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

Solution 1:

(i)\(\displaystyle{36\over 100}\)

On dividing \(36\) by \(100\), we get

Therefore, \(\displaystyle{36\over 100}= 0.36\), which is a terminating decimal.

(ii)\(\displaystyle{1\over 11}\)

On dividing \(1\) by \(11\), we get

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

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

(iii) \(\displaystyle{41\over 8}\)

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

On dividing \(33\) by \(8\), we get \(4.125\)

while dividing \(33\) by \(8\), the remainder is \(0\).

Therefore, \(\displaystyle{4 + { {1\over 8}} = {33\over 8} = 4.125}\), which is a terminating decimal.

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

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, \(\displaystyle{3\over 13}= 0.23076913\), which is a terminating decimal.

(v) \(\displaystyle{2\over 11}\)

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, \(\displaystyle{2\over 11}=0.1818 \dots\)

or \(20.1811\) , which is a non-terminating and recurring decimal.

(vi) \(\displaystyle{329 \over 400}\)

On dividing \(329\) by \(400\), we get

While dividing \(329\) by \(400\), the remainder is \(0\).

Therefore,\(\displaystyle{329\over 400}=0.8225\), which is a terminating decimal.

Question 3: 

Express the following in the form \(\displaystyle {p\over q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).

(i) \(0.\bar{6}\)

(ii)\(0.4\bar{7}\)

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

Solution 3:

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

⇒ \(x = 0.6666\)_____________(1)

Multiply both sides by \(10\),

\(10x = 0.6666 ×10\)

\(10x = 6.6666\)_____________(2)

Subtracting (1) from (2), we get

\(9x = 6\)

\(x =\displaystyle{6\over 9}={2\over 3}\)

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

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

⇒ \( x = 0.47777\)_____________(a)

Multiply both sides by \(10\), we get

\(\displaystyle 10x = 4.7777\)____________(b)

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

\(10x - x =4.39\)

\(x =\displaystyle{43\over 90}\)

Therefore, on converting \(0.47 = \displaystyle{43\over 90}\) in the \(\displaystyle{ p\over q}\) form.

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

\(x=0.001001\) _____________(a)

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

\(1000 × x = 1000 × 0.001001\dots \)

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

Subtract the equation (b) from (a),

\(1000x - x= 1.001001-0.001001\)

\(999x = 1\)

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

Question 4:

Express 0.99999.... in the form \( \displaystyle {p\over q}\) . 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

\(x = \displaystyle{9\over 9}\)

or x = 1

Therefore, on converting 0.99999...= 1 which is in the \(\displaystyle {p\over q}\)  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  \(\displaystyle {1\over 17}\) ? Perform the division to check your answer.

Solution 5:\(\)

We need to find the number of digits in the recurring block of  \(\displaystyle{1\over 17}\).

Let us perform the long division to get the recurring block of \(\displaystyle{1\over 17}\).

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

\(\displaystyle{1\over 17} = 0.0588235294117647 \dots \)

or \(\displaystyle{1\over 17}  = 0.0588235294117647 \),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 \(\displaystyle{5\over 7}\) and \(\displaystyle{9\over 11}\).

Solution 8:

Let us convert \(\displaystyle{5\over 7}\) and \(\displaystyle{9\over 11}\) into decimal form, we get

\(\displaystyle{5\over 7} = 0.714285\dots\) and \(\displaystyle{9\over 11}= 0.818181\dots\)

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 \(\displaystyle{p\over q}\).

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 \(\displaystyle {p\over q}\) form.

While, converting \( {7.478478}\dots \) into \(\displaystyle {p\over q}\) form, we get

\(x =  {7.478478}\)_____________(a)

\(1000x = 7478.478478\)_____________(b)

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

\(999x = 7471\)

\(x=\displaystyle{7471\over 999}\)

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 \(\displaystyle {p\over q}\) form.

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