# NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Ex 1.1) Exercise 1.1

NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 is concerned with the basis of the numeric system. Students may fully understand all of their root-level ideas in Real Numbers Class 10 as they get ready for their Class 10 examinations. These NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 have been created by Extramarks’ professional subject matter experts to help students study for their first-term exams. Students can better understand the ideas because it thoroughly covers all of the key principles.

NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 focuses on key ideas including Euclid’s division lemma, prime numbers, composite numbers, the Fundamental Theorem of Arithmetic, HCF and LCM via the Prime Factorization Method, and irrational numbers, among others.

The NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 are very helpful for students to properly prepare for tests and earn good grades. These NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 are important resources that may help students not only complete the entire curriculum but also conduct in-depth analyses of the courses. On the Extramarks website and mobile application, users can download the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 in pdf format. Some of them are also included in the exercises.

On the Extramarks website and mobile application, students may find NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. To aid students in their board exam preparation, Extramarks’ professional faculty have created these NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. In order to help the students quickly solve the difficulties, experts solved them and shared the NCERT Solutions for Mathematics. They put a lot of effort into making the solutions simple for the students to understand. For each of the answers given to the questions in the exercises of the NCERT books, Extramarks provides a thorough and step-by-step explanation in the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1.

NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1, contains the solutions to the questions from Real Numbers. Here, students are introduced to a variety of crucial ideas that will be helpful to those who plan to continue studying mathematics in the eleventh grade. Students can study for their forthcoming board exam with these NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. The fact that these NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 follow the CBSE Syllabus for 2022–2023 makes them useful.

The Fundamental Theorem of Arithmetic , Revisiting Rational and Irrational Numbers, Decimal Expansions, and Euclid’s division algorithm are the primary subjects covered in the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. The NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 were successfully developed by the professionals at Extramarks based on these ideas. Students who prioritise these ideas will do well in both their class assessments and the Class 10 board exam.

The NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1, contain important questions. These NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 aid students in deftly and effectively resolving issues. They also put a lot of effort into creating mathematics problem solutions that are simple for the students to understand.

Using Euclid’s Division Algorithm, the first exercise in Real Numbers- Exercise 1.1 demonstrates how to divide integers. Each response to a question in the exercises is given a thorough process and a step-by-step explanation. The NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 are always created in accordance with CBSE recommendations in order to properly cover the entire curriculum. These are highly beneficial for getting good test results.

NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1: The first exercise in Chapter 1 of Class 10 Mathematics is Real Numbers. Real Numbers are introduced in Class 9 and are covered in greater detail by studying Euclid’s division algorithm in Class 10. The divisibility of numbers is covered in the exercise. According to Euclid’s division procedure, any positive integer ‘a’ can be divided by another positive integer ‘b’ in such a way that the result will be a smaller positive integer than ‘b’.

Here’s a list of reasons as to why students must adhere to the CBSE pertaining NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 by Extramarks:

- Learners can use these NCERT Solutions to complete and review all of the questions in Exercise 1.1.
- Students can earn additional points by reading the step-by-step solutions provided by the subject-matter experts at Extramarks.
- It aids in achieving high exam math grades.
- It adheres to CBSE standards, which aid in properly preparing the student.
- From the standpoint of the exam, it comprises all the crucial questions.

All students should take the Class 10 board examinations seriously because it is the first board exam they will ever be attempting for the first time. It’s crucial to perform well in each subject. For many students, mathematics seems to be a challenging subject. Consequently, getting good grades in this subject is complicated. At this point, all learners should rely on the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. For most students, mathematics is regarded as the best subject to study. According to the most recent CBSE curriculum and rules, these solutions have been produced by the greatest subject matter specialists in Extramarks. As a result, every response given in these solutions is entirely accurate.

The questions posed in Class 10 Maths Chapter 1 Exercise 1.1 are answered in these NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. These NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 must be used by students who prefer to study independently because they are readily available in PDF format on the Extramarks website and applications. Therefore, it can be downloaded at the convenience of the user. Along with the aforementioned features, these NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 aid in speedy review as well.

Four examples are given in Mathematics Class 10 Exercise 1.1. Both Euclid’s Division Lemma and Division Algorithm serve as its foundation. Students are introduced to real numbers and associated ideas in Chapter 1. These four sample questions are fully discussed prior to Exercise 1.1. Every example clarifies the idea and the kind of questions to anticipate in the subsequent exercise. The examples are helpful in making the idea simple to understand and in giving the students extra practice.

