maths assignment class 9 chapter 1

Search This Blog

Cbse mathematics.

Basic concepts, definitions and formulas of mathematics, mathematics assignments for 9th standard to 10+2 standard, maths study material for 8th, 9th, 10th, 11th, 12th classes, Mathematics lesson plan for classes 8th,10th and 12th standard, Interesting maths riddles and maths magic, Class-wise mathematics study material for students from 8th to 12. A complete resource centre of mathematics for students and teachers

Featured Posts

Maths assignment class viii | quadrilateral, math assignment class ix ch -1| number system, maths assignment   class 9th chapter 1.

MATHEMATICS ASSIGNMENT CLASS IX  NUMBER SYSTEM

 (a) 1⁸ × 3⁰ × 5³ × 2²           Ans; 500,                

(b)  4 -3 × 4⁸ ÷ 4²                Ans; 64

(d) 3 -4 × 3 -5 ÷ 3¹⁰             Ans; 3 -19

Download complete pdf file 

Plese send us solution today is our test

Need solutions toooo

Post a Comment

Breaking news, popular post on this blog, lesson plan maths class 10 | for mathematics teacher.

Lesson Plan Maths Class XII | For Maths Teacher

Lesson Plan Math Class X (Ch- 3) | Pair of Linear Equations in Two Variables

• Assignment 10 15
• Assignment 11 12
• Assignment 12 14
• Assignment 8 8
• Assignment 9 5
• Lesson plan 10 15
• Lesson Plan 12 13
• Lesson Plan 8 10
• Maths 10 20
• Maths 11 21
• Maths 12 17

SUBSCRIBE FOR NEW POSTS

Get new posts by email:.

NCERT Solutions Class 9 Maths Chapter 1 Number Systems

NCERT solutions for class 9 maths chapter 1 number systems consists of an introduction about the number system and the different kinds of numbers in it. The number system has been classified into different types of numbers like natural numbers, whole numbers , integers, rational numbers, irrational numbers , etc. The NCERT solutions class 9 maths chapter 1 covers all the basics of the number system which will be helpful in forming the basic foundation of mathematics.

Class 9 maths chapter 1 number systems will help the students in differentiating between rational and irrational numbers, wherein irrational numbers cannot be expressed in the form of a ratio, and also about real numbers. Class 9 maths NCERT solutions chapter 1 number systems sample exercises can be downloaded from the links below and also you can find some of these in the exercises given below.

• NCERT Solutions Class 9 Maths Chapter 1 Ex 1.1
• NCERT Solutions Class 9 Maths Chapter 1 Ex 1.2
• NCERT Solutions Class 9 Maths Chapter 1 Ex 1.3
• NCERT Solutions Class 9 Maths Chapter 1 Ex 1.4
• NCERT Solutions Class 9 Maths Chapter 1 Ex 1.5
• NCERT Solutions Class 9 Maths Chapter 1 Ex 1.6

NCERT Solutions for Class 9 Maths Chapter 1 PDF

These NCERT solutions for class 9 maths involving the important concepts of real numbers , rational and irrational numbers, are available for free pdf download. The questions involving real numbers and their decimal form, the law of exponents are given below:

☛ Download Class 9 Maths NCERT Solutions Chapter 1 Number Systems

NCERT Class 9 Maths Chapter 1   Download PDF

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

It is advisable for the students to practice the questions in the above links as this will give them better clarity on the kind of numbers and their properties. An exercise-wise detailed analysis of NCERT Solutions Class 9 Maths Chapter 1 number systems is given below for reference.

• Class 9 Maths Chapter 1 Ex 1.1 - 4 Questions
• Class 9 Maths Chapter 1 Ex 1.2 - 4 Questions
• Class 9 Maths Chapter 1 Ex 1.3 - 9 Questions
• Class 9 Maths Chapter 1 Ex 1.4 - 2 Questions
• Class 9 Maths Chapter 1 Ex 1.5 - 5 Questions
• Class 9 Maths Chapter 1 Ex 1.6 - 11 Questions

☛ Download Class 9 Maths Chapter 1 NCERT Book

Topics Covered: The important topics focussed upon are irrational numbers, real numbers, and real numbers when expanded in the decimal form. The class 9 maths NCERT solutions chapter 1 covers the representation of real numbers on a number line, methods to perform operations on real numbers, and laws of exponents when dealing with real numbers.

Total Questions: Class 9 maths chapter 1 Number Systems consists of total 35 questions of which 30 are easy, 2 are moderate and 3 are long answer-type questions.

List of Formulas in NCERT Solutions Class 9 Maths Chapter 1

NCERT solutions class 9 maths chapter 1 covers important facts about the number systems which will help strengthen the math foundation. Like if a number ‘a’ is rational, and ‘b’ represents an irrational number, then ‘a+b’, and ‘a-b’ are irrational numbers, and ‘ab’ and ‘a/b’ are supposed to be irrational numbers, and ‘b’ is not equal to zero. For ‘a’ and ‘b’ positive real numbers the following formula or entities will be true:

• √ab = √a √b
• √(a/b) = √a / √b

Important Questions for Class 9 Maths NCERT Solutions Chapter 1

Video solutions for class 9 maths ncert chapter 1, faqs on ncert solutions class 9 maths chapter 1, do i need to practice all questions provided in ncert solutions class 9 maths number systems.

Practicing the NCERT solutions class 9 maths number systems and exercises on real numbers, rational numbers will help in exploring the number systems in a better way. The NCERT Solutions Class 9 Maths Number Systems will also provide a good insight into the solving of problems.

Why are Class 9 Maths NCERT Solutions Chapter 1 Important?

Since the number systems chapter deals with rational and irrational numbers, real numbers, and their expansion, their decimal form, also covering the law of exponents. Hence, this makes the NCERT solutions class 9 maths important for examinations.

What are the Important Formulas in NCERT Solutions Class 9 Maths Chapter 1?

There are several formulas or entities for positive real numbers which will be helpful in learning mathematics even for higher grades. Like if one wants to rationalize the denominator of 1/ ( √a + b ), then we can multiply and divide by its algebraic conjugate which is √a - b

How Many Questions are there in NCERT Solutions Class 9 Maths Chapter 1 Real Numbers?

The questions in the NCERT Solutions Class 9 Maths Chapter 1 are a great way for learning real numbers. There are around 35 questions dealing with number systems with 25 of them being simple and have straightforward logic, 6 of them are with medium complexity and 4 are elaborative questions.

What are the Important Topics Covered in NCERT Solutions Class 9 Maths Chapter 1?

The NCERT Solutions Class 9 Maths Chapter 1 deal with integers, real numbers, rational and irrational numbers. Apart from these the important topics covered are the real numbers, and what happens when they are expanded in decimal form, the law of exponents in the case of real numbers, how to differentiate between rational and irrational numbers etc.

How CBSE Students can utilize NCERT Solutions Class 9 Maths Chapter 1 effectively?

The students should first practice all the examples to understand the logic and problem solving technique and should try to solve all the exercise questions. The CBSE itself recommends the NCERT Solutions Class 9 Maths for the board exam studies.

• Class 6 Maths
• Class 6 Science
• Class 6 Social Science
• Class 6 English
• Class 7 Maths
• Class 7 Science
• Class 7 Social Science
• Class 7 English
• Class 8 Maths
• Class 8 Science
• Class 8 Social Science
• Class 8 English
• Class 9 Maths
• Class 9 Science
• Class 9 Social Science
• Class 9 English
• Class 10 Maths
• Class 10 Science
• Class 10 Social Science
• Class 10 English
• Class 11 Maths
• Class 11 Computer Science (Python)
• Class 11 English
• Class 12 Maths
• Class 12 English
• Class 12 Economics
• Class 12 Accountancy
• Class 12 Physics
• Class 12 Chemistry
• Class 12 Biology
• Class 12 Computer Science (Python)
• Class 12 Physical Education
• GST and Accounting Course
• Excel Course
• Tally Course
• Finance and CMA Data Course
• Payroll Course

Interesting

• Learn English
• Learn Excel
• Learn Tally
• Learn GST (Goods and Services Tax)
• Learn Accounting and Finance
• GST Tax Invoice Format
• Accounts Tax Practical
• Tally Ledger List
• GSTR 2A - JSON to Excel

Are you in school ? Do you love Teachoo?

We would love to talk to you! Please fill this form so that we can contact you

You are learning...

Chapter 1 Class 9 Number Systems

Click on any of the links below to start learning from Teachoo ...

Get solutions of all NCERT Questions of Chapter 1 Class 9 Number System free at teachoo. Answers to all NCERT Exercises and Examples are solved for your reference. Theory of concepts is also made for your easy understanding

In this chapter, we will learn

• Different Types of numbers like Natural Numbers, Whole numbers, Integers, Rational numbers
• How to find rational numbers between two rational numbers
• What is an irrational number
• Checking if number is irrational or not
• And how to draw an irrational number on the number line
• Then, we will study What a real number is
• And find Decimal expansions - Terminating, Non terminating - repeating, Non terminating Non repeating
• Converting non-terminating repeating numbers into p/q form
• Finding irrational numbers between two numbers
• Representing real numbers on the number line (we use magnification)
• We will learn how to add , subtract and multiply numbers with square root (like 5√2 + 3√3 - 8√2)
• We will learn some identities of numbers with square root (like (√a + √b) 2 )
• How to rationalize numbers
• We will also do questions on Law of Exponents (here, the exponents can also be in fractions)

Click on an NCERT Exercise below to get started. 

Or you can also check the concepts from the Concept wise. Check it out now

Serial order wise

Concept wise.

What's in it?

Hi, it looks like you're using AdBlock :(

Please login to view more pages. it's free :), solve all your doubts with teachoo black.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems are provided here. Our NCERT Maths solutions contain all the questions of the NCERT textbook that are solved and explained beautifully. Here you will get complete NCERT Solutions for Class 9 Maths Chapter 1 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.

Class 9 Maths Chapter 1 Number Systems NCERT Solutions

Below we have provided the solutions of each exercise of the chapter. Go through the links to access the solutions of exercises you want. You should also check out our NCERT Class 9 Solutions for other subjects to score good marks in the exams.

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.1

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.2

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.3

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.4

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.5

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.6

NCERT Solutions for Class 9 Maths Chapter 1 – Topic Discussion

Below we have listed the topics that have been discussed in this chapter. As Number System is one of the important topics in Maths, it has a weightage of 6 marks in class 9 Maths exams. 

• Introduction of 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

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Save my name, email, and website in this browser for the next time I comment.

CBSE NCERT Solutions

NCERT and CBSE Solutions for free

Class 9 Mathematics Assignments

We have provided below free printable Class 9 Mathematics Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 9 Mathematics . These Assignments for Grade 9 Mathematics cover all important topics which can come in your standard 9 tests and examinations. Free printable Assignments for CBSE Class 9 Mathematics , school and class assignments, and practice test papers have been designed by our highly experienced class 9 faculty. You can free download CBSE NCERT printable Assignments for Mathematics Class 9 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Class 9. Students can click on the links below and download all Pdf Assignments for Mathematics class 9 for free. All latest Kendriya Vidyalaya Class 9 Mathematics Assignments with Answers and test papers are given below.

Mathematics Class 9 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 9 Mathematics . Students and teachers can download and save all free Mathematics assignments in Pdf for grade 9th. Our expert faculty have covered Class 9 important questions and answers for Mathematics as per the latest syllabus for the current academic year. All test papers and question banks for Class 9 Mathematics and CBSE Assignments for Mathematics Class 9 will be really helpful for standard 9th students to prepare for the class tests and school examinations. Class 9th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 9 Mathematics Download in Pdf

More assignments for class 9 mathematics.

Advantages of Class 9 Mathematics Assignments

• As we have the best and largest collection of Mathematics assignments for Grade 9, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Mathematics textbooks for Class 9 .
• All Mathematics assignments for Class 9 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 9 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Mathematics chapter wise worksheets and assignments for free in Pdf
• Class 9 Mathematics question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 9 Mathematics students

At https://www.cbsencertsolutions.com, we have provided the biggest database of free assignments for Mathematics Class 9 which you can download in Pdf

We provide here Standard 9 Mathematics chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Mathematics in Grade 9, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

We have provided here topic-wise Mathematics Grade 9 question banks, revision notes and questions for all difficult topics, and other study material.

We have provided the best collection of question bank and practice tests for Class 9 for all subjects. You can download them all and use them offline without the internet.

Related Posts

Class 9 Social Science Economics Assignments

Class 9 Biology Assignments

Class 9 Mathematics Areas of Parallelogram and Triangle Assignments

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.1

Class 9 Maths Exercise 1.1 Solutions Class 9 Maths Exercise 1.1 in Hindi 9th Maths Book Download in PDF Class 9 Maths Chapter 1 Solutions Class 9 Maths NCERT Solutions Class 9 all Subjects Solutions

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.1 Number Systems in Hindi Medium and English Medium modified for academic session 2024-25. The 9th Maths solutions are revised according to new curriculum. All the questions and solutions are updated as per new NCERT textbook issued for 2024-25.

Class 9 Maths Exercise 1.1 Solutions

• Class 9 Maths Exercise 1.1 in English
• Class 9 Maths Exercise 1.1 in Hindi
• 9th Maths Chapter 1 in PDF
• Class 9 Maths Chapter 1 Solutions
• Class 9 Maths NCERT Solutions
• Class 9 all Subjects Solutions

Download (Exercise 1.1) in PDF format otherwise you can use it online. These solutions are related to CBSE as well as Gujrat, MP and UP Board (High School) books. Visit to Discussion Forum to ask your doubts and answer the questions asked by other users.

