On this page you will find Maths RD Sharma Class 9 Number System Exercise 1.4 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 1 for class 9 deals with the topic of triangles and it is one of the most important chapters.

**Download RD Sharma Class 9 Number System Exercise 1.4 Solutions in PDF**

Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 1 Number System Exercise 1.4 below and it will be beneficial for them.

**RD Sharma Class 9 Number System Exercise 1.4 Solutions**

**Q1. Define an irrational number.
**

*Solution:*

An irrational number is a real number which can be written as a decimal but not as a fraction i.e.

it cannot be expressed as a ratio of integers. It cannot be expressed as terminating or repeating

decimal.

**Q2. Explain how an irrational number is differing from rational numbers?
**

*Solution:*An irrational number is a real number which can be written as a decimal but not as a

fraction i.e. it cannot be expressed as a ratio of integers. It cannot be expressed as terminating or

repeating decimal.

For example, $0.10110100$ is an irrational number

A rational number is a real number which can be written as a fraction and as a decimal i.e. it can

be expressed as a ratio of integers. . It can be expressed as terminating or repeating decimal.

For examples,

**Q3. Find, whether the following numbers are rational and irrational
**

**(i) $\sqrt{7}$**

**(ii) $\sqrt{4}$**

**(iii) $2+\sqrt{3}$**

**(iv) $\sqrt{3}+\sqrt{2}$**

**(v) $\sqrt{3}+\sqrt{5}$**

**(vi) $(\sqrt{2}-2)^{2}$**

**(vii) $(2-\sqrt{2})(2+\sqrt{2})$**

**(viii) $(\sqrt{2}+\sqrt{3})^{2}$**

**(ix) $\sqrt{5}-2$**

(x) $\sqrt{23}$

(x) $\sqrt{23}$

**(xi) $\sqrt{225}$**

**(xii) $0.3796$**

**(xiii) 7.478478……**

**(xiv) $1.101001000100001 \ldots \ldots$**

*Solution:*

(i) $\sqrt{7}$ is not a perfect square root so it is an Irrational

number.

(ii) $\sqrt{4}$ is a perfect square root so it is an rational number.

We have,

$\sqrt{4}$ can be expressed in the form of

$\frac{a}{b}$, so it is a rational number. The decimal

representation of $\sqrt{9}$ is $3.0 .3$ is a rational number.

(iii) $2+\sqrt{3}$

Here, 2 is a rational number and $\sqrt{3}$ is an irrational number

So, the sum of a rational and an irrational number is an irrational number.

(iv) $\sqrt{3}+\sqrt{2}$

$\sqrt{3}$ is not a perfect square and it is an irrational number and $\sqrt{2}$ is not a perfect square and is an irrational number. The sum of an irrational number and an irrational number is an irrational number, so $\sqrt{3}+\sqrt{2}$ is an irrational number.

(v) $\sqrt{3}+\sqrt{5}$

$\sqrt{3}$ is not a perfect square and it is an irrational number and $\sqrt{5}$ is not a perfect square and is an irrational number. The sum of an irrational number and an irrational number is an irrational number, so $\sqrt{3}+\sqrt{5}$ is an irrational number.

(vi) $(\sqrt{2}-2)^{2}$

We have, $(\sqrt{2}-2)^{2}$

$=2+4-4 \sqrt{2}$

$=6+4 \sqrt{2}$

6 is a rational number but $4 \sqrt{2}$ is an irrational number.

The sum of a rational number and an irrational number is an irrational number, so

$(\sqrt{2}+\sqrt{4})^{2}$ is an irrational number.

(vii) $(2-\sqrt{2})(2+\sqrt{2})$

We have,

$(2-\sqrt{2})(2+\sqrt{2})=(2)^{2}-(\sqrt{2})^{2}$ $\left[\right.$ Since, $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

$4-2=\frac{2}{1}$

Since, 2 is a rational number.

$(2-\sqrt{2})(2+\sqrt{2})$ is a rational number.

