ (in lowest terms) has a terminating decimal iff $q$ has only 2 and 5 as prime factors (i.e. $q = 2^a5^b$).

Quick worked examples

Example 1 — HCF by Euclid

Find HCF(312, 675).

Solution:

  1. 675 = 312 × 2 + 51
  2. 312 = 51 × 6 + 6
  3. 51 = 6 × 8 + 3
  4. 6 = 3 × 2 + 0

Last non-zero remainder = 3, so HCF = 3.

Example 2 — Prime factorization and LCM/HCF

Find HCF and LCM of 360 and 126.

Solution:

Prime factors:
$360 = 2^3 \times 3^2 \times 5$
$126 = 2 \times 3^2 \times 7$.

HCF = $2^{\min(3,1)} \times 3^{\min(2,2)} = 2^1 \times 3^2 = 18$.
LCM = $2^{\max(3,1)} \times 3^{\max(2,2)} \times 5 \times 7 = 2^3 \times 3^2 \times 5 \times 7 = 2520$.
Check: $360 \times 126 = 18 \times 2520$ (true).

Example 3 — Rational decimal

Express $0.375$ as a fraction.

Solution: $0.375 = \dfrac{375}{1000} = \dfrac{3}{8}$ after simplifying by 125.

10 NCERT-style practice questions (easy → higher)

Each solution is hidden in a toggle. Try the question first, then open the answer.

  1. Q1 (Easy): Use Euclid's division algorithm to find HCF(312, 675).
    Show solution

    675 = 312 × 2 + 51

    312 = 51 × 6 + 6

    51 = 6 × 8 + 3

    6 = 3 × 2 + 0

    HCF = 3.

  2. Q2 (Easy): Express $0.375$ as a rational number in lowest terms.
    Show solution

    $0.375 = \dfrac{375}{1000} = \dfrac{3}{8}$.

  3. Q3 (Easy-Medium): Convert $ \dfrac{7}{40}$ to decimal form.
    Show solution

    $\dfrac{7}{40} = \dfrac{7 \times 25}{40 \times 25} = \dfrac{175}{1000} = 0.175$ (terminating).

  4. Q4 (Medium): Using prime factorization, find HCF and LCM of 360 and 126.
    Show solution

    Prime factors: $360 = 2^3 \times 3^2 \times 5$, $126 = 2 \times 3^2 \times 7$.

    HCF = $2^1 \times 3^2 = 18$.

    LCM = $2^3 \times 3^2 \times 5 \times 7 = 2520$.

  5. Q5 (Medium): Prove that $\sqrt{13}$ is irrational.
    Show solution

    Proof (by contradiction): Suppose $\sqrt{13} = \dfrac{p}{q}$ in lowest terms (p, q integers, q > 0). Then $p^2 = 13 q^2$. So 13 divides $p^2$, hence 13 divides $p$. Let $p = 13k$. Then $p^2 = 169 k^2 = 13 q^2$ ⇒ $q^2 = 13 k^2$ ⇒ 13 divides $q^2$ ⇒ 13 divides $q$.

    So both p and q are divisible by 13, contradicting lowest terms. Hence $\sqrt{13}$ is irrational.

  6. Q6 (Medium-Higher): Show that $0.\overline{142857} = \dfrac{1}{7}$.
    Show solution

    Let $x = 0.142857142857...$ (repeating block length 6). Then $10^6 x = 142857.142857...$.

    Subtract: $10^6 x - x = 142857$ ⇒ $999999 x = 142857$ ⇒ $x = \dfrac{142857}{999999} = \dfrac{1}{7}$ (since 142857 × 7 = 999999).

  7. Q7 (Higher): Prove that $\sqrt[3]{2}$ is irrational.
    Show solution

    Assume $\sqrt[3]{2} = \dfrac{p}{q}$ in lowest terms. Then $p^3 = 2 q^3$.

    So 2 divides $p^3$, so 2 divides p. Let $p = 2k$. Then $8k^3 = 2 q^3$ ⇒ $4k^3 = q^3$.

    Thus 2 divides $q^3$, so 2 divides q. Therefore p and q are both even, contradicting lowest terms. So $\sqrt[3]{2}$ is irrational.

    Alternate clean proof uses unique prime factorization: exponent of 2 on left is multiple of 3, on right it's 1 + multiple of 3, contradiction.

  8. Q8 (Higher): Find HCF(1456, 240) using Euclid's algorithm.
    Show solution

    1456 = 240 × 6 + 16

    240 = 16 × 15 + 0

    HCF = 16.

  9. Q9 (Advanced): Write 1/6 as a decimal. Identify if it terminates or repeats and why.
    Show solution

    $\dfrac{1}{6} = 0.1666\ldots = 0.1\overline{6}$ (repeating). Reason: denominator in lowest terms is 6 = 2 × 3; presence of the prime 3 makes it repeating, not terminating.

  10. Q10 (Advanced): Factorize 2310 into primes and find how many positive divisors it has.
    Show solution

    $2310 = 2 \times 3 \times 5 \times 7 \times 11$ (all primes).

    Each prime exponent is 1, so number of divisors = $(1+1)^5 = 2^5 = 32$.

5 FAQs (simple → higher)

  1. Q: How do I quickly tell if a fraction has a terminating decimal?
    A: Reduce to lowest terms. If denominator has only 2s and 5s as prime factors ($2^a 5^b$), decimal terminates. Else it repeats.
  2. Q: Why does Euclid's algorithm work for HCF?
    A: Because if $a = bq + r$, any common divisor of a and b also divides r, and vice versa. Repeating this reduces the pair of numbers while preserving common divisors until the remainder is zero.
  3. Q: Are all square roots of non-perfect squares irrational?
    A: Yes. If n is not a perfect square, $\sqrt{n}$ is irrational. Proof is by contradiction and uniqueness of prime factorization.
  4. Q: How is the Fundamental Theorem of Arithmetic useful in problems?
    A: It lets you break numbers into primes and compare prime exponents to compute HCF, LCM, number of divisors, etc., reliably and quickly.
  5. Q: Can decimals like 0.999... be equal to 1? How does that fit into real numbers?
    A: Yes. $0.\overline{9} = 1$. The real number system identifies different decimal representations that are equal. This is consistent with limits and rational representation.

Study tips

  • Practice Euclid's algorithm by hand for random pairs — it becomes fast with practice.
  • Learn to spot terminating vs repeating decimals by examining denominator factors.
  • Use prime factorization for LCM/HCF problems — it's reliable and exam-friendly.
  • When you get stuck, paste the problem into our 24-hour AI Math Solver on the FUZY LMS — it explains steps like a teacher.

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