NCERT Solutions Class 9 Maths Chapter 1 Exercise 1.1 of Number Systems is given below to use online or download in PDF for session 2024-25. Download options are given at the top of the page. Go back to Class 9 Maths Chapter 1 main page to see other exercises.

Here we are providing NCERT Solutions Class 9 Maths Chapter 1 Exercise 1.1 to view online without downloading for academic session 2024-25. The option of download are also available on this page. In this way students have choice to download or view it online only. NCERT Books of class 9 are also given in PDF.

1. Are 3, -4 and -3.6 rational numbers? If yes, then express them in the form of p/q, where p is Number Systems and q is natural number. 2. Find six rational numbers between – 2 and 2. 3. Find three rational numbers between 2/7 and 3/7. 4. Are the following statements true or false? Give the reason. Every whole number is an integer. Every rational number is a whole number. Every integer is a rational number. Chose the correct answer in each of the following: 5. Between two rational numbers (A) there is no rational number (B) there is exactly one rational number (C) there are infinitely many rational numbers (D) there are only rational numbers and no irrational numbers 6. The product of any two irrational numbers is (A) always an irrational number (B) always a rational number (C) always an integer (D) sometimes rational, sometimes irrational

Download NCERT Books and Offline apps based on new CBSE Syllabus. Ask your doubts and share your knowledge with your friends and other users through Discussion Forum.

Which questions are important for class test from exercise 1.1 class 9 Maths?

In exercise 1.1 of class 9 Maths, there are four questions and two examples (examples 1, 2). All questions and examples of this exercise are equally important from the exam point of view. Example 1 and question 4 are of the same type. Example 2 and questions 2, 3 are of the same type. From this exercise, one mark and two marks questions can come in the exam.

Which question is the best question for exams in exercise 1.1 of class 9 Maths?

Question 1 (Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?) is the best question of exercise 1.1 of class 9 Maths. This question is very interesting, logical and good.

Is exercise 1.1 of class 9th Maths easy to understand?

Yes, exercise 1.1 of class 9th Maths is very easy to understand. Question 1 of this exercise is a yes or no type question. In Questions 2, 3 and example 2, students have to find rational numbers between any two given rational numbers. Question 4 and example 1 are true or false type questions.

How much time, students need to complete exercise 1.1 of class IX Maths?

Students need a maximum of one day to complete exercise 1.1 (chapter 1) of class 9th Maths. This time also depends on student’s speed, efficiency, capability, and many other factors.

Class 9 Maths Chapter 1 Exercise 1.2 »

Copyright 2024 by Tiwari Academy | A step towards Free Education

• Bihar Board

SRM University

Cbse board result 2024.

• CBSE 10th Result 2024
• CBSE 12th Result 2024
• JAC Board Result 2024
• CG Board Result 2024
• UK Board Result 2024
• Karnataka Board Result 2024
• Shiv Khera Special
• Education News
• Web Stories
• Current Affairs
• नए भारत का नया उत्तर प्रदेश
• School & Boards
• College Admission
• Govt Jobs Alert & Prep
• GK & Aptitude
• Class 9 Maths NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 1 - Number System

Here you can find the ncert solutions for cbse class 9maths chapter 1- number system. it includes a detailed explanation of the answers and covers the easiest methods for solving the questions given in the chapter..

In this article, we are providing the NCERT solutions to Class 9 Maths Chapter 1 - Number System.  All these solutions are available in PDF format which you may access totally free of cost.  Our subject experts have reviewed these NCERT solutions to provide you the error-free content which will make it easy for you to make an effective preparation for the annual exams.

Why you should solve NCERT exercise questions?

Solving the NCERT exercise problems will help you

→ clear all the concepts and formulae you learned in a chapter

→ familiarise with different types of questions that can be asked in exams          

→ get enough practice which is the key to success in the Mathematics exam

→ improve your accuracy and speed

So, to get the desired result in exams, it’s very necessary for students to thoroughly solve the questions given at the end of each chapter of the Class 9 Maths NCERT book.

Some of the questions and solutions are as follows:

Get here latest School , CBSE and Govt Jobs notification in English and Hindi for Sarkari Naukari and Sarkari Result . Download the Jagran Josh Sarkari Naukri App . Check  Board Result 2024  for Class 10 and Class 12 like  CBSE Board Result ,  UP Board Result ,  Bihar Board Result ,  MP Board Result ,  Rajasthan Board Result  and Other States Boards.

• SSC GD Result 2024
• WBBSE Result 2024
• wbbse.wb.gov.in Result 2024
• WB Board 10th Result 2024 Jagran Josh
• WB Madhyamik Toppers 2024
• wbresults.nic.in Result 2024
• West Bengal Madhyamik Supplementary Exam 2024
• CGBSE Result 2024
• AP SSC Results 2024 Manabadi by Jagran Josh
• AP SSC Topper List 2024
• CBSE Class 9 Subjects
• CBSE Class 9 Mathematics
• CBSE Class 9 Study Material
• NCERT Solutions for Class 9
• CBSE Class 9

Latest Education News

ICSE 10th, ISC 12th Result 2024 LIVE: Check CISCE Results Online at results.cisce.org and Latest Updates Here

Who Won Yesterday IPL Match: KKR vs LSG, Match 54, Check All Details and Latest Points Table

(Updated) MI vs SRH Head to Head in IPL: Check Stats, Records and Results

[Official] TN HSE Plus Two Result 2024 Date and Time: Check Tamil Nadu 12th Result Notice and Link at dge.tn.gov.in

TN 12th Public Exam Result 2024 Releasing on May 6, What’s Next and Latest Updates Here

TN 12th Result 2024: TNDGE HSE(+2) Exam Results Tomorrow at 9.30 AM, Check at tnresults.nic.in

[LIVE] TN 12th Result 2024: TNDGE HSC Plus Two (+2) Results to be Released Today at tnresults.nic.in, How to Check Scores Online

Who Won Yesterday IPL Match: CSK vs PBKS, Match 53, Check All Details and Latest Points Table

Purple Cap in IPL 2024: Top Players List with Most Wickets in TATA IPL

[Current] Orange Cap and Purple Cap Holders in IPL 2024

Orange Cap in IPL 2024: Top Players List with Most Runs in TATA IPL

[Today] IPL 2024 Points Table: Team Rankings and Net Run Rate

Top Wicket Keeper In The World: दुनिया के 7 सर्वश्रेष्ठ विकेटकीपर कौन है? एक भारतीय भी शामिल

[New Update] IPL Points Table 2024: आईपीएल 2024 अपडेटेड पॉइंट टेबल यहां देखें, राजस्थान ऑन टॉप

Picture Puzzle IQ Test: Can You Find The Poor LOST PUPPY Hidden In The Park? 21 Seconds Left!

NEET Answer Key 2024 by Aakash Byju's: Download NEET UG Question Paper PDF Here, All Codes (A, B,C, D)

NEET 2024 Question Paper: Download NEET UG Paper PDF Set-wise (Paper Code)

NEET 2024 Answer Key: Download NEET UG Exam Answer Key and Question Paper Solution PDF (All Sets)

Optical Illusion Vision Test: Find the different car in the picture in 5 seconds!

NEET UG Cut Off 2024: Expected Category-Wise Qualifying or Passing Marks

Talk to our experts

1800-120-456-456

• CBSE Class 9 Maths Important Questions for Chapter 1 - Number System

CBSE Class 9 Maths Important Questions Chapter 1 - Number System Free PDF Download

Chapter 1 of Mathematics Class 9 deals with an introduction to various other topics. Those who are planning to pursue a career in mathematics should prepare well for this chapter. Mathematics is the subject to deal with practical life calculations and Class 9 Maths Chapter 1 Important Questions will help set a good base for the students.

Based on these crucial questions, students can prepare for mathematics finals without any hassle. Class 9 is the base to prepare well for 10th boards. Hence students need to master their concepts and utilise their time efficiently. According to CBSE’s basic guidelines, these Important Questions for Class 9 Maths Chapter 1 Number System are prepared.

As a result, students do not need to be concerned or go elsewhere for answers to such inquiries. The first chapter of Mathematics 9th standard is Number System, which deals with whole numbers, rational and irrational numbers, and integers. Read about the topics included in the Crucial Questions for Class 9 Mathematics number system and comprehensive curriculum created by specialists in detail. Vedantu offers students a free PDF download option for all updated CBSE textbook solutions. Topics such as Science, Math, and English will become easier to study if you have access to NCERT Class 9 Science, Math, and other answers that are only accessible on Vedantu.

Download CBSE Class 9 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 9 Maths Important Questions for other chapters:

Important Topics Covered in Class 9 Maths Chapter 1

Introduction to number system

Irrational Number

Real Number and Their Decimal Expansion

Representation of Real Number on Number Line

Operations on Real Number

Laws of Exponents for Real Number

Study Important Questions for Class 9 Maths Chapter 1 – Number Systems

1 Marks Questions

1. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is irrational number.

Ans: We know that the square root of every positive integer will not yield an integer.

We know that \[\sqrt{4}\] is $2$, which is an integer. But, $\sqrt{7}$ or $\sqrt{10}$ will give an irrational number.

Therefore, we conclude that the square root of every positive integer is not an irrational number.

2. Write three numbers whose decimal expansions are non-terminating non-recurring.

Ans: The three numbers that have their expansions as non-terminating on recurring decimals are given below.

0.04004000400004.... 

0.07007000700007....

0.13001300013000013.... 

3. Find three different irrational numbers between the rational numbers $\frac{\text{5}}{\text{11}}$ and $\frac{\text{9}}{\text{11}}$ .

Ans: Let us convert $\frac{5}{11}$ and $\frac{9}{11}$ into decimal form, to get

$\frac{5}{7}=0.714285....and\frac{9}{11}=0.818181....$

Three irrational numbers that lie between $0.714285....$ and $0.818181....$ are:

0.73073007300073...

0.74074007400074.... 

0.76076007600076.... 

4. Which of the following rational numbers have terminating decimal representation?

 $(i)\frac{3}{5}$             

$(ii)\frac{2}{13} $

$(iii)\frac{40}{27}   $    

$(iv)\frac{23}{7}$

Ans: $(i)\frac{3}{5}$

5. How many rational numbers can be found between two distinct rational numbers?

(iv) Infinite

Ans: (iv) Infinite

6. The value of $\left( \text{2+}\sqrt{\text{3}} \right)\left( \text{2-}\sqrt{\text{3}} \right)$ in

(i) $\text{1}$

(ii) $\text{-1}$

(iii) $\text{2}$

(iv) none of these

Ans: (i) $1$

7. ${{\left( \text{27} \right)}^{\text{-2/3}}}$ is equal to

(i) $\text{9}$

(ii) $\text{1/9}$

(iii) $\text{3}$

Ans: (ii) $1/9$

8. Every natural number is

(i) not an integer

(ii) always a whole number

(iii) an irrational number

(iv) not a fraction

Ans: (ii) always a whole number

9. Select the correct statement from the following

(i) $\frac{\text{7}}{\text{9}}\text{}\frac{\text{4}}{\text{5}}$

(ii) $\frac{\text{2}}{\text{6}}\text{}\frac{\text{3}}{\text{9}}$

(iii) $\frac{\text{-2}}{\text{3}}\text{}\frac{\text{-4}}{\text{5}}$

(iv)$\frac{\text{-5}}{\text{7}}\text{}\frac{\text{-3}}{\text{4}}$

Ans: (iii) $\frac{-2}{3}>\frac{-4}{5}$

10. $\text{7}\text{.}\overline{\text{2}}$ is equal to

(i) $\frac{\text{68}}{\text{9}}$

(ii) $\frac{\text{64}}{\text{9}}$

(iii) $\frac{\text{65}}{\text{9}}$

(iv) $\frac{\text{63}}{\text{9}}$

Ans: (iii) $\frac{65}{9}$

11. $\text{0}\text{.83458456}......$ is

(i) an irrational number

(ii) rational number

(iii) a natural number

(iv) a whole number

Ans: (i) an irrational number

12. A terminating decimal is

(i) a natural number

(ii) a rational number

(iii) a whole number

(iv) an integer.