(viii) $(\sqrt{2}+\sqrt{3})^{2}$

We have,

$(\sqrt{2}+\sqrt{3})^{2}=2+2 \sqrt{6}+3=5+\sqrt{6}$ $\left[\right.$ Since,$(a+b)^{2}=a^{2}+2 a b+b^{2}$

The sum of a rational number and an irrational number is an irrational number, so $(\sqrt{2}+\sqrt{3})^{2}$ is an irrational number.

(ix) $\sqrt{5}-2$

The difference of an irrational number and a rational number is an irrational number.

$(\sqrt{5}-2)$ is an irrational number.

(x) $\sqrt{23}$

$\sqrt{23}=4.795831352331 \ldots$

As decimal expansion of this number is non-terminating, non-recurring so it is an irrational number.

(xi) $\sqrt{225}$

$\sqrt{225}=15=\frac{15}{1}$

$\sqrt{225}$ is rational number as it can be represented in $\frac{\mathrm{p}}{\mathrm{q}}$ form.

(xii) $0.3796$

$0.3796$, as decimal expansion of this number is terminating, so it is a rational number.

(xiii) $7.478478 \ldots \ldots$

$7.478478=7.478$, as decimal expansion of this number is non-terminating recurring so it is a rational number.

(xiv) $1.101001000100001 \ldots \ldots$

$1.101001000100001 \ldots \ldots .$ as decimal expansion of this number is non-terminating, non-recurring so it is an irrational number

**Q4. Identify the following as irrational numbers. Give the decimal representation of rational numbers:
**

**(i) $\sqrt{4}$**

**(ii) $3 \times \sqrt{18}$**

**(iii) sqrt1.44**

**(iv) $\sqrt{\frac{9}{27}}$**

**(v) $-\sqrt{64}$**

**(vi) $\sqrt{100}$**

*Solution:*

(i) We have,

$\sqrt{4}$ can be written in the form of

$\frac{p}{q}$. So, it is a rational number. Its decimal

representation is $2.0$

(ii). We have,

$3 \times \sqrt{18}$

$=3 \times \sqrt{2 \times 3 \times 3}$

$=9 \times \sqrt{2}$

Since, the product of a ratios and an irrational is an irrational number.

$9 \times \sqrt{2}$ is an irrational.

$3 \times \sqrt{18}$ is an irrational number.

(iii) We have,

sqrt1. 44

$=\sqrt{\frac{144}{100}}$

$=\frac{12}{10}$

$=1.2$

Every terminating decimal is a rational number, so $1.2$ is a rational number.

Its decimal representation is $1.2$.

(iv) $\sqrt{\frac{9}{27}}$

We have,

$\sqrt{\frac{9}{27}}$8

$=\frac{3}{\sqrt{27}}$

$=\frac{1}{\sqrt{3}}$

Quotient of a rational and an irrational number is irrational numbers so

$\frac{1}{\sqrt{3}}$ is an irrational number.

$\sqrt{\frac{9}{27}}$ is an irrational number.

(v) We have,

$-\sqrt{64}$

$=-8$

$=-\frac{8}{1}$

$=-\frac{8}{1}$ can be expressed in the form of $\frac{\mathrm{a}}{\mathrm{b}}$

so $-\sqrt{64}$ is a rational number.

Its decimal representation is $-8.0$.

(vi) We have,

$\sqrt{100}$

$=10$ can be expressed in the form of $\frac{a}{b}$,

so $\sqrt{100}$ is a rational number

Its decimal representation is $10.0$.

**Q5. In the following equations, find which variables $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ etc. represent rational or irrational numbers:
**

**(i) $x^{2}=5$**

**(ii) $\mathrm{y}^{2}=9$**

**(iii) $z^{2}=0.04$**

**(iv) $\mathrm{u}^{2}=\frac{17}{4}$**

**(v) $\mathrm{v}^{2}=3$**

**(vi) $\mathrm{w}^{2}=27$**

**(vii) $\mathrm{t}^{2}=0.4$**

*Solution:*

(i) We have,

x^{2}=5

Taking square root on both the sides, we get

$x=\sqrt{5}$

$\sqrt{5}$ is not a perfect square root, so it is an irrational number.

