Careers360 Logo
Real Numbers Class 10th Notes - Free NCERT Class 10 Maths Chapter 1 Notes - Download PDF

Real Numbers Class 10th Notes - Free NCERT Class 10 Maths Chapter 1 Notes - Download PDF

Edited By Ramraj Saini | Updated on Apr 14, 2025 07:43 AM IST

Real numbers are defined as numbers that include all rational and irrational numbers that can be represented on the number line, excluding complex numbers. All numbers which we use in our day-to-day life are real numbers. Like, we perform counting operations with natural numbers, yet we measure temperature through integers, express fractions by rational numbers and use irrational numbers to perform operations on square roots and cubes. Real numbers in our notes will help us study Euclid's Division Lemma as well as the Fundamental Theorem of Arithmetic and the concepts of HCF along with LCM through factorization and properties of rational and irrational numbers.

This Story also Contains
  1. NCERT Class 10 Chapter 1 Real Numbers: Notes
  2. Euclid’s Division Lemma:
  3. The Fundamental Theorem of Arithmetic:
  4. HCF And LCM by Prime Factorization
  5. Exploring Again the Irrational Numbers
  6. Exploring Again the Rational Numbers and Their Decimal Expansions
  7. Class 10 Chapter Wise Notes
Real Numbers Class 10th Notes - Free NCERT Class 10 Maths Chapter 1 Notes - Download PDF
Real Numbers Class 10th Notes - Free NCERT Class 10 Maths Chapter 1 Notes - Download PDF

The NCERT class 10th maths notes comprehensively cover all the topics, sub-topics, short tricks, formulae, and other important key points. The notes in this article are based on real numbers and the fundamental theorem of arithmetic, rational, and irrational numbers. CBSE Class 10 chapter Real Numbers also includes Euclid's division lemma and Euclid’s division algorithm. These topics can be easily downloaded from the Class 10 Maths chapter 1 notes PDF. The students must practice the NCERT examples and check the NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers. Apart from this, the use of NCERT class 10th maths notes provides you with the ability to examine mathematical concepts rigorously. Also, after studying the NCERT Notes, you will have access to short summaries that help you to remember information fast.

Background wave

NCERT Class 10 Chapter 1 Real Numbers: Notes

Real numbers: These numbers are defined as those that can be represented on the number line. They are further divided into two parts such as rational and irrational numbers. They can be represented as R. The real numbers chart is given below for a better understanding.

Types of Numbers:

There are mainly two types of numbers, such as real numbers and Imaginary numbers. For better understanding, look at the chart given below.

1744089434080

1. Natural Numbers: These are the numbers that start from the count of one till infinity. The natural numbers can be represented by “N”.
Example: N = {1, 2, 3, 4, 5, 6……}

2. Whole Numbers: These include all natural numbers but start from the count of zero. The whole numbers can be represented by “W”.
Example: W = { 0, 1, 2, 3, 4, 5, 6…...}

3. Integers: These are the numbers that include both positive and negative numbers along with zero.
Example: {…-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5……}

4. Rational numbers: These are the numbers that are any real
number and can be expressed in the form pq of R where the denominator q≠0. It is represented by "Q". It is noted that every integer number is a rational number.
Example: 12, 34, −78, 0.3333..., etc.

5. Irrational numbers: These are the numbers that cannot be expressed in the form pq of R where the denominator q≠0.
Example: √2,√3, π, 0.34343..., etc.

6. Prime numbers: The prime numbers are those numbers that can not be divided by any number other than 1 and the number itself.
Example: 2,3, 5, 7, 11, etc.

7. Composite numbers: The composite numbers are those numbers that can be divided by at least one of the numbers other than 1 and itself.
Example: 4, 6, 8, 9, 12, etc.

8. Co-primes: The co-primes are those numbers whose HCF is 1.
Example: (13, 21) (19, 23) etc.

Euclid’s Division Lemma:

According to Euclid’s division Lemma, for two positive integers, a and b, there exists a unique integer such that a=bq+r, where 0≤r<b. Where a is a dividend, b is a divisor, q is a quotient, and r is the remainder. The lemma is always equivalent to: Dividend = Divisor × Quotient + Remainder.

Usage of Euclid's Division Lemma

Euclid’s Division Lemma can be used in finding the highest common factor of any two positive integers.