Students do best in mathematics when they have ample practice, are familiar with all the concepts, and are aware of every question. The idea is to take note of all the formulas and practice more questions. It’s crucial for them to practice previous year’s question papers as they get ready for the test so they can familiarise themselves with the questions and learn time management skills. In the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 PDF that is available on the Extramarks website and mobile application, they can practice questions from previous years. In-depth responses to questions are also provided to help students grasp the idea.

**NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Ex 1.1) Exercise 1.1 **

According to the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 the foundation of Real Numbers is based on the Euclid’s Division algorithm. Finding the qualities of numbers based on the algorithm is the subject of the questions in this exercise. It also covers the subject of utilising the division lemma to get the HCF of positive integers.

In the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 by Extramarks, word problems are used to answer the majority of the questions.

Euclid’s division lemma or algorithm has numerous uses because it aids in determining the characteristics of numbers. Euclid’s Division Algorithm can occasionally be used for all integers except zero, even though it is only described for positive integers. Five questions make up this practice, three of which are short answers and two of which are long answers.

Students can conveniently download the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 in PDF format from the Extramarks website and mobile application.

Below are some tips for students to ace the exams with the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 by Extramarks:

Through the NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1 of the questions in Real Numbers, the students will learn how to use Euclid’s division lemma. The only thing to keep in mind is that children should adhere strictly to the procedures when computing the HCF using this lemma. They risk getting the wrong outcomes if they veer off course. Additionally, students might use a quicker approach for determining the HCF, such as prime factorization, to check their answers.

The division lemma is a mathematical concept that is covered in the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. It states that “for any positive integers a and b, we can find unique integers q and r such that a= bq+r.” If you look closely, the division where an is the dividend, b is the divisor, q is the quotient, and r is the remainder can be seen as a repetition of this theorem. Students can solve all related issues using the thorough proof provided in the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1.

**Access NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers **

A student in Class 10 finds it difficult to balance so many topics and assignments as they advance through the grades. Because of this, the Extramarks teachers have created thorough NCERT Solutions for Class 10 Math Chapter 1 Exercise 1.1 of Real Numbers. The answers adhere to the revised CBSE curriculum. Extramarks’ subject matter specialists have conducted an in-depth study to create NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 that are simple for students to understand. Students can improve their understanding of real numbers at the fundamental level by studying these NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 by Extramarks, enabling them to tackle challenging issues independently.

Students can easily get the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 in PDF format from the Extramarks website or mobile application if they need assistance with all the questions in Chapter 1 of Class 10 Mathematics. Users can store these solutions on their device and access them when they are not connected to the internet. They can even print out the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 as an additional method for quickly reviewing key concepts and formulas. Students can get the NCERT Solutions for Class 10 Science on Extramarks if they’re looking for them.

One of the key areas of Mathematics is Real Numbers, and in the Class 10 (Unit – Number Systems) Mathematical Board Exams, Real Numbers are worth six points. There are typically 3 questions from this chapter on average. In the board exam from the previous year, this chapter was the subject of three questions (2021).

one of the three part A questions (1 marks).

one of the three part B questions (2 marks).

one of the three part C questions (3 marks).

This chapter discusses

The Euclid Algorithm for Division

Reviewing Rational and Irrational Numbers in The Fundamental Theorem of Arithmetic

Expansions in Decimals

List of Exercises for Mathematics Chapter 1 in Class 10:

Exercise 1.1 Solutions Question 5 ( 4 long, 1 short)

Exercise 1.2 Solutions Question 7 ( 4 long, 3 short)

Exercise 1.3 Solutions 3 Question ( 3 short)

Exercise 1.4 Solutions 3 Question ( 3 short)

Real numbers are covered in greater detail in Class 10 after being introduced in Class 9. NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 have answers to all of the complex problems present in it. Real numbers and their uses are covered in this chapter. According to Euclid’s algorithm for division, any positive integer a can be divided by another positive integer b such that the result will be a smaller positive integer than b. The Fundamental Theorem of Arithmetic, however, only applies to the multiplication of positive integers.

In section 1.1 of the chapter, real numbers are introduced. Sections 1.2 and 1.3 then cover two crucial subjects.

