NCERT Solutions for Class 9 Maths Chapter 1: Number System - Exercise 1.5

Exercise (1.5)

1. Classify the Following Numbers As Rational or Irrational:

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

Ans: The given number is $2-\sqrt{5}$.

Here, $\sqrt{5}=2.236.....$ and it is a non-repeating and non-terminating irrational number.

Therefore, substituting the value of $\sqrt{5}$ gives

$2-\sqrt{5}=2-2.236.....$

$=-0.236.....$, which is an irrational number.

So, $2-\sqrt{5}$ is an irrational number.

(ii) $\left(  {3+}\sqrt{ {23}} \right) {-}\left( \sqrt{ {23}} \right)$

Ans: The given number is $\left( 3+\sqrt{23} \right)-\left( \sqrt{23} \right)$.

The number can be written as

$\begin{align} & \left( 3+\sqrt{23} \right)-\sqrt{23}=3+\sqrt{23}-\sqrt{23} \\ & =3 \\ \end{align}$

$=\dfrac{3}{1}$, which is in the $\dfrac{p}{q}$ form and so, it is a rational number.

Hence, the number $\left( 3+\sqrt{23} \right)-\sqrt{23}$ is a rational number.

(iii) $\dfrac{ {2}\sqrt{ {7}}}{ {7}\sqrt{ {7}}}$

Ans: The given number is $\dfrac{2\sqrt{7}}{7\sqrt{7}}$.

The number can be written as

$\dfrac{2\sqrt{7}}{7\sqrt{7}}=\dfrac{2}{7}$, which is in the $\dfrac{p}{q}$  form and so, it is a rational number.

Hence, the number  $\dfrac{2\sqrt{7}}{7\sqrt{7}}$ is a rational number.

(iv) $\dfrac{ {1}}{\sqrt{ {2}}}$

Ans: The given number is $\dfrac{1}{\sqrt{2}}$.

It is known that, $\sqrt{2}=1.414.....$ and it is a non-repeating and non-terminating irrational number.

Hence, the number $\dfrac{1}{\sqrt{2}}$ is an irrational number.

(v) $ {2\pi }$

Ans: The given number is $2\pi $.

It is known that, $\pi =3.1415$ and it is an irrational number.

Now remember that, Rational $\times $ Irrational = Irrational.

Hence, $2\pi $ is also an irrational number.

2. Simplify Each of the of the Following Expressions:

(i) $\left(  {3+}\sqrt{ {3}} \right)\left(  {2+}\sqrt{ {2}} \right)$

Ans: The given number is $\left( 3+\sqrt{3} \right)\left( 2+\sqrt{2} \right)$.

By calculating the multiplication, it can be written as

$\left( 3+\sqrt{3} \right)\left( 2+\sqrt{2} \right)=3\left( 2+\sqrt{2} \right)+\sqrt{3}\left( 2+\sqrt{2} \right)$.

$=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$.

(ii) $\left(  {3+}\sqrt{ {3}} \right)\left(  {3-}\sqrt{ {3}} \right)$

Ans: The given number is $\left( 3+\sqrt{3} \right)\left( 3-\sqrt{3} \right)$.

By applying the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, the number can be written as

$\left( 3+\sqrt{3} \right)\left( 3-\sqrt{3} \right)={{3}^{2}}-{{\left( \sqrt{3} \right)}^{2}}=9-3=6$.

(iii)  ${{\left( \sqrt{ {5}} {+}\sqrt{ {2}} \right)}^{ {2}}}$

Ans: The given number is ${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}$.

Applying the formula ${{\left( a+b \right)}^{2}}={{a}^{2+}}2ab+{{b}^{2}}$, the number can be written as

${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}={{\left( \sqrt{5} \right)}^{2}}+2\sqrt{5}\sqrt{2}+{{\left( \sqrt{2} \right)}^{2}}$

 $=5+2\sqrt{10}+2$

 $=7+2\sqrt{10}$.

