Home/
CBSE/
Class 10/
Mathematics/
Study Notes/
Real Numbers

Class 10 . Mathematics . Mathematics . Real Numbers

Chapter 1: Real Numbers

Ch 1

Mathematics

Class 10

Lecture 1

Introduction

Reference video · Lecture 1Introduction

Chapter Roadmap

Fundamental Theorem of Arithmetic

Definition

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 prime factors.

Uniqueness of prime factorisation

Writing the prime factors in ascending order gives one unique factorisation. Grouping equal primes gives a product of prime powers.

$$x=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k},\quad p_1<p_2<\cdots<p_k$$

Key points and applications

  • Prime factors can repeat; for example, \(4=2\times2\) and \(253=11\times23\).
  • Prime powers are a compact way to write repeated factors, such as \(32760=2^3\times3^2\times5\times7\times13\).
  • Applications include checking divisibility, finding HCF and LCM, proving certain roots irrational, and analysing rational-number decimal expansions from the denominator’s prime factors.
  • Prime supply is infinite, so primes and their products generate infinitely many positive integers.
A factor tree showing the prime factorisation of 32760.
A factor tree showing the prime factorisation of 32760.

Prime factorisation

  1. Break the number into factors until every factor is prime.
  2. Write all prime factors in ascending order, then combine repeated primes as powers.
Worked example

Find the prime factorisation of 32760.

  1. Factorise 32760 by splitting it into convenient factors.
    $$32760=3276\times10$$
  2. Continue factoring each part until only primes remain.
    $$3276=2^2\times3^2\times7\times13,\quad 10=2\times5$$
  3. Collect equal prime factors as powers.
    $$32760=2^3\times3^2\times5\times7\times13$$
Answer: \(32760=2^3\times3^2\times5\times7\times13\)

Using Prime Factors to Find HCF and LCM

Definition

Prime-factorisation method — A method for finding HCF and LCM by comparing the prime powers in the numbers’ factorizations.

HCF and LCM from prime factors

For the HCF, take the smallest exponent of each prime common to all the numbers. For the LCM, take the greatest exponent of every prime appearing in any of them.

$$\operatorname{HCF}=\prod p^{\text{smallest common exponent}},\quad \operatorname{LCM}=\prod p^{\text{greatest exponent}}$$
HCF–LCM relation for two positive integers
$$\operatorname{HCF}(a,b)\times\operatorname{LCM}(a,b)=a\times b$$

Use this to find either the HCF or LCM once the other is known.

Find HCF and LCM by prime factorisation

  1. Write each number as a product of prime powers.
  2. For the HCF, multiply the smallest powers of the primes common to every number.
  3. For the LCM, multiply the greatest powers of all primes present.
Worked example

Find the HCF and LCM of 6 and 20 using prime factorisation.

  1. Write 6 and 20 as prime powers.
    $$6=2\times3,\quad 20=2^2\times5$$
  2. The only shared prime is 2; use its smallest exponent for the HCF.
    $$\operatorname{HCF}(6,20)=2^1=2$$
  3. Use the greatest exponent of every prime present for the LCM.
    $$\operatorname{LCM}(6,20)=2^2\times3\times5=60$$
  4. Check the two-number product relation.
    $$2\times60=6\times20=120$$
Answer: \(\operatorname{HCF}(6,20)=2,\quad \operatorname{LCM}(6,20)=60\)

Important applications and limits

  • Ending in zero: A number ending in \(0\) is divisible by \(5\); its prime factorisation must therefore contain \(5\).
  • Powers of 4: Since \(4^n=2^{2n}\) contains only the prime \(2\), no natural-number power of \(4\) ends in \(0\).
  • Two numbers: The HCF–LCM product relation applies to any pair of positive integers.
  • Three or more numbers: In general, the product of all the numbers is not equal to their HCF multiplied by their LCM.
⚠ Common mistake

Do not use the smallest exponent of every prime for the HCF: only primes common to all numbers contribute. The HCF–LCM product identity is guaranteed for two numbers, not generally for three or more.

Practice

Can \(4^n\) end in the digit \(0\) for any natural number \(n\)?

Hint: A number ending in \(0\) must have \(5\) as a prime factor.
Key step: \(4^n=2^{2n}\) has no prime factor \(5\).
Solution: No. \(4^n\) contains only the prime factor \(2\), so it is not divisible by \(5\) and cannot end in \(0\).
Practice

Find the HCF and LCM of \(96\) and \(404\) using prime factorisation.

