(i) \(\sqrt { 10 } \) is an irrational number ( since 10 is not a perfect square number).
(ii) \(\sqrt { 49 } =7=\frac { 7 }{ 1 } \) a rational number(since 49 is a perfect square number).
(iii) 0.025 is a rational number (since it is a terminating decimal).
(iv) \(0.7\overline { 6 } \) = 0.7666…. is a rational number ( since it is a non – terminating and recurring decimal expansion).
(v) 2.505500555…. is an irrational number (since it is a non – terminating and non–recurring decimal).
(vi) \(\frac { \sqrt { 2 } }{ 2 } =\frac { \sqrt { 2 } }{ \sqrt { 2 } \times \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \) is an irrational number ( since 2 is not a perfect square number).

To represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line we make 11 markings each being equal distance \(\frac { 1 }{ 11 } \) on the left of 0.
The point A represents \(\left( -\frac { 2 }{ 11 } \right) \) , the point B represents\(\left( -\frac { 2 }{ 11 } \right) \) and the point C represents \(\left( -\frac { 9 }{ 11 } \right) \)

Relational 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 \(\frac { 7 }{ 20 } ,\frac { 17 }{ 40 } and\frac { 37 }{ 80 } \)

Let us convert the given rational numbers having the same denominators.
L.C.M of 8 and 6 is 24.
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
Now the rational numbers between \(-\frac { 202 }{ 24 } and\frac { 15 }{ 24 } are-\frac { 19 }{ 24 } ,-\frac { 18 }{ 24 } ,-\frac { 17 }{ 24 } ,....,\frac { 0 }{ 24 } ,\frac { 1 }{ 24 } ,\frac { 22 }{ 24 } ,...,\frac { 14 }{ 24 } \)
we can take any seven of them \(\frac { 1 }{ 24 } ,\frac { 2 }{ 24 } ,\frac { 3 }{ 24 } ,\frac { 4 }{ 24 } ,\frac { 5 }{ 24 } ,\frac { 6 }{ 24 } ,\frac { 7 }{ 24 } \)

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

\(\sqrt { 5 } \) = \({ 5 }^{ \frac { 1 }{ 2 } }={ 5 }^{ \frac { 6 }{ 12 } }=\sqrt [ 12 ]{ { 5 }^{ 6 } } =\sqrt [ 12 ]{ 15625 } \)

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

\({7\over 16}={7\over 2^4}={7\over 2^4\times 5^6}\)
\(\therefore {7\over 16}\) has a terminating decimal expansion.

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

(500000)⁵ \(\times\) (3000)³ 
\( =\left(5.0 \times 10^{5}\right)^{3} \times\left(3.0 \times 10^{3}\right)^{3} \\ =(5.0)^{2} \times\left(10^{5}\right)^{2} \times(3.0)^{3} \times\left(10^{3}\right)^{3} \\ =25 \times 10^{10} \times 27 \times 10^{9}=675 \times 10^{19} \\ =675.0 \times 1019=6.75 \times 102 \times 1019=6.75 \times 10^{21} \)

\(3\sqrt { 162 } \times 7\sqrt { 50 } \times 6\sqrt { 98 } \) = \(\left( 3\times 9\sqrt { 2 } \times 7\times 5\sqrt { 2 } \times 6\times 7\sqrt { 2 } \right) \)
= \(3 \times 7 \times 6 \times 9 \times 5 \times 7 \times \sqrt{2} \times \sqrt{2} \times \sqrt{2}\)  = 79380\(\sqrt { 2 } \)

\(9\sqrt { 7 } -6\sqrt { 7 } =\left( 9-6 \right) \sqrt { 7 } =3\sqrt { 7 } \) The answer is irrational.

\(\sqrt [ 4 ]{ 324 } \) =\(\sqrt [ 4 ]{ 81\times 4 } =\sqrt [ 4 ]{ { 3 }^{ 4 }\times 4 } =\sqrt [ 4 ]{ { 3 }^{ 4 } } \times \sqrt [ 4 ]{ 4 } \)     \(\left[ \because \sqrt [ n ]{ a^{ n } } \times \sqrt [ n ]{ b } =\sqrt [ n ]{ ab } \right] \)
=  \(3\times \sqrt [ 4 ]{ 4 } \)                                                                     \(\left[ \because \sqrt [ n ]{ a^{ n } } =a \right] \)
order = 4; radicand = 4; Coefficient = 3

The order of the surds \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 4 ]{ 7 } \) and \(\sqrt [ 2 ]{ 6 } \) 3, 4, 2 L.C.M of 3, 4, 2 = 12
\(\sqrt [ 3 ]{ 5 } ={ 5 }^{ \frac { 1 }{ 3 } }={ 5 }^{ \frac { 4 }{ 12 } }=\left( 625 \right) ^{ \frac { 1 }{ 12 } }\)
\(\sqrt [ 4 ]{ 7 } ={ 7 }^{ \frac { 1 }{ 4 } }=7^{ \frac { 3 }{ 12 } }=\left( 343 \right) ^{ \frac { 1 }{ 12 } }\)
\(\sqrt [ 2 ]{ 6 } ={ 6 }^{ \frac { 1 }{ 2 } }=7^{ \frac { 6 }{ 12 } }=\left( 46656 \right) ^{ \frac { 1 }{ 12 } }\)
The order of the surds \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 4 ]{ 7 } ,\sqrt [ 2 ]{ 6 } \)  is \(\left( 343 \right) ^{ \frac { 1 }{ 12 } }<\left( 625 \right) ^{ \frac { 1 }{ 12 } }<\left( 46656 \right) ^{ \frac { 1 }{ 12 } }\) that is \(\sqrt [ 4 ]{ 7 } <\sqrt [ 3 ]{ 5 } <\sqrt [ 2 ]{ 6 } \)