Exercise 1.3 (Revised) - Chapter 1 - Number Systems - Ncert Solutions class 9 - Maths

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Chapter 1 - Number Systems | NCERT Solutions Class 9 Maths

Ex 1.3 Question 1.

Write the following in decimal form and say what kind of decimal expansion each has: $\sqrt{5}$
(i) $\frac{36}{100}$
(ii) $\frac{1}{11}$
(iii) $4 \frac{1}{8}$
(iv) $\frac{3}{13}$
(v) $\frac{2}{11}$
(vi) $\frac{329}{400}$

Answer.

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

On dividing 36 by 100 , we get

Therefore, we conclude that $\frac{36}{100}=0.36$, which is a terminating decimal.
(ii) $\frac{1}{11}$

On dividing 1 by 11 , we get

We can observe that while dividing 1 by 11 , we got the remainder as 1 , which will continue to be 1 . recurring decimal.
(iii) $4 \frac{1}{8}=\frac{33}{8}$

On dividing 33 by 8 , we get

We can observe that while dividing 33 by 8 , we got the remainder as 0 .

Therefore, we conclude that $4 \frac{1}{8}=\frac{33}{8}=4.125$, which is a terminating decimal.
(iv) $\frac{3}{13}$

On dividing 3 by 13 , we get

We can observe that while dividing 3 by 13 we got the remainder as 3 , which will continue to be 3 after carrying out 6 continuous divisions.

Therefore, we conclude that $\frac{3}{13}=0.230769 \ldots .$. or $\frac{3}{13}=0 . \overline{230769}$, which is a non-terminating decimal and recurring decimal.
(v) $\frac{2}{11}$

On dividing 2 by 11 , we get

We can observe that while dividing 2 by 11 , first we got the remainder as 2 and then 9 , which will continue to be 2 and 9 alternately.

Therefore, we conclude that $\frac{2}{11}=0.1818 \ldots \ldots$ or $\frac{2}{11}=0 \overline{18}$, which is a non-terminating decimal and recurring decimal.
(vi) $\frac{329}{400}$

On dividing 329 by 400 , we get

$\text { We can observe that while dividing } 329 \text { by } 400 \text {, we got the remainder as } 0 \text {. }$

Ex 1.3 Question 2.

You know that $\frac{1}{7}=0.142857 \ldots \ldots$. 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?
[Hint: Study the remainders while finding the value of $\frac{1}{7}$ carefully.]
Answer.

We are given that $\frac{1}{7}=0 \overline{142857}$ or $\frac{1}{7}=0.142857 \ldots .$.

We need to find the values of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}$ and $\frac{6}{7}$, without performing long division.
We know that, $\frac{2}{7}=\frac{3}{7}=\frac{4}{7}=\frac{5}{7}$ and $\frac{6}{7}$ can be rewritten as $2 \times \frac{1}{7}, 3 \times \frac{1}{7}, 4 \times \frac{1}{7}, 5 \times \frac{1}{7}$ and $6 \times \frac{1}{7}$.

On substituting value of $\frac{1}{7}$ as $0.142857 . . .$. , we get
$
\begin{aligned}
& 2 \times \frac{1}{7}=2 \times 0.142857 \ldots \ldots=0.285714 \ldots \ldots \ldots . \\
& 3 \times \frac{1}{7}=3 \times 0.142857 \ldots \ldots=0.428571 \\
& 4 \times \frac{1}{7}=4 \times 0.142857 \ldots=0.571428 \\
& 5 \times \frac{1}{7}=5 \times 0.142857 \ldots \ldots=0.714285 \\
& 6 \times \frac{1}{7}=6 \times 0.142857 \ldots \ldots=0.857142 \\
&
\end{aligned}
$

Therefore, we conclude that, we can predict the values of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}$ and $\frac{6}{7}$, without performing long division, to get
$
\frac{2}{7}=0 . \overline{285714}, \frac{3}{7}=0 . \overline{428571}, \frac{4}{7}=0 . \overline{571428}, \frac{5}{7}=0 . \overline{714285} \text {, and } \frac{6}{7}=0 . \overline{857142}
$

Ex 1.1 Question 3

 $\text {Express the following in the form } \bar{q} \text {, where } p \text { and } q \text { are integers and } q \neq 0 \text {. }$

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

Answer.
(i) Let $x=0 . \overline{6} \Rightarrow x=0.6666 \ldots \ldots .$. (a)

We need to multiply both sides by 10 to get
$
10 x=6.6666 \ldots .
$

We need to subtract (a)from (b), to get

We can also write $9 x=6$ as $\quad x=\frac{6}{9}$ or $\quad x=\frac{2}{3}$.

Therefore, on converting $0 . \overline{6}$ in the $\frac{p}{q}$ form, we get the answer as $\frac{2}{3}$.
(ii) Let $x=0.4 \overline{7} \Rightarrow x=0.47777$ (a)

We need to multiply both sides by 10 to get
$
10 x=4.7777
$

We need to subtract $(a)$ from $(b)$, to get

We can also write $9 x=4.3$ as $x=\frac{4.3}{9}$ or $x=\frac{43}{90}$.
(iii) Let $x=0 . \overline{001} \Rightarrow x=0.001001$. $\qquad$ (a)

We need to multiply both sides by 1000 to get

$
1000 x=1.001001 \ldots .
$

We need to subtract $(a)$ from $(b)$, to get

We can also write $999 x=1$ as $x=\frac{1}{999}$

Therefore, on converting $0 . \overline{001}$ in the $\frac{p}{q}$ form, we get the answer as $\frac{1}{999}$.

