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# NCERT Solutions for Class 9 Maths Exercise 1.3 Chapter 1 Number System

Last updated date: 09th Aug 2024
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## NCERT Solutions for Maths Chapter 1 Exercise 1.3 Class 9

Study easily from NCERT Solutions Class 9 Maths Chapter 1 Number Systems Exercise 1.3, which has been created by our subject matter experts. When solving the practice problems, the students can refer to the NCERT Ex 1.3 Class 9 Maths Solutions PDF. Real numbers and their decimal expansion are covered in Number Systems' third exercise. The answers provide an extensive and systematic breakdown of every response to the questions in the Class 9 NCERT textbook exercises. As a result, using these strategies can greatly improve exam scores.

Table of Content
1. NCERT Solutions for Maths Chapter 1 Exercise 1.3 Class 9
2. Glance of NCERT Solutions for Class 9 Maths Chapter 1  Number Systems - Exercise 1.3 | Vedantu
3. Topics Covered in Class 9 Maths Chapter 1 Exercise 1.3
4. Access NCERT Solutions for Ex 1.3 Class 9 Maths Chapter 1 – Number System
4.11. Write the following in decimal form and say what kind of decimal expansion each has:
5. NCERT Solutions for Class 9 Maths Chapter 1 Other Exercises
6. Chapter-Specific NCERT Solutions for Class 9 Maths
FAQs

## Glance of NCERT Solutions for Class 9 Maths Chapter 1  Number Systems - Exercise 1.3 | Vedantu

• The chapter focuses on the types of numbers in the number system, including natural numbers, integers, rational numbers, and irrational numbers, emphasizing their properties and differences.

• It introduces real numbers as a combination of rational and irrational numbers, exploring their behaviors and characteristics.

• For rational numbers, the decimal expansion is either terminating or non-terminating and recurring.

• For an irrational number, the decimal expansion is non-terminating and non-recurring.

• A significant part of the chapter deals with decimal expansions of rational and irrational numbers, teaching how to distinguish between terminating and non-terminating decimals.

• This chapter of Chapter 1 Maths Ex 1.3 Class 9 helps students to understand the Number Systems and its types .

• There are links to video tutorials explaining chapter 1 Ex 1.3 Class 9 Number Systems for better understanding.

• There are five exercises (16 fully solved questions) in Class 9 Math chapter 1 Exercise 1.3 Number Systems.

## Topics Covered in Class 9 Maths Chapter 1 Exercise 1.3

• Decimal Expansions: This exercise explores how real numbers can be expressed in decimal form. You'll encounter two main types:

• Terminating decimals.

• Non-terminating decimals.

• Classifying Decimals: The exercise equips you to differentiate between rational and irrational numbers based on their decimal expansions.

• Converting Terminating Decimals: This exercise may involve converting terminating decimals into equivalent fractions with a denominator of 10.

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## Access NCERT Solutions for Ex 1.3 Class 9 Maths Chapter 1 – Number System

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

(i) $\dfrac{ {36}}{ {100}}$

Ans: Divide $36$ by $100$.

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

So, $\dfrac{36}{100}=0.36$ and it is a terminating decimal number.

(ii) $\dfrac{ {1}}{ {11}}$

Ans: Divide $1$ by $11$.

$\begin{matrix} &{0.0909..}\\ 11&{\overline{)\;1\quad}}\\ &\underline{-0\quad}\\ &10\\ &\underline{\;\;-0\quad}\\ &\;\;100\\ &\underline{\;-99}\\ &\quad10\\ &\quad\underline{\;\;-0\quad}\\ &\;\;100\\ &\underline{\;-99}\\ &\quad1\quad\\ \end{matrix}$

It is noticed that while dividing $1$ by $11$, in the quotient $09$ is repeated.

So, $\dfrac{1}{11}=0.0909.....$ or

$\dfrac{1}{11}=0.\overline{09}$

and it is a non-terminating and recurring decimal number.

