Exercise 1.3 Number Systems NCERT Solutions Class 9

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Chapter 1 Ex.1.3 Question 1

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

i) \(\begin{align}\frac{36}{100}\end{align}\)

ii) \(\begin{align}\frac{1}{11}\end{align}\)

iii)  \(\begin{align}4 \frac{1}{8}\end{align}\)

iv)\(\begin{align}\frac{3}{13}\end{align}\)

v) \(\begin{align}\frac{2}{11}\end{align}\)

vi) \(\begin{align}\frac{329}{400}\end{align}\)

Solution

Video Solution

Steps:

(i) \(\begin{align}\frac{{36}}{{100}}=0.36\end{align}\)

Terminating decimal.

(ii) \(\begin{align}\frac{1}{11}\end{align}\)

The remainder \(1\) keeps repeating. \(\begin{align}\frac{1}{{11}} = 0.0909\end{align}\) and can be written as

\(\begin{align}\frac{1}{{11}} = 0.\overline {09} \end{align}\)

Non-terminating recurring decimal.

 (iii)  \(\begin{align}4 \frac{1}{8}=\frac{33}{8}\end{align}\)

\(\begin{align}4\frac{1}{8} = 4.125\end{align}\)

Terminating decimal (\(∵\) The remainder is zero)

(iv) \(\begin{align}\frac{3}{13}=0.23076923\end{align}\)

\(\because\) We find the block of numbers \(230769\) keep repeating.

This is non-terminating recurring decimal and is written as:

\(\begin{align}\,\frac{3}{{13}} = 0.\overline {230769} \end{align}\)

(v) \(\begin{align}\frac{2}{{11}} = 0.1818\end{align}\)

Here we find the block of numbers \(18\) keep repeating. Hence this is a non-terminating recurring decimal and is written as:

\(\begin{align}~\frac{2}{11}=0.\overline{18}\end{align}\)

(vi) \(\begin{align}\frac{{329}}{{400}} = \frac{{329}}{{4 \times 100}}\end{align}\)

\(\begin{align}{\frac{82.25}{100}} \\ {=0.8225}\end{align}\)

Terminating decimal (∵ The remainder is zero)

Chapter 1 Ex.1.3 Question 2

You know that \(\begin{align}\frac{1}{7} = \overline {0.142857} .\end{align}\) Can you predict what the decimal expansions of \(\begin{align}\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}\end{align}\) are, without actually doing the long division? If so, how?

 

Solution

Video Solution

What is Known?

The decimal expansion of \(\begin{align}\,\frac{1}{7}\end{align}\).

What is Unknown?

The decimal expansions of \(\begin{align}\,\,\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}.\end{align}\)

Steps:

\(\begin{align}\frac{1}{7}=0 . \overline{142587}\end{align}\)

This is a non-terminating recurring decimal.

We can use this to find the decimal expansion of \(\begin{align}\frac{2}{7},\,\,\frac{3}{7},\,\frac{4}{7},\,\frac{5}{7},\frac{6}{7}.\end{align}\)

To write the decimal expansion for:

i) \(\begin{align}\frac{2}{7}: \end{align}\) We observe that we get \(2\) as remainder after the second step in the above division.

 Hence,we start writing the quotient after the second decimal place and we get \(\begin{align}\frac{2}{7} = 0.\overline {285714} \end{align}\)

ii) \(\begin{align}\frac{3}{7}:\end{align}\) \(3\) is the remainder after the first step.

\(\begin{align}\rm{Hence}\; \frac{3}{7}=0.\overline {428571}\end{align}\)

iii) \(\begin{align}\frac{4}{7}:\end{align}\) \(4\) is the remainder at the \(4^{th}\) step.

 \(\begin{align}\text{Hence }\frac{4}{7}=0.\overline {571428} \end{align}\)

iv) \(\begin{align}\frac{5}{7}:\end{align}\)\(5\) is the remainder at the \(5^{th}\) step

\(\begin{align}\text{Hence }\frac{5}{7} = 0.\overline {714285} \end{align}\)

v)\(\begin{align}\frac{6}{7}:\end{align}\) \(6\) is the remainder after the \(3^{rd}\) step.