Ex 1.3 Question 4.

Express $0.99999 \ldots$ in the form $\frac{p}{q}$. Are you surprised by your answer? Discuss why the answer makes sense with your teacher and classmates.

Answer.

Let $x=0.99999$ $\qquad$ (a)

We need to multiply both sides by 10 to get
$
10 x=9.9999
$
We need to subtract $(a)$ from $(b)$, to get
$
\begin{aligned}
10 x & =9.99999 \ldots . . \\
-x & =0.99999 \ldots \\
\hline 9 x & =9
\end{aligned}
$

We can also write $9 x=9$ as $\quad x=\frac{9}{9}$ or $x=1$.

Therefore, on converting $0.99999 \ldots .$. in the $\frac{p}{q}$ form, we get the answer as 1 .

Yes, at a glance we are surprised at our answer.

Yes, at a glance we are surprised at our answer.

But the answer makes sense when we observe that 0.9999 $\qquad$ goes on forever. SO there is not gap between 1 and 0.9999 $\qquad$ and hence they are equal.

Ex 1.3 Question 5.

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

Answer.

We need to find the number of digits in the recurring block of $\frac{1}{17}$.

Let us perform the long division to get the recurring block of $\frac{1}{17}$.

We need to divide 1 by 17 , to get

We can observe that while dividing 1 by 17 we got the remainder as 1 , which will continue to be 1 after carrying out 16 continuous divisions.

Therefore, we conclude that $\frac{1}{17}$ or $\frac{1}{17}=0 . \overline{0588235294117647}$ $\qquad$ $\frac{1}{17}=0.0588235294117647$ $\qquad$ Therefore, we conclude that $\frac{1}{17}=0.0588235294117647 \ldots .$. or $\frac{1}{17}=0 . \overline{0588235294117647}$, which is a nonterminating decimal and recurring decimal.

Ex 1.3 Question 6.

Look at several examples of rational numbers in the form $\frac{p}{q}(q \neq 0)$, 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?

Answer. 

Let us consider the examples of the form $\frac{p}{q}$ that are terminating decimals.
$
\begin{aligned}
& \frac{5}{2}=2.5 \\
& \frac{5}{4}=1.25 \\
& \frac{2}{5}=0.4 \\
& \frac{2}{10}=0.2 \\
& \frac{5}{16}=0.3125
\end{aligned}
$

We can observe that the denominators of the above rational numbers have powers of 2,5 or both.

Therefore, we can conclude that the property, which $q$ must satisfy in $\frac{p}{q}$, so that the rational number $\frac{p}{q}$ is a terminating decimal is that $q$ must have powers of 2,5 or both.

Ex 1.3 Question 7.

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

Answer.

The three numbers that have their expansions as non-terminating on recurring decimal are given below.
$
\begin{aligned}
& 0.04004000400004 \ldots \\
& 0.07007000700007 \ldots . \\
& 0.013001300013000013 \ldots .
\end{aligned}
$

Ex 1.3 Question 8.

Find three different irrational numbers between the rational numbers $\frac{5}{11}$ and $\frac{9}{11}$.

Answer.

Let us convert $\frac{5}{7}$ and $\frac{9}{11}$ into decimal form, to get
$
\frac{5}{7}=0.714285 \ldots \text { and } \frac{9}{11}=0.818181 \ldots .
$

Three irrational numbers that lie between $0.714285 \ldots$ and $0.818181 \ldots$ are:
$0.73073007300073 \ldots \ldots$
$0.74074007400074 \ldots \ldots$
$0.76076007600076 \ldots$

Ex 1.3 Question 9.

Classify the following numbers as rational or irrational:
(i) 23
(ii) 225
(iii) 0.3796
(iv) $7.478478 \ldots$
(v) $1.101001000100001 \ldots$

Answer.

(i) $\sqrt{23}$

We know that on finding the square root of 23 , we will not get an integer.

Therefore, we conclude that $\sqrt{23}$ is an irrational number.
(ii) $\sqrt{225}$

We know that on finding the square root of 225 , we get 15 , which is an integer.

Therefore, we conclude that $\sqrt{225}$ is a rational number.

(iii) 0.3796

We know that 0.3796 can be converted into $\frac{p}{q}$.

While, converting 0.3796 into $\frac{p}{q}$ form, we get
$
0.3796=\frac{3796}{10000}
$

The rational number $\frac{3796}{10000}$ can be converted into lowest fractions, to get $\frac{949}{2500}$.

We can observe that 0.3796 can be converted into a rational number.

Therefore, we conclude that 0.3796 is a rational number.
(iv) $7.478478 \ldots$.

We know that $7.478478 \ldots$ is a non-terminating recurring decimal, which can be converted into $\frac{p}{q}$ form.

While, converting 7.478478.... into $\frac{\underline{p}}{q}$ form, we get
$x=7.478478 \ldots$.
....(a)
$1000 x=7478.478478$ $\qquad$ (b)

While, subtracting (a) from (b), we get
$
\begin{aligned}
1000 x & =7478.478478 \ldots \\
-x & =\quad 7.478478 \ldots \\
\hline 999 x & =7471
\end{aligned}
$

We know that $999 x=7471$ can also be written as $x=\frac{7471}{999}$.

Therefore, we conclude that $7.478478 \ldots$ is a rational number.
(v) $1.101001000100001 \ldots$.

We can observe that the number $1.101001000100001 \ldots$ is a non-terminating on recurring decimal.
$\underline{p}$
We know that non-terminating and non-recurring decimals cannot be converted into $q$ form. Therefore, we conclude that $1.101001000100001 . .$. is an irrational number.