(iv)  $\left( \sqrt{ {5}}-\sqrt{ {2}} \right)\left( \sqrt{ {5}} {+}\sqrt{ {2}} \right)$

Ans: The given number is $\left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}+\sqrt{2} \right)$.

Applying the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, the number can be expressed as

$\left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}+\sqrt{2} \right)={{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}$

 $\begin{align} & =5-2 \\ & =3. \\ \end{align}$

3. Recall that, $ {\pi }$ is defined as the ratio of the circumference (say $ {c}$) of a circle to its diameter (say $ {d}$). That is, $ {\pi =}\dfrac{ {c}}{ {d}}$ .This seems to contradict the fact that $ {\pi }$ is irrational. How will you resolve this contradiction?

Ans: It is known that, $\pi =\dfrac{22}{7}$, which is a rational number. But, note that this value of $\pi $ is an approximation.

On dividing $22$ by $7$, the quotient $3.14...$ is a non-recurring and non-terminating number. Therefore, it is an irrational number.

In order of increasing accuracy, approximate fractions are

$\dfrac{22}{7}$, $\dfrac{333}{106}$, $\dfrac{355}{113}$, $\dfrac{52163}{16604}$, $\dfrac{103993}{33102}$, and \[\dfrac{245850922}{78256779}\].

Each of the above quotients has the value $3.14...$, which is a non-recurring and non-terminating number.

Thus, $\pi $ is irrational.

So, either circumference $\left( c \right)$ or diameter $\left( d \right)$ or both should be irrational numbers.

Hence, it is concluded that there is no contradiction regarding the value of $\pi $ and it is made out that the value of $\pi $ is irrational.

4. Represent $\sqrt{ {9} {.3}}$ on the number line.

Ans: Follow the procedure given below to represent the number $\sqrt{9.3}$.

• First, mark the distance $9.3$ units from a fixed-point $A$ on the number line to get a point $B$. Then $AB=9.3$ units.

• Secondly, from the point $B$ mark a distance of $1$ unit and denote the ending point as $C$.

• Thirdly, locate the midpoint of $AC$ and denote as $O$.

• Fourthly, draw a semi-circle to the centre $O$ with the radius $OC=5.15$ units. Then 

$\begin{align} & AC=AB+BC \\ & =9.3+1 \\ & =10.3 \\ \end{align}$

So, $OC=\dfrac{AC}{2}=\dfrac{10.3}{2}=5.15$.

• Finally, draw a perpendicular line at $B$ and draw an arc to the centre $B$ and then let it meet at the semicircle $AC$ at $D$ as given in the diagram below.

5. Rationalize the denominators of the following:

(i) $\dfrac{ {1}}{\sqrt{ {7}}}$

Ans: The given number is $\dfrac{1}{\sqrt{7}}$.

Multiplying and dividing by $\sqrt{7}$ to the number gives

$\dfrac{1}{\sqrt{7}}\times \dfrac{\sqrt{7}}{\sqrt{7}}=\dfrac{\sqrt{7}}{7}$.

(ii) $\dfrac{ {1}}{\sqrt{ {7}} {-}\sqrt{ {6}}}$

Ans: The given number is $\dfrac{1}{\sqrt{7}-\sqrt{6}}$.

Multiplying and dividing by $\sqrt{7}+\sqrt{6}$ to the number gives

$\dfrac{1}{\sqrt{7}-\sqrt{6}}\times \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\dfrac{\sqrt{7}+\sqrt{6}}{\left( \sqrt{7}-\sqrt{6} \right)\left( \sqrt{7}+\sqrt{6} \right)}$

Now, applying the formula $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ to the denominator gives

$\begin{align} & \dfrac{1}{\sqrt{7}-\sqrt{6}}=\dfrac{\sqrt{7}+\sqrt{6}}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}} \\ & =\dfrac{\sqrt{7}+\sqrt{6}}{7-6} \\ & =\dfrac{\sqrt{7}+\sqrt{6}}{1}. \\ \end{align}$

(iii) $\dfrac{ {1}}{\sqrt{ {5}} {+}\sqrt{ {2}}}$

Ans: The given number is $\dfrac{1}{\sqrt{5}+\sqrt{2}}$.