\(\begin{align}\text{Hence }\frac{6}{7} = 0.\overline {857142} \end{align}\)

Chapter 1 Ex.1.3 Question 3

Express the following in the form of \(\begin{align}\frac{\mathrm{p}}{\mathrm{q}},\end{align}\) where \(p\) and \(q\) are integers and \(q≠ 0\).

(i) \(\begin{align}0.\overline 6\end{align}\)

(ii) \(\begin{align}0.4\overline 7\end{align}\)

(iii) \(\begin{align} 0.\overline {001} .\end{align}\)

 

Solution

Video Solution

Steps:

\(\begin{aligned} \text { (i) } &0 . \overline{6} \\ & \text { Let } \mathrm{x} =0 . \overline{6} \\&\qquad \mathrm{x} =0.666 \ldots \qquad \qquad \dots(1) \end{aligned}\)

Since one digit is repeating. We should multiply both sides of (\(1\)) by \(10\). We get,

\[\begin{align}10\,{\rm{ x}} &= {\rm{ }}6.666...\\&= {\rm{ }}6{\rm{ }} + {\rm{ }}0.666...\\10\;{\rm{ x}}& = {\rm{ }}6{\rm{ }} + {\rm{ }}x\\10{\rm{\,x }}{\rm{ }}-{\rm{x }}{\rm{ }} &= {\rm{ }}6\\9\,{\rm{x }} &= {\rm{ }}6\\\rm{x} &= \frac{6}{9}\\&= \frac{2}{3}\\\end{align}\]

Hence\(\begin{align}0.\overline 6 = \frac{2}{3}\end{align}\)

\(\begin{align}{\rm{(ii}})\quad 0.4\overline 7 \end{align}\)

\(\begin{align}\text{Let x = 0.4777} \qquad…… (1) \end{align}\)

Here the repetition starts after the first decimal place and one digit is repeated.

\(\begin{align}\text{10 x} = 4.777 \qquad…... (2)\end{align}\)

\(\begin{align}(2) – (1)\; \text{gives}\end{align}\)

\[\begin{align} 10 \mathrm{x}-\mathrm{x} &=4.777 \ldots-0.4777 \ldots \\ 9 \mathrm{x} &=4.3 \\ 9 \mathrm{x} &=\frac{43}{10} \\ ∴ \mathrm{x} &=\frac{43}{90} \end{align}\]

Hence \(\begin{align}0.4 \overline{7}=\frac{43}{90}\end{align}\)

\(\begin{align}\rm(iii) \quad 0.\overline {001} \end{align}\)

\(\begin{align}\text{Let x }= 0.001001…… \qquad(1)\end{align}\)

Since \(3\) digits are repeated multiply both the sides of (\(1\)) by \(1000\)

\[\begin{align}1000\;\mathrm{x}&=1.001001 \\ 1000\;\mathrm{x}&=1+0.001001 \\ 1000 \;\mathrm{x}&=1+\mathrm{x} \\ 1000\;\mathrm{x}-\mathrm{x}&=1 \\ 999 \;\mathrm{x}&=1 \\ ∴ \;\mathrm{x}&=\frac{1}{999}\end{align}\]

Chapter 1 Ex.1.3 Question 4

Express \(\begin{align} 0.99999…\end{align}\) in the form of \(\begin{align}\frac{p}{q}\end{align}\). Are you surprised with your Answer? With your teacher and classmates discuss why the answer makes sense?

 

Solution

Video Solution

Steps:

Let  \(\begin{align}\text{ x }= 0.99999…...\qquad (1)\end{align}\)

Since one digit is repeated. We should multiply both the sides of \((1)\) by \(10\)

\[\begin{align}10\;\mathrm{x}&=9.9999 \\ 10\;\mathrm{x}&=9+0.9999 \\ 10\;\mathrm{x}&=9+\mathrm{x} \\ 10 \;\mathrm{x}-\mathrm{x}&=9 \\ 9\; \mathrm{x}&=9 \\ \mathrm{x}&=1\end{align}\]

Hence \(0.99999 = 1\)

Chapter 1 Ex.1.3 Question 5

What can be the maximum number of digits be in the repeating block of digits in decimal expansion of \(\begin{align}\,\frac{1}{{17}}\end{align}\)? Perform the division to check your Answer.