(iii)  ${4}\dfrac{ {1}}{ {8}}$

Ans: $4\dfrac{1}{8}=4+\dfrac{1}{8}=\dfrac{32+1}{8}=\dfrac{33}{8}$

Divide $33$ by $8$.

$\begin{matrix} &{4.125}\\ 8&{\overline{)\;33\quad}}\\ &\underline{-32\quad}\\ &10\\ &\underline{\;\;-8\quad}\\ &\;\;20\\ &\underline{-16}\\ &\;\quad\;40\\ &\quad\underline{\quad-40\quad}\\ &\quad\underline{\quad 0\quad}\\ \end{matrix}$

Notice that, after dividing $33$ by $8$, the remainder is found as $0$.

So, $4\dfrac{1}{8}=4.125$ and it is a terminating decimal number.

(iv)  $\dfrac{ {3}}{ {13}}$

Ans: Divide $3$ by $13$.

$\begin{matrix} &{0.230769}\\ 13&{\overline{)\;3\quad}}\\ &\underline{-0\quad}\\ &30\\ &\underline{\;-26\quad}\\ &\;40\\ &\underline{-39\quad}\\ &\;\quad\;10\\ &\quad\underline{\quad-0\quad}\\ &\quad{\quad100}\\ &\quad\quad\underline{-91\quad}\\ &\quad\quad90\\ &\quad\quad\underline{-78\quad}\\ &\quad\quad 120\\ &\quad \quad\underline{-117\quad}\\ &\quad\quad\underline{\quad 3\quad} \end{matrix}$

It is observed that while dividing $3$ by $13$, the remainder is found as $3$ and that is repeated after each $6$ continuous divisions.

So, $\dfrac{3}{13}=0.230769.......$ or

$\dfrac{3}{13}=0.\overline{230769}$

and it is a non-terminating and recurring decimal number.

(v)   $\dfrac{ {2}}{ {11}}$

Ans: Divide $2$ by $11$.

$\begin{matrix} &{0.1818}\\ 11&{\overline{)\;2\quad}}\\ &\underline{-0\quad}\\ &20\\ &\underline{\;-11\quad}\\ &\;90\\ &\underline{-88\;}\\ &\;\quad\;20\\ &\quad\underline{\quad-11\quad}\\ &\quad{\quad90}\\ &\quad\underline{-88}\\ &\quad2\\ \end{matrix}$

It can be noticed that while dividing $2$ by $11$, the remainder is obtained as $2$ and then $9$, and these two numbers are repeated infinitely as remainders.

So, $\dfrac{2}{11}=0.1818.....$ or

$\dfrac{2}{11}=0.\overline{18}$

and it is a non-terminating and recurring decimal number.

(vi) $\dfrac{ {329}}{ {400}}$

Ans: Divide $329$ by $400$.

$\begin{matrix} &{0.8225}\\ 400&{\overline{)\;329\quad}}\\ &\underline{-0\quad}\\ &3290\\ &\underline{\;-3200\quad}\\ &\;900\\ &\underline{-800\;}\\ &\quad\;1000\\ &\quad\underline{\quad-800\quad}\\ &\quad{\quad2000}\\ &\quad\underline{\quad-2000\quad}\\ &\quad\underline{\quad 0 \quad}\\ \end{matrix}$

It can be seen that while dividing $329$ by $400$, the remainder is obtained as $0$.

So, $\dfrac{329}{400}=0.8225$ and is a terminating decimal number.

2. You know that $\dfrac{ {1}}{ {7}} {=0} {.142857}...$. Can you predict what the decimal expansions of $\dfrac{ {2}}{ {7}} {,}\dfrac{ {3}}{ {7}} {,}\dfrac{ {4}}{ {7}} {,}\dfrac{ {5}}{ {7}} {,}\dfrac{ {6}}{ {7}}$  are, without actually doing the long division? If so, how?