Multiplying and dividing by $\sqrt{5}-\sqrt{2}$ to the number gives

$\dfrac{1}{\sqrt{5}+\sqrt{2}}\times \dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{2}}{\left( \sqrt{5}+\sqrt{2} \right)\left( \sqrt{5}-\sqrt{2} \right)}$

Now, applying the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$  to the denominator gives

$\begin{align} & \dfrac{1}{\sqrt{5}+\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{2}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \\ & =\dfrac{\sqrt{5}-\sqrt{2}}{5-2} \\ & =\dfrac{\sqrt{5}-\sqrt{2}}{3}. \\ \end{align}$

(iv) $\dfrac{ {1}}{\sqrt{ {7}} {-2}}$

Ans: The given number is $\dfrac{1}{\sqrt{7}-2}$.

Multiplying and dividing by $\sqrt{7}+2$ to the number gives

$\dfrac{1}{\sqrt{7}-2}=\dfrac{\sqrt{7}+2}{\left( \sqrt{7}-2 \right)\left( \sqrt{7}+2 \right)}\\$.

Now, applying the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ to the denominator gives

$\begin{align} & \dfrac{1}{\sqrt{7}-2}=\dfrac{\sqrt{7}+2}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( 2 \right)}^{2}}} \\ & =\dfrac{\sqrt{7}+2}{7-4} \\ & =\dfrac{\sqrt{7}+2}{3}. \\ \end{align}$

Important Concepts Discussed in Class 9 Maths Chapter 1

• Number systems

• Irrational Numbers

• Real Numbers And Their Decimal Expansions

• Representing Real Numbers On The Number Line

• Operations On Real Numbers

• Laws Of Exponents For Real Numbers

Class 9 Maths Chapter 1 - Number System

Class 9 Maths Chapter 1 is about the properties of whole numbers, natural numbers, irrational numbers, and prime numbers, etc. This is one of the easiest and scoring chapters of the syllabus but only when you learn the basic concepts. In exercise 1.5, you will encounter problems in which you need to find the difference between rational and irrational numbers. Here, at Vedantu, you will come across the complete solutions which will give you a solution to every question for this exercise in a precise manner. 

In this exercise, you will be given numbers in the complex form like exponential and square root. You need to simplify those expressions to discover whether it is a rational number or irrational number.  

Here, with Vedantu’s notes for Class 9 Maths Chapter 1 Exercise 1.5, you will be able to develop your basic concepts regarding the numbers and their characteristics.

Below mentioned are some of the types of problems so that you can get a gist of exercise 1.5. These are the questions for which you will get the complete solution in Vedantu CBSE Class 9 Maths Chapter 1 Exercise 1.5 Solutions. 

Flip through the types of problems asked in the chapter to get the solution. 

Rational Numbers and Irrational Numbers 

Here, in the solution of Class 9 Maths Chapter 1 Exercise 1.5, you will go through the stepwise procedure to solve problems based on the classification of numbers into rational and irrational numbers. You will go through the steps to convert the square roots into decimals so that you can find a solution to the given problems. You can solve our quizzes to get perfection for these types of problems. Our quizzes include every type of questions to boost your confidence.

Simplify of Expression 

Class 9 Maths Chapter 1 Exercise 1.5 includes the simplification of complex expression which includes square roots and whole numbers. You can solve these problems by simple methods given in Class 9 Maths Chapter 1 Exercise 1.5 solutions.  You can read the procedure and short cuts to solve these problems and save your time. If any doubt arises while solving or understanding the solution, then you can ask our experts about this.  

Word Problem

In word problems from chapter 1 exercise 1.5, you will encounter calculations of arithmetic expressions. Our experts will teach you about the precise calculation you need to do for solving these problems. You can download the solutions from our website for free. These solutions are available in the form of PDF.