(ii) We have,

$=y^{2}=9$

$=3$

$=\frac{3}{1}$ can be expressed in the form of $\frac{\mathrm{a}}{\mathrm{b}}$, so it a rational number.

(iii)

$\mathrm{Z}^{2}=0.04$

Taking square root on the both sides, we get

$z=0.2$

$\frac{2}{10}$ can be expressed in the form of $\frac{\mathrm{a}}{\mathrm{b}}$, so it is a rational number.

(iv) We have,

$\mathrm{u}^{2}=\frac{17}{4}$

Taking square root on both sides, we get,

$\mathrm{u}=\sqrt{\frac{17}{4}}$

$\mathrm{u}=\frac{\sqrt{17}}{2}$

Quotient of an irrational and a rational number is irrational, so u is an Irrational number.

(v) We have,

$\mathrm{V}^{2}=3$

Taking square root on both sides, we get,

$v=\sqrt{3}$

$\sqrt{3}$ is not a perfect square root, so $v$ is irrational number.

(vi) We have,

$\mathrm{W}^{2}=27$

Taking square root on both the sides, we get,

$w=3 \sqrt{3}$

Product of a irrational and an irrational is an irrational number. So $\mathrm{w}$ is an irrational number.

(vii) We have,

$\mathrm{t}^{2}=0.4$

Taking square root on both sides, we get,

$t=\sqrt{\frac{4}{10}}$

$t=\frac{2}{\sqrt{10}}$

Since, quotient of a rational and an Irrational number is irrational

number. $\mathrm{t}^{2}=0.4$ is an irrational number.

**Q6. Give an example of each, of two irrational numbers whose:
**

**(i) Difference in a rational number.**

**(ii) Difference in an irrational number.**

**(iii) Sum in a rational number.**

**(iv) Sum is an irrational number.**

**(v) Product in a rational number.**

**(vi) Product in an irrational number.**

**(vii) Quotient in a rational number.**

**(viii) Quotient in an irrational number.**

*Solution:*

(i) $\sqrt{2}$ is an irrational number.

Now, $\sqrt{2}-\sqrt{2}=0$

0 is the rational number.

(ii) Let two irrational numbers are $3 \sqrt{2}$ and $\sqrt{2}$.

$3 \sqrt{2}-\sqrt{2}=2 \sqrt{2}$

$5 \sqrt{6}$ is the rational number.

(iii) $\sqrt{11}$ is an irrational number.

Now, $\sqrt{11}+(-\sqrt{11})=0$

0 is the rational number.

(iv) Let two irrational numbers are $4 \sqrt{6}$ and $\sqrt{6}$

$4 \sqrt{6}+\sqrt{6}$

$5 \sqrt{6}$ is the rational number.

(iv) Let two Irrational numbers are $7 \sqrt{5}$ and

$\sqrt{5}$

Now, $7 \sqrt{5} \times \sqrt{5}$

$=7 \times 5$

$=35$ is the rational number.

(v) Let two irrational numbers are $\sqrt{8}$ and $\sqrt{8}$.

Now, $\sqrt{8} \times \sqrt{8}$

8 is the rational number.

(vi) Let two irrational numbers are $4 \sqrt{6}$ and $\sqrt{6}$

Now, $\frac{4 \sqrt{6}}{\sqrt{6}}$

$=4$ is the rational number

(vii) Let two irrational numbers are $3 \sqrt{7}$ and $\sqrt{7}$

Now, 3 is the rational number.

(viii) Let two irrational numbers are $\sqrt{8}$ and $\sqrt{2}$

Now $\sqrt{2}$ is an rational number.

