NCERT Solutions Class 9 Maths Chapter 1 Number Systems

Number System is the first chapter of Class 9 Maths NCERT Book. It covers various topics, including an introduction to number systems, irrational numbers, real numbers and their decimal expansions, operations on real numbers, and the laws of exponents for real numbers and their applications.

The chapter begins with an introduction to Number Systems in section 1.1, followed by significant topics in sections 1.2, 1.3, 1.4, and 1.5.

1.0Download Class 9 Science Chapter 1 NCERT Solutions PDF Online

ALLEN'S Experts lucidly curated the solutions to improve the students' problem-solving abilities. For a more precise idea about Number Systems NCERT solutions, students can download the below NCERT Solutions for Class 9 Maths chapter 1 pdf solution.

2.0NCERT Solutions Class 9 Maths Chapter 1 : Important Topics

Irrational Numbers: Numbers that can't be written as p/q.

Real Numbers and their Decimal Expansions: This section includes the decimal expansions of real numbers and how to distinguish between rational and irrational numbers.

Number Line: Representing Real Numbers on the Number Line.

Operations on Real Numbers: Here, you explore operations like addition, subtraction, multiplication and division on irrational numbers.

Laws of Exponents for Real Numbers: Use these laws of exponents to solve the questions.

3.0Class 9 Maths Chapter 1 Number Systems: Exercise Solutions

There are five exercises in chapter 1 (Number systems) of class 9 Maths. Students can find the split below:

Explore Number Systems and learn how to solve various problems only on NCERT Solutions For Class 9 Maths.

4.0Advantages of Class 9 Maths Chapter 1 NCERT Solutions – Number Systems

1. Foundation for Advanced Math: It forms the basis for understanding more complex

mathematical concepts in higher classes, including algebra, calculus, and trigonometry.

2. Real-Life Applications: Helps in solving real-world problems like calculating areas and volumes and understanding financial transactions.

3. Understanding Different Types of Numbers: Students learn about rational, irrational, and real numbers, enabling them to classify and use numbers effectively in various mathematical problems.

4. Preparation for Competitive Exams: A strong grasp of number systems is crucial for excelling in competitive exams like NTSE, Olympiads, and even entrance exams like JEE and NEET.

5. Building a Strong Mathematical Base: It provides a solid foundation for other topics in mathematics, making it easier to understand future concepts and chapters.

By mastering this chapter, students not only excel academically but also develop skills that are useful in everyday life and future studies.

5.0NCERT Questions with Solutions for Class 9 Maths Chapter 1 - Detailed Solutions

Exercise: 1.1

1. Is zero a rational number? Can you write it in the form $\mathrm{p}/\mathrm{q}$, where p and q are integers and $q\ne 0$ ? Sol. Yes, zero is a rational number. We can write zero in the form $\mathrm{p}/\mathrm{q}$ where p and q are integers and $q\ne 0$. So, 0 can be written as $\frac{0}{1}=\frac{0}{2}=\frac{0}{3}$ etc.
2. Find six rational numbers between 3 and 4 . Sol. First rational number between 3 and 4 is $=\frac{3+4}{2}=\frac{7}{2}$ Similarly, other numbers are $\frac{3+\frac{7}{2}}{2}=\frac{13}{4}$ $\frac{3+\frac{13}{4}}{2}=\frac{25}{8}$ $\frac{3+\frac{25}{8}}{2}=\frac{49}{16}$ $\frac{3+\frac{49}{16}}{2}=\frac{97}{32}$ $\frac{\frac{97}{32}+3}{2}=\frac{193}{64}$ So, numbers are $\frac{7}{2},\frac{13}{4},\frac{25}{8},\frac{49}{16},\frac{97}{32},\frac{193}{64}$
3. Find five rational numbers between $3/5$ and $4/5$. Sol. Let $\frac{3}{5}\frac{(\mathrm{n}+1)}{\mathrm{n}+1}=\frac{3}{5}\times \frac{6}{6}=\frac{18}{30}$ $\frac{4}{5}\frac{(\mathrm{n}+1)}{\mathrm{n}+1}=\frac{4}{5}\times \frac{6}{6}=\frac{24}{30}$ So, required rational numbers are $\frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30}$
4. State whether the following statements are true or false? Give reasons for your answers. (i) Every natural number is a whole number. (ii) Every integer is a whole number. (iii) Every rational number is a whole number. Sol. (i) True, the collection of whole numbers contains all natural numbers. (ii) False, -2 is not a whole number. (iii) False, $\frac{1}{2}$ is an integer but a rational number but not a whole number.