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

Ans: Note that,  $\dfrac{2}{7},\dfrac{3}{7},\dfrac{4}{7},\dfrac{5}{7}$ and $\dfrac{6}{7}$ can be rewritten as $2\times \dfrac{1}{7},\text{ 3}\times \dfrac{1}{7},\text{ 4}\times \dfrac{1}{7},\text{ 5}\times \dfrac{1}{7},$ and $6\times \dfrac{1}{7}$

Substituting the value of $\dfrac{1}{7}=0.142857$ , gives

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

$3\times \dfrac{1}{7}=3\times 0.428571...$$\text{=}\,\text{0}\text{.428571}...$

$4\times \dfrac{1}{7}=4\times 0.142857...$$\text{=}\,\text{0}\text{.571428}...$

$5\times \dfrac{1}{7}=5\times 0.71425...$  $\text{=}\,\text{0}\text{.714285}...$

$6\times \dfrac{1}{7}=6\times 0.142857...$$\text{=}\,\text{0}\text{.857142}...$

So, the values of $\dfrac{2}{7},\text{ }\dfrac{3}{7},\text{ }\dfrac{4}{7},\text{ }\dfrac{5}{7}$ and $\dfrac{6}{7}$ obtained without performing long division are

$\dfrac{2}{7}=0.\overline{285714}$

$\dfrac{3}{7}=0.\overline{428571}$

$\dfrac{4}{7}=0.\overline{571428}$

$\dfrac{5}{7}=0.\overline{714285}$

$\dfrac{6}{7}=0.\overline{857142}$

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

(i) ${0} {.}\overline{ {6}}$

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

$\Rightarrow x=0.6666$                                                   ….… (1)

Multiplying both sides of the equation (1) by $10$, gives

$10x=0.6666\times 10$

$10x=6.6666$…..                 …… (2)

Subtracting the equation $\left( 1 \right)$ from $\left( 2 \right)$, gives

\begin{align} & 10x=6.6666..... \\ & \underline{-x=0.6666.....} \\ & 9x=6 \\ & 9x=6 \\ & x=\dfrac{6}{9}=\dfrac{2}{3} \\ \end{align}

So, the decimal number becomes

$0.\overline{6}=\dfrac{2}{3}$  and it is in the required  $\dfrac{p}{q}$ form.

(ii) ${0} {.}\overline{ {47}}$

Ans: Let  $x=0.\overline{47}$

$\text{ }\Rightarrow x=0.47777.....$                                             ……(a)

Multiplying both sides of the equation (a) by $10$, gives

$10x=4.7777.....$         ……(b)

Subtracting the equation $\left( a \right)$ from $\left( b \right)$, gives

\begin{align} & 10x=4.7777..... \\ & \underline{-x=0.4777.....} \\ & 9x=4.3 \\ \end{align}

Therefore,

\begin{align} & \,\,\,\,\,\,x=\dfrac{4.3}{9}\times \dfrac{10}{10} \\ & \Rightarrow x=\dfrac{43}{90} \\ \end{align}

So, the decimal number becomes

$0.\overline{47}=\dfrac{43}{90}$  and it is in the required $\dfrac{p}{q}$ form.

(iii) ${0} {.}\overline{ {001}}$

Ans: Let $x=0.\overline{001}\Rightarrow$           …… (1)

Since the number of recurring decimal number is $3$, so multiplying both sides of the equation (1) by $1000$, gives

$1000\times x=1000\times 0.001001.....$ …… (2)

Subtracting the equation (1) from (2) gives

\begin{align} & 1000x=1.001001..... \\ & \underline{\text{ }-x=0.001001.....} \\ & 999x=1 \\ \end{align}

$\Rightarrow x=\dfrac{1}{999}$

Hence, the decimal number becomes

$0.\overline{001}=\dfrac{1}{999}$ and it is in the $\dfrac{p}{q}$ form.

4. Express ${0} {.99999}.....$ in the form of $\dfrac{ {p}}{ {q}}$ . Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense.

Ans:

Let $x=0.99999.....$                                                             ....... (a)

Multiplying by $10$ both sides of the equation (a), gives

$10x=9.9999.....$                                                             …… (b)

Now, subtracting the equation (a) from (b), gives

\begin{align} & 10x=9.99999..... \\ & \underline{\,-x=0.99999.....} \\ & \,\,9x=9 \\ \end{align}

$\Rightarrow x=\dfrac{9}{9}$

$\Rightarrow x=1$.