Representation of Numbers on Number Line 

In this chapter, you have to deal with the questions where you will be given different numbers in square root form and you need to represent them on a number line. Representation of a number is always a tough task for students but at the same time, it is scoring. Therefore, to solve these types of questions you should have knowledge about the construction of semi-circles. This technique is clearly taught by our experts in the Class 9 Maths Chapter 1 Exercise 1.5 solutions. You can learn this in our live sessions too by the Vendatu’s experts.

Questions on Rationalization 

Problems related to rationalization are one of the most important questions in this chapter, Number System. These are very simple once you will get the method of solving them. This type of question will able to collect good marks in the exams. In these questions, there are square root denominators and these can be rationalized. You can find the stepwise solution for it in our solution for Class 9 Maths Chapter 1 Exercise 1.5. You can revise the important points before start solving the questions through our revision notes for you.

You can find examples for terminating and non terminating decimals in this CBSE Class 9 Maths Chapter 1 Exercise 1.5 Solutions. 

In further exercises of chapter Number System Class 9, you will find out the questions based on a number line and irrational and rational numbers. Therefore, it is important to clear your all doubts in this exercise by solving all the questions with perfection. We also provide recorded video lectures through which a student can learn the concepts and methods to solve the question asked in Class 9th Maths Chapter 1.

Our professionals have created solutions for the students which can be easily learned by them. As our experts are highly qualified, therefore, we are able to give students simple tips and tricks to solve the questions in less time. 

Class 9 Maths Chapter 1

Maths has always been an interesting subject which can give a boost to your overall percentage. It gives you the strength to keep yourself different from others; therefore, we at Vedantu give you unique solutions to learn the concept easily. These will help you to give your best in the final examination. 

In Maths Chapter 1 class 9, you will know about integer, whole number and natural number in detail. Integers are those numbers that can’t have a decimal but the whole value in both positive and negative form. Whole numbers do not include fractions and negative terms. Natural numbers are always positive and can be represented on a number line which starts from 1 and ends at infinity. You will get the complete knowledge about different types of numbers with examples in our NCERT Solutions for Class 9 Maths Chapter 1.

In these solutions, you will know how these numbers like integers, whole numbers, and natural numbers can be represented on the number line. With the number line representation, you will able to learn how you can solve difficult expressions and convert them into a simple one. In these solutions, we have also included six laws for exponents to solve the problems.  Here you will get perfect knowledge about these numbers so that you will be able to solve advanced math problems in the entire maths syllabus.

We, at Vedantu, give you a one-stop solution for NCERT Maths Class 9 Chapter 1. In this solution, you can find each and every easy solution for every question. 

Download NCERT Solution Class 9 Maths PDF to score better in your upcoming examination.

Benefits of Opting Vedantu for Class 9 Maths Chapter 1 Exercise 1.5

Vedantu always efficiently teaches students. Whether the students are struggling with Maths problems of exercise 1.1 or 1.6, we have covered solutions for every exercise. We offer students the best study material, in which they can find step by step solution so that they can understand it in a better way. Here, the solutions are drafted by some of the top faculty and subject experts. These experts know how the solutions can be made simple for the students. The simpler solution will take you less time, and you will be able to remember them for a longer time.

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Our CBSE Class 9 Maths Chapter 1 Exercise 1.5 Solutions are created on the basis of NCERT guidelines. Therefore, we are able to cover every question and provide a solution for that. We are here to provide you with NCERT Solutions for every grade so that in every grade you will get the best marks.

Importance of Rational, Irrational and Real Numbers in Our Daily Life

Rational, irrational and real numbers are very important in our day to day life. These numbers makes the daily life maths easier. Most important use of these numbers are in physics and engineering. We need these numbers to measure quantities, calculating taxes or interests, and also measuring portions. 

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclid's Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

NCERT Solution Class 9 Maths of Chapter 1 All Exercises