\(\sqrt [ 3 ]{ 2 } =\sqrt [ 15 ]{ { 2 }^{ 5 } } =\sqrt [ 15 ]{ 32 } \)
\(\sqrt [ 5 ]{ 3 } =\sqrt [ 15 ]{ 3^{ 3 } } =\sqrt [ 15 ]{ 27 } \)
\(\sqrt [ 15 ]{ 32 } >\sqrt [ 5 ]{ 3 } \)
Therefore, \(\sqrt [ 3 ]{ 2 } >\sqrt [ 5 ]{ 3 } \)

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.

i) Draw a line and mark a point A on it.
ii) Mark a point B such that AB = 9.3 cm.
iii) Mark a point C on this line such that BC = 1 unit
iv) Find the midpoint of AC by drawing perpendicular bisector of AC and let it be O
v) With O as center and OC = OA as radius, draw a semicircle
vi) Draw a line BD, which is perpendicular to AB at B.
vii) Now BD = \(\sqrt { 9.3 } \)  which can be marked in the number line as the value of  BE = BD =\(\sqrt { 9.3 } \)

Rational between \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 5 } \) = \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
=\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
= \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
Rational numbers between\(\frac { 1 }{ 2 } \) and \(\frac { 7 }{ 20 } \)
=\(\frac { 1 }{ 2 } \left( \frac { 10+7 }{ 20 } \right) \)
=\(\frac { 17 }{ 40 } \)
Rational number between \(\frac { 1 }{ 2 } \) and \(\frac { 7 }{ 20 } \) =\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } \times \frac { 17 }{ 40 } \right) \)
=\(\frac { 1 }{ 2 } \left( \frac { 20+17 }{ 40 } \right) \)
=\(\frac { 37 }{ 80 } \)
Thus the rational numbers are \(\frac { 7 }{ 20 } ,\frac { 17 }{ 40 } and\frac { 37 }{ 80 } \)

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

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

\(\frac { 1 }{ 4 } \) = 0.25 and \(\frac { 1 }{ 3 } \) = 0.3333….. = \(0.\bar3\)
In between 0.25 and \(0.\bar3\), there are infinitely many irrational numbers.
Four irrational numbers between 0.25 and \(0.\bar3\) are
0.2601001000100001…..
0.2701001000100001…..
0.2801001000100001…..
0.3101001000100001…..

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

(i) \(2\over 3\) = 0.666... = \(0.\overline 6\)

(ii) \(47\over 99\) = 0.4747... = \(0.\overline {47}\)

(iii) \(-{16\over 45}\) = 0.355.... = \(0.3\overline{5}\)

(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.

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

\({ 625 }^{ \frac { 3 }{ 4 } }\) =\(\left( \sqrt [ 4 ]{ 625 } \right) ^{ 3 }=\left( \sqrt [ 4 ]{ { 5 }^{ 4 } } \right) ^{ 3 }={ 5 }^{ 3 }=5\times 5\times 5=125\)

10^-4 = \(\cfrac { 1 }{ { 10 }^{ 4 } } =\cfrac { 1 }{ 1000 } =0.0001\)

(a) 0.750750075000750....
(b) 0.767076700767000767 ...
(c) 0.78080078008000780 ...

Let x = 0.474747 ....... \(\rightarrow\) (1)
100x = 47.4747 ......... \(\rightarrow\) (2)

 \(x={47\over 99}\)
\(\therefore 0.\overline {47}={47\over 99}\)