So, the decimal number becomes

$0.99999...=\dfrac{1}{1}$ which is in the $\dfrac{p}{q}$ form.

Yes, for a moment we are amazed by our answer, but when we observe that $0.9999.........$ is extending infinitely, then the answer makes sense.

Therefore, there is no difference between $1$ and $0.9999.........$ and hence these two numbers are equal.

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

Ans: Here the number of digits in the recurring block of $\dfrac{1}{17}$ is to be determined. So, let us calculate the long division to obtain the recurring block of $\dfrac{1}{17}$. Dividing $1$ by $17$ gives

Thus, it is noticed that while dividing $1$ by $17$, we found $16$ number of digits in the

repeating block of decimal expansion that will continue to be $1$ after going through $16$ continuous divisions.

Hence, it is concluded that $\dfrac{1}{17}=0.0588235294117647.....$ or

$\dfrac{1}{17}=0.\overline{0588235294117647}$ and it is a recurring and non-terminating decimal number.

6. Look at several examples of rational numbers in the form $\dfrac{ {p}}{ {q}}\left( {q}\ne {0} \right)$, 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 such rational numbers $\dfrac{5}{2},\dfrac{5}{4},\dfrac{2}{5},\dfrac{2}{10},\dfrac{5}{16}$ of the form $\dfrac{p}{q}$ which have terminating decimal representations.

\begin{align} & \dfrac{5}{2}=2.5 \\ & \dfrac{5}{4}=1.25 \\ & \dfrac{2}{5}=0.4 \\ & \dfrac{2}{10}=0.2 \\ & \dfrac{5}{16}=0.3125 \\ \end{align}

In each of the above examples, it can be noticed that the denominators of the rational numbers have powers of $2,5$ or both.

So, $q$ must satisfy the form either ${{2}^{m}}$, or ${{5}^{n}}$, or  both ${{2}^{m}}\times {{5}^{n}}$ (where $m=0,1,2,3.....$ and $n=0,1,2,3.....$) in the form of $\dfrac{p}{q}$.

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

Ans: All the irrational numbers are non-terminating and non-recurring, because irrational numbers do not have any representations of the form of $\dfrac{p}{q}$ $\left( q\ne 0 \right)$, where $p$ and $q$are integers. For example:

$\sqrt{2}=1.41421.....$,

$\sqrt{3}=1.73205...$

$\sqrt{7}=2.645751....$

are the numbers whose decimal representations are non-terminating and non-recurring.

8. Find any three irrational numbers between the rational numbers $\dfrac{ {5}}{ {7}}$ and $\dfrac{ {9}}{ {11}}$.

Ans: Converting  $\dfrac{5}{7}$and $\dfrac{9}{11}$ into the decimal form gives

$\dfrac{5}{7}=0.714285.....$ and

$\dfrac{9}{11}=0.818181.....$

Therefore, $3$ irrational numbers that are contained between $0.714285......$ and $0.818181.....$

are:

\begin{align} & 0.73073007300073...... \\ & 0.74074007400074...... \\ & 0.76076007600076...... \\ \end{align}

Hence, three irrational numbers between the rational numbers $\dfrac{5}{7}$ and $\dfrac{9}{11}$ are

\begin{align} & 0.73073007300073...... \\ & 0.74074007400074...... \\ & 0.76076007600076...... \\ \end{align}

9. Classify the following numbers as rational or irrational:

(i) $\sqrt{ {23}}$

Ans: The following diagram reminds us of the distinctions among the types of rational and irrational numbers.

After evaluating the square root gives

$\sqrt{23}=4.795831.....$ , which is an irrational number.

(ii) $\sqrt{ {225}}$

Ans: After evaluating the square root gives

$\sqrt{225}=15$, which is a rational number.

That is, $\sqrt{225}$ is a rational number.

(iii) ${0} {.3796}$

Ans: The given number is $0.3796$. It is terminating decimal.