Ans: (ii) a rational number

13. The $\frac{\text{p}}{\text{q}}$ form of the number $\text{0}\text{.8}$ is

(i) $\frac{\text{8}}{\text{10}}$

(ii) $\frac{\text{8}}{\text{100}}$

(iii) $\frac{\text{1}}{\text{8}}$

(iv) $\text{1}$

Ans: (i) $\frac{8}{10}$

14. The value of $\sqrt[\text{3}]{\text{1000}}$ is

Ans: (ii) $10$

15. The sum of rational and an irrational number

(i) may be natural

(ii) may be irrational

(iii) is always irrational

(iv) is always rational

Ans: (iii) is always rational

16. The rational number not lying between $\frac{\text{3}}{\text{5}}$ and $\frac{\text{2}}{\text{3}}$ is

(i) $\frac{\text{49}}{\text{75}}$

(ii) $\frac{\text{50}}{\text{75}}$

(iii) \[\frac{\text{47}}{\text{75}}\]

(iv) $\frac{\text{46}}{\text{75}}$

Ans: (B) $\frac{50}{75}$

17. $\text{0}\text{.12}\overline{\text{3}}$ is equal to

(i) $\frac{\text{122}}{\text{90}}$

(ii) $\frac{\text{122}}{\text{100}}$

(iii) $\frac{\text{122}}{\text{99}}$

(iv) None of these

Ans: (a) $\frac{122}{990}$

18. The number ${{\left( \text{1+}\sqrt{\text{3}} \right)}^{\text{2}}}$ is

(a) natural number

(b) irrational number

(c) rational number

(d) integer

Ans: (b) irrational number

19. The simplest form of $\sqrt{\text{600}}$ is

(i) $\text{10}\sqrt{\text{60}}$

(ii) $\text{100}\sqrt{\text{6}}$

(iii) $\text{20}\sqrt{\text{3}}$

(iv) $\text{10}\sqrt{\text{6}}$

Ans: (D) $10\sqrt{6}$

20. The value of $\text{0}\text{.}\overline{\text{23}}\text{+0}\text{.}\overline{\text{22}}$ is

(i) $\text{0}\text{.4}\overline{\text{5}}$

(ii) $\text{0}\text{.4}\overline{\text{4}}$

(iii) $\text{0}\text{.}\overline{\text{45}}$

(iv) $\text{0}\text{.}\overline{\text{44}}$

Ans: (A) $0.\overline{23}=0.232323....$

$0.\overline{22}=0.222222....$

$0.\overline{23}+0.\overline{22}=0.454545....$

$=0.\overline{45}$

21. The value of ${{\text{2}}^{\frac{\text{1}}{\text{3}}}}\text{ }\!\!\times\!\!\text{ }{{\text{2}}^{\text{-}\frac{\text{4}}{\text{3}}}}$ is

(i) $\text{2}$

(ii) $\frac{\text{1}}{\text{2}}$

Ans: (B) ${{2}^{\frac{1}{3}}}\times {{2}^{-\frac{4}{3}}}={{2}^{\frac{1}{3}-\frac{4}{3}}}={{2}^{\frac{1-4}{3}}}={{2}^{-\frac{3}{3}}}$

22. \[\text{16}\sqrt{\text{13}}\text{ }\!\!\div\!\!\text{ 9}\sqrt{\text{52}}\] is equal to

(i) $\frac{\text{3}}{\text{9}}$

(ii) $\frac{\text{9}}{\text{8}}$

(iii) \[\frac{\text{8}}{\text{9}}\]

Ans: $16\sqrt{13}\div 9\sqrt{52}$

$\frac{16\sqrt{13}}{9\sqrt{52}}=\frac{16}{9}\sqrt{\frac{13}{52}}=\frac{8}{9}$

23. $\sqrt{\text{8}}$ is an

(i) natural number

(iii) integer

(iv) irrational number

Ans: (D) $\sqrt{8}$ is an irrational number

$\therefore \sqrt{4\times 2}=2\sqrt{2}$

2 Marks Questions

1. Is zero a rational number? Can you write it in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$?

Ans: Consider the definition of a rational number. A rational number is the one that can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.

Zero can be written as $\frac{0}{1},\frac{0}{2},\frac{0}{3},\frac{0}{4},\frac{0}{5}......$

So, we arrive at the conclusion that $0$ can be written in the form $\frac{p}{q}$, where $q$is any integer.

Therefore, zero is a rational number.

2. Find six rational numbers between $3$ and $4$.

Ans: We know that there are infinite rational numbers between any two numbers.

A rational number is the one that can be written in the form of $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.

We know that the numbers $3.1,3.2,3.3,3.4,3.5$ and $3.6$ all lie between $3$ and $4$.

We need to rewrite the numbers $3.1,3.2,3.3,3.4,3.5$ and $3.6$ in $\frac{p}{q}$ form to get the rational numbers between $3$ and $4$.

So, after converting we get $\frac{32}{10},\frac{32}{10},\frac{33}{10},\frac{34}{10},\frac{35}{10},$ and $\frac{36}{10},$ into lowest fractions.

On converting the fractions into lowest fractions, we get $\frac{16}{5},\frac{17}{5},\frac{7}{2}$ and $\frac{18}{5}$.

Therefore, six rational numbers between $3$ and $4$are \[\frac{31}{10},\frac{16}{5},\frac{33}{10},\frac{17}{5},\frac{7}{2}\] and $\frac{18}{5}$.

3. Find five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$.

We know that the numbers $\frac{3}{5}$ and $\frac{4}{5}$ can also be written as $0.6\text{ and }0.8$.

We can conclude that the numbers$0.61,0.62,0.63,0.64$ and $0.65$ in \[\frac{p}{q}\] form to get the rational numbers between $3\text{ and }4$.

So, after converting, we get $\frac{61}{100},\frac{62}{100},\frac{63}{100},\frac{64}{100}\text{ and }\frac{65}{100}$.

We can further convert the rational numbers $\frac{62}{100},\frac{64}{100}\text{ and }\frac{65}{100}$ into lowest fractions.

On converting the fractions, we get $\frac{31}{50},\frac{16}{25}\text{ and }\frac{13}{20}$.

Therefore, six rational numbers between $3\text{ and }4$ are $\frac{61}{100},\frac{31}{50},\frac{63}{100},\frac{16}{50}\text{ and }\frac{13}{50}$.

4. Show how $\sqrt{5}$ can be represented on the number line.

Ans: According to Pythagoras theorem, we can conclude that

${{\left( \sqrt{5} \right)}^{2}}={{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}$.

We need to draw a line segment $AB\text{ of }1$unit on the number line. Then draw a straight line segment $BC\text{ of }2$ units. Then join the points $C$ and $A$, to form a line segment $BC$.

Then draw the arc $ACD$, to get the number $\sqrt{5}$ on the number line.

5. You know that $\frac{1}{7}=0.142857....$. Can you predict what the decimal expansion of $\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}$ are, without actually doing the long division? If so, how?

(Hint: Study the remainder while finding the value of $\frac{1}{7}$ carefully.)

Ans: We are given that $\frac{1}{7}=0.\overline{142857}$ or $\frac{1}{7}=0.142857....$

We need to find the value of \[\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7}\text{ and }\frac{6}{7}\], without performing long division.

We know that \[\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7}\text{ and }\frac{6}{7}\] can be rewritten as

$2\times \frac{1}{7},3\times \frac{1}{7},4\times \frac{1}{7},5\times \frac{1}{7}\text{ and }6\times \frac{1}{7}$.

On substituting value of $\frac{1}{7}$ as $0.142857....$, we get 

$2\times \frac{1}{7}=2\times 0.142857....=0.285714.... $

$3\times \frac{1}{7}=3\times 0.142857....=0.428571.... $ 

$4\times \frac{1}{7}=4\times 0.142857....=0.571428....$ 

$5\times \frac{1}{7}=5\times 0.142857....=0.714285.... $ 

 $6\times \frac{1}{7}=6\times 0.142857....=0.857142.... $ 

Therefore, we conclude that, we can predict the values of \[\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7}\text{ and }\frac{6}{7}\], without performing long division, to get 

\[\frac{2}{7}=0.\overline{285714},\frac{3}{7}=0.\overline{428571},\frac{4}{7}=0.\overline{571428},\frac{5}{7}=0.\overline{714285},\frac{6}{7}=0.\overline{857142}\]

6. Express $0.99999....$in the form $\frac{p}{q}$. Are you surprised by your answer? Discuss why the answer makes sense with your teacher and classmates.

Ans: Let $x=0.99999....\text{ }......(a)$

We need to multiply both sides by $10$ to get 

$10x=9.9999....\text{ }......(b)$

We need to subtract $(a)\text{ from }(b)$, to get

10x=9.99999.... 

x=0.99999.... 

_____________

We can also write $9x=9\text{ as }x=\frac{9}{9}\text{ or }x=1$.

Therefore, on converting $0.99999....$ in the $\frac{p}{q}$ form, we get the answer as $1$.

Yes, at a glance we are surprised at our answer. But the answer makes sense when we observe that $0.99999....$ goes on forever. So there is no gap between $1$  and $0.9999....$ and hence they are equal.

7. Visualize $3.765$ on the number line using successive magnification.

Ans: We know that the number $3.765$ will lie between $3.764\text{ and }3.766$.

We know that the number $3.764$and $3.766$ will lie between $3.76\text{ and }3.77$.

We know that the number $3.76\text{ and }3.77$. will lie between $3.7\text{ and }3.8$.

We know that the number  $3.7\text{ and }3.8$ will lie between $3\text{ and }~4$.

Therefore, we can conclude that we need to use the successive magnification, after locating numbers $3\text{ and }~4$ on the number line

(Image will be uploaded soon)

8. Visualize $4.\overline{26}$ on the number line, upto $4$decimal places.

Ans: We know that the number $4.\overline{26}$ can also be written as$4.262....$ .

We know that the number $4.262....$ will lie between $4.261\text{ and }4.263$.

We know that the number  $4.261\text{ and }4.263$ will lie between $4.26\text{ and }4.27$.

We know that the number $4.26\text{ and }4.27$ will lie between $4.2\text{ and }4.3$.

We know that the number $4.2\text{ and }4.3$ will lie between $4\text{ and }5$.

Therefore, we can conclude that we need to use the successive magnification, after locating numbers $4\text{ and }5$ on the number line.

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

Ans: We know that when we measure the length of the line or a figure by using a scaleneory device, we do not get an exact measurement. In fact, we get an approximate rational value. So, we are not able to realize that either the circumference ($c$) or diameter ($d$) of a circle is irrational.

Therefore, we can conclude that as such there is not any contradiction regarding the value of $\pi $ and we realize that the value of $\pi $ is irrational.

10. Represent $9.3$ on the number line.

Ans: Mark the distance $9.3$ units from a fixed point $A$ on a given line to obtain a point $B$ such that $AB=9.3$ units. From $B$ mark a distance of $1$ unit and call the new point as $C$. Find themid-point of $AC$ and call that point as $O$. Draw a semi-circle with centre $O$ and radius $OC=5.15$units. Draw a line perpendicular to $AC$ passing through $B$ cutting the semi-circle at $D$.

Then $BD=\sqrt{9.3}$.

11. Find (i) ${{64}^{\frac{1}{5}}}$ (ii) ${{32}^{\frac{1}{5}}}$ (iii) ${{125}^{\frac{1}{3}}}$

(i) ${{64}^{\frac{1}{2~}}}$

We know that${{a}^{\frac{1}{n}}}=\sqrt[n]{a},\text{ where }a>0$

We conclude that  ${{64}^{\frac{1}{2~}}}$can also be written as $\sqrt[2]{64}=\sqrt[2]{8\times 8}$

$\sqrt[2]{64}=\sqrt[2]{8\times 8}$$=8$

Therefore, the value of  ${{64}^{\frac{1}{2~}}}$will be $8$.

(ii) ${{32}^{\frac{1}{5}}}$

We conclude that  ${{32}^{\frac{1}{5}}}$can also be written as $\sqrt[5]{32}=\sqrt[5]{2\times 2\times 2\times 2\times 2}$

$\sqrt[5]{32}=\sqrt[5]{2\times 2\times 2\times 2\times 2}=2$

Therefore, the value of  ${{32}^{\frac{1}{5}}}$will be $2$.

(iii) ${{125}^{\frac{1}{3}}}$

We conclude that ${{125}^{\frac{1}{3}}}$can also be written as $\sqrt[3]{125}=\sqrt[3]{5\times 5\times 5}$

$\sqrt[3]{125}=\sqrt[3]{5\times 5\times 5}=5$

Therefore, the value of ${{125}^{\frac{1}{3}}}$will be $5$.

12. Simplify $\sqrt[3]{2}\times \sqrt[4]{3}$

Ans: $\sqrt[3]{2}\times \sqrt[4]{3}$

${{2}^{\frac{1}{3}}}\times {{3}^{\frac{1}{4}}}$

The LCM of $3\text{ and }4\text{ is }12$

$\therefore {{2}^{\frac{1}{3}}}={{2}^{\frac{4}{12}}}={{\left( {{2}^{4}} \right)}^{\frac{1}{12}}}={{16}^{\frac{1}{12}}} $

${{3}^{\frac{1}{4}}}={{3}^{\frac{3}{12}}}={{\left( {{3}^{3}} \right)}^{\frac{1}{12}}}={{27}^{\frac{1}{12}}} $

${{2}^{\frac{1}{3}}}\times {{3}^{\frac{1}{4}}}={{16}^{\frac{1}{12}}}\times {{27}^{\frac{1}{12}}}={{\left( 16\times 27 \right)}^{\frac{1}{12}}} $

$={{\left( 432 \right)}^{\frac{1}{12}}} $

13. Find the two rational numbers between$\frac{1}{2}$ and $\frac{1}{3}$.

Ans: First rational number between $\frac{1}{2}$ and $\frac{1}{3}$

$=\frac{1}{2}\left[ \frac{1}{2}+\frac{1}{3} \right]\Rightarrow \frac{1}{2}\left[ \frac{3+2}{6} \right]\Rightarrow \frac{5}{12} $

$ =\frac{1}{2},\frac{5}{12}\text{ and }\frac{1}{3} $

Second rational number between $\frac{1}{2}$ and $\frac{1}{3}$

$=\frac{1}{2}\left[ \frac{1}{2}+\frac{5}{12} \right]\Rightarrow \frac{1}{2}\left[ \frac{6+5}{12} \right]\Rightarrow \frac{11}{24}$

$=\frac{5}{12}\text{ and }\frac{11}{24}$ are two rational numbers between $\frac{1}{2}$ and $\frac{1}{3}$.

14. Find two rational numbers between $2$ and $3$.

Ans: Irrational numbers between $2$ and $3$ is $\sqrt{2\times 3}=\sqrt{6}$

Irrational number between $2$ and $3$ is $\sqrt{6}$.

