Real Number System

The product of a rational and an irrational number is irrational. And Difference of rational and irrational is irrational. Therefore,  $3\sqrt 2 -1$ is an irrational number.

If 'a' is an irrational number, then $\frac{a}{a} = \frac{1}{1}$, which is a rational number.

Irrational numbers have non-terminating, non-repeating decimal expansion. The decimal expansion is 0.132013200132000132..…, which is non-terminating and non-repeating.

$\frac{\sqrt 5 \times \sqrt 2}{\sqrt {10}} = \frac{\sqrt{10}}{\sqrt{10}}$= 1, so it is a rational expression.

$\sqrt 6 \times \sqrt 3 = \sqrt {18}$= 3$\sqrt 2$ is an irrational expression.

If the decimal expansion of a number is non-terminating and non-repeating, then it is an irrational number. So, 0.1234569873215... is non-terminating and non-repeating, and hence, it is an irrational number. 

$\frac{- \sqrt {2}}{\sqrt {2}} = -1$is not an irrational number.

$\sqrt 2 \times \sqrt 8 = \sqrt {16} = 4$is an integer and all the integers are rational numbers. So, $\sqrt 2 \times \sqrt 8$ is a rational number as well as an integer.

$\dfrac{-18}{5} = -3.6 $
Therefore, it lies between $-3$ and $-4.$

$0.43 = \dfrac {43}{100}$ (As it is expressed a fraction.)
Therefore, $A$ is the correct answer.

$\displaystyle 2.\overline{8768}=2+0.\overline{8768}$
= $\displaystyle 2+\frac{8768-8}{9990}=2+\frac{8760}{9990}$
= $\displaystyle 2+\frac{292}{333}=2\frac{292}{333}$

Infinite number of rational numbers exist between any two distinct rational numbers. We know that a rational number is a number which can be written in the form of $\frac { p }{ q } $ where p and q are integers and q $\neq $0.

Between any two real number, there is always many rational number.

$6^{2} = 36$

$2^{4} =16$