**Q7. Give two rational numbers lying between $0.232332333233332$ and $0.212112111211112$.
**

*Solution:*Let $\mathrm{a}=0.212112111211112$

And, $\mathrm{b}=0.232332333233332 \ldots$

Clearly, $a<b$ because in the second decimal place a has digit 1 and $b$ has digit 3 If we

consider rational numbers in which the second decimal place has the digit 2 , then they will lie between $\mathrm{a}$ and $\mathrm{b}$.

Let. $x=0.22$

$y=0.22112211 \ldots$… Then $a<x<y<b$

Hence, $x$, and $y$ are required rational numbers.

**Q8. Give two rational numbers lying between $0.515115111511115$ and $0.5353353335$
**

*Solution:*Let, $a=0.515115111511115 \ldots$

And, $b=0.5353353335 .$

We observe that in the second decimal place a has digit 1 and $\mathrm{b}$ has digit 3 , therefore, $a<b$.

So If we consider rational numbers

$x=0.52$

$y=0.52062062 \ldots$

We find that,

$a<x<y<b$

Hence $x$ and $y$ are required rational numbers.

**Q9. Find one irrational number between $0.2101$ and $0.2222 \ldots=0 . \overline{2}$
**

*Solution:*

Let, $a=0.2101$ and,

$b=0.2222 \ldots$

We observe that in the second decimal place a has digit 1 and $\mathrm{b}$ has digit 2 , therefore $a<b$ in the third decimal place a has digit 0 .

So, if we consider irrational numbers

$x=0.211011001100011 \ldots$

We find that $a<x<b$

Hence $\mathrm{x}$ is required irrational number.

**Q10. Find a rational number and also an irrational number lying between the numbers $0.3030030003 \ldots$ and $0.3010010001$…
**

*Solution:*Let,

$\mathrm{a}=0.3010010001$ and,

$b=0.3030030003 \ldots$

We observe that in the third decimal place a has digit 1 and $\mathrm{b}$ has digit

3 , therefore $a<b$ in the third decimal place a has digit $1 .$ So, if we

consider rational and irrational numbers

$x=0.302$

$y=0.302002000200002 \ldots . .$

We find that $a<x<b$ and, $a<y<b$.

Hence, $x$ and $y$ are required rational and irrational numbers respectively.

**Q11. Find two irrational numbers between $0.5$ and $0.55 .$
**

*Solution:*Let $a=0.5=0.50$ and $b=0.55$

We observe that in the second decimal place a has digit 0 and $\mathrm{b}$ has digit

5, therefore $\mathrm{a}<0$ so, if we consider irrational numbers

$x=0.51051005100051 \ldots$

$y=0.530535305353530 \ldots$

We find that $a<x<y<b$

Hence $x$ and $y$ are required irrational numbers.

**Q12. Find two irrational numbers lying between $0.1$ and $0.12$.
**

*Solution:*

Let $a=0.1=0.10$

And $b=0.12$

We observe that In the second decimal place a has digit 0 and $\mathrm{b}$ has digit 2 .

Therefore, $a<b$.

So, if we consider irrational numbers

$x=0.1101101100011 \ldots y=0.111011110111110 \ldots$ We find that $a<x<y<0$

Hence, $x$ and $y$ are required irrational numbers.

**Q13. Prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.**

**If possible, let $\sqrt{3}+\sqrt{5}$ be a rational number equal to $x$.
**

*Solution:*

Then,

$x=\sqrt{3}+\sqrt{5}$

$\mathrm{x}^{2}=(\sqrt{3}+\sqrt{5})^{2}$

$x^{2}=8+2 \sqrt{15}$

$\frac{x^{2}-8}{2}=\sqrt{15}$

Now, $\sqrt{\frac{x^{2}-8}{2}}$ is rational

$\sqrt{15}$ is rational

Thus, we arrive at a contradiction.

Hence, $\sqrt{3}+\sqrt{5}$ is an irrational number.