$\sqrt{2\times \sqrt{6}}={{2}^{\frac{1}{2}}}\times {{6}^{\frac{1}{4}}}={{2}^{2\times \frac{1}{4}}}\times {{6}^{\frac{1}{4}}} $

$ ={{\left( {{2}^{2}} \right)}^{\frac{1}{4}}}\times {{6}^{\frac{1}{4}}}={{4}^{\frac{1}{4}}}\times {{6}^{\frac{1}{4}}}={{\left( 24 \right)}^{\frac{1}{4}}}=\sqrt[4]{24} $

$\sqrt{6}\text{ and }\sqrt{24}$ are two rational numbers between $2$ and $3$.

15. Multiply $\left( 3-\sqrt{5} \right)$ by $\left( 6+\sqrt{2} \right)$.

Ans: $\left( 3-\sqrt{5} \right)$$\left( 6+\sqrt{2} \right)$

$  =3\left( 6-\sqrt{2} \right)-\sqrt{5}\left( 6+\sqrt{2} \right) $

$=18+3\sqrt{2}-6\sqrt{5}-\sqrt{5}\times \sqrt{2} $

$  =18+3\sqrt{2}-6\sqrt{5}-\sqrt{10} $

16. Evaluate (i) $\sqrt[3]{125}$ (ii) $\sqrt[4]{1250}$

(i) $\sqrt[3]{125}$$={{\left( 5\times 5\times 5 \right)}^{\frac{1}{3}}}={{\left( {{5}^{3}} \right)}^{\frac{1}{3}}}=5$

(ii) $\sqrt[4]{1250}$$\begin{align}

$={{\left( 2\times 5\times 5\times 5\times 5 \right)}^{\frac{1}{4}}}={{\left( 2\times {{5}^{4}} \right)}^{\frac{1}{4}}} $ 

$={{2}^{\frac{1}{4}}}\times {{\left( {{5}^{4}} \right)}^{\frac{1}{4}}}=5\times \sqrt[4]{2} $

17. Find rationalizing factor of $\sqrt{300}$.

Ans: $\sqrt{300}=\sqrt{2\times 2\times 3\times 5\times 5}$

$  =\sqrt{{{2}^{2}}\times 3\times {{5}^{2}}} $

$ =2\times 5\sqrt{3}=10\sqrt{3} $

Rationalizing factor is $\sqrt{3}$

18. Rationalizing the denominator $\frac{1}{\sqrt{5}+\sqrt{2}}$ and subtract it from $\sqrt{5}-\sqrt{2}$.

Ans: $\frac{1}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$=\frac{\sqrt{5}-\sqrt{2}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}$

Difference between $\left( \sqrt{5}-\sqrt{2} \right)\text{ and }\left( \sqrt{5}-\frac{\sqrt{2}}{3} \right)$

$=\sqrt{5}-\sqrt{2}-\left( \frac{\sqrt{5}-\sqrt{2}}{3} \right) $

$ =\sqrt{5}-\sqrt{2}-\frac{\sqrt{5}}{3}+\frac{\sqrt{2}}{3} $

$=\left( \sqrt{5}-\frac{\sqrt{5}}{3} \right)-\left( \sqrt{2}-\frac{\sqrt{2}}{3} \right) $

$=\frac{2\sqrt{5}}{3}-\frac{2\sqrt{2}}{3}=\frac{2}{3}\left( \sqrt{5}-\sqrt{2} \right) $

19. Show that $\sqrt{7}-3$ is irrational.

Ans: Suppose $\sqrt{7}-3$ is rational

Let $\sqrt{7}-3=x$ ($x$ is a rational number)

$\sqrt{7}=x+3$

$x$ is a rational number $3$ is also a rational number

$\therefore x+3$ is a rational number

But is $\sqrt{7}$ irrational number which is contradiction

$\therefore \sqrt{7}-3$ is an irrational number.

20. Find two rational numbers between $7$ and $5$.

Ans: First rational number $=\frac{1}{2}\left[ 7+5 \right]=\frac{12}{2}=6$

Second rational number $=\frac{1}{2}\left[ 7+6 \right]=\frac{1}{2}\times 13=\frac{13}{2}$

Two rational numbers between $7\text{ and }5\text{ are }6\text{ and }\frac{13}{2}$.

21. Show that $5+\sqrt{2}$ is not a rational number.

Ans: Let $5+\sqrt{2}$ is a rational number.

Say $5+\sqrt{2=x}$ i.e., $\sqrt{2}=x-5$

$x$ is a rational number $5$ is also rational number

$\therefore x-5$ is also a rational number.

But $\sqrt{2}$ is irrational number which is a contradiction

$\therefore 5+\sqrt{2}$ is an irrational number.

22. Simplify ${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}$.

Ans: ${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}={{\left( \sqrt{5} \right)}^{2}}+{{\left( \sqrt{2} \right)}^{2}}+2\sqrt{5}\times \sqrt{2}=5+2+2\sqrt{10}=7+2\sqrt{2}$

23. Evaluate $\frac{{{11}^{\frac{5}{2}}}}{{{11}^{\frac{3}{2}}}}$.

Ans: $\frac{{{11}^{\frac{5}{2}}}}{{{11}^{\frac{3}{2}}}}={{11}^{\frac{5}{2}-\frac{3}{2}}}\left[ \because \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \right]$

$ ={{11}^{\frac{5-3}{2}}}={{11}^{\frac{2}{2}}} $

$=11 $ 

24. Find four rational numbers between $\frac{3}{7}$ and $\frac{4}{7}$.

$\frac{3}{7}\times \frac{10}{10}=\frac{30}{70}\text{ and }\frac{4}{7}\times \frac{10}{10}=\frac{40}{70}$

Take any four rational numbers between $\frac{30}{70}\text{ and }\frac{40}{70}$ i.e., rational numbers between  $\frac{3}{7}$ and $\frac{4}{7}$ are $\frac{31}{70},\frac{32}{70},\frac{33}{70},\frac{34}{70},\frac{35}{70}$

25. Write the following in decimal form (i) $\frac{36}{100}$ (ii) $\frac{2}{11}$

(i) $\frac{36}{100}=0.36$

(ii) $\frac{2}{11}=0.\overline{18}$

26. Express $2.417\overline{8}$ in the form $\frac{a}{b}$

Ans: $x=2.4\overline{178}$

$10x=24.\overline{178}$$......(1)$$[\text{Multiplying both sides by }10]$

$10x=24.178178178.... $

$1000\times 10x=1000\times 24.178178178....$Multiplying both sides by 1000

$10,000x=24178.178178.... $

$ 10000x=24178.\overline{178}\text{ }......(2) $

Subtracting $(1)\text{ from }(2)$

$10,000x-x=24178.\overline{178}-24.\overline{178} $

$9990x=24154 $

$x=\frac{24154}{9990} $

$ 2.4\overline{178}=\frac{24154}{9990}+\frac{12077}{4995} $

27. Multiply $\sqrt{3}$ by $\sqrt[3]{5}$.

Ans: $\sqrt{3}\text{ and }\sqrt[3]{5}$

Or ${{3}^{\frac{1}{2}}}\text{ and }{{5}^{\frac{1}{3}}}$

$LCM\text{ of }2\text{ and }3\text{ is }6 $

${{3}^{\frac{1}{2}}}={{3}^{\frac{1}{2}\times \frac{3}{3}}}={{\left( {{3}^{3}} \right)}^{\frac{1}{6}}}={{\left( 27 \right)}^{\frac{1}{6}}} $

 ${{5}^{\frac{1}{3}}}={{5}^{\frac{1}{3}\times \frac{2}{2}}}={{\left( {{5}^{2}} \right)}^{\frac{1}{6}}}={{\left( 25 \right)}^{\frac{1}{6}}} $

 $\sqrt{3}\times \sqrt[3]{5}={{\left( 27 \right)}^{\frac{1}{6}}}\times {{\left( 25 \right)}^{\frac{1}{6}}}={{\left( 27\times 25 \right)}^{\frac{1}{6}}} $

$ ={{675}^{\frac{1}{6}}}=\sqrt[6]{675} $

28. Find the value of $\frac{\sqrt{2}+\sqrt{5}}{\sqrt{5}}$ if $\sqrt{5}=2.236$ and $\sqrt{10}=3.162$.

Ans: $\frac{\sqrt{2}+\sqrt{5}}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{10}+5}{5}=\frac{8.162}{5}=1.6324$

29. Convert $0.\overline{25}$ into rational number.

Ans: Let \[x=0.\overline{25}\]                            ......(i)

$x=0.252525....$

Multiply both sides by 100

$100x=25.252525....$

$100x=25.\overline{25}$                                                  ......(ii)

Subtract (i) from (ii)

$100x-x=25.\overline{25}-0.\overline{25} $

 $x=\frac{25}{99} $

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

Ans: By multiplying each terms in the given product we have,

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

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

 $=18+9\sqrt{6}+4\sqrt{6}+12 $

 $ =30+\left( 9+4 \right)\sqrt{6} $

 $=30+13\sqrt{6} $

31. Simplify $\frac{{{9}^{\frac{3}{2}}}\times {{9}^{-\frac{4}{2}}}}{{{9}^{\frac{1}{2}}}}$.

Ans: By using the formulas of exponents with same base we get,

$\frac{{{9}^{\frac{3}{2}}}\times {{9}^{-\frac{4}{2}}}}{{{9}^{\frac{1}{2}}}}=\frac{{{9}^{\frac{3}{2}-\frac{4}{2}}}}{{{9}^{\frac{1}{2}}}}\left[ {{a}^{m}}.{{a}^{n}}={{a}^{m-n}} \right]$

$\frac{{{9}^{-\frac{1}{2}}}}{{{9}^{\frac{1}{2}}}}=\frac{1}{{{9}^{\frac{1}{2}+\frac{1}{2}}}}\left[ {{a}^{-m}}=\frac{1}{{{a}^{m}}} \right] $

$ =\frac{1}{{{9}^{\frac{2}{2}}}}=\frac{1}{9} $

3 Marks Questions

1. State whether the following statements are true or false. Give

reasons for your answers.

i. Every natural number is a whole number.

Separately, consider whole numbers and natural numbers.

We know that the whole number series is 0,1,2,3,4,5....

We know that the natural number series is 0,1,2,3,4,5....

As a result, every number in the natural number series may be found in the whole number series.

Therefore, we can safely conclude that any natural number is a whole number.

ii. Every integer is a whole number.

Ans: Separately, consider whole numbers and integers.

We know that integers are those numbers that can be written in the form of $\frac{p}{q}$ where q=1.

In the case of an integer series, we now have.... 4,3,2,1,0,1,2,3,4....

We can conclude that all whole number series numbers belong to the integer series.

However, the whole number series does not contain every number of integer series.

As a result, we can conclude that no integer is a whole number.

iii. Every rational number is a whole number.

Ans: Separately, consider whole numbers and rational numbers.

We know that integers are those numbers that can be written in the form of $\frac{p}{q}$ where $q\ne 0$.

We know that every number of whole number series can be written in the form of $\frac{p}{q}$ as $\frac{0}{1},\frac{1}{1},\frac{2}{1},\frac{3}{1},\frac{4}{1},\frac{5}{1}...$

We conclude that every number of the whole number series is a rational number. 

But, every rational number does not appear in the whole number series.

2. State whether the following statements are true or false. Justify your answers.

i. Every irrational number is a real number.

Ans: Separately, consider irrational numbers and real numbers.

We know that irrational numbers are the numbers that cannot be converted in the form $\frac{p}{q}$, where p and q are integers and $q\ne 0$.

A real number is made up of both rational and irrational numbers, as we all know.

As a result, we might conclude that any irrational number is, in fact, a real number.

ii. Every point on the number line is of the form $\sqrt{m}$, where m is a natural number.

Ans: Consider a number line. We know that we can express both negative and positive numbers on a number line.

We know that when we take the square root of any number, we cannot receive a negative value.

Therefore, we conclude that not every number point on the number line is of the form $\sqrt{m}$, where m is a natural number.

iii. Every real number is an irrational number.

As a result, we can deduce that any irrational number is actually a real number. However, not every real number is irrational.

Therefore, we conclude that every real number is not a rational number.

3. Express the following in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q\ne 0$.

i. $0.\overline{6}$

Let $x=0.\overline{6}$

$\Rightarrow x=0.6666$            ......(a)

Multiplying both sides by 10 we get

$10x=6.6666$             ......(b)

We need to subtract (a) from (b), to get

We can also write $9x=6$ as $x=\frac{6}{9}$ or $x=\frac{2}{3}$.

Therefore, on converting $0.\overline{6}$ in the $\frac{p}{q}$ form, we get the answer as $\frac{2}{3}$.

ii. $0.4\overline{7}$

Ans: Let $x=0.4\overline{7}\Rightarrow x=0.47777$                     ......(a)

Multiplying both sides by 10 we get 

$10x=4.7777$                                            ......(b)

We can also write $9x=4.3$ as $x=\frac{4.3}{9}$ or $x=\frac{43}{90}$

Therefore, on converting $0.4\overline{7}$ in the $\frac{p}{q}$ form, we get the answer as $\frac{43}{90}$.

iii. $0.\overline{001}$

Ans: Let $x=0.\overline{001}\Rightarrow x=0.001001$                     ......(a)

Multiplying both sides by 1000 we get 

$1000x=1.001001$                                         ......(b)

We can also write $999x=1$ as $x=\frac{1}{999}$

Therefore, on converting $0.\overline{001}$ in the $\frac{p}{q}$ form, we get the answer as $\frac{1}{999}$.

4. What can the maximum number of digits be in the recurring block of digits in the decimal expansion of $\frac{1}{17}$ ? Perform the division to check your answer.