Five questions are based on Theorem 1.1, also known as Euclid’s Division Lemma, in the Euclid’s Division Algorithm section.

The Arithmetic Fundamental Theorem: Through the solutions to the seven problems in Exercise 1.2, explore the applications of this topic, which discusses the multiplication of positive integers.

The following subjects are thereafter covered; they were first covered in Class 9.

Reviewing Rational and Irrational Numbers: In this section, the answers to three problems from Exercise 1.3 that also employ the subject of the previous Exercise 1 are provided.

- Decimal Expansions: It investigates when a rational number’s decimal expansion is ending and when it is repeating. In Exercise 1.4, there are a total of 3 problems with sub-parts.

**Important Topics under NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Ex 1.1) Exercise 1.1 **

The very basic yet crucial subject of mathematics is introduced in the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. Real numbers are introduced in the first section of the chapter on real numbers for class 10, followed by two extremely significant topics: Euclid’s Division Algorithm and The Fundamental Theorem of Arithmetic. Numerous practical and scientific applications can be made of this basic arithmetic theorem. These are also used in other relevant fields. Therefore, Extramarks’ NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 on Real Numbers will undoubtedly be helpful for building a solid mathematical foundation that will support subsequent higher education as well.

This chapter aids students in comprehending the basic theorem of mathematics, which has several practical and scientific applications. The Real numbers that are covered in this chapter are also used in other related subjects. Therefore, Extramarks’ NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 will undoubtedly be helpful for building a solid foundation in mathematics that will support subsequent higher education as well. Major categories of real numbers include rational and irrational numbers. With the use of suitable examples and applications, theorems will be clarified.

Here’s a detailed study of the Sub-topics:

Introduction:

The learner will understand the domain of real numbers and their associated applications in this chapter. Some highly significant positive number features are presented in this chapter.

Euclid’s Division Lemma:

In this chapter, the learner will learn how to use Euclid’s approach to find the two positive integers’ highest common factor (HCF).

The Fundamental Theorem of Arithmetic:

The student will discover that each composite number has a distinct way to be stated as a product of primes. This property is a fundamental theorem of mathematics.

Revisiting Irrational Numbers:

The irrational numbers are redefined in this chapter. A few pertinent examples will make the subject simpler to comprehend. They can be demonstrated by the use of contradiction.

Revisiting Rational Numbers and Their Decimal Expansions:

The idea of rational numbers will be covered again by the learner utilising both fraction expression and decimal expansions. This is due to the fact that every rational number’s decimal expansion either terminates or repeatedly does not terminate.

**Importance of Real Numbers**

Real numbers have always been crucial in mathematics, and knowing how they are classified helps us understand how they are used. There are an endless number of real numbers, which are all the numbers on the number line. They are significant because they can provide pupils with more details about the issue they are examining. Additionally, there are imaginary numbers; this subject will be covered later in this chapter. In fact, some numbers can point students in the direction of more mathematical definitions or formulas.

When kids first start learning about numbers or are taught numbers, they start counting at 1. As an illustration, 1, 2, 3, 4,… These are classified as natural numbers or counting numbers.

Children quickly pick up the concept of “none” when counting something for which there are no items to count after learning how to count. The quantity of no objects is designated by the number zero. To construct the set of whole numbers, this number is added to the counting numbers: 0, 1, 2, 3, and so on.

When they try to subtract a greater number from a smaller number, they find that some of the numbers are not positive. It is possible to represent a debt as a negative number. The set of integers consists of the whole numbers and their antipodes (negatives): -3, -2, -1, 0, 1, 2, 3, and so on.

Children then look at numbers that are components of a whole. These figures, which might be fractional or decimal, indicate parts of the total or a mix of wholes and parts of the whole. Because each of these can be expressed as a ratio, they are referred to as “rational numbers.” All integers are considered to be rational numbers.

The last group of numbers is known as “irrational numbers” because they are so unusual that they cannot be expressed as ratios. They are taken into consideration because they are crucial to mathematics.

**Exercise 1.1**

Real numbers are introduced in the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. Real numbers and complex numbers (real numbers plus imaginary numbers) are the two categories of numbers. In the number system, real numbers are essentially a blend of both rational and irrational numbers. These real numbers have been subjected to a variety of arithmetic operations in NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1. The real numbers are frequently defined in Exercise 1.1 of Class 10 Mathematics as the result of the union of both rational and irrational numbers. They are represented by the letter “R” and can be positive or negative. This category includes all natural integers, decimals, and fractions. This NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 by Extramarks provides clear explanations of the real number fundamentals.

