Real Numbers

Notes of chapter: Real Numbers are presented below. Indepth notes along with worksheets and NCERT Solutions for Class 10.


(1) Real numbers-

Real numbers are collection of rational and irrational numbers.

(2) Euclid’s Division Lemma-

We know-

It is just equivalent to division equation

Dividend = Divisor × Quotient + Remainder

But a formal statement is given by Euclid which is known as Euclid’s Division Lemma.

Given positive integers a and b, there exist whole numbers q and r satisfying

a = bq + r, 0 ≤ r < b.

Here, q and r are quotient and remainder.

a = dividend and

b = divisor

Eg:- Find out the integers q and r for positive integers 4, 19.


Euclid’s Division Lemma

a = bq + r, 0 ≤ r < b.


19 = 4 × 4 + 3

(4 goes into 19 up to 4) (remainder is 3)

4 = 3 × 1 + 1

3 = 1 × 3 + 0

(3) Euclid’s Division Algorithm

This theorem is based on the Euclid’s Division Lemma.

According to this algorithm-

Certain steps should be follow to get HCF of two positive integers c and d with c>d.

Steps to follow-

Step 1- Apply Euclid’s Division Lemma , to c and d to find out whole numbers q and r. Such that c = dq + r, 0 ≤ r < d.

Step 2- If r = 0, d is the HCF of c and d

If r ≠ 0, apply lemma to d and r.

Step 3- Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

This algorithm works because HCF (c,d) = HCF (d,r)

Where, the symbol HCF (c, d) denotes the HCF of c and d etc.

Eg:- Use Euclid’s algorithm to find the HCF of 455 and 42.

Step 1- Apply Euclid’s Division Lemma, to c and d

455 = 42 × 10 + 35

Step 2- If r ≠ 0, apply lemma to d and r.

Here, r ≠ 0, apply Euclid’s Division Lemma

42 = 35 × 1 + 7

Again apply Euclid’s Division Lemma

35 = 7  5 + 0

Remainder is zero.

Divisor is 7.

Therefore, 7 is HCF of 455 and 42.

Utility of Euclid’s Division Algorithm –

(i) Euclid’s Algorithm is very useful for calculating HCF of very large numbers.

(ii) It is one of the earliest examples of an algorithm that a computer had been programmed to carry out.

(4)Fundamental theorem of Arithmetic-

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.

The prime factorisation of a natural number is unique, except for the order of its factors.

Given a composite number x, we factorise it as x = p1p2… pn,

Where p1 , p2… pn are prime numbers written in ascending order.


120 = 2 × 2 × 2 × 3 × 5

= 23  3  5 (order is ascending)

Therefore, factorisation is unique.

(5) Revisiting Irrational Numbers-

Eg:- √2, √3, π, 0.10110111011110….

Theorem 1

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


Let the prime factorisation of a = p1p2………,

Where p1,p2……… are primes, not necessarily distinct.


a2 = (p1p2……… (p1p2………

= p12 p22………..pn2.

p divides a2 (Given)

Therefore, according to fundamental theorem of arithmetic p is one of the prime factors of a2.

By using uniqueness part of the fundamental theorem of arithmetic

Only prime factors of a2 are p1, p2,………,

p is one of p1, p2,………

Now, since a = p1p2………,

Therefore, p divides a.

Theorem 2- ∠2 is irrational.


Let us assume that ∠2 is rational.

Suppose a and b have a common factor other than 1.

Let, a and b are coprime numbers.

So, b√2 = a

2b2 = a2 (squaring both sides)…(1)

Therefore, 2 divides a2.

So, 2 divides a. (Theorem 1-If p divides a2, then p divides a)

a = 2c…(2)

Put value of a in equation 1

2b2 = (2c)2

2b2 = 4c2

b2 = 2 c2

Therefore, 2 divides b2.

So, 2 divides b. (Theorem 1- If p divides a2, then p divides a)

Therefore, a and b have at least 2 as a common factor.

But, a and b have no common factor other than 1 because both are coprime numbers. Therefore, it contradicts.

Hence, √2  is irrational.

Eg: – Show that 5 – √3 is irrational.


Let us assume that 5 – √3 is rational.

We can find coprime a and b, where b ≠ 0,

But, this contradicts the fact that √3 is irrational number.

Therefore, our assumption that 5 -√3 is rational is incorrect.

So, we conclude that 5 – √3 is irrational.

Revisiting Rational Numbers and their Decimal Expansion-

Clearly remainder is not zero and denominator is not of the form 2n5m. But after some time decimal expansion is repeating. Therefore it is non – terminating recurring decimal expansion.

Helping Topics

NCERT Solutions Class 9

Worksheet Class 9