Ans: The number of digits in the recurring block of $\frac{1}{17}$ must be determined.

To acquire the repeating block of $\frac{1}{17}$ we'll use long division.

We need to divide 1 by 17, to get 0.0588235294117647.... and we got the remainder as 1, which will continue to be 1 after carrying out 16 continuous divisions.

Therefore, we conclude that

\[\frac{1}{17}=0.0588235294117647\] or \[\frac{1}{17}=0.\overline{0588235294117647}\] which is a non-terminating decimal and recurring decimal.

5. Look at several examples of rational numbers in the form $\frac{p}{q}\left( q\ne 0 \right)$ where $p$ and $q$ are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property $q$ must satisfy?

Ans: Let us consider the examples of the form $\frac{p}{q}$ that are terminating decimals .

$ \frac{5}{2}=2.5 $ 

$ \frac{5}{4}=1.25 $ 

$ \frac{2}{5}=0.4 $

$  \frac{5}{16}=0.3125 $ 

It can be observed that the denominators of the above rational numbers have powers of 2,5 or both.

Therefore, we can conclude that property, which $q$ must satisfy in $\frac{p}{q}$ , so that the rational number $\frac{p}{q}$ is a terminating decimal is that q must have powers of 2,5 or both.

6. Classify the following numbers as rational or irrational:

i. $2-\sqrt{5}$

Ans: $2-\sqrt{5}$

We know that $\sqrt{5}=2.236....$, which is an irrational number.

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

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

As a result, we can deduce that $2-\sqrt{5}$ is an irrational number.

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

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

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

As a result, we can deduce that $\left( 3+\sqrt{23} \right)-\sqrt{23}$ is a rational number.

iii. $\frac{2\sqrt{7}}{7\sqrt{7}}$

Ans: $\frac{2\sqrt{7}}{7\sqrt{7}}$

We can cancel $\sqrt{7}$ in the numerator and denominator to get $\frac{2\sqrt{7}}{7\sqrt{7}}=\frac{2}{7}$, because $\sqrt{7}$ is a common number in both the numerator and denominator.

iv. $\frac{1}{\sqrt{2}}$

Ans: $\frac{1}{\sqrt{2}}$

We know that $\sqrt{2}=1.4142...$, which is an irrational number.

$\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

$=\frac{1.4142...}{2}=0.707...$ which is also an irrational number.

As a result, we can deduce that $\frac{1}{\sqrt{2}}$ is an irrational number.

Ans: $2\pi $

We know that $\pi =3.1415....,$ which is an irrational number.

We can conclude that $2\pi $ will also be an irrational number.

As a result, we can deduce that $2\pi $ is an irrational number.

7. Simplify each of the following expression:

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

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

Applying distributive law,

\[(3+3\sqrt{3})(2+\sqrt{2})=3(2+\sqrt{2})\sqrt{3}(2+\sqrt{2})\]

\[=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}\]

ii. \[\left( 3+3\sqrt{3} \right)3-\sqrt{3}\]

Ans: $(3+3\sqrt{3})(3-\sqrt{3})$

$  (3+3\sqrt{3})(3-\sqrt{3})=(3-\sqrt{3})+\sqrt{3}(3-\sqrt{3}) $ 

$ =9-3\sqrt{3}+3\sqrt{3}-3 $

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

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

Applying the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

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

$ =5+2\sqrt{10}+2 $

$=7+2\sqrt{10}$

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

Ans: $\left( 5+\sqrt{2} \right)\left( 5+\sqrt{2} \right)$

Applying the formula $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$

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

8. Find 

i. ${{9}^{\frac{3}{2}}}$

Ans: We know that ${{a}^{\frac{1}{n}}}=\sqrt[n]{a},a>0$

As a result, we can deduce that ${{9}^{\frac{3}{2}}}$ can also be written as

$\sqrt[2]{{{\left( 9 \right)}^{3}}}=\sqrt[2]{9\times 9\times 9}=\sqrt[2]{3\times 3\times 3\times 3\times 3\times 3}$

$=3\times 3\times 3 $

Therefore, the value of ${{9}^{\frac{3}{2}}}$ will be $27$ .

ii. ${{32}^{\frac{2}{5}}}$

As a result, we can deduce that ${{32}^{\frac{2}{5}}}$ can also be written as

$ \sqrt[5]{{{\left( 32 \right)}^{2}}}=\sqrt[5]{\left( 2\times 2\times 2\times 2\times 2 \right)\left( 2\times 2\times 2\times 2\times 2 \right)} $

$=2\times 2 $ 

Therefore, the value of ${{32}^{\frac{2}{5}}}$ will be $4$.

iii. ${{16}^{\frac{3}{4}}}$

As a result, we can deduce that ${{16}^{\frac{3}{4}}}$ can also be written as 

$\sqrt[4]{{{\left( 16 \right)}^{3}}}=\sqrt[4]{\left( 2\times 2\times 2\times 2 \right)\left( 2\times 2\times 2\times 2 \right)\left( 2\times 2\times 2\times 2 \right)} $

$ =2\times 2\times 2 $

 & =8 $

Therefore, the value of ${{16}^{\frac{3}{4}}}$ will be $8$ .

iv. ${{125}^{-\frac{1}{3}}}$

Ans: We know that ${{a}^{-n}}=\frac{1}{{{a}^{n}}}$

As a result, we can deduce that ${{125}^{-\frac{1}{3}}}$ can also be written as $\frac{1}{{{125}^{\frac{1}{3}}}},or{{\left( \frac{1}{125} \right)}^{\frac{1}{3}}}$

We know that ${{a}^{\frac{1}{n}}}=\sqrt[n]{a},a>0$

$ \sqrt[3]{\frac{1}{125}}=\sqrt[3]{\left( \frac{1}{5}\times \frac{1}{5}\times \frac{1}{5} \right)} $

$=\frac{1}{5} $

Therefore, the value of ${{125}^{-\frac{1}{3}}}$ will be $\frac{1}{5}$.

9. Simplify

i. ${{2}^{\frac{2}{3}}}{{.2}^{\frac{1}{5}}}$

Ans: We know that ${{a}^{m}}.{{a}^{n}}={{a}^{\left( m+n \right)}}$

As a result, we can deduce that ${{2}^{\frac{2}{3}}}{{.2}^{\frac{1}{5}}}={{\left( 2 \right)}^{\frac{2}{3}+\frac{1}{5}}}$

${{2}^{\frac{2}{3}}}{{.2}^{\frac{1}{5}}}=\left( 2 \right)\frac{10+3}{15}={{\left( 2 \right)}^{\frac{13}{15}}}$

Therefore, the value of ${{2}^{\frac{2}{3}}}{{.2}^{\frac{1}{5}}}$ will be ${{\left( 2 \right)}^{\frac{13}{15}}}$.

ii. ${{\left( {{3}^{\frac{1}{3}}} \right)}^{7}}$

As a result, we can deduce that ${{\left( {{3}^{\frac{1}{3}}} \right)}^{7}}$ can also be written as ${{3}^{\frac{7}{3}}}$

iii. $\frac{{{11}^{\frac{1}{2}}}}{{{11}^{\frac{1}{4}}}}$

Ans: We know that $\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{\left( m-n \right)}}$

As a result, we can deduce that  $\begin{align}

$ \frac{{{11}^{\frac{1}{2}}}}{{{11}^{\frac{1}{4}}}}={{11}^{\frac{1}{2}}}-{{11}^{\frac{1}{4}}} $

$={{11}^{\frac{2-1}{4}}}={{11}^{\frac{1}{4}}} $

Therefore, the value of  $\frac{{{11}^{\frac{1}{2}}}}{{{11}^{\frac{1}{4}}}}$ will be ${{11}^{\frac{1}{4}}}$.

iv. ${{7}^{\frac{1}{2}}}{{.8}^{\frac{1}{2}}}$

Ans: We know that ${{a}^{m}}.{{b}^{m}}={{\left( a\times b \right)}^{m}}$

As a result, we can deduce that ${{7}^{\frac{1}{2}}}{{.8}^{\frac{1}{2}}}={{\left( 7\times 8 \right)}^{\frac{1}{2}}}.$

${{7}^{\frac{1}{2}}}{{.8}^{\frac{1}{2}}}={{\left( 7\times 8 \right)}^{\frac{1}{2}}}={{\left( 56 \right)}^{\frac{1}{2}}}.$

Therefore, the value of ${{7}^{\frac{1}{2}}}{{.8}^{\frac{1}{2}}}$ will be ${{\left( 56 \right)}^{\frac{1}{2}}}$.

10. Express $0.8888....$ in the form $\frac{p}{q}$.

Ans: Let us assume that the given decimal as,

$x=0.\overline{8}......\left( 1 \right)$

$10x=10\times 0.8888$ (Multiply both sides by 10)

$ 10x=8.8888 $

$10x=8.\overline{8}.....\left( 2 \right) $

$10x-x=8.\overline{8}-0.\overline{8}$ (Subtracting (1) from (2))

$x=\frac{8}{9} $

11.  Simplify by rationalizing denominator $\frac{7+3\sqrt{5}}{7-3\sqrt{5}}$.

Ans: We are given the fraction to rationalize. By rationalizing the denominator we get,

$\frac{7+3\sqrt{5}}{7-3\sqrt{5}}=\frac{7+3\sqrt{5}}{7-3\sqrt{5}}\times \frac{7+3\sqrt{5}}{7+3\sqrt{5}}$ 

$=\frac{{{\left( 7+3\sqrt{5} \right)}^{2}}}{{{7}^{2}}-{{\left( 3\sqrt{5} \right)}^{2}}} $

$ =\frac{{{7}^{2}}+{{\left( 3\sqrt{5} \right)}^{2}}+2\times 7\times 3\sqrt{5}}{49-{{3}^{2}}\times 5} $

$=\frac{49+9\times 5+42\sqrt{5}}{49-45} $

$=\frac{49+45+42\sqrt{5}}{4} $

$ =\frac{94+42\sqrt{5}}{4} $ 

$ =\frac{94}{4}+\frac{42}{4}\sqrt{5} $

$ =\frac{47}{2}+\frac{21}{2}\sqrt{5} $

12 . Simplify ${{\left\{ {{\left[ {{625}^{-}}^{\frac{1}{2}} \right]}^{-\frac{1}{4}}} \right\}}^{2}}$.

Ans:  Let us take the given expression to simplify and using the exponents formulas we get,

  \[{{\left\{ {{\left[ {{625}^{-}}^{\frac{1}{2}} \right]}^{-\frac{1}{4}}} \right\}}^{2}}\]

$ ={{\left\{ {{\left( \frac{1}{{{625}^{\frac{1}{2}}}} \right)}^{-\frac{1}{4}}} \right\}}^{2}} $

$={{\left\{ {{\left( \frac{1}{{{\left( {{25}^{2}} \right)}^{\frac{1}{2}}}} \right)}^{-\frac{1}{4}}} \right\}}^{2}}\left( {{a}^{-m}}=\frac{1}{{{a}^{m}}} \right) $

 $ =\left\{ {{\left( \frac{1}{25} \right)}^{-\frac{1}{4}\times 2}} \right\} $

$=\left( \frac{1}{{{25}^{-\frac{1}{2}}}} \right)=\frac{1}{{{\left( {{5}^{2}} \right)}^{-\frac{1}{2}}}}=\frac{1}{{{5}^{-1}}}=5 $

13. Visualize 3.76 on the number line using successive magnification.

14. Prove that $\frac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\frac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\frac{1}{1+{{x}^{a-c}}+{{x}^{b-c}}}=1$

Ans: We are asked to prove the expression,

$\frac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\frac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\frac{1}{1+{{x}^{a-c}}+{{x}^{b-c}}}=1$

Let us take the LHS of the given expression that is,

$LHS=\frac{1}{1+{{x}^{b}}.{{x}^{-a}}+{{x}^{c}}.{{x}^{-a}}}+\frac{1}{1+{{x}^{a}}.{{x}^{-b}}+{{x}^{c}}.{{x}^{-b}}}+\frac{1}{1+{{x}^{a}}.{{x}^{-c}}+{{x}^{b}}.{{x}^{-c}}} $

$=\frac{1}{{{x}^{-a}}.{{x}^{a}}+{{x}^{b}}.{{x}^{-a}}+{{x}^{c}}.{{x}^{-a}}}+\frac{1}{{{x}^{b}}.{{x}^{-b}}+{{x}^{a}}.{{x}^{-b}}+{{x}^{c}}.{{x}^{-b}}}+\frac{1}{{{x}^{c}}.{{x}^{-c}}+{{x}^{a}}.{{x}^{-c}}+{{x}^{b}}.{{x}^{-c}}} $

$ =\frac{1}{{{x}^{-a}}\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}+\frac{1}{{{x}^{-b}}\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}+\frac{1}{{{x}^{-c}}\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)} $

$=\frac{{{x}^{a}}}{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}+\frac{{{x}^{b}}}{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}+\frac{{{x}^{c}}}{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)} $ $ =\frac{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}=1 $

15. Represent $\sqrt{3}$ on number line.

Ans: Consider a number line $\text{OD}$ such that the construction to form two triangles is done as shown below.

Take $OA=AB=1$ unit.

