NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers

NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers

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CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26

Updated on 29 Apr 2025, 04:47 PM IST

Understanding the decimal representation of real numbers becomes essential for mathematics because real numbers contain all rational and irrational elements. The behavior of these numbers in decimal format is addressed in this exercise. We use this method to identify non-terminating decimals while understanding how rational numbers form repeating patterns. Learning these concepts establishes our ability to correctly identify numbers while linking theoretical number concepts to their actual decimal expressions.

This Story also Contains

  1. NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2
  2. Access Solutions of Real Numbers Class 10 Chapter 1 Exercise 1.2
  3. Topics covered in Chapter 1, Real Numbers: Exercise 1.2
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers
Ex - 1.3

Students obtain maximum benefit from these concepts when they consult with NCERT Solutions. The NCERT Books related solutions provide step-by-step explanations to simplify difficult proofs. Students who refer to these educational resources can develop fundamental number theory knowledge that helps them succeed in advanced mathematics studies as well as competitive exams.

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2

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Access Solutions of Real Numbers Class 10 Chapter 1 Exercise 1.2

Q1 Prove that $\sqrt 5$ is irrational.

Answer:

Let us assume $\sqrt{5}$ is rational, which means it can be written in the form $\frac{p}{q}$ where p and q are co-primes and $q\neq 0$

$\\\sqrt{5}=\frac{p}{q}$

Squaring both sides, we obtain

$\\\left ( \sqrt{5} \right )^{2}=\left (\frac{p}{q} \right )^{2}\\$

$5=\frac{p^{2}}{q^{2}}\\$

$p^{2}=5q^{2}$

From the above equation, we can see that p2 is divisible by 5, therefore, p will also be divisible by 5, as 5 is a prime number. $(i)$

Therefore, p can be written as 5r

p = 5r

p2 = (5r)2

5q2 = 25r2

q2 = 5r2

From the above equation, we can see that q2 is divisible by 5, Therefore, q will also be divisible by 5 as 5 is a prime number. $(ii)$

From (i) and (ii), we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that $\sqrt{5}$ is rational was wrong. Hence proved that $\sqrt{5}$ is irrational.

Q2 Prove that $3 + 2 \sqrt 5$ is irrational.

Answer:

Let us assume $3 + 2 \sqrt 5$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\3+2\sqrt{5}=\frac{p}{q}$

$2\sqrt{5}=\frac{p}{q}-3$

$\sqrt{5}=\frac{p-3q}{2q}$

As p and q are integers $\frac{p-3q}{2q}\\$ would be rational, which contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $3 + 2 \sqrt 5$ is rational was wrong. Therefore $3 + 2 \sqrt 5$ is irrational.

Q3 Prove that the following are irrationals :

(i) $\frac{1}{\sqrt 2}$

Answer:

Let us assume $\frac{1}{\sqrt{2}}$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\frac{1}{\sqrt{2}}=\frac{p}{q}$

$\sqrt{2}=\frac{q}{p}$

Since p and q are co-prime integers $\frac{q}{p}$ will be rational, which contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $\frac{1}{\sqrt{2}}$ is rational was wrong. Therefore $\frac{1}{\sqrt{2}}$ is irrational.

Q3 (2) Prove that the following are irrationals :

(ii) $7 \sqrt 5$

Answer:

Let us assume $7 \sqrt 5$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$7\sqrt{5}=\frac{p}{q}$

$\sqrt{5}=\frac{p}{7q}$

As p and q are integers $\frac{p}{7q}\\$ would be rational, which contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $7 \sqrt 5$ is rational was wrong. Therefore $7 \sqrt 5$ is irrational.

Q3 (3) Prove that the following are irrationals : $6 + \sqrt 2$

Answer:

Let us assume $6 + \sqrt 2$ is rational, this means it can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$6+\sqrt{2}=\frac{p}{q}$

$\sqrt{2}=\frac{p}{q}-6$

$\sqrt{2}=\frac{p-6q}{q}$

As p and q are integers $\frac{p-6q}{q}$ would be rational, which contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $6 + \sqrt 2$ is rational was wrong. Therefore $6 + \sqrt 2$ is irrational.

Also Read-

Topics covered in Chapter 1, Real Numbers: Exercise 1.2

1. Irrational Numbers: When expressed in the state of integers these numbers become impossible to rationalize themselves. The goal of this exercise is to identify particular numbers which prove to be irrational while developing verification methods.

2. Proof by Contradiction: The method makes an assumption that results in a contradiction to show the initial assumption is wrong. Through this approach, it becomes possible to determine that √5 represents an irrational number.

3. Properties of Rational and Irrational Numbers: Operations of addition and multiplication reveal how numbers become rational or irrational when performed.

4. Application of Prime Factorization: Prime factorization functions as an investigational tool to determine both the divisibility aspects as well as properties of numbers which lead to irrationality proofs.

5. Logical Reasoning: A necessary ability for mathematics students should be mastering logical argumentation because it extends beyond mathematical applications.


Also see-

NCERT Solutions of Class 10 Subject Wise

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

NCERT Exemplar Solutions of Class 10 Subject Wise

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

Frequently Asked Questions (FAQs)

Q: What is the sum and difference of rational and irrational numbers, rational or irrational?
A:

Sum and difference of a rational and irrational number is irrational.

Q: State the theorem “Fundamental theorem of Arithmetic”.
A:

“Fundamental theorem of Arithmetic” given in the Class 10 Maths chapter 1 states that “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”. 

Q: Which technique is used to prove root(2) irrational?
A:

The proof is based on a most common technique called ‘proof by contradiction.

Q: Is this exercise important for board exams?
A:

important in board exams, you can check previous year papers for better understanding.


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