NCERT Solutions for Class 10 Maths Chapter 1 - Exercise

1. Prove that $\sqrt{5}$ is irrational.

Ans: We have to prove that $\sqrt{5}$ is irrational.

We will use a contradiction method to prove it.

Let $\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $\sqrt{5}=\dfrac{a}{b}$

Now, squaring both sides, we get

${{\left( \sqrt{5} \right)}^{2}}={{\left( \dfrac{a}{b} \right)}^{2}}$

\[\Rightarrow 5=\dfrac{{{a}^{2}}}{{{b}^{2}}}\] 

$\Rightarrow {{a}^{2}}=5{{b}^{2}}$ …….(1)

If ${{a}^{2}}$ is divisible by $5$ then $a$ is also divisible by $5$.

Let $a=5k$, where, $k$ is any integer.

Again squaring both sides, we get

\[\Rightarrow {{a}^{2}}={{\left( 5k \right)}^{2}}\] 

Substitute the value in eq. (1), we get

\[\Rightarrow {{\left( 5k \right)}^{2}}=5{{b}^{2}}\]

$\Rightarrow {{b}^{2}}=5{{k}^{2}}$ …..(2)

If ${{b}^{2}}$ is divisible by $5$ then $b$ is also divisible by $5$.

From, eq. (1) and (2), we can conclude that $a$ and $b$ have $5$ as a common factor.

This contradicts our assumption.

Therefore, we can say that $\sqrt{5}$ is irrational.

Hence proved.

2. Prove that $3+2\sqrt{5}$ is irrational.

Ans: We have to prove that $3+2\sqrt{5}$ is irrational.

We will use a contradiction method to prove it.

Let $3+2\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $3+2\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow 2\sqrt{5}=\dfrac{a}{b}-3$ 

$\Rightarrow \sqrt{5}=\dfrac{1}{2}\left( \dfrac{a}{b}-3 \right)$ ……..(1)

From eq. (1) we can say that $\dfrac{1}{2}\left( \dfrac{a}{b}-3 \right)$ is rational so $\sqrt{5}$ must be rational.

But this contradicts the fact that $\sqrt{5}$ is irrational. Hence the assumption is false.

Therefore, we can say that $3+2\sqrt{5}$ is irrational.

Hence proved.

3. Prove that following are irrationals:

(i) $\dfrac{1}{\sqrt{2}}$ 

Ans: We have to prove that $\dfrac{1}{\sqrt{2}}$ is irrational.

We will use a contradiction method to prove it.

Let $\dfrac{1}{\sqrt{2}}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}$

$\Rightarrow \sqrt{2}=\dfrac{b}{a}$ ………..(1)

From eq. (1) we can say that $\dfrac{b}{a}$ is rational so $\sqrt{2}$ must be rational.

But this contradicts the fact that $\sqrt{2}$ is irrational. Hence the assumption is false.

Therefore, we can say that $\dfrac{1}{\sqrt{2}}$ is irrational.

Hence proved.

(ii) $7\sqrt{5}$ 

Ans: We have to prove that $7\sqrt{5}$ is irrational.

We will use a contradiction method to prove it.

Let $7\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $7\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow \sqrt{5}=\dfrac{a}{7b}$ ………..(1)

From eq. (1) we can say that $\dfrac{a}{7b}$ is rational so $\sqrt{5}$ must be rational.

But this contradicts the fact that $\sqrt{5}$ is irrational. Hence the assumption is false.

Therefore, we can say that $7\sqrt{5}$ is irrational.

Hence proved.

(iii) $6+\sqrt{2}$ 

Ans: We have to prove that $6+\sqrt{2}$ is irrational.

We will use a contradiction method to prove it.

Let $6+\sqrt{2}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $6+\sqrt{2}=\dfrac{a}{b}$

$\Rightarrow \sqrt{2}=\dfrac{a}{b}-6$ ………..(1)

From eq. (1) we can say that $\dfrac{a}{b}-6$ is rational so $\sqrt{2}$ must be rational.

But this contradicts the fact that $\sqrt{2}$ is irrational. Hence the assumption is false.

Therefore, we can say that $6+\sqrt{2}$ is irrational.

Hence proved.

