(i) 5.348 lies between 5 and 6.

Steps of construction:
1. Divide the distance between 5 and 6 into 10 equal intervals.
2. Mark the point 5.3 which is the sixth from the left of 6 and 3 from the right of 5.
3. 5.34 lies between 5.30 and 5.40. Divide the distance into 10 equal intervals.
4. Mark the point 5.34 which is sixth from the left of 5.40
5. 5.348 lies between 5.34 and 5.35. Divide the distance into 10 equal intervals.
6. Mark a point 5.348 which is second from the left of 5.350 and seventh from the right of 5.340.
(ii) \(6.\overline{4}\) up to 3 decimal places.
\(6.\overline{4}\) = 6.4444 .......
\(6.\overline{4}\) = 6.444 (correct to 3 decimal places)
The number lies between 6 and 7.

Steps of construction:
1. Divide the distance between 6 and 7 into 10 equal intervals.
2. Mark the point 6.4 which is the sixth from the left of 7 and fourth from the right of 6.
3. 6.44 lies between 6.4 and 6.5. Divide the distance into 10 equal intervals.
4. Mark the point 6.44 which is sixth from the left of 6.5 and fourth from the right of 6.40.
5. Mark the point 6.444 which is sixth from the left of 6.450 and fourth from the right of 6.440.
(iii) \(4.\overline{73}\) = 4.737373 ......
= 4.737374 (correct to 4 decimal places 4.7374 lies between 4 and 4)

Steps of construction:
1. Divide the distance between 4 and 5 into 10 equal parts.
2. Mark the point 4.7 which is third from the left of 5 and seventh from the right of 4.
3. 4.73 lies between 4.7 and 4.8. Divide the distance into 10 equal intervals.
4. Mark the point 4.73 which is seventh from the left of 4.80 and third from the left of 4.70.
5. 4.737 lies between 4.73 and 4.74. Divide the distance into 10 equal intervals.
6. Mark the point 4.737 which is third from the left of 4.740 and seventh from the right of 4.730.
7. 4.7374 lies between 4.737 and 4.738. Divide the distance into 10 equal intervals.
8. Mark the point 4.7374 which is sixth from the left of 4.7380 and fourth from the right of 4.7370.

\({6\over 7}=0.\overline {857142}\)
\({12\over 13}=0.\overline{923076}\)

The two irrational numbers are  0.866142 ...and 0.903076.......

(60000000)³ = \(\left(6.0 \times 10^{7}\right)^{4}=(6.04)^{4} \times\left(10^{7}\right)^{4}\)
\( =1296 \times 10^{28}=1.296 \times 10^{3} \times 10^{28} \\ =1.296 \times 10^{31} \) 

\(\sqrt { 44 } +\sqrt { 99 } -\sqrt { 275 } \) = \(\sqrt { 4\times 11 } +\sqrt { 9\times 11 } -\sqrt { 11\times 25 } \)
= \(\left( 2\sqrt { 11 } +3\sqrt { 11 } \right) -5\sqrt { 11 } =5\sqrt { 11 } -5\sqrt { 11 } =0\)

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
\( x=\frac{17197.936}{999} \\ x=\frac{17197936}{999000} \\ x=\frac{5681}{330} \)

Let x = 0.242424..... \(\rightarrow\) (1)
100 x = 24.2424 ...... \(\rightarrow\) (2)
(2) - (1) \(\Rightarrow\) 100x - x = 24.2424 
99x = 24
x = \(24\over 99\) (or) \(8\over 33\)
x = \(8\over 33\)

A rational number between \(\frac { 1 }{ 2 } \) and \(\frac { 2 }{ 3 } \) = \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 2 }{ 3 } \right) =\frac { 1 }{ 2 } \left( \frac { 3+4 }{ 6 } \right) =\frac { 1 }{ 2 } \left( \frac { 7 }{ 6 } \right) =\frac { 7 }{ 12 } \)
A rational number between \(\frac { 1 }{ 2 } \) and \(\frac { 7 }{ 12 } \) = \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 7 }{ 12 } \right) =\frac { 1 }{ 2 } \left( \frac { 6+7 }{ 12 } \right) =\frac { 1 }{ 2 } \left( \frac { 13 }{ 12 } \right) =\frac { 13 }{ 24 } \)
Hence two rational numbers between \(\frac { 1 }{ 2 } \) and \(\frac { 2 }{ 3 } \) are \(\frac { 7 }{ 12 } \) and \(\frac { 13 }{ 24 } \) (of course, there are many more!)
There is an interesting result that could help you to write instantly rational numbers between any two given rational numbers.

Two rational numbers between the given two irrational numbers are 0.5152 and 0.5352

No. \(\sqrt{2}\) = 1.4142135 .... it is an irrational number. It has non-terminating and non-repeating decimal representation.

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) 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 \(x=\frac { 563 }{ 990 } \)

We know that \(\frac { 1 }{ 3 } =0.\overline { 3 } \)
Therefore \(\frac { 1 }{ 27 } =\frac { 1 }{ 9 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 9 } \times 0 333... = 0.037037..\ =\ 0.\overline { 037 } \)
Also, \(\frac { 59 }{ 27 } =2\frac { 5 }{ 27 } =2+\frac { 5 }{ 27 } =2+\left( 5\times \frac { 1 }{ 27 } \right) \)
\(=2+(5\times 0.\overline { 037 } )=2+(5\times 0.037037037..)\)
\(=2+0.185185...=2.185185..\quad =2.\overline { 185 } \)

The rational numbers between \(-7\over 11\) and \(2\over 11\) is
\(\frac {- 6 }{ 11 } , \frac { -5 }{ 11 } , \frac { -4 }{ 11 },\frac {- 3 }{ 11 },\frac {- 2 }{ 11 },\frac {- 1 }{ 11 } , 0, \frac { 1 }{ 11 } \)