Exercise 1.1 from Class 10 Math Chapter 1: The set of real numbers in this exercise includes both whole and natural numbers, integers, rational and irrational numbers, and integers and rational numbers. Along with the identity and distribution properties, real numbers also adhere to the commutative and associative qualities. Euclid’s division Lemma and technique are mostly used in the NCERT solutions for Class 10 Math Exercise 1.1. Exercise 1.1 in Class 10 Math presents five significant questions pertaining to Euclid’s division lemma and algorithm.

**NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1**

Benefits of the NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1:

- All significant exam-related questions are contained in NCERT Class 10 Maths Chapter 1 Exercise 1.1, and all of the questions have been changed from that chapter’s first exercise.
- The introduction to real numbers and Euclid’s Division Lemma and Division Algorithm, which are key ideas in the chapter, are the foundations of NCERT Book Exercise 1.1 for Class 10 Math.
- The greatest resource for resolving Class 10 Math chapter 1 exercise 1.1 is generally regarded as the NCERT syllabus solutions.

There are 15 chapters in all in the NCERT Mathematics textbook for class 10. Algebra, geometry, trigonometry, statistics, and probability are all covered in Class 10 Mathematics 45.

From Exercise 1.1, students can anticipate questions worth 2 or 3 marks. There might occasionally be no questions. However, there is still a very high likelihood that you will have questions after this activity.

**NCERT Solutions for Class 10 Maths Chapter 1 Exercises**

- Chronological order wise

- Exercise 1.1
- Exercise 1.2
- Exercise 1.3
- Exercise 1.4
- Examples
- Case Based Questions (MCQ)
- MCQs from NCERT Examples
- Previous Year MCQ (Mathematics Standard)

- Concept wise

- Introduction
- HCF using Euclid’s Division Algorithm
- Euclid’s Division Algorithm – Proving
- Prime Factorization
- LCM/HCF
- Decimal Expansion
- Irrational numbers

**Q.1** Use Euclid’s division algorithm to find the HCF of:

- 135 and 225
- 196 and 38220
- 867 and 255

**Ans.**

$\begin{array}{l}\left(\text{i}\right)\text{135 and 225}\\ \text{Step 1:}\\ \text{Since 225}>\text{135},\text{we apply the division lemma}\\ \text{to 225 and 135 to get}\\ \text{225}=\text{135}\times \text{1}+\text{9}0\\ \\ \text{Step 2:}\\ \text{Since remainder 9}0\ne 0,\text{we apply the division lemma}\\ \text{to 135 and 9}0\text{to get}\\ \text{135}=\text{9}0\times \text{1}+\text{45}\\ \\ \text{Step 3:}\\ \text{We consider the new divisor 9}0\text{and new remainder 45},\\ \text{and apply the division lemma to get}\\ \text{9}0=\text{2}\times \text{45}+0\\ \text{The remainder has now become zero},\text{so the procedure stops}.\\ \text{Since the divisor at this stage is 45},\\ \text{the HCF of 135 and 225 is 45}.\\ \left(\text{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}196 and 3822}0\\ \text{Step 1:}\\ \text{Since 3822}0>\text{196},\text{we apply the division lemma}\\ \text{to 3822}0\text{and 196 to get}\\ \text{3822}0=\text{196}\times \text{195}+0\\ \text{Since the remainder is zero},\text{the procedure stops}.\\ \text{The divisor at this stage is 196.}\\ \text{Therefore},\text{HCF of 196 and 3822}0\text{is 196}.\\ \left(\text{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}867 and 255}\\ \text{Step 1:}\\ \text{Since 867}>\text{255},\text{we apply the division lemma}\\ \text{to 867 and 255 to get}\\ \text{867}=\text{255}\times \text{3}+\text{1}0\text{2}\\ \\ \text{Step 2:}\\ \text{Since remainder 1}0\text{2}\ne 0,\\ \text{we apply the division lemma to 255 and 1}0\text{2 to get}\\ \text{255}=\text{1}0\text{2}\times \text{2}+\text{51}\\ \\ \text{Step 3:}\\ \text{We consider the new divisor 1}0\text{2 and new remainder 51},\\ \text{and apply the division lemma to get}\\ \text{1}0\text{2}=\text{51}\times \text{2}+0\\ \text{Since the remainder is zero},\text{the procedure stops}.\\ \text{The divisor at this stage is 51.}\\ \text{Therefore},\text{HCF of 867 and 255 is 51}.\end{array}$

