# Number Systems : Exercise 1.3 (Mathematics NCERT Class 9th)

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

(i) ${{36}\over{100}}$
(ii) ${1\over{11}}$

(iii) $4{1 \over 8}$

(iv) ${3\over{13}}$

(v) ${2\over{11}}$
(vi) ${{329} \over {400}}$
Sol.

(i)    ${{36} \over {100}}$ in decimal form -
${{36} \over {100}} = 0.36,$, terminating decimal.
(ii)   By long division,we have Therefore ${1 \over {11}} = 0.090909 ..... = 0.\overline {09} ,$ non- terminating and repeating decimal.
(iii)   $4{1 \over 8} = {{4 \times 8 + 1} \over 8} = {{33} \over 8}$ By long division , we have Therefore ${{33} \over 8} = 4.125\,$, terminating decimal.
(iv)  By long division, we have Therefore ${3 \over {13}} = 0.23076923.. = 0.\overline {230769} \,$,
non- terminating and repeating decimal.
(v)   By long division we have Therefore ${2 \over {11}} = 0.181818.. = 0.\overline {18} \,$,
non- terminating and repeating decimal.
(vi)  By long division, we have Therefore ${{329} \over {400}} = 0.8225$, terminating decimal.

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

Yes. All of the above will have repeating decimals which are permutations of 1, 4, 2, 8, 5, 7.
For example, here To find ${2 \over 7}$, locate when the remainder becomes 2 and respective quotient (here it is 2), then write the new quotient beginning from there (the arrows in the above division) using the repeating digits 1, 4, 2, 8, 5, 7. Therefore ${2 \over 7} = 0.\overline {285714}$
Similarly,      ${{\rm{3}} \over {\rm{7}}} = 0.\overline {428571} ,\,{4 \over 7} = 0.\overline {571428}$
${{\rm{5}} \over {\rm{7}}} = 0.\overline {714285} ,\,{6 \over 7} = 0.\overline {857142}$

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

(i) $0.\overline 6$
(ii) $0.4\overline 7$
(iii) $0.\overline {001}$
Sol.

(i)    Let $x = 0.\overline 6$
Then, x = 0.666 ... ... (1) Here , we have only one repeating digit.
So, we multiply both sides of (1) by 10 to get
10x = 6.66.... ... (2)

Subtracting (1) from (2) we get
10x – x = (6.66...) – (0.66...) $\Rightarrow$ 9x = 6 $\Rightarrow$ $x = {2 \over 3}$
Hence, $0.\overline 6 = {2 \over 3}$

(ii)   Let $x = 0.4\overline 7$
Clearly, there is just one digit
on the right side of the decimal point which is without bar.
So, we multiply both sides by 10.
So that only the repeating decimal is left on the right side of the decimal point.

Therefore $10\,x = 4.\overline 7$
$\Rightarrow$ $10\,x = 4 + 0.\overline 7$
$\Rightarrow$ $10\,x = 4 + {7 \over 9}$
For example : $0.\overline 7 = {7 \over 9},0.\overline {35} = {{35} \over {99}}$ etc.
$\Rightarrow$ $10x = {{4 \times 9 + 7} \over 9}$ $\Rightarrow$ $10x = {{43} \over 9}$

$\Rightarrow$ $x = {{43} \over {90}}$
Hence $0.4\overline 7 = {{43} \over {90}}$
ALITER
Let $x = 0.4\overline 7 = 0.4777...$
Therefore 10x = 4.777... ... (1)
and 100x = 47.777... ... (2)
Subtracting (1) from (2) we get
100 x – 10 x = (47.777 ... ) – (4.777... )
$\Rightarrow$ 90x = 43 $\Rightarrow$ $x = {{43} \over {90}}$
Hence, $0.4\overline 7 = {{43} \over {90}}$
(iii)   Let $x = 0.\overline {001}$

$\Rightarrow$ x = 0.001001001 .... ... (1)
Here , we have three repeating digits after the decimal point. So we multiply (1) by ${10^3} = 1000$,
we get
1000x = 1.001001... ... (2)

Subtracting (1) from (2) we get -
1000 x – x = (1.001001...) – (0.001001...)
$\Rightarrow$ 999x = 1 $\Rightarrow$ $x = {1 \over {999}}$
Hence, $0.\overline {001} = {1 \over {999}}$

Q.4      Express 0.99999... in the form ${p \over q}$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Sol.

