(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).

(i) \({7\over 128}={7\over 2^7}\)

\(\therefore \frac{7}{128}\) has terminating decimal expansion.
(ii) \({21\over 15}={7\over 5}={7\over 5^1}\)
\(\therefore {21\over 15}\) has terminating decimal expansion.
(iii) \(4{9\over 35}={149\over 35}\)
  \(={149\over 5\times 7}\) (it is not in the form of \(\frac{P}{2^m\times 5^n}\))
\(\therefore 4\frac{9}{35}\) has a non-terminating recurring decimal expansion.
(iv) \({219\over 2200}={219\over 2^3\times 5^2\times 11}\) (It is not it the form of \({P\over 2^m\times 5^n}\))

\(\therefore {219\over 2200}\) has a non-terminating recurring decimal expansion.

(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 5.3 and 5.4. 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 form the right of 5.340.
(ii) \(6.\overline{4}\) upto 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 6.44 and 6.45. 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.

(i) \(2\over 7\) = 0.2857142....
= \(0.\overline {285714}\)

Non-terminating and recurring decimal expansion.
(ii) \(-5\frac { 3 }{ 11 } \) = -5.272....
= \(-5.\overline { 27 } \)   

Non-terminating and recurring decimal expansion.
(iii) \(\frac { 22 }{ 3 } \) = 7.333...
= \(7.\overline { 3 } \)    

Non-terminating and recurring decimal expansion.
(vi) \(\frac{327}{200}=\frac{327}{2\times 100}\)
= \(\frac{3.27}{2}\)
= 1.635

Terminating decimal expansion.

(i) Yes \( \sqrt{3}+\sqrt{3} =2 \sqrt{3} =\sqrt{2^{2} \times 3} =\sqrt{12} \) pure sured
(ii) Yes \(2 \sqrt{5}-\sqrt{5}=\sqrt{5}\) pure sured
(iii) Yes \(\sqrt {5}\times\sqrt {7}=\sqrt {35}\), a pure sured
(iv) Yes, the quotient of two surds can be a pure surd. \(\frac{\sqrt{5}}{\sqrt{2}}=\sqrt{\frac{5}{2}}\)

(i) \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 9 ]{ 4 } ,\sqrt [ 6 ]{ 3 } \)
\(5^{\frac{1}{3}}\)
ஃ The order of the surds \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 9 ]{ 4 } ,\sqrt [ 6 ]{ 3 } \) are 3,9,6
\(4^{\frac{1}{9}}\)
\(3^{\frac{1}{6}}\) l.c.m of 3,9,6 is 18
ஃ \(\frac{1}{3}=\frac{1\times6}{3\times6}=\frac{6}{18}\)
\(\frac{1}{9}=\frac{1\times2}{9\times2}=\frac{2}{18};\frac{1}{6}=\frac{1\times3}{6\times3}=\frac{3}{18}\)
\((5^{\frac{1}{3}})=5^{\frac{6}{18}}=(15625)^{\frac{1}{18}}\)
\((4^{\frac{1}{9}})=4^{\frac{2}{18}}=(4^2)^{\frac{1}{18}}=16^{\frac{1}{18}}\)
\((3^{\frac{1}{6}})=3^{\frac{3}{18}}=(3^3)^{\frac{1}{18}}=27^{\frac{1}{18}}\)
ஃ The descending order of \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 9 ]{ 4 } ,\sqrt [ 6 ]{ 3 } \) is \((15625)^{\frac{1}{18}}>(27)^{\frac{1}{18}}>16^{\frac{1}{18}}\) i.e., \(\sqrt { \sqrt { 3 } } >\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } >\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } \)
(ii) \(\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } ,\sqrt { \sqrt { 3 } } \)
The order of the surds \(\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } ,\sqrt[2] { \sqrt [2]{ 3 } } \)  are 6, 12, 4
l.c.m of 6,12,4 is 12
\(\sqrt[2]{\sqrt[3]{5}}=5^{\frac{1}{6}}=5^{\frac{1\times2}{6\times2}}=5^{\frac{2}{12}}=(5^2)^{\frac{1}{12}}=25^{\frac{1}{12}}\)
\(\sqrt[3]{\sqrt[4]{\sqrt{7}}}=7^{\frac{1}{12}};\sqrt{\sqrt{3}}=3^{\frac{1}{4}}=3^{\frac{1\times3}{4\times3}}=3^{\frac{3}{12}}=(3^3)^{\frac{1}{12}}=27^{\frac{1}{12}}\)
ஃ The ascending order of the surds
\(\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } ,\sqrt { \sqrt { 3 } } \) is \(7^{\frac{1}{12}}<25^{\frac{1}{2}}<27^{\frac{1}{2}}\), that is \(\sqrt[3]{\sqrt[4]{7}}<\sqrt[2]{\sqrt[3]{5}}<\sqrt{\sqrt{3}}\)

(i) \(\sqrt{3}\times\sqrt{5}\times\sqrt{2}=\sqrt{3\times5\times2}=\sqrt{30}\)
(ii) \(\sqrt{35}\div\sqrt{7}=\sqrt{\frac{35}{7}}=\sqrt{5}\)
(iii) \(\sqrt[3]{27}\times\sqrt[3]{8}\times\sqrt[3]{125}=\sqrt[3]{27\times8\times125}=\sqrt[3]{3^3\times2^3\times5^3}=3\times2\times5=30\)
(iv) \((7\sqrt{a}-5\sqrt{b})(7\sqrt{a}+5\sqrt{b})=(7\sqrt{a})^2-(5\sqrt{b})^2=49a-25b\) = 21\(\sqrt 3\)
(v) \([\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}]\div\sqrt{\frac{16}{81}}\)

=\([\sqrt{\frac{15^2}{27^2}}-\sqrt{\frac{5^2}{12^2}}]\times\sqrt{\frac{9^2}{4^2}}\)
=\((\frac{15}{27}-\frac{5}{12})\times\frac{9}{4}=(\frac{5}{9}-\frac{5}{12})\times\frac{9}{4}\)
=\((\frac{20-15}{36})\times\frac{9}{4}=\frac{5}{ ̶3̶6̶}\times\frac{ ̶9̶}{4}=\frac{5}{16}\)

\(\sqrt[5]{100000}=(100000)^{\frac{1}{5}}=(10^{ ̶5̶})^{\frac{1}{ ̶5̶}}=10\)

\(\sqrt[5]{243}=243^{\frac{1}{5}}=(3^5)^{\frac{1}{5}}=3^{ ̶5̶\times\frac{1}{ ̶5̶}}=3\)

\(\sqrt[5]{32}=32^{\frac{1}{5}}=(2^5)^{\frac{1}{5}}=2^{ ̶5̶\times\frac{1}{ ̶5̶}}=2\)