Exercise : 1.2

1. State whether the following statements are true or false? Justify your answers. (i) Every irrational number is a real number. (ii) Every point on the number line is of the form $\sqrt{\mathrm{m}}$, where m is a natural number. (iii) Every real number is an irrational number. Sol. (i) True, since collection of real numbers consists of rationals and irrationals. (ii) False, because no negative number can be the square root of any natural number. (iii) False, 2 is real but not irrational.
2. Are the square roots of all positive integer's irrational? If not, give an example of the square root of a number that is a rational number. Sol. No, $\sqrt{4}=2$ is a rational number.
3. Show how $\sqrt{5}$ can be represented on the number line. Sol. $\sqrt{5}$ on Number line. OABC is unit square. So, $\mathrm{OB}=\sqrt{1^2+1^2}=\sqrt{2}$ $\mathrm{OD}=\sqrt{(\sqrt{2}{)}^2+1}=\sqrt{3}$ $\mathrm{OE}=\sqrt{(\sqrt{3}{)}^2+1}=2$ $\mathrm{OF}=\sqrt{(2{)}^2+1}=\sqrt{5}$

• Using compass we can cut arc with centre O and radius $=\mathrm{OF}$ on number line. ON is required result.

Exercise : 1.3

1. Write the following in decimal form and say what kind of decimal expansion each has : (i) $\frac{36}{100}$ (ii) $\frac{1}{11}$ (iii) $4\frac{1}{8}$ (iv) $\frac{3}{13}$ (v) $\frac{2}{11}$ (vi) $\frac{329}{400}$ Sol. (i) $\frac{36}{100}=0.36$ (Terminating) (ii) $\frac{1}{11}=0.090909$..... (Non-Terminating Repeating) 11 1 1.00000 0.090909 .... $\frac{-99}{100}$ $\frac{99}{100}$ $\frac{99}{1}$ (iii) $4\frac{1}{8}=\frac{33}{8}=4.125$ (Terminating decimal) (iv) $\frac{3}{13}=0.230769230769$...... $=0.\overline{230769}$ (Non-Terminating repeating) (v) $\frac{2}{11}=0.1818\dots \dots$. $=0.\overline{18}$ (Non-Terminating repeating) (vi) $\frac{329}{400}$

• $329.0000(0.8225$ $\frac{3200}{900}$ $\frac{800}{1000}$ $\frac{800}{2000}$ $\frac{2000}{x}$ $\frac{329}{400}=0.8225$ (Terminating)
• You know that $\frac{1}{7}=0.\overline{142857}$. Can you predict what the decimal expansions 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? Sol. Yes, we can predict decimal explain without actually doing long division method as $\frac{2}{7}=2\times \frac{1}{7}=2\times 0.\overline{142857}=0.\overline{285714}$ $\frac{3}{7}=3\times \frac{1}{7}=3\times 0.\overline{142857}=0.\overline{428571}$ $\frac{4}{7}=4\times \frac{1}{7}=4\times 0.\overline{142857}=0.\overline{571428}$ $\frac{5}{7}=5\times \frac{1}{7}=5\times 0.\overline{142857}=0.\overline{714285}$ $\frac{6}{7}=6\times \frac{1}{7}=6\times 0.\overline{142857}=0.\overline{857142}$
• Express the following in the form $p/q$, where p and q are integers and $\mathrm{q}\ne 0$. (i) $0.\overline{6}$ (ii) $0.4\overline{7}$ (iii) $0.\overline{001}$ Sol. (i) Let $x=0.6666$.... Multiplying both the sides by 10 $10\mathrm{x}=6.666$. Subtract (1) from (2) $10\mathrm{x}-\mathrm{x}=(\mathrm{6.6666....})-.(0.6666\dots ..$. $\Rightarrow 9\mathrm{x}=6\Rightarrow \mathrm{x}=\frac{6}{9}=\frac{2}{3}$ (ii) Let $\mathrm{x}=0.4\overline{7}=.4777\dots$ Multiply both sides by 10 $10\mathrm{x}=4.\overline{7}$ Multiply both sides by 10 $100\mathrm{x}=47.\overline{7}$ Subtract (1) from (2) $90\mathrm{x}=43$ $x=\frac{43}{90}$ (iii) Let $\mathrm{x}=0.\overline{001}=0.001001001$ Multiply both sides by 1000 $1000\mathrm{x}=1.\overline{001}$ Subtract (1) from (2) $999x=1$ $x=\frac{1}{999}$
• Express 0.99999..... in the form $\frac{p}{q}$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Sol. Let $\mathrm{x}=0.999$.... Multiply both sides by 10 we get $10\mathrm{x}=9.99\dots$... Subtract (1) from (2) $$\begin{gathered} 9x=9 \\ \Rightarrow x=1 \\ 0.9999\dots =1=\frac{1}{1} \\ \therefore p=1,q=1 \end{gathered}$$
• What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$ ? Perform the division to check your answer. Sol. Maximum number of digits in the repeating block of digits in decimal expansion of $\frac{1}{17}$ can be 16 .