Let x = 0.9999 ... ... (1)
Here , we have only one repeating digit. So, we multiply both sides of (1) by 10 to get.
10x = 9.999... ... (2)
Subtracting (1) from (2) we get
10 x – x = (9.999...) – (0.999...)

$\Rightarrow$ 9x = 9 $\Rightarrow$ x = 1
Hence, 0.9999 ... = 1
Since , 0.9999... goes on forever. So, there is no gap between 1 and 0.9999... and hence they are equal.

Q.5     What can the maximum number of digits be in the repeating block of digits in the decimal expansion of ${1 \over {17}}?$ Perform the division to check your answer.
Sol.         We have Thus, ${1 \over {17}} = 0.\overline {588235294117647}$
Therefore the maximum number of digits in the quotient while computing ${1 \over {17}}$ are 15.

Q.6       Look at several examples of rational numbers in the form ${p \over q}\left( {q \ne 0} \right)$, 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?
Sol.

Consider several rational numbers in the form ${p \over q}\left( {q \ne 0} \right)$, where p and q are integers with common factors other than 1 and having terminating decimal representation.
Let the various such rational numbers be ${1 \over 2},{1 \over 4},{7 \over 8},{{37} \over {25}},$
${8 \over {125}},{{17} \over {20}},{{31} \over {16}}\,\,etc.$
In all cases, we think of the natural number which when multiplied  by their respective denominators gives 10 or a power of 10.
${1 \over 2} = {{1 \times 5} \over {2 \times 5}} = {5 \over {10}} = 0.5$ [Since 2 × 5 = 10]
${1 \over 4} = {{1 \times 25} \over {4 \times 25}} = {{25} \over {100}} = 0.25$ [Since 4 ×25 = 100]
${7 \over 8} = {{7 \times 125} \over {8 \times 125}} = {{875} \over {1000}} = 0.875$ [Since 8 × 125 = 1000]
${{37} \over {25}} = {{37 \times 4} \over {25 \times 4}} = {{128} \over {100}} = 1.28$ [Since 25 × 4 = 100]
${8 \over {125}} = {{8 \times 8} \over {125 \times 8}} = {{64} \over {1000}} = 0.064$ [Since 125 × 8 = 1000]
${{17} \over {20}} = {{17 \times 5} \over {20 \times 5}} = {{85} \over {100}} = 0.85$ [Since 20 × 5 = 100]
${{31} \over {16}} = {{31 \times 625} \over {16 \times 625}} = {{19375} \over {10000}} = 1.9375$ [Since 16 × 625 = 10000]
We have seen that those rational numbers whose denominators when multiplied by a suitable integer produce a power of 10 are expressible in the finite decimal form. But this can always be done only when the denominator of the given rational number has either 2 or 5 or both of them as the only prime factors. Thus, we obtain the following property :
If the denominator of a rational number in standard form has no prime factors other than 2 or 5, then and only then it can be represented as a terminating decimal.

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

Three numbers whose decimal representations are non- terminating non- recurring are
$\sqrt 2 ,\,\sqrt 3 \,\,and\,\,\sqrt 5$ or we can say 0.100100010001..., 0.20200200020002... and .003000300003.

Q.8       Find three different irrational numbers between the rational numbers ${5 \over 7}and{9 \over {11}}$.
Sol.          We have Therefore ${5 \over 7} = 0.\overline {714285\,} \,and\,\,{9 \over {11}} = 0.\overline {81}$
Thus , three different irrational numbers between ${5 \over 7}and{9 \over {11}}$ are.
0.75075007500075000075.... , 0.767076700767000... and 0.80800800080000...

Q.9          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) $\sqrt {23}$ is an irrational number as 23 is not a perfect square.
(ii) $\sqrt {225} = \sqrt {3 \times 3 \times 5 \times 5}$ $= 3 \times 5$
$= 15$
Thus, 15 is a rational number.
(iii) 0.3796 is a rational number as it is terminating decimal.
(iv) 7.478478 .... is non- terminating but repeating , so, it is a rational number.
(v) 1.101001000100001 ... is non - terminating and non- repeating so, it is an irrational number.

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