So, $0.3796$ is a rational number.

(iv) ${7} {.478478}$

Ans: The given number is $7.478478\ldots .$

It is a non-terminating and recurring decimal that can be written in the $\dfrac{p}{q}$ form.

Let      $x=7.478478\ldots .$                                   ……(a)

Multiplying the equation (a) both sides by $100$ gives

$\Rightarrow 1000x=7478.478478.....$                                               ……(b)

Subtracting the equation (a) from (b), gives

\begin{align} & 1000x=7478.478478.... \\ & \underline{\text{ }-x=\text{ }7.478478\ldots .} \\ & 999x=7471 \\ & 999x=7471 \\ & \text{ }x=\dfrac{7471}{999} \\ \end{align}

Therefore, $7.478478.....=\dfrac{7471}{999}$, which is in the form of $\dfrac{p}{q}$

So, $7.478478...$ is a rational number.

(v) ${1} {.101001000100001}.....$

Ans: The given number is $1.101001000100001....$

It can be clearly seen that the number $1.101001000100001....$ is a non-terminating and non-recurring decimal and it is known that non-terminating non-recurring decimals cannot be written in the form of $\dfrac{p}{q}$.

Hence, the number $1.101001000100001....$ is an irrational number.

## Conclusion

The number system-related NCERT Solutions for Class 9, Chapter 1, Exercise 1.3, give an excellent foundation in fundamental mathematical topics, such as real and irrational numbers. Understanding and identifying between various numbers is the main objective of this particular activity, which is essential for comprehending higher-level math in later grades. As this is a foundational exercise, students should concentrate on developing their capacity to differentiate between and work with rational and irrational numbers because they are skills that are often expanded upon in other math areas. Gaining a thorough understanding of these ideas can also help you solve increasingly challenging mathematical and geometrical issues.

## NCERT Solutions for Class 9 Maths Chapter 1 Other Exercises

 Chapter 1 - Number System Other Exercises in PDF Format Exercise 1.1 4 Questions & Solutions Exercise 1.2 4 Questions & Solutions Exercise 1.4 5 Questions & Solutions Exercise 1.5 3 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 9 Maths Exercise 1.3 Chapter 1 Number System

1. Give an overview of the chapter Number system.

In this Chapter, you will get to understand about the different types of numbers along with their varied characteristics. The chapter primarily deals with irrational numbers, real numbers and their decimal expansion. It is followed by representing real numbers on the numbers, operating on real numbers, and lastly laws of exponents for real numbers.

You will also be able to learn about rational numbers and irrational numbers with their properties. The chapter also introduces you to:

• Classification of expressions into rational or irrational numbers

• Simplification of expressions

• Number Line representation

• Rationalization

2. Give a brief on decimal and decimal classification.

Decimal is a very interesting and fun part of the Number System. Decimal fractions were first introduced and used by Chinese at the end of the 4th century and then spread to the Middle East and from there to Europe. Decimals are used in our daily life without consciousness. For instance – counting of money, filling fuel into our vehicle or while measuring our weight. Decimals can never be whole numbers. Few examples: 1.8, 000.23 etc.

Decimals classifications: Terminating decimal fractions are 17/4= 4.25, 21/5 = 4.2 and so on.

Non - terminating decimal fractions are 16/3 = 5.33333 , 15.35353535 etc.

Integer: All the numbers that does not have decimals in them are known as integers. Thus, -9, 4, 1476 etc. Do you know? All integers are whole numbers including the negative numbers.

3. What does this chapter mainly focuses on?

The chapter mainly focuses on different types of numbers. They are:

• Complex Number

• Imaginary Number

• Real Number

• Rational Number

• Irrational Numbers

• Integers

• Whole Numbers

• Natural Numbers

• Natural Numbers - any of the given numbers that are used for counting purpose, starting from one, is considered a natural number.

• Whole Numbers - the total union set of all the Natural numbers which includes zero are the set of whole numbers.

• Integers - the set of all the whole numbers including their negative terms is called the set of integers.

