(i) 
The decimal point is to be moved six places to the left. Therefore n = 6.
(ii) 
The decimal point is to be moved two places to the right. Therefore n = −2
(iii) 
The decimal point is to be moved seven places to the left. Therefore n = 7

(i) 3\(\sqrt{7}\) + 5\(\sqrt{7}\) = (3 + 5)\(\sqrt{7}\) = 8\(\sqrt{7}\). The answer is irrational.
(ii) 7\(\sqrt{5}\) - 4\(\sqrt{5}\) = (7 - 4)\(\sqrt{5}\) = 3\(\sqrt{5}\). The answer is irrational.

\(\sqrt [ 3 ]{ (1024)^{ -2 } } =\left[ \sqrt [ 3 ]{ ({ 2 }^{ 3 }\times { 2 }^{ 3 }\times { 2 }^{ 3 }\times 2 } )^{ -2 } \right] \)
=\(\left[ \sqrt [ 3 ]{ ({ 2 }^{ 3 }\times { 2 }^{ 3 }\times { 2 }^{ 3 }\times 2 } )^{ -2 } \right] \)  [Laws of radicals - (i)]
=\(\left[ \sqrt [ 3 ]{ 2^{ 3 } } \times \sqrt [ 3 ]{ { 2 }^{ 3 } } \times \sqrt [ 3 ]{ { 2 }^{ 3 } } \times \sqrt [ 3 ]{ { 2 } } \right] ^{ -2 }\) [Laws of radicals – (ii)]
= \(\left[ 2\times 2\times 2\times \sqrt [ 3 ]{ 2 } \right] ^{ -2 }\) [Laws of radicals – (i)]
=\(\left[ 8\times \sqrt [ 3 ]{ 2 } \right] ^{ -2 }=\left[ \frac { 1 }{ 8 } \right] ^{ 2 }\times \left( \frac { 1 }{ \sqrt [ 3 ]{ 2 } } \right) ^{ 2 }\)
=\(\frac { 1 }{ 64 } \sqrt [ 3 ]{ \frac { 1 }{ 4 } } \)
order = 3 ; radicand = \(\frac{1}{4}\); coefficient = \(\frac{1}{64}\)
(These results can also be obtained using index notation).

(i) \( \sqrt{3} =3^{\frac{1}{2}} =3^{\frac{6}{12}} =\sqrt[12]{3^{6}} =\sqrt[12]{729} \)
(ii) \( \sqrt[4]{3} =3^{\frac{1}{4}} =3^{\frac{3}{12}} =\sqrt[12]{3^{3}} =\sqrt[12]{27} \)
(iii) \( \sqrt[3]{3} =3^{\frac{1}{3}} =3^{\frac{4}{12}} =\sqrt[12]{3^{4}} =\sqrt[12]{81} \)
The last row has surds of same order.

(i) The world population is nearly
7000,000,000 = 7.0 \(\times\) 10⁹
(ii) One light year means the distance
9460528400000000 km = 9.4605284 \(\times\) 10¹⁵ km
(iii) Mass of an electron is
0.000 000 000 000 000 000 000 000 000 00091093822 kg = 9.1093822 \(\times\) 10^-31 kg