• Look at several examples of rational numbers in the form $p/q(q\ne 0)$, 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? Sol. There is a property that q must satisfy rational number of form $\frac{p}{q}(q\ne 0)$ where $\mathrm{p},\mathrm{q}$ are integers with no common factors other than 1 having terminating decimal representation (expansions) is that the prime factorisation of $q$ has only powers of 2 or powers of 5 or both (i.e., $q$ must be of the form $2^m\times 5^n$ ). Here $m,n$ are whole numbers.
• Write three numbers whose decimal expansions are non-terminating nonrecurring. Sol. 0.01001000100001... 0.202002000200002... 0.003000300003...
• Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$. Sol. $75.000000$ (0.714285...

• Thus, $\frac{5}{7}=0.\overline{714285}$
  
  Thus, $\frac{9}{11}=0.\overline{81}$ Three different irrational numbers between $\frac{5}{7}$ and $\frac{9}{11}$ are taken as 0.750750075000750000... 0.780780078000780000... 0.80800800080000800000...
• Classify the following numbers as rational or irrational : (i) $\sqrt{23}$ (ii) $\sqrt{225}$ (iii) 0.3796 (iv) 7.478478 ...... (v) 1.101001000100001...... Sol. (i) $\sqrt{23}=$ Irrational number (ii) $\sqrt{225}=15=$ Rational number (iii) 0.3796 decimal expansion is terminating $\Rightarrow .3796=$ Rational number (iv) 7.478478... $=7.\overline{478}$ which is non-terminating recurring. = Rational number (v) 1.101001000100001..... decimal expansion is non terminating and non-repeating. = Irrational number

Exercise: 1.4

1. Visualise 3.765 on the number line, using successive magnification. Sol. $\mathrm{n}=3.765$

• Visualise $4.\overline{26}$ on the number line, up to 4 decimal places. Sol. $\mathrm{n}=4.\overline{26}$ So, $\mathrm{n}=4.2626$ (upto 4 decimal places)

Exercise 1.5

• Classify the following numbers as 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) $2\pi$ Sol. (i) 2 is a rational number and $\sqrt{5}$ is an irrational number. $\therefore 2-\sqrt{5}$ is an irrational number. (ii) $(3+\sqrt{23})-\sqrt{23}\Rightarrow (3+\sqrt{23})-\sqrt{23}$ $=3$ is a rational number. (iii) $\frac{2\sqrt{7}}{7\sqrt{7}}=\frac{2}{7}$ is a rational number. (iv) $\frac{1}{\sqrt{2}}$ $\because 1$ is a rational number and $\sqrt{2}$ is an irrational number. So, is an irrational number. (v) $2\pi$ $\because 2$ is a rational number and $\pi$ is an irrational number. So, $2\pi$ is an irrational number.
• Simplify each of the following expressions : (i) $(3+\sqrt{3})(2+\sqrt{2})$ (ii) $(3+\sqrt{3})(3-\sqrt{3})$ (iii) $(\sqrt{5}+\sqrt{2}{)}^2$ (iv) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$ Sol. (i) $(3+\sqrt{3})(2+\sqrt{2})=3(2+\sqrt{2})+\sqrt{3}(2+\sqrt{2})$ $=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$ (ii) $(3+\sqrt{3})(3-\sqrt{3})=(3{)}^2-(\sqrt{3}{)}^2$ $=9-3=6$ (iii) $(\sqrt{5}+\sqrt{2}{)}^2$ $$\begin{gathered} =(\sqrt{5}{)}^2+2\sqrt{10}+(\sqrt{2}{)}^2 \\ =7+2\sqrt{10} \end{gathered}$$ (iv) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=5-2=3$
• Recall, $\pi$ is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, $\pi =c/d$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction? Sol. There is no contradiction. When we measure a length with a scale or any other device, we only get an approximate rational value. Therefore, we may not realise that c is irrational.
• Represent $\sqrt{9.3}$ on the number line. Sol.