• Rational Numbers - any number which can be written as a ratio of two natural numbers is known as a rational number.

• Irrational Numbers - any number which cannot be written in the form of a ratio of two natural numbers is known as an irrational number.

4. How will Vedantu’s NCERT solutions of class 9 maths chapter 1 number system will help us?

Solutions to the questions are prepared by our science experts as per the latest guidelines provided by the CBSE and NCERT. We have drafted the solutions in a step by step way to make your preparation simple and easy to understand. We have given all the relevant exams and diagrams to the questions with real-life examples.

These solutions are in a systematic way which helps you in building up a strong base of all the concepts. We make sure that all the topics and sub-topics are covered from every chapter and we also have designed these solutions in a way to make your learning process more fun, fascinating and enjoyable.

5. Are the Vedantu NCERT Solutions for Chapter 1 of Class 9 Maths helpful?

Vedantu is the best website for students to get solutions for their NCERT subjects. These solutions are really helpful for the students because they are very accurate according to the NCERT pattern. Also, they help students prepare for examinations in a better way. Solutions of Chapter 1 of Class 9 Maths by Vedantu offers comprehensive solutions, making learning much easier for students.

6. Where can I download solutions for Chapter 1 of Class 9 Maths?

7. How can I score good marks in Chapter 1 of Class 9 Maths?

You can score good marks in the Chapter 1 of Class 9 Maths in the following ways-

• Do not be in doubt regarding any concept. Concepts are very important when it comes to Maths.

• Learn the formulas of each chapter very thoroughly. Formulae are the key to solving all kinds of Math problems.

• Solve each and every question of the exercises at the end of each chapter. The more you practise the better you can get at Maths.

• Refer to the solutions for a better understanding of a chapter. The solution will contribute to a better understanding of each concept.

8. How many chapters are there in Class 9 Maths apart from Chapter 1?

The NCERT Class 9 Maths has a total of 13 chapters. All these chapters are a combination of the Basic Number Systems, Algebra, Geometry, And Mensuration. All of these 13 chapters are equally important for us to learn. For all of these chapters, you can get free solutions on Vedantu. You can download the chapter-wise solutions from Vedantu in PDF format free of cost.

9. What are the important chapters in Class 9 Maths apart from Chapter 1?

All the 13 chapters in Class 9 Maths are important. Do not leave out a single chapter when you are preparing for your examinations. All the chapters carry a certain amount of weightage in your exams. Also, leaving out even a single chapter will make you struggle to understand other chapters. So studying all the chapters is important. Other than that, you will get questions from all of the chapters as a part of your final exam.

10. Are there any tricky questions in Exercise 1.3 Class 9 Maths?

Students often face difficulties in visually representing irrational numbers due to their lack of a terminating decimal expansion. To tackle these problems, they need to understand successive approximations or known lengths to approximate the position of irrational numbers. Practice with examples and visualization can reinforce the concept and improve accuracy in plotting. Mastering this skill enhances spatial reasoning and precision in mathematical drawing, a crucial skill for higher-level math and sciences.

11. What should I focus on in Class 9 Maths Exercise 1.3 to perform well in exams?

To excel in exams on Class 9 Maths Ex 1.3, focus primarily on understanding irrational numbers, recognizing how they differ from rational numbers, and why they're unique. Practice plotting these numbers on a number line, which can involve estimating their position between known values. This visualization is crucial for grasping the abstract nature of these numbers. Also, practice solving different types of problems that use irrational numbers, as this will help solidify your understanding and improve your problem-solving skills. Download free pdf of Class 9 Maths Chapter 1 Ex 1.3

12. Can I skip class 9 maths 1.3 if I am short on time?

It's not advisable to skip class 9 maths ch 1 ex 1.3, even if you're short on time. This exercise introduces important concepts about irrational numbers and their placement on the number line, foundational knowledge that's vital for understanding more complex mathematical topics in future studies. If time is tight, try to at least grasp the key concepts and work through a couple of problems to familiarize yourself with the basics. Skipping it entirely could leave a gap in your understanding, making it harder to grasp related concepts later on.