And $\angle A=90{}^\circ $

In $\Delta OAB$, by using the Pythagorean theorem we get,

$O{{B}^{2}}={{1}^{2}}+{{1}^{2}}$

$O{{B}^{2}}=2 $

$ OB=\sqrt{2}$ 

Now from triangle $\text{ }\!\!\Delta\!\!\text{ OBD}$, using the Pythagorean theorem we get,

$O{{D}^{2}}=O{{B}^{2}}+B{{D}^{2}} $

$ O{{D}^{2}}={{\left( \sqrt{2} \right)}^{2}}+{{\left( 1 \right)}^{1}} $

$O{{D}^{2}}=2+1=3 $

$OD=\sqrt{3} $

Now, if the point $\text{O}$ is $0$ units then the point $\text{D}$ represents $\sqrt{3}$units.

16. Simplify ${{\left( 3\sqrt{2}+2\sqrt{3} \right)}^{2}}{{\left( 3\sqrt{2}-2\sqrt{3} \right)}^{2}}$.

Ans: We are given the expression as,

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

Now, by regrouping the terms in the above expression we have,

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

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

$=\left[ {{\left( 3\sqrt{2} \right)}^{2}}-{{\left( 2\sqrt{3} \right)}^{2}} \right]\left[ {{\left( 3\sqrt{2} \right)}^{2}}-{{\left( 2\sqrt{3} \right)}^{2}} \right] $

$ =\left[ 9\times 2-4\times 3 \right]\left[ 9\times 2-4\times 3 \right] $

$ =\left[ 18-12 \right]\left[ 18-12 \right] $ 

$=6\times 6=36 $

17. Express $2.\overline{4178}$ in the form $\frac{p}{q}$.

Ans: Let $\frac{p}{q}=2.\overline{4178}$

$\frac{p}{q}=2.4178178178$

Multiply by 10

$10\frac{p}{q}=24.178178$

Multiply by 1000

$10000\frac{p}{q}=1000\times 24.178178 $ 

$1000\frac{p}{q}-\frac{p}{q}=24178.178178-14.178178 $

$9999\frac{p}{q}=24154 $

$\frac{p}{q}=\frac{24154}{9999} $

18. Simplify ${{\left( 27 \right)}^{-\frac{2}{3}}}\div {{9}^{\frac{1}{2}}}{{.3}^{-\frac{3}{2}}}$.

Ans: ${{\left( 27 \right)}^{-\frac{2}{3}}}\div {{9}^{\frac{1}{2}}}{{.3}^{-\frac{3}{2}}}$

$ =\frac{{{\left( 3\times 3\times 3 \right)}^{-\frac{2}{3}}}\times {{3}^{\frac{3}{2}}}}{{{\left( 3\times 3 \right)}^{\frac{1}{2}}}}\left[ {{a}^{-m}}=\frac{1}{{{a}^{m}}} \right] $

$ =\frac{{{\left( {{3}^{3}} \right)}^{-\frac{2}{3}}}\times {{3}^{\frac{3}{2}}}}{{{\left( {{3}^{2}} \right)}^{\frac{1}{2}}}} $

$=\frac{{{3}^{\frac{3}{2}-2}}}{3}=\frac{{{3}^{-\frac{1}{3}}}}{3} $

$=\frac{1}{{{3}^{\frac{4}{3}}}}=\frac{1}{\sqrt[3]{81}} $

19. Find three rational numbers between $2.\overline{2}$ and $2.\overline{3}.$

Ans: The irrational numbers are the numbers that do not end after the decimal point nor repeat its numbers in a sequence. 

Representing the given numbers in decimal form we have,

$ 2.\overline{2}=2.222222222...... $

$ 2.\overline{3}=2.333333333....... $ 

So any numbers between these two numbers that do not end nor repeat in any sequence gives the required irrational numbers.

Three rational numbers between $2.\overline{2}$ and $2.\overline{3}$ are $2.222341365....$, $2.28945187364....$ and $2.2321453269....$

20. Give an example of two irrational numbers whose

i. Sum is a rational number

Ans: The required two irrational numbers are $2+\sqrt{2}$ and $2-\sqrt{2}$

Sum $2+\sqrt{2}+2-\sqrt{2}=4$ which is a rational number.

ii. Product is a rational number

Ans: The required two irrational numbers are $3\sqrt{2}$ and $6\sqrt{2}$

Product $3\sqrt{2}\times 6\sqrt{2}=18\times 2=36$ which is rational.

iii. Quotient is a rational number

Ans: The required two irrational numbers are $2\sqrt{125}$ and $3\sqrt{5}$

Quotient $\frac{2\sqrt{125}}{3\sqrt{5}}=\frac{2}{3}\sqrt{\frac{125}{5}}=\frac{2}{3}\times 5=\frac{10}{3}$

21 . If $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$, find the value of $\frac{5}{\sqrt{2}+\sqrt{3}}$.

Ans: First let us take the given expression and by rationalizing the denominator we get,

$\frac{5}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$ 

$\frac{5\left( \sqrt{2}-\sqrt{3} \right)}{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}} $

$=\frac{5\left( \sqrt{2}-\sqrt{3} \right)}{2-3} $

Now, substituting the required values of irrational numbers we get,

$=-5\left[ 1.414-1.732 \right] $

$  =-5\times -0.318 $

$  =1.59 $ 

22. Visualize 2.4646 on the number line using successive magnification.

23. Rationalizing the denominator of $\frac{1}{4+2\sqrt{3}}$.

Ans: First let us take the given expression and rationalizing the denominator by multiplying the numerator and denominator with its conjugate we get,

$\frac{1}{4+2\sqrt{3}}=\frac{1}{4+2\sqrt{3}}\times \frac{4-2\sqrt{3}}{4+2\sqrt{3}} $

$ =\frac{4-2\sqrt{3}}{{{\left( 4 \right)}^{2}}-{{\left( 2\sqrt{3} \right)}^{2}}} $

$=\frac{4-2\sqrt{3}}{16-{{\left( 2\sqrt{3} \right)}^{2}}} $ 

$=\frac{4-2\sqrt{3}}{16-12} $ 

$  =\frac{4-2\sqrt{3}}{4} $

$ =\frac{2\left( 2-\sqrt{3} \right)}{4} $

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

24. Visualize the representation of $5.3\overline{7}$ on the number line up to 3 decimal places.

Ans: The representation of $5.3\overline{7}$ on the number line is given below:

25. Show that \[5\sqrt{2}\] is not a rational number.

Ans: Let us assume that \[5\sqrt{2}\] is a rational number.

Take \[x=5\sqrt{2}\] , with \[x\]being rational as well.

\[x=5\sqrt{2}\]

\[\Rightarrow \frac{x}{5}=\sqrt{2}\]

Let us compare the terms in LHS and RHS.

In LHS, we have\[\frac{x}{5}\] , with \[x\] and $5$ being rational numbers (Here \[x\] is rational, based on our assumption). So \[\frac{x}{5}\] is a rational number.

In RHS, we have$\sqrt{2}$, which is not a rational number, but an irrational number.

This is a contradiction, i.e. $LHS\ne RHS$.

So, we can conclude that  \[5\sqrt{2}\] is not a rational number.

26. Simplify \[3\sqrt[3]{250}+7\sqrt[3]{16}-4\sqrt[3]{54}\].

Ans: Let us first find the cube roots of given numbers to their simplest forms by using the prime factorization then we get,

\[3\sqrt[3]{250}+7\sqrt[3]{16}-4\sqrt[3]{54}=3\sqrt[3]{5\times 5\times 5\times 2}+7\sqrt[3]{2\times 2\times 2\times 2}-4\sqrt[3]{3\times 3\times 3\times 2}\]

\[=\left( 3\times 5\sqrt[3]{2} \right)+\left( 7\times 2\sqrt[3]{2} \right)-\left( 4\times 3\sqrt[3]{2} \right)\]

\[=\left( 15\sqrt[3]{2} \right)+\left( 14\sqrt[3]{2} \right)-\left( 12\sqrt[3]{2} \right)\]

\[=\left( 15+14-12 \right)\sqrt[3]{2}\]

\[=17\sqrt[3]{2}\]

Thus, we get \[3\sqrt[3]{250}+7\sqrt[3]{16}-4\sqrt[3]{54}=17\sqrt[3]{2}\]

27. Simplify \[3\sqrt{48}-\frac{5}{2}\sqrt{\frac{1}{3}}+4\sqrt{3}\].

Ans: Let us first find the square roots of given numbers to their simplest forms by using the prime factorization then we get,

\[3\sqrt{48}-\frac{5}{2}\sqrt{\frac{1}{3}}+4\sqrt{3}=\left( 3\sqrt{2\times 2\times 2\times 2\times 3} \right)-\left[ \frac{5}{2}\left( \sqrt{\frac{1}{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \right) \right]+\left( 4\sqrt{3} \right)\]

\[=\left( 3\times 2\times 2\sqrt{3} \right)-\left[ \frac{5}{2}\left( \frac{\sqrt{3}}{3} \right) \right]+\left( 4\sqrt{3} \right)\]

\[=\left( 12\sqrt{3} \right)-\left( \frac{5\sqrt{3}}{6} \right)+\left( 4\sqrt{3} \right)\]

\[=\left( 12-\frac{5}{6}+4 \right)\sqrt{3}\]

\[=\left( \frac{72-5+24}{6} \right)\sqrt{3}\]

\[=\frac{91}{6}\sqrt{3}\]

Thus, we get \[3\sqrt{48}-\frac{5}{2}\sqrt{\frac{1}{3}}+4\sqrt{3}=\frac{91}{6}\sqrt{3}\]

28. If $\frac{1}{7}=0.\overline{142857}$. Find the value of $\frac{2}{7},\frac{3}{7},\frac{4}{7}$

Ans: It is given that – $\frac{1}{7}=0.\overline{142857}$

(i) $\frac{2}{7}=2\times \frac{1}{7}$

$=2\times 0.\overline{142857}$

$=0.\overline{285714}$

$\Rightarrow \frac{2}{7}=0.\overline{285714}$

(ii) $\frac{3}{7}=3\times \frac{1}{7}$

$=3\times 0.\overline{142857}$

$=0.\overline{428571}$

$\Rightarrow \frac{3}{7}=0.\overline{428571}$

(iii) $\frac{4}{7}=4\times \frac{1}{7}$

$=4\times 0.\overline{142857}$

$=0.\overline{571428}$

$\Rightarrow \frac{4}{7}=0.\overline{571428}$

29. Find $6$ rational numbers between $\frac{6}{5}$ and $\frac{7}{5}$

Ans: It is possible to divide the interval between $\frac{6}{5}$ and $\frac{7}{5}$ into $10$ equal parts.

Then we will have – $\frac{6}{5},\frac{6.1}{5},\frac{6.2}{5},\frac{6.3}{5},\frac{6.4}{5},\frac{6.5}{5},\frac{6.6}{5},\frac{6.7}{5},\frac{6.8}{5},\frac{6.9}{5},\frac{7}{5}$

i.e. $\frac{60}{50},\frac{61}{50},\frac{62}{50},\frac{63}{50},\frac{64}{50},\frac{65}{50},\frac{66}{50},\frac{67}{50},\frac{68}{50},\frac{69}{50},\frac{70}{50}$

From these fractions, it is possible to choose $6$ rational numbers between $\frac{6}{5}$ and $\frac{7}{5}$

Thus , $6$ rational numbers between $\frac{6}{5}$ and $\frac{7}{5}$ are $\frac{61}{50},\frac{62}{50},\frac{63}{50},\frac{64}{50},\frac{65}{50},\frac{66}{50}$

30. Show how $\sqrt{4}$ can be represented on the number line.

Ans: Take $AB=OA=1\text{ }unit$ on a number line.

Also, $\angle A={{90}^{\circ }}$

In $\vartriangle OAB$, apply Pythagoras Theorem,

$\therefore O{{A}^{2}}+A{{B}^{2}}=O{{B}^{2}}$

$\Rightarrow O{{B}^{2}}={{1}^{2}}+{{1}^{2}}$

$\Rightarrow O{{B}^{2}}=1+1$

$\Rightarrow O{{B}^{2}}=2$

$\Rightarrow OB=\sqrt{2}$

Now, draw $OB=O{{A}_{1}}=\sqrt{2}$

And, ${{A}_{1}}{{B}_{1}}=1\text{ unit}$ with$\angle {{A}_{1}}={{90}^{\circ }}$

In \[\vartriangle O{{A}_{1}}{{B}_{1}}\], apply Pythagoras Theorem,

$\therefore O{{A}_{1}}^{2}+{{A}_{1}}{{B}_{1}}^{2}=O{{B}_{1}}^{2}$

$\Rightarrow O{{B}_{1}}^{2}={{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}$

$\Rightarrow O{{B}_{1}}^{2}=2+1$

$\Rightarrow O{{B}_{1}}^{2}=3$

$\Rightarrow O{{B}_{1}}=\sqrt{3}$

Now, draw $O{{B}_{1}}=O{{A}_{2}}=\sqrt{3}$

And, \[{{A}_{2}}{{B}_{2}}=1\text{ unit}\] with$\angle {{A}_{2}}={{90}^{\circ }}$

In \[\vartriangle O{{A}_{2}}{{B}_{2}}\], apply Pythagoras Theorem,

$\therefore O{{A}_{2}}^{2}+{{A}_{2}}{{B}_{2}}^{2}=O{{B}_{2}}^{2}$

$\Rightarrow O{{B}_{2}}^{2}={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}$

$\Rightarrow O{{B}_{2}}^{2}=3+1$

$\Rightarrow O{{B}_{2}}^{2}=4$

$\Rightarrow O{{B}_{2}}=\sqrt{4}$

Now, draw $O{{B}_{2}}=O{{A}_{3}}=\sqrt{4}$

Thus line segment $O{{A}_{3}}=\sqrt{4}$

Short Answer Questions (4 Marks)

1. Write the following in decimal form and say what kind of decimal expansion each has:

i. $\frac{36}{100}$

Ans: Performing long division of $36$ by $100$

$\begin{matrix} &{0.36}\\ 100&{\overline{)\;36\quad}}\\ &\underline{-0\quad}\\ &360\\ &\underline{-300\quad}\\ &\;\;600\\ &\underline{-600}\\ &\underline{\quad 0 \;\;} \end{matrix}$

Thus, $\frac{36}{100}=0.36$ - this is a terminating decimal.

ii. $\frac{1}{11}$

Ans: Performing long division of $1$ by $11$

$\begin{matrix} {} & 0.0909.. \\ 11 & \overline{)\text{ }1\text{ }} \\ {} & \underline{-0} \\ {} & 10 \\ {} & \underline{-0} \\ {} & 100 \\ {} & \underline{-99} \\ {} & 10 \\ {} & \underline{-0} \\ {} & 100 \\ {} & \underline{-99} \\ {} & 1 \\ \end{matrix}$

It can be seen that performing further division will produce a reminder of $1$ continuously.