 

Solution

Video Solution

Steps:

Let us perform the division \(\begin{align}1÷ 17\end{align}\)

There are \(16\) digits in the repeating block of the decimal expansion of \(\begin{align}\frac{1}{{17}}\end{align}\).

Chapter 1 Ex.1.3 Question 6

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

 

Solution

Video Solution

Steps:

We shall look at some examples of rational numbers in the form of \(\begin{align}\frac{{\rm{p}}}{{\rm{q}}}\end{align}\) \(\begin{align}(q \ne 0)\end{align}\) where decimal representations are terminating.

\(\begin{align}\frac{2}{5} &= 0.4 \qquad \qquad\frac{3}{{100}} = 0.03\\\\\frac{{27}}{{16}} & = 1.6875 \qquad \quad \frac{{33}}{{50}} = 0.66\end{align}\)

We observed that the denominators of above rational numbers are in the form of \(\begin{align}{2^a} \times {5^b}\end{align}\) Where, \(a\) and \(b\) are whole numbers.

Hence if \(q\) is in the form \(\begin{align}{2^a} \times {5^b}\end{align}\) then \(\begin{align}\frac{{\rm{p}}}{{\rm{q}}}\end{align}\) is a terminating decimal.

Chapter 1 Ex.1.3 Question 7

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

 

Solution

Video Solution

Steps:

(i) \(0.212212221…\)

(ii) \(0.03003000300003… \)

(iii) \(0.825882588825… \ldots\)

Chapter 1 Ex.1.3 Question 8

Find three irrational numbers between the rational numbers \(\begin{align}\frac{5}{7}\end{align}\) and \(\begin{align}\,\frac{9}{{11}}\end{align}\).

 

Solution

Video Solution

Steps:

Let us find the decimal expansion of \(\begin{align}\frac{5}{7}\end{align}\) and \(\begin{align}\frac{9}{{11}}\end{align}\) .

We can write \(3\) irrational numbers between them as follows:

(i) \(\begin{align}0.723350980011234...\end{align}\)

(ii) \(\begin{align} 0.77653472899423…\end{align}\)

(iii) \(\begin{align}0.80983435253947…\end{align}\)

Chapter 1 Ex.1.3 Question 9

Classify the following numbers as rational or irrational:

(i) \(\begin{align}\sqrt{23}\end{align}\)

(ii) \(\begin{align} \sqrt{225}\end{align}\)

(iii) \(\begin{align} 0.3796\end{align}\)

iii) \(\begin{align}0.3796\end{align}\)

iv) \(\begin{align}7.478478 \ldots . . . .=7. \overline{478}\end{align}\)

v) \(\begin{align}1.101001000100001\end{align}\)

 

Solution

Video Solution

Steps:

i) \(\begin{align}\sqrt {23} = \frac{{\sqrt {23} }}{1} = \frac{p}{q},\end{align}\) but \(p\) is not an integer.

Hence \(\begin{align}{\sqrt{23}}\end{align}\) is an irrational number.

ii) \(\begin{align} \sqrt {225} = \frac{{15}}{1} = \frac{p}{q},\end{align}\) where \(p\) and \(q\) are integers. \(\begin{align}q \ne 0.\end{align}\)

Hence \(\begin{align}\sqrt {225} \end{align}\) is a rational number.

iii) \(\begin{align}0.3796\end{align}\)

\(\begin{align}0.3796\end{align}\) is a rational number . Because it is a terminating decimal number.

iv) \(\begin{align}7.478478 \ldots . . . .=7. \overline{478}\end{align}\)

It is an rational number. As It is a non-terminating recurring decimal.

v) \(\begin{align}1.101001000100001\end{align}\)

It is an irrational number because it is a non-terminating and non-recurring decimal. 

  
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