Maths Syllabus of CBSE Class 10

The Mathematics syllabus of Class 10 contains some crucial part of Mathematics such as Geometry, Algebra, Arithmetic, Trigonometry, Statistics, etc. There are 15 chapters in Class 10 in Mathematics. The chapters are real numbers, polynomials, pair of linear equations in two variables, Arithmetic progressions, quadratic equations, triangles, introduction to Trigonometry, coordinate Geometry, applications of Trigonometry, constructions, circles, the area related to circles, surface areas and volumes, probability, Statistics. There are chapters from different parts of Mathematics.

CBSE Class 10 Mathematics Chapter 1

Central Board of Secondary Education (CBSE) has included some vital concepts of Arithmetic in the Class 10 Mathematics syllabus. Chapter 1 of Class 10 Mathematics is one of them. This chapter contains all about real numbers. Real numbers are a strong conceptual topic of Mathematics. From this chapter, the students will get to know the definition, properties and classification of the real number. This chapter is also an essential part of the mathematical number system. In this chapter, some examples and property proof is included for the students. By solving those examples, the students can learn the real number system easily. The property proofs will be beneficial for them. The students should practise the exercises of the textbook. They will get a good topic insight from Class 10 maths ex 1.3 NCERT solutions and it will be efficient for your practice.

NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3

Class 10 maths ex 1.3 is an important practice set for the students. This exercise contains some testing problems of real numbers. By practising those problems, the students will get a clear understanding of the properties and classification of real numbers. The examples are much efficient to clear the conceptual part. The solutions of this exercise are provided by NCERT. Class 10 maths ex 1.3 NCERT solutions are available online for free. The students can download it for their convenience.

Necessity of NCERT Solutions Class 10 Maths Ch 1 Ex 1.3

Practising exercises of individual Mathematics chapters is essential for the Class 10 board exam. The students should practise the exercises of specific chapters sincerely. If they get solutions for those exercises after solving, that will be beneficial for the students. That is why Class 10 maths ex 1.3 NCERT solutions are necessary. It increases the confidence of the students.

Irrational Numbers

Any number which cannot be represented in the form of p/q (where p and q are integers and q≠0.) is an irrational number. Examples √2,π, e, and so on.

Interesting Results of Number Theory

• If a number p (a prime number) divides a2, then p divides a. Example: 3 divides 62 i.e 36, which implies that 3 divides 6.

• The addition or subtraction of a rational and an irrational number always results in an irrational number.

• The multiplication value of a non-zero rational number and an irrational number is always irrational.

• The quotient obtained when division operation is performed between a non-zero rational number and an irrational number is always irrational.

• √m is irrational when ‘m’ is a prime. For example, 11 is a prime number and √11 is irrational which can be proved by the method of “Proof by contradiction”.

Proof by Contradiction

In a contradiction method, we start with an assumption which is contrary to what we are required to prove. Using a series of logical deductions from this assumption of contradiction, we will reach a mathematical inconsistency (error) – which enables us to conclude that our assumption of contradiction was incorrect.

Let’s suppose, we are going to prove √2 is an irrational number through the method of contradiction. Firstly, we assume the contradiction that √2 is a rational number and So it can be written in the form of a/b, where a and b are two co-prime numbers and n ≠ 0. By observing, we will find that there exist no coprime integers a and b for √2, so our assumption was wrong. 

Solved Examples

1. Prove that, √5 is an international number.

Solution:

Suppose, √5 is rational. Therefore, there are two integers a and b, where a/b=√5.

Suppose, the common factor of a and b is other than 1. So, we can assume them as co-prime numbers by dividing by the common factor.

a = b√5

Or,  5b2 = a2

Therefore, a2 is divisible by 5 as well as a is divisible by 5.

Let, a=5k, where k is an integer

Now, (5k)2=5b2

Or, b2 = 5k2

So, b2 and b both are divisible by 5.

Hence, we can say that a and b have common factor 5, which contradicts the co-prime fact.

Therefore, √5 cannot be expressed in p/q form. √5 is an irrational number (proved).

2. 3+2√5 is irrational - prove this.

Solution:

Suppose, 3+2√5 is a rational number.

That means, there are two integers called a and b such that 

a/b=3+2√5

Or, a/b - 3 = 2√5

Or, √5 = 1/2 (a/b-3)

As a and b are integers and 1/2 (a/b - 3) will also be rational. So, √5 is rational.

This fact contradicts that √5 is irrational.

Hence, our consumption is false and √5 is an irrational number.

NCERT Solutions for Class 10 Maths Chapter 1 Exercises