Hint: Factorise both numbers, then use the smallest common and greatest overall prime powers.
Key step: \(96=2^5\times3\) and \(404=2^2\times101\).
Solution: The HCF is \(2^2=4\). The LCM is \(2^5\times3\times101=9696\).
Practice

Find the HCF and LCM of \(6\), \(72\) and \(120\) by prime factorisation.

Hint: Use the smallest powers common to all three for the HCF, and the greatest powers present for the LCM.
Key step: \(6=2\times3\), \(72=2^3\times3^2\), and \(120=2^3\times3\times5\).
Solution: The HCF is \(2\times3=6\). The LCM is \(2^3\times3^2\times5=360\). The product of the three numbers is not \(6\times360\).

Quick recap

  • Euclid’s division algorithm writes \(a=bq+r\) with \(0\le r<b\) and helps find HCF.
  • Every composite number has a unique prime factorisation, except for the order of its factors.
  • The HCF uses the smallest powers of primes common to all numbers; the LCM uses the greatest powers of all primes present.
  • For two positive integers, \(\operatorname{HCF}(a,b)\times\operatorname{LCM}(a,b)=a\times b\).
  • A power of \(4\) cannot end in \(0\) because its prime factorisation contains no factor \(5\).
  • Prime factorisation can help prove irrationality and determine the decimal form of rational numbers.
Lecture 2

Revisiting Irrational Numbers

Reference video · Lecture 2Revisiting Irrational Numbers

Irrational Numbers and Prime Factorisation

Definition

Irrational number — A real number is irrational if it cannot be written as \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q\ne 0\).

Examples and real numbers

  • Examples: \(\sqrt{2},\sqrt{3},\sqrt{15},\pi,\frac{\sqrt{2}}{\sqrt{3}}\) are irrational.
  • Decimal example: \(0.10110111011110\ldots\) is non-terminating and non-repeating.
  • Real numbers: Rational and irrational numbers together form the real numbers.
Fundamental Theorem of Arithmetic

Every composite number has a prime factorisation, unique apart from the order of its prime factors.

A Prime Dividing a Square

Prime divisor property

If \(p\) is prime and \(p\mid a^2\), then \(p\mid a\), for a positive integer \(a\). Prime factorisation shows that the prime factors of \(a^2\) come from those of \(a\).

Prime factorisation of a square
$$a^2=p_1^2p_2^2\cdots p_n^2$$

If \(a=p_1p_2\cdots p_n\), then each prime factor appears twice in \(a^2\).

Proving Square Roots of Primes Are Irrational

Square root of a prime

For every prime \(p\), \(\sqrt{p}\) is irrational. If we assume \(\sqrt{p}=\frac{a}{b}\) in lowest terms, then \(a^2=pb^2\) forces \(p\) to divide both \(a\) and \(b\), contradicting that they are coprime.

Proof by contradiction for \(\sqrt{p}\)

  1. Assume the opposite and write the root as a reduced fraction, with \(b\ne0\).
    $$\sqrt{p}=\frac{a}{b}$$
  2. Square and rearrange; the prime divisor property implies \(p\mid a\).
    $$a^2=pb^2$$
  3. Write \(a=pc\) and substitute; this implies \(p\mid b\) as well.
    $$b^2=pc^2$$
  4. Since \(a\) and \(b\) share the factor \(p\), this contradicts their being coprime; conclude the root is irrational.
Worked example

Prove that \(\sqrt{3}\) is irrational.

  1. Assume, for contradiction, that \(\sqrt{3}=\frac{a}{b}\) in lowest terms, where \(b\ne0\).
    $$\sqrt{3}=\frac{a}{b}$$
  2. Squaring gives \(3b^2=a^2\). Thus \(3\mid a^2\), so \(3\mid a\); write \(a=3c\).
    $$3b^2=a^2$$
  3. Substitute \(a=3c\); then \(3\mid b^2\), so \(3\mid b\).
    $$3b^2=9c^2\quad\Rightarrow\quad b^2=3c^2$$
  4. Both \(a\) and \(b\) are divisible by \(3\), contradicting that they are coprime.
Answer: Therefore, \(\sqrt{3}\) is irrational.
⚠ Common mistake

When assuming a number is rational for a contradiction proof, write it as a fraction in lowest terms before deriving divisibility.