**Q.2** Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

**Ans.**

$\begin{array}{l}\text{Let a is a positive integer and}\mathrm{b}=\text{6.}\\ \text{By Euclid}\u2019\text{s algorithm},\text{we get}\\ \mathrm{a}=\text{6}\mathrm{q}+\mathrm{r}\text{\hspace{0.17em}\hspace{0.17em}for some integer}\mathrm{q}\ge 0,\\ \text{and}0\le \mathrm{r}<\text{6}.\\ \text{Therefore},\\ \mathrm{a}=\text{6}\mathrm{q}\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em} 6}\mathrm{q}+\text{1 \hspace{0.17em}or\hspace{0.17em} 6}\mathrm{q}+\text{2 or \hspace{0.17em}6}\mathrm{q}+\text{3 or \hspace{0.17em}6}\mathrm{q}+\text{4 \hspace{0.17em}or \hspace{0.17em}6}\mathrm{q}+\text{5}\\ \\ \text{Also},\text{6}\mathrm{q}+\text{1}=\text{2}\times \text{3}\mathrm{q}+\text{1}=\text{2}{\mathrm{k}}_{\text{1}}+\text{1},\text{}\\ \text{where}{\mathrm{k}}_{\text{1}}=3\mathrm{q}\text{.}\\ \text{6}\mathrm{q}+\text{3}=(\text{6}\mathrm{q}+\text{2})+\text{1}=\text{2}(\text{3}\mathrm{q}+\text{1})+\text{1}=\text{2}{\mathrm{k}}_{\text{2}}+\text{1},\\ \text{where}{\mathrm{k}}_{\text{2}}=3\mathrm{q}+1.\\ \text{6}\mathrm{q}+\text{5}=(\text{6}\mathrm{q}+\text{4})+\text{1}=\text{2}(\text{3}\mathrm{q}+\text{2})+\text{1}=\text{2}{\mathrm{k}}_{\text{3}}+\text{1},\text{}\\ \text{where}{\mathrm{k}}_{\text{3}}=3\mathrm{q}+2.\\ \text{Clearly},\text{6}\mathrm{q}+\text{1},\text{6}\mathrm{q}+\text{3},\text{6}\mathrm{q}+\text{5 are of the form 2}\mathrm{k}+\text{1},\\ \text{where}\mathrm{k}\text{\hspace{0.17em}\hspace{0.17em}is an integer}.\\ \\ \text{Therefore},\text{6}\mathrm{q}+\text{1},\text{6}\mathrm{q}+\text{3},\text{6}\mathrm{q}+\text{5 are not divisible by 2}\\ \text{and so these represent odd numbers.}\\ \\ \text{Thus},\text{any odd integer can be expressed in the form}\\ \text{6}\mathrm{q}+\text{1},\text{or \hspace{0.17em}6}\mathrm{q}+\text{3},\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em} 6}\mathrm{q}+\text{5.}\end{array}$

**Q.3** An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

**Ans.**

$\begin{array}{l}\text{HCF of 616 and 32 will give the maximum number of columns in}\\ \text{which each group can march}.\\ \text{By Euclid}\u2019\text{s algorithm, we get}\\ \text{616}=\text{32}\times \text{19}+\text{8}\\ \text{and}\\ \text{32}=\text{8}\times \text{4}+0\\ \text{The HCF}(\text{616},\text{32})\text{is 8}.\\ \text{Therefore},\text{each group can march in 8 columns}.\end{array}$

