# Real Numbers (Chapter 1, Exercise 1.3) – Class 10 Maths Solutions NCERT

This page contains Exercise 1.3, Chapter 1 of Class 10 Mathematics under NCERT Syllabus. The Chapter 1 of Class 10 Mathematics is about the Real Numbers. This part contains the Solutions of Questions based on Irrational Numbers which the part of Chapter of Class 10.

The solutions of all the three questions of Exercise 1.3 is given below.

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

Solution:

Let $\sqrt{5}$ is a rational number and $\sqrt{5}= \frac{a}{b}$, where $a$ and $b$ are coprime.

$\therefore b\times \sqrt{5}=a$

Squaring both sides,

$\Rightarrow b^2\times 5=a^2\,\,\,\,\,……(i)$

$\Rightarrow b^2=\frac{a^2}{5}$

This means that $5$ divides $a^2$. Therefore, $5$ divides $a$ also and we can write,

$a=5c$, where $c$ is some integer.

Now, from Equation $(i)$

$\Rightarrow 5\times b^2=a^2$

$\Rightarrow 5\times b^2=(5c)^2$

$\Rightarrow 5\times b^2=25c^2$

$\Rightarrow b^2=5c^2$

$\Rightarrow \frac{b^2}{5}=c^2$

This means that $5$ divides $b^2$ and hence $5$ divides $b$ also.

Therefore, $5$ is a common factor of $a$ and $b$. This contradicts the fact that $a$ and $b$ are coprime.

The contradiction is due to the incorrect assumption of $\sqrt{5}$ as rational number.

$\therefore \sqrt{5}$ is irrational.

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

Solution:

Let us assume that $3+2\sqrt{5}$ is a rational number.

$\therefore 3+2\sqrt{5}=\frac{a}{b}$, where $a$ and $b$ are coprime.

Now,

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

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

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

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

Since, $a,\, b,\, 2$ and $3$ are integers, $\frac{a-3b}{2b}$ is also an integer and hence rational. This means that $\sqrt{5}$ is also rational. This contradicts the fact that $\sqrt{5}$ is irrational.

The contradiction is due to the incorrect assumption that $3+2\sqrt{5}$ is a rational number. $\therefore 3+2\sqrt{5}$ is irrational.

## Question 3: Prove that the following are irrationals:

### i. $\frac{1}{\sqrt{2}}$

Solution:

Let us assume that $\frac{1}{\sqrt{2}}$ is rational number.

$\therefore \frac{1}{\sqrt{2}}=\frac{a}{b}$, where $a$ and $b$ are coprime.

Now,

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

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

Since $a$ and $b$ are integers, $\frac{b}{a}$ is also an integer and hence rational. This means that $\sqrt{2}$ is also rational.

This contradicts the fact that $\sqrt{2}$ is irrational. The contradiction is due to the incorrect assumption that $\frac{1}{\sqrt{2}}$ is rational.

$\therefore \frac{1}{\sqrt{2}}$ is irrational.

### ii. $7\sqrt{5}$

Solution:

Let us assume that $7\sqrt{5}$ is rational number.

$\therefore 7\sqrt{5}=\frac{a}{b}$, where $a$ and $b$ are coprime.

Now,

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

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

Since $a$, $b$ and $7$ are integers, $\frac{a}{7b}$ is also an integer and hence rational. This means that $\sqrt{5}$ is also rational.

This contradicts the fact that $\sqrt{5}$ is irrational. The contradiction is due to the incorrect assumption that $7\sqrt{5}$ is rational.

$\therefore 7\sqrt{5}$ is irrational.

### iii.$6+\sqrt{2}$

Solution:

Let us assume that $6+\sqrt{2}$ is rational number.

$\therefore 6+\sqrt{2}=\frac{a}{b}$, where $a$ and $b$ are coprime.

Now,

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

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

Since $a$, $b$ and $6$ are integers, $\frac{a}{b}-6$ is also an integer and hence rational. This means that $\sqrt{2}$ is also rational.

This contradicts the fact that $\sqrt{2}$ is irrational. The contradiction is due to the incorrect assumption that $6+\sqrt{2}$ is rational.

$\therefore 6+\sqrt{2}$ is irrational.

I hope the solutions of Class 10, Exercise 1.3, Chapter 1 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 1.3, feel free to contact me at [email protected] or [email protected] or fill the form here.

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