class 9 maths chapter 1 exercise 1.3

Question 1:

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

(i) ${\displaystyle \frac{36}{100}}$  (ii) ${\displaystyle \frac{1}{11}}$ (iii) ${\displaystyle \frac{41}{8}}$ (iv) ${\displaystyle \frac{3}{13}}$ (v)${\displaystyle \frac{2}{11}}$(vi) ${\displaystyle \frac{329}{400}}$

Solution 1:

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

On dividing $36$ by $100$, we get

Therefore, ${\displaystyle \frac{36}{100}=0.36}$, which is a terminating decimal.

(ii)${\displaystyle \frac{1}{11}}$

On dividing $1$ by $11$, we get

We observe that while dividing $1$ by $11$, the quotient = $0.09$ is repeated.

Therefore, ${\displaystyle \frac{1}{11}=0.0909\dots }$ or ${\displaystyle \frac{1}{11}=0.09}$, which is a non-terminating and recurring decimal.

(iii) ${\displaystyle \frac{41}{8}}$

$=4+{\displaystyle \frac{1}{8}=\frac{{\displaystyle (32+1)}}{8}=\frac{33}{8}}$

On dividing $33$ by $8$, we get $4.125$

while dividing $33$ by $8$, the remainder is $0$.

Therefore, ${\displaystyle 4+\frac{1}{8}=\frac{33}{8}=4.125}$, which is a terminating decimal.

(iv) ${\displaystyle \frac{3}{13}}$

On dividing $3$ by $13$, we get $0.230769$

while dividing $3$ by $13$ the remainder is $3$, which will continue to be $3$ after carrying out continuous divisions.

Therefore,$0.23076913\dots$ or $0.23076913$, which is a non-terminating and recurring decimal.

Therefore, ${\displaystyle \frac{3}{13}=0.23076913}$, which is a terminating decimal.

(v) ${\displaystyle \frac{2}{11}}$

On dividing $2$ by $11$, we get

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

Therefore, ${\displaystyle \frac{2}{11}=0.1818\dots }$

or $20.1811$ , which is a non-terminating and recurring decimal.

(vi) ${\displaystyle \frac{329}{400}}$

On dividing $329$ by $400$, we get

While dividing $329$ by $400$, the remainder is $0$.

Therefore,${\displaystyle \frac{329}{400}=0.8225}$, which is a terminating decimal.

Question 3: 

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

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

(ii)$0.4\overline{7}$

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

Solution 3:

(i). Let ${\displaystyle x=0.\overline{6}}$

⇒ $x=0.6666$_____________(1)

Multiply both sides by $10$,

$10x=0.6666\times 10$

$10x=6.6666$_____________(2)

Subtracting (1) from (2), we get

$9x=6$

$x={\displaystyle \frac{6}{9}=\frac{2}{3}}$

Therefore, on converting $0.6={\displaystyle \frac{2}{3}}$,which is in the ${\displaystyle \frac{p}{q}}$ form.

(ii).Let ${\displaystyle x=0.4\overline{7}}$

⇒ $x=0.47777$_____________(a)

Multiply both sides by $10$, we get

${\displaystyle 10x=4.7777}$____________(b)

Subtract the equation (a) from (b),we get

$10x-x=4.39$

$x={\displaystyle \frac{43}{90}}$

Therefore, on converting $0.47={\displaystyle \frac{43}{90}}$ in the ${\displaystyle \frac{p}{q}}$ form.

(iii)Let $x=0.\overline{001}$

$x=0.001001$ _____________(a)

multiply both sides by $1000)$ because the number of recurring decimal number is $3)$

$1000\times x=1000\times 0.001001\dots$

So, $1000x=1.001001$_____________(b)

Subtract the equation (b) from (a),

$1000x-x=1.001001-0.001001$

$999x=1$

Therefore, on converting $0.\overline{001}={\displaystyle \frac{1}{999}}$ in the ${\displaystyle \frac{p}{q}}$ form.

Question 4:

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

Solution 4:

Let x = 0.99999_____________(a)

We need to multiply by 10 on both sides, we get

10x = 9.9999_____________(b)

Subtract the equation (a) from (b), to get

10x - x = 9.9999 - 0.9999

9x = 9

$x={\displaystyle \frac{9}{9}}$

or x = 1

Therefore, on converting 0.99999...= 1 which is in the ${\displaystyle \frac{p}{q}}$  form,

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

But the answer makes sense when we observe that 0.9999$\dots$ goes on forever.

So, there is no gap between 1 and 0.9999$\dots$ and hence they are equal.

Question 5:

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

Solution 5:$$

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

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

We need to divide $1$ by $17$, to get

We can observe that while dividing $1$ by $17$ we get $16$  number of digits in the repeating block of decimal expansion which will continue to be $1$ after carrying out $16$ continuous divisions.

Therefore, we conclude that

${\displaystyle \frac{1}{17}=0.0588235294117647\dots }$

or ${\displaystyle \frac{1}{17}=0.0588235294117647}$,which is a non-terminating and recurring decimal.

Question 7:

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

Solution 7:

All irrational numbers are non-terminating and non-recurring.

Examples: 

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

$\sqrt{3}=1.73205\dots$

$\sqrt{7}=2.645751\dots$

Question 8: 

Find three different irrational numbers between the rational numbers ${\displaystyle \frac{5}{7}}$ and ${\displaystyle \frac{9}{11}}$.

Solution 8:

Let us convert ${\displaystyle \frac{5}{7}}$ and ${\displaystyle \frac{9}{11}}$ into decimal form, we get

${\displaystyle \frac{5}{7}=0.714285\dots }$ and ${\displaystyle \frac{9}{11}=0.818181\dots }$

Three irrational numbers that lie between $0.714285\dots$ and $0.818181\dots$ are:

$0.73073007300073\dots$

$0.74074007400074\dots$

$0.76076007600076\dots$

Irrational numbers cannot be written in the form of ${\displaystyle \frac{p}{q}}$.

Question 9:

Classify the following numbers as rational or irrational:

(i)$$$\sqrt{23}$ 

(ii)$\sqrt{225}$ 

(iii)$0.3796$ 

(iv)$7.478478\dots$ 

(v)$1.101001000100001\dots$

Solution 9:

(i)$\sqrt{23}$

$\sqrt{23}=4.795831\dots$

It is an irrational number

(ii) $\sqrt{225}=15$

Therefore $\sqrt{225}$ is a rational number.

(iii)$0.3796$

It is terminating decimal. Therefore, it is rational number

(iv)$7.478478\dots$.

The given number $7.478478\dots$ is a non-terminating recurring decimal, which can be converted into ${\displaystyle \frac{p}{q}}$ form.

While, converting $7.478478\dots$ into ${\displaystyle \frac{p}{q}}$ form, we get

$x=7.478478$_____________(a)

$1000x=7478.478478$_____________(b)

While, subtracting (a) from (b), we get

$999x=7471$

$x={\displaystyle \frac{7471}{999}}$

Therefore, $7.478478\dots$ is a rational number.

(v)$1.101001000100001\dots$

We can observe that the number $1.101001000100001\dots$is a non-terminating non-recurring decimal.

Thus, non-terminating and non-recurring decimals cannot be converted into ${\displaystyle \frac{p}{q}}$ form.

Therefore, we conclude that $1.101001000100001\dots$ is an irrational number.