**Q.4** Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q+ 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

**Ans.**

$\begin{array}{l}\text{Let}\mathrm{a}\text{be a positive integer and}\mathrm{b}=\text{3}.\\ \text{By Euclid\u2019s Algorithm,}\\ \mathrm{a}=\text{3}\mathrm{q}+\mathrm{r}\text{for some integer}\mathrm{q}\ge 0\text{and}0\le \mathrm{r}<\text{3.}\\ \text{The possible remainders are 0, 1 and 2. Therefore},\\ \mathrm{a}\text{can be 3}\mathrm{q}\text{or 3}\mathrm{q}+\text{1 or 3}\mathrm{q}+\text{2.}\\ \text{Thus,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}=9{\mathrm{q}}^{2}{\text{or (3q+1)}}^{2}\text{or (}3\mathrm{q}+2{)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9{\mathrm{q}}^{2}\text{or}(9{\mathrm{q}}^{2}+6\mathrm{q}+1)\text{or (}9{\mathrm{q}}^{2}+12\mathrm{q}+4)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\times \left(3{\mathrm{q}}^{2}\right)\text{or 3}(3{\mathrm{q}}^{2}+2\mathrm{q})+1\text{or 3(}3{\mathrm{q}}^{2}+4\mathrm{q}+1)+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3{\mathrm{k}}_{1}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or 3}{\mathrm{k}}_{2}+1\text{or 3}{\mathrm{k}}_{3}+1\\ \text{where}{\mathrm{k}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{k}}_{2}\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{\mathrm{k}}_{3}\text{are some positive integers.}\\ \text{Hence, square of any positive integer is either of the form}\\ \text{3m or 3m + 1 for some integer m.}\end{array}$

**Q.5** Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+ 8.

**Ans.**

$\begin{array}{l}\text{Let a be a positive integer and}\mathrm{b}=3\\ \mathrm{a}=3\mathrm{q}+\mathrm{r}\text{, where q}\ge \text{0 and}0\le \mathrm{r}<3\\ \text{Therefore,}\\ \mathrm{a}=3\mathrm{q}\text{or}3\mathrm{q}+1\text{or}3\mathrm{q}+2\text{}\\ \text{Therefore, an integer can be represented by these three forms.}\\ \mathbf{Case}\text{}\mathbf{1}:\text{When}\mathrm{a}\text{}=\text{}3\mathrm{q},\text{\hspace{0.17em}then}\\ {\mathrm{a}}^{3}={\left(3\mathrm{q}\right)}^{3}=27{\mathrm{q}}^{3}=9\left(3{\mathrm{q}}^{3}\right)=9\mathrm{m}\\ \text{where m is an integer and}\mathrm{m}=3{\mathrm{q}}^{3}.\\ \\ \mathbf{Case}\text{}\mathbf{2}:\text{When}\mathrm{a}=\text{3}\mathrm{q}+\text{1},\text{then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{3}={(\text{3}\mathrm{q}+\text{1})}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow {\mathrm{a}}^{3}=\text{27}{\mathrm{q}}^{3}+\text{27}{\mathrm{q}}^{2}+\text{9}\mathrm{q}+\text{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow {\mathrm{a}}^{3}=\text{9}(\text{3}{\mathrm{q}}^{3}+\text{3}{\mathrm{q}}^{2}+\mathrm{q})+\text{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow {\mathrm{a}}^{3}=\text{9}\mathrm{m}+\text{1}\\ \text{Where}\mathrm{m}\text{is an integer such that}\mathrm{m}=\text{3}{\mathrm{q}}^{3}+\text{3}{\mathrm{q}}^{2}+\text{}\mathrm{q}\\ \\ \mathbf{Case}\text{}\mathbf{3}:\text{When}\mathrm{a}=\text{3}\mathrm{q}+\text{2},\text{then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{3}=\text{}{(\text{3}\mathrm{q}+\text{2})}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow {\mathrm{a}}^{3}=\text{27}{\mathrm{q}}^{3}+\text{54}{\mathrm{q}}^{2}+\text{36}\mathrm{q}+\text{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow {\mathrm{a}}^{3}=\text{9}(\text{3}{\mathrm{q}}^{3}+\text{6}{\mathrm{q}}^{2}+\text{4}\mathrm{q})+\text{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow {\mathrm{a}}^{3}=\text{9}\mathrm{m}+\text{8}\\ \text{Where}\mathrm{m}\text{is an integer such that}\mathrm{m}=\text{3}{\mathrm{q}}^{3}+\text{6}{\mathrm{q}}^{2}+\text{4}\mathrm{q}\\ \text{Therefore},\text{the cube of any positive integer is of the form}\\ \text{9}\mathrm{m},\text{9}\mathrm{m}+\text{1},\text{or 9}\mathrm{m}+\text{8}.\end{array}$

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