\(1\over 11\) = 0.0909 ......... = \(0.\overline{09}\)
\(\therefore \frac{1}{33}=\frac{1}{3}\times \frac{1}{11}\)
 \(=\frac{1}{3}\times 0.0909.......\)
= 0.0303........ = \(0.\overline{03}\)
Also \(\frac{71}{33}=2\frac{5}{33}=2+\frac{5}{33}\)
\(=2+5\times \frac{1}{33}\)
= 2 + 5 \(\times\) \(0.\overline {03}\)
= 2 + (5 \(\times\) 0.030303 .......)
= 2 + 0.151515 .......
= 2 + \(0.\overline {15}\)
= \(2.\overline {15}\)

\(\sqrt{2}=1.414\)
\(\sqrt{3}=1.732\)
The two irrational numbers between \(\sqrt{2}\) and \(\sqrt{3}\) are 1.514 and 1.632

Thus, by division method \(\sqrt { 3 } \) = 1.7320508…
It is found that the square root of every positive non perfect square number is an irrational number  \(\sqrt { 2 } ,\sqrt { 3 } ,\sqrt { 5 } ,\sqrt { 6 } ,\sqrt { 7 } \)..... are all irrational numbers.

Let x = 17.2151515 ......  (1)
Here period of decimal is 3, multiply equation (1) by 1000
1000 x = 17215.151515
(2) - (1)
999x = 17197.936
\( x=\frac{17197.936}{999} \\ x=\frac{17197936}{999000} \\ x=\frac{5681}{330} \)

0.86 = \({8\over 10}+{6\over 100}={80+6\over 100}\)
        = \(86\over 100\) (or) \(43\over 50\)

Yes, it is possible to write.

(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) \(\frac { -4 }{ 11 } \)

Thus we see that \(\frac { -4 }{ 11 } =-0.\overline { 36 } \)
(ii) \(\frac { 11 }{ 75 } \)

Thus we see that \(\frac { 11 }{ 75 } =0.14\overline { 6 } \)
A rational number can be expressed by 
(i) either a terminating
(ii) or a non-terminating and recurring (repeating) decimal expansion. 
The converse of this statement is also true. That is, if the decimal expansion of a number is terminating or non-terminating and recurring, then the number is a rational number.