(i) \( (300000)^{2} \times(20000)^{4}\)
\(=\left(3.0 \times 10^{5}\right)^{2} \times\left(2.0 \times 10^{4}\right)^{4} \)
\(=3^{2} \times 10^{10} \times 2^{4} \times 10^{16} \)
\(=9 \times 16 \times 10^{10+16} \)
\(=144 \times 10^{26} \)
\(=1.44 \times 10^{28} \)
(ii) (0.000001)¹¹ ÷ (0.005)³
\( =\left(1.0 \times 10^{-6}\right)^{11} \div\left(5.0 \times 10^{-3}\right)^{3} \)
\( =\frac{1.0 \times 10^{-66}}{125.0 \times 10^{9}} \)
\( =\frac{1000.0 \times 10^{-69}}{125.0 \times 10^{-9}} \)
\( =8.0 \times 10^{-69} \times 10^{9} \)
\( =8.0 \times 10^{-60} \)
(iii) \(\left\{(0.00003)^{6} \times(0.00005)^{4}\right\} \div\left\{(0.009)^{3} \times(0.05)^{2}\right\}\)
=\(\frac{(3.0\times10^{-5})^6\times(5.0\times10^{-5})^4}{(9.0\times10^{-3})^3\times(5.0\times10^{-2})^2}\)
=\(\frac{3^6\times10^{-30}\times5^4\times10^{20}}{9^3\times10^{-9}\times5^{2}\times10^{-4}}=\frac{3^6\times5^4\times10^{-30-20}}{(3^{2})^3\times10^{-9-4}\times5^{2}}=\frac{ ̶3̶^6̶\times5^4\times10^{-50}}{ ̶3̶^6̶\times5^2\times10^{-13}}\)
\( =5^{4-2} \times 10^{-50+13} \)
\(=5^{2} \times 10^{-37} \)
\(=25 \times 10^{-37} \)
\(=2.5 \times 10^{1} \times 10^{-37}\)
\(=2.5 \times 10^{-36} \)

\(\frac { \sqrt { 5 } }{ \sqrt { 6 } +2 } -\frac { \sqrt { 5 } }{ \sqrt { 6 } -2 } =\frac{\sqrt{5}(\sqrt{6}-2)-\sqrt{5}(\sqrt{6}+2)}{(\sqrt{6}+2)(\sqrt{6}-2)}\)
\(\frac{ ̶​̶​̶​̶​̶​̶​̶​̶√̶3̶0̶-2\sqrt{5}- ̶​̶​̶​̶​̶​̶​̶​̶√̶3̶0̶-2\sqrt{5}}{\sqrt{6}^2-2^2}=\frac{-4\sqrt{5}}{6-4}=\frac{ ̶4̶\sqrt{5}}{ ̶2̶}=-2\sqrt{5}\)

\( =\frac{\sqrt{3 \times 16}+\sqrt{2 \times 16}}{\sqrt{3 \times 9}-\sqrt{2 \times 9}} \)
\( =\frac{\sqrt{3 \times 4^{2}}+\sqrt{2 \times 4^{2}}}{\sqrt{3 \times 3^{2}}-\sqrt{2 \times 3^{2}}} \)
\( =\frac{4 \sqrt{3}+4 \sqrt{2}}{3 \sqrt{3}-3 \sqrt{2}} \)
\( =\frac{4 \sqrt{3}+\sqrt{2}}{3 \sqrt{3}-\sqrt{2}}\)
\( =\frac{4(\sqrt{3}+\sqrt{2})}{3(\sqrt{3}-\sqrt{2})} \)
\( =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
\( =\frac{4}{3}\left(\frac{(\sqrt{3}+\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}\right)\)
\( =\frac{4}{3}\left(\frac{3+2 \sqrt{6}+2}{3-2}\right) \)
\( =\frac{4}{3}\left(\frac{5+2 \sqrt{6}}{1}\right)=\frac{4}{3}(5+2 \sqrt{6}) \)

(i) Yes, the sum of two surds is a rational number.
Example \( 3 \sqrt{4}+2 \sqrt{4} =5 \sqrt{4} =\sqrt{5^{2} \times 4}=\sqrt{25 \times 4} =\sqrt{100}=10 \) rational number.
(ii) ) Yes, the difference of two surds is a rational number
Example  \(2 \sqrt{9}-\sqrt{9}=\sqrt{9}=3\), a rational number
(iii) ) Yes, the product of two surds is a rational number.
Example \(\sqrt{2} \times \sqrt{2}=\sqrt{4}=2\), a rational number
(iv) ) Yes, the quotient of two surds is a rational number.
 \(\sqrt{75} \div \sqrt{3}=\sqrt{\frac{75}{3}}=\sqrt{25}=5 \), a rational number

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