NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers Exercise 1.3: Detailed Guide for Assam State Board Syllabus 2025-2026 (English Medium)

Free Solutions for Assam State Board Class 10 Maths Chapter 1 Exercise 1.3 in English Medium

 

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3 – Digital Pipal Academy

Digital Pipal Academy offers detailed and easy-to-follow NCERT Solutions for Class 10 Maths Chapter 1, Exercise 1.3, covered under the Assam State Board Syllabus 2025-2026. This exercise focuses on the Fundamental Theorem of Arithmetic and its applications. Our expert faculty provides step-by-step explanations to ensure students understand the concepts thoroughly. These solutions strictly follow NCERT guidelines, helping students prepare effectively and score better in their exams.

Real Number	Solution Link
Exercise 1.1 (Real Number)	Click Here
Exercise 1.2 (Real Number)	Click Here
Exercise 1.3 (Real Number)	Click Here
Exercise 1.4 (Real Number)	Click Here

 

Exercise 1.3

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

Solution:
Let, if possible, $\sqrt{5}$ is a rational number.

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

where $p, q \in \mathbb{Z}$, $q \neq 0$, and $p, q$ are coprime.

$$p^2 = 5q^2$$

Since 5 divides $p^2$, it follows that 5 divides $p$.

Let

$$p = 5m$$

for some integer $m$.

$$(5m)^2 = 5q^2$$ $$25m^2 = 5q^2$$ $$q^2 = 5m^2$$

Since 5 divides $q^2$, it follows that 5 divides $q$.

Thus, both $p$ and $q$ are divisible by 5, contradicting that they are coprime.

Hence, our assumption was wrong, proving that $\sqrt{5}$ is irrational.

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

Solution:
Let, if possible, $3 + 2\sqrt{5}$ is a rational number.

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

where $a, b \in \mathbb{Z}$, $b \neq 0$, and $a, b$ are coprime.

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

Since $\frac{a}{b} - 3$ is a rational number, it follows that $2\sqrt{5}$ is rational.

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

Since the right-hand side is rational, this contradicts the fact that $\sqrt{5}$ is irrational.

Thus, our assumption was wrong, proving that $3 + 2\sqrt{5}$ is irrational.

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3. Prove that the following are irrational:

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

Solution:
Let, if possible, $\frac{1}{\sqrt{2}}$ is rational.

$$\frac{1}{\sqrt{2}} = \frac{a}{b}$$

where $a, b \in \mathbb{Z}$, $b \neq 0$.

$$\sqrt{2} = \frac{b}{a}$$

Since the right-hand side is rational, this contradicts the fact that $\sqrt{2}$ is irrational.

Thus, our assumption was wrong, proving that $\frac{1}{\sqrt{2}}$ is irrational.

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(ii) $7\sqrt{5}$

Solution:
Let, if possible, $7\sqrt{5}$ is rational.

$$7\sqrt{5} = \frac{a}{b}$$

where $a, b \in \mathbb{Z}$, $b \neq 0$.

$$\sqrt{5} = \frac{a}{7b}$$

Since the right-hand side is rational, this contradicts the fact that $\sqrt{5}$ is irrational.

Thus, our assumption was wrong, proving that $7\sqrt{5}$ is irrational.

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(iii) $6 + \sqrt{2}$

Solution:
Let, if possible, $6 + \sqrt{2}$ is rational.

$$6 + \sqrt{2} = \frac{a}{b}$$

where $a, b \in \mathbb{Z}$, $b \neq 0$.

$$\sqrt{2} = \frac{a}{b} - 6$$

Since $\frac{a}{b} - 6$ is rational, it follows that $\sqrt{2}$ is rational.

This contradicts the fact that $\sqrt{2}$ is irrational.

Thus, our assumption was wrong, proving that $6 + \sqrt{2}$ is irrational.

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Sudev Chandra Das (B.Sc. Mathematics)

Hi! I'm Sudev Chandra Das, Founder of Digital Pipal Academy. I've dedicated myself to guiding students toward better education. I believe, 'Success comes from preparation, hard work, and learning from failure.' Let’s embark on a journey of growth and digital excellence together!.

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