How do you simplify $\dfrac{2}{{\sqrt 5 }}$ ?

Hint: First of all, we will have to know the significance of rationalizing the denominator which simply means to simplify the calculation. Now, doing rationalization of fraction having irrational part in denominator is as shown below:
$\dfrac{{a + \sqrt b }}{{a - \sqrt b }} \times \dfrac{{a + \sqrt b }}{{a + \sqrt b }} = \dfrac{{{a^2} + b + 2a\sqrt b }}{{{a^2} - b}}$
Here the conjugate of denominator, that is, algebraic sign has been changed between rational number and irrational part and it has been multiplied by numerator as well as denominator.

Complete step-by-step solution:
Given expression is $\dfrac{2}{{\sqrt 5 }}$ .
The conjugate of the denominator is $\sqrt 5 $ itself as there is no rational part added or subtracted in this question.
On multiplying the numerator and denominator by $\sqrt 5 $ , we get as follows: $\dfrac{2}{{\sqrt 5 }} = \dfrac{2}{{\sqrt 5 }} \times \dfrac{{\sqrt 5 }}{{\sqrt 5 }} = \dfrac{{2\sqrt 5 }}{{\sqrt 5 \times \sqrt 5 }} = \dfrac{{2\sqrt 5 }}{5}$
Because we know $\sqrt a \times \sqrt a = a$
Hence $\dfrac{2}{{\sqrt 5 }} = \dfrac{{2\sqrt 5 }}{5}$

Therefore, $\dfrac{{2\sqrt 5 }}{5}$ is the desired answer.

Note: Be careful while finding the conjugate of the denominator of a given fraction. Also take care of the calculation part because there is a chance that due to the sign or any other reason you may get the incorrect answer of the given question. Remember the way of rationalizing a fraction having irrational denominators which have already shown in above.
We need to remember a most important concept is if the denominator is of the form $(a + b)$ we take $(a - b)$ as rationalizing factor and if denominator is of the form $(a - b)$ and we take $(a + b)$ as a rationalizing factor. Thereafter, use property $(a - b)(a + b) = {a^2} - {b^2}$ .
At the end, the denominator is $1$ which is a rational number and hence our answer is correct.
We, generally, rationalize the expression to get the real number that can be easily represented on the number line, as the irrational number is not easy to find or represent on the number line.