The steps are included to find HCF with the help of Euclid’s Division Lemma.

  • Let the two positive integers be a and b, where “a” is greater than “b”.

Now, applying Euclid’s division to these numbers a and b, also find the two numbers q and r.

  • In this step, checking the value of r, if it is found to be 0, then b is the HCF of these numbers.

  • Continuing this process until we get the value 0. Reaching the value of 0 makes the divisor b the HCF of a and b.

NEET/JEE Offline Coaching
Get up to 90% Scholarship on your NEET/JEE preparation from India’s Leading Coaching Institutes like Aakash, ALLEN, Sri Chaitanya & Others.
Apply Now

The Fundamental Theorem of Arithmetic:

The fundamental theorem of arithmetic states that composite numbers can be expressed as the product of prime numbers.

1744089545697

In the above-given arrangement, the highlighted numbers are the prime numbers. Thus, the prime factors are: 5, 7, 11 and 13.

Statement: Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factor occurs.
For example:
33=3×11
21=3×7
6=2×3

HCF And LCM by Prime Factorization

HCF: HCF is the highest common factor, also known as the greatest common divisor (GCD). It is defined as the greatest number that divides each of the given numbers without leaving any remainder, as the highest common factor (HCF) of two or more given numbers.
The HCF can be calculated using two methods such as:

1) Prime factorisation: In this method, we express the given numbers as the products of their respective prime factors. After that, we select the common prime factors from the product of both numbers
Example: To find the H.C.F of 28 and 35
Prime factors of 28 = 2 × 2 × 7
Prime factors of 35 = 7 × 5
The factor common to 28 and 35 is 7, which in turn is the HCF of 28 and 35.

(2) Euclid’s division algorithm: It is defined as the multiple use of Euclid’s division lemma to find the HCF of two numbers.

Example: To find the HCF of 16 and 58.

1744089592196


As shown in the figure above, the HCF of 16 and 58 is 2.

LCM: It is the least common multiple, in which the least number exactly divisible by each one of the given numbers is called the LCM of the numbers.
Example: To find the LCM (Least Common Multiple) of 87 and 145
87 = 3 × 29
145 = 5 × 29
The common prime factor is 29
The uncommon prime factors are 3 for 87 and 5 for 145
Thus, the LCM of 87 and 145 = 3 × 5 × 29, which is 435.

Product of Two Numbers = Product of their LCM and HCF.

For any two positive integers X and Y,
X × Y = HCF × LCM
Example: For 77 and 99, the HCF is 11 and the LCM is 693
77 × 99 = 7623
11 × 693 = 7623
Therefore,77 × 99 = 11 × 693

Exploring Again the Irrational Numbers

Irrational Numbers: An irrational number is any number that cannot be expressed in the form of pq (where p and q are integers and q≠0.).
Examples: √5,π etc.

Let p be a prime number. If p divides x2, then p divides x where x is a positive integer.

Example: 3 divides 92, which is equal to 81, which implies that 3 divides 9.
Point to remember:
1) The subtraction or addition of irrational numbers with rational numbers always produces an irrational result.
2) When multiplying or dividing a non-zero irrational number by a rational number produces an irrational outcome is produced.
3) The square root of prime numbers exists as an irrational value. For example, 5 is a prime number, and √5 is irrational. The validity of the above statement depends on Proof by contradiction methods, where we need to execute the following steps to check whether a given statement is correct or incorrect.

Proof by Contradiction: In this method of proof by contradiction, we have to check whether a given statement is correct or incorrect.
(I) We start our explanation with the premise that the statement is true.
(II) By reaching an inconsistent result, we prove that our initial assumption was incorrect.
Example: Prove that √5 is irrational.
Now, let’s assume that √5 is rational.
Since it is rational √5 can be expressed as
√5 = ab, where a and b are co-prime Integers, b≠0.
On squaring, a2b2 = 5
⇒ a2 = 5b2.
Thus, 5 divides a. Then, there exists a number c such that a = 5c. Then, a2 = 25c2. Thus, 5b2 = 25c2 or b2 = 5c2.
Therefore, 5 divides b. The number 5 acts as a common factor in both the expressions of a and b. The proof contradicts our assumption about a and b being coprime integers.

