A rational number can be written in the form of $\frac{p}{q}$ where $p$ and $q$ are integers and $q$ is not equal to zero. Rational numbers have their decimal expansions as either terminating or non-terminating but recurring. 
$\pi$ = 3.14159265358..., i.e., the digits after decimal point do not terminate and not recurring. Hence it is not rational
$\sqrt{2}$ is approximated to 1.414213562373095..., i.e., it cannot be expressed as a recurring and non-terminating decimal and hence is not rational.
$\frac{2}{0}$ is not defined and for a rational number, the denominator should be non-zero.
4.8 on the other hand, satisfies all the properties of rational numbers.

We rationalise the denominator by dividing and multiplying the number by $\sqrt{2}+\sqrt{3}$.
So, we have 

 $\frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{(\sqrt{3}+\sqrt{2}{)}^2}{{\sqrt{3}}^2-{\sqrt{2}}^2}$
$=\frac{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2+2\sqrt{6}}{3-2}$
[$\text{Using the identity}(a+b{)}^2=a^2+b^2+2ab$]
$=5+2\sqrt{6}$.

Multiplication of two rational numbers is always a rational number.
This is because every rational number can be expressed as a $\frac{p}{q}$ form. Therefore, multiplying two rational numbers would result in multiplying two numbers in their $\frac{p}{q}$ forms, with the denominators not equal to 0, in each case. This would result in another number which is also in its $\frac{p}{q}$ form, with both the numerator and denominator being integers.
The sum of two irrational numbers need not always be irrational. 
For example, consider $2+\sqrt{3}$ and $2-\sqrt{3}$, both of which are irrational numbers. Adding these gives us 4, which is a rational number.
Irrational numbers are real numbers too. Since we know that the number line constitutes all the real numbers, irrational numbers form part of it too.
The sum or difference of a rational and an irrational number is always irrational. 

We know that ${\left(a^m\right)}^n=a^{mn}$ .
${\left(256\right)}^{\frac{-1}{2}}={\left(2^8\right)}^{\frac{-1}{2}}\text{(Since}256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^8\text{)}$
$=2^{(8\times (\frac{-1}{2}\left)\right)}$
$=2^{-4}$
$=\frac{1}{16}$

Every irrational number is a real number but real numbers comprise of rational as well as irrational numbers. Moreover, any point on a number line represents a definite real number.

Natural numbers are all the counting numbers starting from 1, i.e. {1, 2, 3, ...}
Therefore, statement: '1 is smallest natural number'; is correct.
Whole numbers are all counting numbers and zero, i.e. {0, 1, 2, 3, ...}
Therefore, statement: '1 is smallest whole number'; is wrong.
Integers include the set of natural numbers, their negatives and 0; i.e. {...-3, -2, -1, 0, 1, 2, 3, ...}. Thus, -1 is the greatest negative integer, not the smallest one.
0 is included among integers.
Hence, the statement is correct.

Given: perpendicular sides of a right angled triangle are $\sqrt{6}$ and $\sqrt{3}$.

Using Pythagoras theorem,

Hypotenuse = $\sqrt{{\sqrt{6}}^2+{\sqrt{3}}^2}$

                    = $\sqrt{6+3}$

                   = $\sqrt{9}$  cm

                  =  3 cm

We know the identity:
$a^2-b^2=(a+b)(a-b)$
$(3+\sqrt{5})(3-\sqrt{5})=(3^2-(\sqrt{5}{)}^2=9-5=4$
 
Hence, $(3+\sqrt{5})(3-\sqrt{5})=4$, which is a rational number.

To convert 0.45 into a fraction:
$0.45=\frac{45}{100}$ 
[Multiplying 0.45 with $\frac{100}{100}$]
Dividing the numerator and the denominator by 5, we get
$\frac{45}{100}=\frac{9}{20}$ 

$\therefore 0.45=\frac{9}{20}$