Thus, $\frac{1}{11}=0.09090...$ i.e. $\frac{1}{11}=0.\overline{09}$, this is a non-terminating, but recurring decimal.

iii. $4\frac{1}{8}$

Ans: First convert the mixed fraction into an improper fraction –

$4\frac{1}{8}=\frac{(4\times 8)+1}{8}=\frac{33}{8}$

Performing long division of $33$ by $8$

$\begin{matrix} {} & 4.125 \\ 8 & \overline{\left){\text{ }33\text{ }}\right.} \\ {} & \underline{-32} \\ {} & 10 \\ {} & \underline{-8} \\ {} & 20 \\ {} & \underline{-16} \\ {} & 40 \\ {} & \underline{-40} \\ {} & 0 \\ \end{matrix}$

Thus, $4\frac{1}{8}=4.125$ - this is a terminating decimal.

iv. $\frac{3}{13}$

Ans: Performing long division of $3$ by $13$

$\begin{matrix} {} & 0.230769.. \\ 13 & \overline{\left){\text{ }3\text{ }}\right.} \\ {} & \underline{-0} \\ {} & 30 \\ {} & \underline{-26} \\ {} & 40 \\ {} & \underline{-39} \\ {} & 10 \\ {} & \underline{-0} \\ {} & 100 \\ {} & \underline{-91} \\ {} & 90 \\ {} & \underline{-78} \\ {} & 120 \\ {} & \underline{-117} \\ {} & 3 \\ \end{matrix}$

It can be seen that performing further division will produce a reminder of $3$ periodically, after every six divisions.

Thus, $\frac{3}{13}=0.230769...$ i.e. \[\frac{3}{13}=0.\overline{230769}\], this is a non-terminating, but recurring decimal.

v. $\frac{2}{11}$

Ans: Performing long division of $2$ by $11$

$\begin{matrix} {} & 0.1818.. \\ 11 & \overline{)\text{ 2 }} \\ {} & \underline{-0} \\ {} & 20 \\ {} & \underline{-11} \\ {} & 90 \\ {} & \underline{-88} \\ {} & 20 \\ {} & \underline{-11} \\ {} & 90 \\ {} & \underline{-88} \\ {} & 2 \\ \end{matrix}$

It can be seen that performing further division will produce a reminder of $2$followed by $9$ alternatively.

Thus, $\frac{2}{11}=0.181818...$ i.e. $\frac{2}{11}=0.\overline{18}$this is a non-terminating, but recurring decimal.

vi. $\frac{329}{400}$

Ans: Performing long division of $33$ by $8$

$\begin{matrix} {} & 0.8225 \\ 400 & \overline{)\text{ 329 }} \\ {} & \underline{-0} \\ {} & 3290 \\ {} & \underline{-3200} \\ {} & 900 \\ {} & \underline{-800} \\ {} & 1000 \\ {} & \underline{-800} \\ {} & 2000 \\ {} & \underline{-2000} \\ {} & 0 \\ \end{matrix}$

Thus, $\frac{329}{400}=0.8225$ - this is a terminating decimal.

2. Classify the following as rational or irrational:

i. $\sqrt{23}$

Ans: It is known that the root of $23$ will produce a non-terminating and non-recurring decimal number (it is not a perfect square value), also it cannot be represented as a fraction. Thus we can say that $\sqrt{23}$ is an irrational number.

ii. $\sqrt{225}$

Ans: It is known that $\sqrt{225}=15$, which is an integer.

Thus $\sqrt{225}$ is a rational number.

iii. $0.3796$

Ans: Here, $0.3796$ is a terminating decimal number, and also it can be expressed as a fraction.

i.e. $0.3796=\frac{3796}{10000}=\frac{949}{2500}$

Thus $0.3796$ is a rational number.

iv. $7.478478...$

Ans: Here, $7.478478...$ is a non-terminating, but recurring decimal number, and also it can be expressed as a fraction.

i.e. $7.478478...=7.\overline{487}$

Converting it into fraction 

If $x=7.478478...\text{          (1)}$

Then $1000x=7478.478478...\text{          (2)}$

Subtract equations $(2)-(1)$

$ 1000x=7478.478478... $

$  \underline{-\text{      }x=\text{      }7.478478...} $

$ \text{  }999x=7471 $

Now, $999x=7471$

$\Rightarrow x=\frac{7471}{999}$

i.e. $7.\overline{478}=\frac{7471}{999}$

Thus $7.478478...$ is a rational number.

v. $1.101001000100001...$

Ans: Here,  $1.101001000100001...$ is a non-terminating and non-recurring decimal number and also it cannot be represented as a fraction. Thus we can say that $1.101001000100001...$ is an irrational number.

3. Rationalize the denominator of the following:

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

Ans: In order to rationalize the denominator, we multiply and divide $\frac{1}{\sqrt{7}}$ by $\sqrt{7}$

$\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$

Rationalizing the denominator of  $\frac{1}{\sqrt{7}}$ produces $\frac{\sqrt{7}}{7}$.

ii. $\frac{1}{\sqrt{7}-\sqrt{6}}$

Ans: In order to rationalize the denominator, we multiply and divide $\frac{1}{\sqrt{7}-\sqrt{6}}$ by $\sqrt{7}+\sqrt{6}$

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

Using the identity - \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

$=\frac{\sqrt{7}+\sqrt{6}}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}}$

$=\frac{\sqrt{7}+\sqrt{6}}{7-6}$

$=\frac{\sqrt{7}+\sqrt{6}}{1}$

$\Rightarrow \frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{7}+\sqrt{6}$

Rationalizing the denominator of  $\frac{1}{\sqrt{7}-\sqrt{6}}$ produces $\sqrt{7}+\sqrt{6}$.

iii. $\frac{1}{\sqrt{5}+\sqrt{2}}$

Ans: In order to rationalize the denominator, we multiply and divide $\frac{1}{\sqrt{5}+\sqrt{2}}$ by $\sqrt{5}-\sqrt{2}$

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

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

$=\frac{\sqrt{5}-\sqrt{2}}{5-2}$

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

$\Rightarrow \frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{3}$

Rationalizing the denominator of  $\frac{1}{\sqrt{5}+\sqrt{2}}$ produces $\frac{\sqrt{5}-\sqrt{2}}{3}$.

iv. $\frac{1}{\sqrt{7}-2}$

Ans: In order to rationalize the denominator, we multiply and divide $\frac{1}{\sqrt{7}-2}$ by $\sqrt{7}+2$

$\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}=\frac{\sqrt{7}+2}{\left( \sqrt{7}-2 \right)\left( \sqrt{7}+2 \right)}$

$=\frac{\sqrt{7}+2}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( 2 \right)}^{2}}}$

$=\frac{\sqrt{7}+2}{7-4}$

$=\frac{\sqrt{7}+2}{3}$

$\Rightarrow \frac{1}{\sqrt{7}+2}=\frac{\sqrt{7}+2}{3}$

Rationalizing the denominator of  $\frac{1}{\sqrt{7}-2}$ produces $\frac{\sqrt{7}+2}{3}$.

Long Answer Questions (5 Marks)

It can be seen that performing further divisions will produce a reminder of $3$ periodically, after every six divisions.

Thus, $\frac{329}{400}=s0.8225$ - this is a terminating decimal.

4. If $\sqrt{5}=2.236$ and $\sqrt{3}=1.732$. Find the value of $\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{7}{\sqrt{5}-\sqrt{3}}$.

Ans: It is given that – 

$\sqrt{5}=2.236$

$\sqrt{3}=1.732$

Now, $\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{7}{\sqrt{5}-\sqrt{3}}$

$\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{7}{\sqrt{5}-\sqrt{3}}=\left[ \frac{2}{\left( \sqrt{5}+\sqrt{3} \right)}\times \frac{\left( \sqrt{5}-\sqrt{3} \right)}{\left( \sqrt{5}-\sqrt{3} \right)} \right]+\left[ \frac{7}{\left( \sqrt{5}-\sqrt{3} \right)}\times \frac{\left( \sqrt{5}+\sqrt{3} \right)}{\left( \sqrt{5}+\sqrt{3} \right)} \right]$

$=\left[ \frac{2\left( \sqrt{5}-\sqrt{3} \right)}{\left( \sqrt{5}+\sqrt{3} \right)\left( \sqrt{5}-\sqrt{3} \right)} \right]+\left[ \frac{7\left( \sqrt{5}+\sqrt{3} \right)}{\left( \sqrt{5}-\sqrt{3} \right)\left( \sqrt{5}+\sqrt{3} \right)} \right]$

$=\left[ \frac{\left( 2\sqrt{5}-2\sqrt{3} \right)+\left( 7\sqrt{5}+7\sqrt{3} \right)}{\left( \sqrt{5}+\sqrt{3} \right)\left( \sqrt{5}-\sqrt{3} \right)} \right]$

$=\left[ \frac{2\sqrt{5}-2\sqrt{3}+7\sqrt{5}+7\sqrt{3}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}} \right]$

$=\left[ \frac{(2+7)\sqrt{5}+(7-2)\sqrt{3}}{5-3} \right]$

$=\left[ \frac{9\sqrt{5}+5\sqrt{3}}{2} \right]$

Since, $\sqrt{5}=2.236$ and $\sqrt{3}=1.732$

$=\left[ \frac{(9\times 2.236)+(5\times 1.732)}{2} \right]$

$=\left[ \frac{20.124+8.66}{2} \right]$

$=\left[ \frac{28.784}{2} \right]$

Thus, $\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{7}{\sqrt{5}-\sqrt{3}}=14.392$

5. Find the value of $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{7}{\sqrt{5}-\sqrt{2}}$, if $\sqrt{5}=2.236$ and $\sqrt{2}=1.414$.

$\sqrt{2}=1.414$

Now, $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{7}{\sqrt{5}-\sqrt{2}}$

$\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{7}{\sqrt{5}-\sqrt{2}}=\left[ \frac{3}{\left( \sqrt{5}+\sqrt{2} \right)}\times \frac{\left( \sqrt{5}-\sqrt{2} \right)}{\left( \sqrt{5}-\sqrt{2} \right)} \right]+\left[ \frac{7}{\left( \sqrt{5}-\sqrt{2} \right)}\times \frac{\left( \sqrt{5}+\sqrt{2} \right)}{\left( \sqrt{5}+\sqrt{2} \right)} \right]$

$=\left[ \frac{3\left( \sqrt{5}-\sqrt{2} \right)}{\left( \sqrt{5}+\sqrt{2} \right)\left( \sqrt{5}-\sqrt{2} \right)} \right]+\left[ \frac{7\left( \sqrt{5}+\sqrt{2} \right)}{\left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}+\sqrt{2} \right)} \right]$

$=\left[ \frac{\left( 3\sqrt{5}-3\sqrt{2} \right)+\left( 7\sqrt{5}+7\sqrt{2} \right)}{\left( \sqrt{5}+\sqrt{2} \right)\left( \sqrt{5}-\sqrt{2} \right)} \right]$

$=\left[ \frac{3\sqrt{5}-3\sqrt{2}+7\sqrt{5}+7\sqrt{2}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \right]$

$=\left[ \frac{(3+7)\sqrt{5}+(7-3)\sqrt{2}}{5-2} \right]$

$=\left[ \frac{10\sqrt{5}+4\sqrt{2}}{3} \right]$

Since, $\sqrt{5}=2.236$ and $\sqrt{2}=1.414$

$=\left[ \frac{(10\times 2.236)+(4\times 1.414)}{3} \right]$

$=\left[ \frac{22.36+5.656}{3} \right]$

$=\left[ \frac{28.016}{3} \right]$

Thus, $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{7}{\sqrt{5}-\sqrt{2}}=\frac{28.016}{3}$

6. Simplify $\frac{2+\sqrt{5}}{2-\sqrt{5}}+\frac{2-\sqrt{5}}{2+\sqrt{5}}$

Ans: $\frac{2+\sqrt{5}}{2-\sqrt{5}}+\frac{2-\sqrt{5}}{2+\sqrt{5}}$

\[\frac{2+\sqrt{5}}{2-\sqrt{5}}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=\left[ \frac{2+\sqrt{5}}{2-\sqrt{5}}\times \frac{\left( 2+\sqrt{5} \right)}{\left( 2+\sqrt{5} \right)} \right]+\left[ \frac{2-\sqrt{5}}{2+\sqrt{5}}\times \frac{\left( 2-\sqrt{5} \right)}{\left( 2-\sqrt{5} \right)} \right]\]

\[=\left[ \frac{\left( 2+\sqrt{5} \right)\left( 2+\sqrt{5} \right)}{\left( 2-\sqrt{5} \right)\left( 2+\sqrt{5} \right)} \right]+\left[ \frac{\left( 2-\sqrt{5} \right)\left( 2-\sqrt{5} \right)}{\left( 2+\sqrt{5} \right)\left( 2-\sqrt{5} \right)} \right]\]

\[=\left[ \frac{{{\left( 2+\sqrt{5} \right)}^{2}}+{{\left( 2-\sqrt{5} \right)}^{2}}}{\left( 2-\sqrt{5} \right)\left( 2+\sqrt{5} \right)} \right]\]

