# RD Sharma Class 9 Number System Exercise 1.4 Solutions

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) $\sqrt{23}$
(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.

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