• Let $\ell$ be the number line. Draw a line segment $\mathrm{AB}=9.3$ units and BC $=1$ unit. Find the mid point 0 of $AC$. Draw a semicircle with centre 0 and radius OA or OC. Draw $\mathrm{BD}\perp \mathrm{AC}$ intersecting the semicircle at D . Then, $\mathrm{BD}=\sqrt{9.3}$ units. Now, with centre B and radius BD , draw an arc intersecting the number line $l$ at $P$. Hence, $\mathrm{BD}=\mathrm{BP}=\sqrt{9.3}$
• Rationalise the denominators of the following : (i) $\frac{1}{\sqrt{7}}$ (ii) $\frac{1}{\sqrt{7}-\sqrt{6}}$ (iii) $\frac{1}{\sqrt{5}+\sqrt{2}}$ (iv) $\frac{1}{\sqrt{7}-2}$ Sol. (i) $\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$ (ii) $\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$ $=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\frac{\sqrt{7}+\sqrt{6}}{1}=\sqrt{7}+\sqrt{6}$ (iii) $\frac{1}{\sqrt{5}+\sqrt{2}}$ $\frac{1}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{3}$ (iv) $\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}$ $=\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}$

Exercise : 1.6

1. Find: (i) $(64{)}^{1/2}$ (ii) $32^{1/5}$ (iii) $125^{1/3}$ Sol. (i) $(64{)}^{1/2}={\left(8^2\right)}^{1/2}=\left(8^{2\times \frac{1}{2}}\right)=8^1=8$ (ii) $32^{1/5}={\left(2^5\right)}^{1/5}=\left(2^{5\times \frac{1}{5}}\right)=2^1=2$ (iii) $(125{)}^{1/3}={\left(5^3\right)}^{1/3}=5^{3\times \frac{1}{3}}=5$
2. Find: (i) $9^{3/2}$ (ii) $32^{2/5}$ (iii) $16^{3/4}$ (iv) $125^{-1/3}$ Sol. (i) $9^{\frac{3}{2}}={\left(9^{\frac{1}{2}}\right)}^3=(3{)}^3=27$ (ii) $32^{\frac{2}{5}}={\left(2^5\right)}^{\frac{2}{5}}=2^{5\times \frac{2}{5}}=2^2=4$ (iii) $16^{3/4}={\left(2^4\right)}^{3/4}=2^3=8$ (iv) $125^{-1/3}={\left(5^3\right)}^{-1/3}=5^{-1}=1/5$
3. Simplify : (i) $2^{2/3}\cdot 2^{1/5}$ (ii) ${\left(\frac{1}{3^3}\right)}^7$ (iii) $\frac{11^{1/2}}{11^{1/4}}$ (iv) $7^{1/2}\cdot 8^{1/2}$ Sol. (i) $2^{\frac{2}{3}}\cdot 2^{\frac{1}{5}}=2^{\frac{2}{3}+\frac{1}{5}}=2^{\frac{10+3}{15}}=2^{\frac{13}{15}}$ (ii) ${\left(\frac{1}{3^3}\right)}^7=\frac{1^7}{{\left(3^3\right)}^7}=\frac{1}{3^{21}}=3^{-21}$ (iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}=11^{\frac{1}{2}-\frac{1}{4}}=11^{\frac{1}{4}}=\sqrt[4]{11}$ (iv) $7^{\frac{1}{2}}.8^{\frac{1}{2}}=(7\times 8{)}^{1/2}=(56{)}^{1/2}$