Three irrational numbers between 0.12 and 0.13 are 0.12010010001…, 0.12040040004…, 0.12070070007…

The square root of every positive but not a perfect square number is an irrational number.
Example: (i) \(\sqrt{18}\) (ii) \(\sqrt{75}\)

\(\frac{22}{7}\) = 3.142857142 ...... and \(\pi\) = 3.141592653289 ........
It is an irrational number.
We usually take  \(\pi\)  as  \(\frac{22}{7}\) (a rational number only Butupto 2 decimal places).
But it is not exactly equal to  \(\frac{22}{7},\) it is approximate  value.

Yes, it is possible to write.

(i) \(3\over 4\) = 0.75

(ii) \(5\over 8\) = 0.625

(iii) \(9\over 25\) = 0.36

Let x = \(0.\overline { 9 } \) =  0.99999… (1)
(Multiply equation (1) by 10)
10 x = 9.99999… (2)
Subtract (1) from (2) 
9x = 9 or x = 1
Thus, \(0.\overline { 9 } \) = 1

(i) Given that a = 2  + \(\sqrt { 3 } \) , b = 2 - \(\sqrt { 2 } \) 
x = a + b  = ( 2 + \(\sqrt { 3 } \)) + (2 - \(\sqrt { 3 } \)) = 4 a rational number
y = a - b = (2 +  \(\sqrt { 3 } \) ) - (2 - \(\sqrt { 3 } \)) = 2\(\sqrt { 3 } \)  an irrational number
(ii) Given that a  = \(\sqrt { 2 } \) + 7 , b = \(\sqrt { 2 } \) - 7
x = a + b = (\(\sqrt { 2 } \) + 7) + (\(\sqrt { 2 } \) - 7) = 2\(\sqrt { 2 } \) an irrational number.
y = a -b  = (\(\sqrt { 2 } \) + 7 ) - (\(\sqrt { 2 } \) - 7) = 14 a rational number
(iii) Given that a = \(\sqrt { 75 } \) , b = \(\sqrt { 3 } \)
x = ab = \(\sqrt { 75 } \times \sqrt { 3 } =\sqrt { 75\times 3 } =\sqrt { 5\times 5\times 3\times 3 } =5\times 3=15\) a rational number
\(y=\frac { a }{ b } =\frac { \sqrt { 75 } }{ \sqrt { 3 } } =\sqrt { \frac { 75 }{ 3 } } =\sqrt { 25 } =5\), rational number
(iv) Given that a = \(\sqrt { 18 } \), b = \(\sqrt { 3 } \)
\(x=ab=\sqrt { 18 } \times \sqrt { 3 } =\sqrt { 18\times 3 } =\sqrt { 6\times 3\times 3 } =3\sqrt { 6 } ,\) an irrational number,
\(y=\frac { a }{ b } =\frac { \sqrt { 18 } }{ \sqrt { 3 } } =\sqrt { \frac { 18 }{ 3 } } =\sqrt { 6 } \) an irrational number

(i) Let \(x=0.\overline { 3 } \)= 0.3333   (1)
(Here period of decimal is 1, multiply equation (1) by 10)
10x = 3.3333....    (2)
(2) - (1) : 9x = 3  or \(x=\frac { 1 }{ 3 } \)
(ii) Let \(x=2.\overline { 124 } \) = 2.124124124…            (1)
(Here period of decimal is 3, multiply equation (1) by 1000)
1000 x = 2124.124124124… (2)
(2)–(1): 999 x = 2122  \(x=\frac { 2122 }{ 999 } \)
(iii) Let x = \(0.4\overline { 5 } \) = 0.45555…              (1)
(Here the repeating decimal digit is 5, which is the second digit after the decimal point, multiply equation (1) by 10)
10 x = 4.5555…  (2)
(Now period of decimal is 1, multiply equation (2) by 10)
100 x = 45.5555…              (3)
(3) – (2): 90 x = 41 or \(x=\frac { 41 }{ 90 } \)
(iv) Let x = \(0.5\overline { 68 } \)=  0.5686868…  (1)
(Here the repeating decimal digit is 68, which is the second digit after the decimal point, so multiply equation (1) by 10)
10 x = 5.686868…                (2)
(Now period of decimal is 2, multiply equation (2) by 100)
1000 x = 568.686868… (3)
3) – (2): 990 x = 563 or \(x=\frac { 563 }{ 990 } \)

(i)  \(\frac { 1 }{ 4 } \) and \(1\over 5\)
Take LCM = 200
\( \frac{1}{4} \times \frac{50}{50}=\frac{50}{200} \)
\( \frac{1}{5} \times \frac{40}{40}=\frac{40}{200}\)
The numbers are \(\frac{41}{200}, \frac{42}{200} \ldots \ldots \ldots \frac{49}{200}\)
(ii) 0.1 and 0.11
Change decimal into fraction \(0.1=\frac{1}{10} \text { and } 0.11=\frac{11}{100}\)
Take LCM = 100
\(\frac{1}{10} \times \frac{10}{10}=\frac{10}{100}, \frac{11}{100} \times \frac{1}{1}=\frac{11}{100}\)
Take LCM = 600
\(\frac{10}{100} \times \frac{6}{6}=\frac{60}{600} \text { and } \frac{11}{100} \times \frac{6}{6}=\frac{66}{600}\)
A rationalnumberbetween and \(\frac{60}{600} \text { and } \frac{66}{600}\)
are \(\frac{61}{600}, \frac{62}{600}, \frac{63}{600}, \frac{64}{600}, \frac{65}{600}\)
(iii) -1 and -2
Take LCM = 10 
\(\frac{-1}{1} \text { and } \frac{-2}{1}\)
\(\frac{-1}{1} \times \frac{10}{10}=\frac{-10}{10} \text { and } \frac{-2}{1} \times \frac{10}{10}=\frac{-20}{10}\)
The Numbers are \(\frac{-11}{10}, \frac{-12}{10} \ldots \ldots \ldots \ldots . \frac{-19}{10}\)

\(\sqrt [ 3 ]{ 128 } \) = \(\sqrt [ 3 ]{ 2\times 2\times 2\times 2\times 2\times 2\times 2 } =4\sqrt [ 3 ]{ 2 } \)

\(\sqrt { 27 } \) =\(\sqrt { 3\times 3\times 3 } =\sqrt [ 3 ]{ 3 } \)

\(\left( 5\sqrt { 125 } \right) ^{ 7 }\) = \({ 125 }^{ \frac { 7 }{ 5 } }\)

243 = 3 \(\times\) 3 \(\times\) 3 \(\times\) 3 \(\times\) 3 = 3⁵

\({17\over 200}={17\over 2^3\times 5^2}\)
\(\therefore {17\over 200}\) has a terminating decimal expansion.

Let x = 0.2454545 ....... \(\rightarrow\) (1)
10x = 2.454545 ........... \(\rightarrow\) (2)
1000x = 245.4545 ........ \(\rightarrow\) (3)

\(x={243\over 990}\)  (or)  \({27\over 110}\)
\(\therefore 0.2\overline{45}={27\over 110}\)