Operations with Rational and Irrational Numbers

Rational–irrational operation properties

The sum or difference of a rational and an irrational number is irrational. The product of a non-zero rational and an irrational number is irrational; dividing an irrational number by a non-zero rational also gives an irrational number.

Show an expression is irrational

  1. Assume the expression is rational.
  2. Rearrange or divide by a non-zero rational to make a known irrational number rational.
  3. The contradiction proves that the original expression is irrational.

Illustrations

  • Difference: If \(5-\sqrt{3}\) were rational, then \(\sqrt{3}=5-(5-\sqrt{3})\) would be rational, a contradiction.
  • Product: If \(3\sqrt{2}\) were rational, then dividing by \(3\) would make \(\sqrt{2}\) rational, a contradiction.

HCF and LCM of Three Numbers

  • Three numbers: For positive integers \(p,q,r\), the product of their HCF and LCM is not generally equal to \(pqr\).
  • Example: For \(2,4,8\), the HCF is \(2\) and the LCM is \(8\), so their product is \(16\), whereas \(pqr=64\).
Practice

Prove that \(\sqrt{2}\) is irrational.

Hint: Assume \(\sqrt{2}=\frac{a}{b}\) in lowest terms and use the prime divisor property.
Key step: From \(a^2=2b^2\), show that \(2\) divides both \(a\) and \(b\).
Solution: If \(\sqrt{2}=\frac{a}{b}\), then \(a^2=2b^2\). Thus \(2\mid a\); write \(a=2c\). Substitution gives \(b^2=2c^2\), so \(2\mid b\). This contradicts coprimality; hence \(\sqrt{2}\) is irrational.
Practice

Show that \(3\sqrt{2}\) is irrational.

Hint: Assume the expression is rational and divide by \(3\).
Key step: If \(3\sqrt{2}=\frac{a}{b}\), then \(\sqrt{2}=\frac{a}{3b}\) would be rational.
Solution: Assuming \(3\sqrt{2}=\frac{a}{b}\) gives \(\sqrt{2}=\frac{a}{3b}\), which is rational. This contradicts that \(\sqrt{2}\) is irrational; therefore \(3\sqrt{2}\) is irrational.
Practice

Show that \(5-\sqrt{3}\) is irrational.

Hint: Assume the difference is rational and isolate \(\sqrt{3}\).
Key step: A rational value for \(5-\sqrt{3}\) would make \(\sqrt{3}\) rational.
Solution: If \(5-\sqrt{3}\) were rational, then \(\sqrt{3}=5-(5-\sqrt{3})\) would be rational, contradicting the irrationality of \(\sqrt{3}\). Thus \(5-\sqrt{3}\) is irrational.
★ Remember
  • Prime-root result: \(\sqrt{p}\) is irrational whenever \(p\) is prime.
  • Contradiction pattern: Assume a reduced fraction, force a prime to divide both numerator and denominator, then reject the assumption.
  • Operation shortcut: Assume the target expression is rational and rearrange to contradict a known irrationality.

Quick recap

  • An irrational number cannot be expressed as a ratio of integers with a non-zero denominator.
  • Unique prime factorisation implies that if a prime divides a square, it divides the original integer.
  • Assuming \(\sqrt{p}\) is a reduced fraction forces the prime \(p\) to divide both numerator and denominator.
  • Therefore, \(\sqrt{2}\) and \(\sqrt{3}\) are irrational; the proof applies to every prime \(p\).
  • Adding or subtracting a rational number from an irrational number gives an irrational number.
  • Multiplying an irrational number by a non-zero rational number, or dividing it by one, gives an irrational number.
  • For three positive integers, HCF times LCM is not generally equal to their product.

Formula sheet

HCF–LCM relation for two positive integers
$$\operatorname{HCF}(a,b)\times\operatorname{LCM}(a,b)=a\times b$$
Prime factorisation of a square
$$a^2=p_1^2p_2^2\cdots p_n^2$$

Mathematics — ALL CHAPTERS

1

Real Numbers

2

Polynomials

3

Pair Of Linear Equations In Two Variables

4

Quadratic Equations

5

Arithmetic Progressions

6

Triangles

7

Coordinate Geometry

8

Introduction To Trigonometry

9

Some Applications Of Trigonometry

10

Circles

11

Areas Related To Circles

12

Surface areas And Volumes

13

Statistics 

14

Probability