Using the identities –

\[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

\[{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]

\[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]

\[=\left[ \frac{\left( {{\left( 2 \right)}^{2}}+{{\left( \sqrt{5} \right)}^{2}}+\left( 2\times 2\times \sqrt{5} \right) \right)+\left( {{\left( 2 \right)}^{2}}+{{\left( \sqrt{5} \right)}^{2}}-\left( 2\times 2\times \sqrt{5} \right) \right)}{{{\left( 2 \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}} \right]\]

\[=\left[ \frac{\left( 4+5+\left( 4\sqrt{5} \right) \right)+\left( 4+5-\left( 4\sqrt{5} \right) \right)}{4-5} \right]\]

$=\left[ \frac{9+9}{-1} \right]$

$=\left[ \frac{18}{-1} \right]$

Thus, $\frac{2+\sqrt{5}}{2-\sqrt{5}}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=\left( -18 \right)$

7. Find a and b, if $\frac{3-\sqrt{6}}{3+2\sqrt{6}}=a\sqrt{6}-b$

Ans: $\frac{3-\sqrt{6}}{3+2\sqrt{6}}=a\sqrt{6}-b$

Here, 

$LHS=\frac{3-\sqrt{6}}{3+2\sqrt{6}}$

$RHS=a\sqrt{6}-b$

Start by rationalizing the denominator in LHS

In order to rationalize the denominator, we multiply and divide $\frac{3-\sqrt{6}}{3+2\sqrt{6}}$ by $3+2\sqrt{6}$

$\frac{3-\sqrt{6}}{3+2\sqrt{6}}\times \frac{3-2\sqrt{6}}{3-2\sqrt{6}}=\frac{\left( 3-\sqrt{6} \right)\left( 3-2\sqrt{6} \right)}{\left( 3+2\sqrt{6} \right)\left( 3-2\sqrt{6} \right)}$

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

$=\frac{\left( 9 \right)-\left( 6\sqrt{6} \right)-\left( 3\sqrt{6} \right)+\left( 12 \right)}{9-24}$

$=\frac{\left( 21 \right)-\left( 9\sqrt{6} \right)}{-15}$

$=\frac{\left( 21 \right)}{-15}-\frac{\left( 9\sqrt{6} \right)}{-15}$

They are all divisible by $3$

$=-\frac{7}{5}+\frac{\left( 3\sqrt{6} \right)}{5}$

Thus, $LHS=\frac{3}{5}\sqrt{6}-\frac{7}{5}$

Comparing with RHS, we get – 

Thus, 

$a=\frac{3}{5}$

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

Important Questions for Class 9 Maths Chapter 1 - Free PDF Download

Class 9 is like beginning to your academics career, which is right before board 10th. Thus students need to be very serious regarding their studies during preparation. No matter what we learn in class 9, it is important to clarify your concepts better. Hence, clearing Number System concepts will help students further apply in electronics physics and higher maths. Thus it is better to build a good base in mathematics with these Important Questions Maths Class 9 Chapter 1.

Vedantu provides a free PDF to download for Class 9 Chapter 1 Important Questions such that students can prepare well according to the CBSE syllabus. CBSE is strict to its pattern and follows the same throughout the question paper set. Students need to understand these guidelines and find solutions with a proper explanation. This free PDF online will surely help students master their concepts and build a Number System base. This PDF covers all the important concepts in the form of question example to learn how to implement them during exams. Thus PDF proves to be magical for those who are weak in mathematics as they also get solutions to the concepts covered in the back exercise.

Number System Class 9 Important Questions

Before you begin practising Class 9 Chapter 1 Maths Important Questions, you need to know the different topics and subtopics to cover in the chapter. Chapter 1 of Mathematics Class 9 covers a total of 6 exercises with a small introduction of the number system, number lines, defining real numbers, natural numbers, whole numbers, rational, and irrational numbers. Also, students become familiar with the concepts of addition, subtraction, division, and multiplication of the real numbers. The last topic in the chapter will be covering the law of exponents in the real numbers.

Below are the mentioned section-wise topics and concepts that a student to prepare through important questions.

Exercise 1.1

Under exercise 1.1, students will become familiar with the basic understanding of rational and irrational numbers. There is also a revision on whole numbers, real number, integers, and natural numbers and definition. They will even know how to represent a number in the form of p/q, where q is not equal to 0.

Exercise 1.2

Further moving to in-depth study about rational and irrational numbers, there are questions on justification with true and false. Also, students become familiar with the concept of representing rational and irrational numbers on the number line. A new concept of constructing a square root spiral comes in exercise 1.2.

Exercise 1.3

Here the student will know how to represent fractions into decimal form and find if it is terminating or non-terminating. Thus the concepts of terminating and non-terminating fractions will further help to identify it is a rational or irrational number.

Exercise 1.4

A new concept of representing decimal expansion on the number line is introduced through exercise 1.4 mathematics class 9. Here students will learn about magnifying the number to the maximum requirement and representing it on the number line. These decimal places can be either terminating or non-terminating. Hence there are two different concepts in number line representation of decimal expansions.

Exercise 1.5

Now comes the basic calculations of different rational and irrational numbers. It includes addition, multiplication, subtraction, and division of the rational and irrational numbers. Questions will be based on such concepts, and thus you have to simplify the statement accordingly. Also, students become familiar with the concept of rationalising.

Exercise 1.6

Here students will learn to solve questions based with a number having power in fractional form. Also, it might cover the basic addition and subtraction of powers for in-depth conceptual and extra knowledge.

These are the six exercises which will be covered under Chapter 1 Maths Class 9 Important Questions. Hence students can prepare questions according to the concepts discussed above.

Chapter 1 Maths Class 9 Important Questions

According to the syllabus mentioned above exercise-wise, below are some important questions covered to let students prepare well for important questions for class 9 maths number system. These exercise-wise solutions will let students master each concept in detail. Below are some of the questions that are usually picked to set question paper as prepared by CBSE.

State if zero is a rational number. Justify your statement by representing it in p/q where q is not equal to 0 and p and q are both integers.

Find 5 different rational numbers between 5 and 6. Mention each step in detail.

Find out 5 different rational numbers between 12/11 and 10/11.

Justify your statement for the following terms stating true or false.

Rational numbers are real.

A number line having representation in the form of √m has m as a natural number.

A real number is always an irrational number.

Check if true or false. Square roots of all positive numbers will be irrational. Explain your answer statement with the help of an example.

Try to represent √5 on the number line.

Represent the following in the form of decimal expansion:

(i) 36/100  (ii) 1/11  (iii) 4⅛  (iv) 3/13  (v) 2/11  (vi) 329/400

Turn 2/7, 13/7, 4/7, 5/7, 6/7

into decimal expansions without actually doing any long division calculations. Here 1/7= 0.142857.

Express the following as fractional form p/q where q is not 0.

(i) \[0.\overline{6}\]

(ii) \[0.4 \overline{7}\]

(iii) \[0. \overline{001}\]

Represent 2.675 on the number line with number line magnification.

How will you visualise 6.2626…... on the number line up to 4 decimal places?

State if the numbers are rational or irrational.

(i) \[2 - \sqrt{5}\]  (ii)  \[(3 + \sqrt{23}) - \sqrt{23}\]  (iii) \[\frac{2\sqrt{7}}{7\sqrt{7}}\]  (iv) \[\frac{1}{\sqrt{2}}\]  (v) \[2 \pi\]

Simplify the following terms:

(i) \[(3 + \sqrt{3}) ( 2 + \sqrt{2})\] (ii) \[(3 + \sqrt{3}) (3 + \sqrt{3})\]

(iii) \[(\sqrt{5} + \sqrt{2})^{2}\] (iv)\[(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})\]

Rationalise the following terms with their denominators:

(i) \[\frac{1}{7}\] (ii) \[\frac{1}{\sqrt{7} - \sqrt{6}}\] (iii) \[\frac{1}{\sqrt{5} + \sqrt{2}}\] (iv) \[\frac{1}{\sqrt{7} - 2}\]

(i) \[64^{1}{2}\] (ii) \[32^{1}{5}\] (iii) \[125^{1}{3}\]

Class 9 Maths Chapter 1 Extra Questions

Find three rational numbers between $\frac{1}{3}$ and $\frac{1}{2}$. 

Express 0.4323232 in the form of  $\frac{a}{b}$ where a and b are integers and b 0.

Simplify and find the value of $(729)^{1/6}$ .

Rationalise the denominator 1 9 + 5 + 6 .

Find 6 rational numbers between 4 and 6.

Simplify $\sqrt[3]{2}$+$\sqrt[4]{3}$ and $\sqrt{5}+\sqrt{2}$ .

Locate $\sqrt{5}$ on the number line.

Visualise the representation of 4.26 on the number line upto 3 decimal places.

Is 2 - 5   a rational number or irrational number?

Convert 0.45 into rational numbers.

Benefits of Important Questions for Class 9 Maths Number System

Number System Class 9 Important Questions PDF is prepared by Vedantu experts and is free of cost. Students can also schedule additional problems to prepare in-depth to clarify their concepts. These questions turn to be magical for those weak in mathematics or do not show interest in it. The questions are prepared precisely accordingly to CBSE guidelines for their question paper pattern such that students need not search them anywhere else.

Below are the mentioned reasons why students should refer to the Important Question of Maths Class 9 Chapter 1:

Students are free to access these important questions PDF. They need not pay for any study material on Vedantu website.

These Number System Class 9 Important Questions are helpful during exams and help students clarify their concepts of homework questions.

These crucial questions are available in PDF format, which can be easily downloaded through the website. Thus students need not unnecessarily waste their precious time in finding solutions to problems.

Students can print these PDF questions and solutions, which reduces the stress of preparing through soft copies.

All the questions and solutions to the questions are prepared according to CBSE guidelines. Thus it will help students to know question paper pattern.

There is a total of 100 marks for the Class 9 Mathematics paper where 20 marks are for internal assessment, and rest 80 is for the written exam. The marks weightage for Chapter 1 Maths Class 9 is 8 out of 80. Rest 72 marks are for rest of the syllabus of Class 9 Mathematics.

There are a total of 26 sums to be solved according to NCERT book for Chapter 1 Of Class 9 Maths in its six exercises. However, there are other Class 9 Maths Chapter 1 Important Questions prepared by the experts and solutions to each of them provided. Thus, students can build a good base in mathematics through this PDF of important questions available with Vedantu for free. According to exercise-wise, some questions describe different questions and situations that students generally encounter while sitting in the exam. Thus students need to struggle to find essential concepts in the chapter and solutions to each of them.

Important Related Links for CBSE Class 9 

FAQs on CBSE Class 9 Maths Important Questions for Chapter 1 - Number System

1. What number of questions are there in each exercise of Chapter 1 of Class 9 Maths?

Real numbers are used in the first chapter of Class 9 Mathematics. There are a total of six workouts. In Chapter 1, students will learn about several real number concepts. There are four questions in Exercise 1.1, four questions in Exercise 1.2, nine questions in Exercise 1.3, two questions in Exercise 1.4, five questions in Exercise 1.5, and three questions in Exercise 1.6. Students must rehearse all NCERT problems offered by Vedantu in order to understand Chapter 1 of Class 9 maths.

2. Is Chapter 1 of Class 9 Maths difficult to solve?

Students of Class 9 should understand the concepts of real numbers given in Chapter 1 of Class 9 Maths. It is not difficult to solve if students understand the basic concepts of real numbers. Students can take help from the NCERT solutions Class 9 Maths Chapter 1 given at Vedantu app and website. All concepts related to the real numbers are explained in a simple way for quick understanding of the students. 

3. Why are Class 9 Maths NCERT Solutions of Chapter 1 important?

Students should rehearse Class 9 Mathematics NCERT Answers for Chapter 1 to comprehend and practise real-world questions. The NCERT Answers are essential since students may face similar questions in their examinations. With the aid of these questions, students may simply tackle the actual numbers problems in the exam. Students may access complete NCERT Answers for Class 9 Mathematics Chapter 1 for free on the Vedantu platform.

4. What are five rational numbers between ⅖ and ⅗?

Students can find five or more rational numbers between the two given rational numbers easily. They can follow the given steps to find five rational numbers between ⅖ and ⅗. We have to multiply the numerator and denominator of the given rational numbers with the same number.

⅖ x 6/6 = 12/30

⅗ x 6/6= 18/30

Now, we can write five rational numbers between 12/30 and 18/30. Thus, the five rational numbers are 13/30, 14/30, 15/30, 16/30, 17/30.

5. What are the main topics covered in Chapter 1 of Class 9 Maths?

In Chapter 1 of Class 9 Mathematics, students will study about rational and irrational numbers. Students will learn about irrational numbers, how to represent real numbers on a number line, how to express real numbers in decimal form, different operations on real numbers, and real-world exponent rules. To comprehend the various principles of rational numbers in Class 9 Chapter 1, students should practise all NCERT problems offered in the textbook.

CBSE Class 9 Maths Important Questions

Cbse study materials.