Exploring Again the Rational Numbers and Their Decimal Expansions

Rational Numbers: These are the numbers that are any real
number and can be expressed in the form pq of R where the denominator q≠0. It is represented by "Q". It is noted that every integer number is a rational number.
Example: 12, 34, −78, 0.3333..., etc.

Terminating Decimals: Terminating decimals are those decimals which end at a specific point. It is also known as finite decimals.
Example: 0.87, 82.25, 9.527, etc.

Non-terminating Decimals: are those decimals where the digits after the decimal point are repeated continuously and infinitely.
Example: 0.142857142857….., 0.3333…., 0.111… etc.

Non-terminating decimals are further divided into parts :

a) Recurring – In this decimal fraction number or group of numbers is repeated indefinitely.
Example: 0.142857142857…

b) Non-recurring – In this, a decimal number continues without any repetition pattern of digits..
Example: 3.1415926535…

Determine if a Given Rational Number is Terminating or Not

If xy is a rational number, and they are satisfies the following conditions, then its decimal expansion would terminate:
a) The HCF of x and y should be 1.
b) y can be represented as a prime factorisation of 2 and 5, i.e. y = 2a × 5b, where either a or b, or both, can = 0.
If the prime factorisation of y contains any number other than 2 or 5, then the decimal expansion of that number will be recurring

Example:
150 = 0.02 is a terminating decimal, as the HCF of 1 and 50 is 1, and the denominator (50) can be expressed as 2 × 52.

17 = 0.1428571 is a recurring decimal as the HCF of 1 and 7 is 1, and the denominator (7) is equal to 71.

Class 10 Chapter Wise Notes

Students must download the notes below for each chapter to ace the topics.


NCERT Exemplar Solutions for Class 10

Students must check the NCERT Exemplar solutions for class 10 of Mathematics and Science Subjects.

NCERT Solutions for Class 10

Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.

NCERT Books and Syllabus

To learn about the NCERT books and syllabus, read the following articles and get a direct link to download them.

Frequently Asked Questions (FAQs)

1. What are real numbers in Class 10 Maths?

Real numbers are defined as numbers that include all rational and irrational numbers that can be represented on the number line, excluding complex numbers. All numbers which we use in our day-to-day life are real numbers. Like, we perform counting operations with natural numbers, yet we measure temperature through integers, express fractions by rational numbers and use irrational numbers to perform operations on square roots and cubes. 

2. What are the properties of real numbers?

Real numbers' primary properties are:

Closure Property: Any real number is always the sum, difference, and product of two real numbers.

Commutative property: For all real numbers, the commutative property states that a + b = b + a and a × b = b × a.

Associative properties: (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c) are examples of associative properties.

Distributive property: According to the distributive property, a × (b + c) equals a × b + a × c.

Identity Property: The multiplicative identity is 1 and the additive identity is 0.

Inverse property: The Inverse Property states that 1/a is the multiplicative inverse of a (a ≠ 0) and -a is the additive inverse of a.

3. What is the difference between rational and irrational numbers?

Rational numbers, such as 1/2, 3, -5, and 0.75, can be written as p/q, where p and q are integers and q ≠ 0. They are either repeated or terminating in their decimal representation.

Irrational numbers have non-terminating, non-repeating decimal expansions (such as √2, π, and e) and cannot be represented as p/q.

4. What is Euclid’s Division Lemma?

According to Euclid’s division Lemma, for two positive integers, a and b, there exists a unique integer such that a = bq + r, where 0 ≤ r < b. Where a is a dividend, b is a divisor, q is a quotient, and r is the remainder. The lemma is always equivalent to: Dividend = Divisor × Quotient + Remainder.

5. How to find HCF using Euclid’s algorithm?

The steps are included to find HCF with the help of Euclid’s Division Lemma.

1) Let the two positive integers be a and b, where “a” is greater than “b”.
Now, applying Euclid’s division to these numbers a and b, also find the two numbers q and r.
2) In this step, checking the value of r, if it is found to be 0, then b is the HCF of these numbers.
3) Continuing this process until we get the value 0. Reaching the value of 0 makes the divisor b the HCF of a and b.

Articles

Upcoming School Exams

Application Date:24 March,2025 - 23 April,2025

Admit Card Date:25 March,2025 - 21 April,2025

Admit Card Date:25 March,2025 - 17 April,2025

View All School Exams

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top