Maths 1 Marks Question

We know that $a ^ { m } \cdot b ^ { m } = ( a \times b ) ^ { m }$.

So, $7 ^ { \frac { 1 } { 2 } } \cdot 8 ^ { \frac { 1 } { 2 } } = ( 7 \times 8 ) ^ { \frac { 1 } { 2 } }$

$= ( 56 ) ^ { \frac { 1 } { 2 } }$

Therefore, the value of $7 ^ { \frac { 1 } { 2 } } \cdot 8 ^ { \frac { 1 } { 2 } }$will be $( 56 ) ^ { \frac { 1 } { 2 } }$.

$2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=2^{\frac{2}{3}+\frac{1}{5}}$ {by $a^{m} \cdot a^{n}=a^{(m+n)}$} 
$2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=2^{\left(\frac{2 \times 5+1 \times 3}{15}\right)}$$=2^{\frac{13}{15}}$

3.142678 = $\frac{3142678}{1000000}$, and hence is a rational number.
Now, let us consider the case when the decimal expansion is non-terminating recurring.

$13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}}=(13 \times 17)^{\frac{1}{5}}=221^{\frac{1}{5}}$

$\frac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}=7^{\left(\frac{1}{5}-\frac{1}{3}\right)}=7^{\frac{3-5}{15}}=7^{\frac{-2}{15}}$

$\left(3^{\frac{1}{5}}\right)^{4}=3^{\frac{4}{5}}$

We know that  $a ^ { m } \cdot a ^ { n } = a ^ { ( m +n) }$.

So, $2 ^ { \frac { 2 } { 3 } } \cdot 2 ^ { \frac { 1 } { 3 } } = ( 2 ) ^ { \frac { 2 } { 3 } + \frac { 1 } { 3 } }$

$= ( 2 ) ^ { \frac { 2+1} { 3} }\\ = ( 2 ) ^ { \frac { 3 } { 3 } }\\=(2)^1\\=2$

Therefore, the value of  $2 ^ { \frac { 2 } { 3 } } \cdot 2 ^ { \frac { 1 } { 3 } }$ will be  $2$.

       $=\frac{2-\sqrt{3}}{4-3}$ [using a² - b² = (a+b)(a-b)]

      $=2-\sqrt{3}$

$=\frac{\sqrt{2}}{2} $     (since $\sqrt{2} \cdot \sqrt{2}$ = 2)
In this form, it is easy to locate $\frac{1}{\sqrt{2}}$ on the number line. It is halfway between 0 and $\sqrt{2}$