How would you simplify $\dfrac{2}{\sqrt{3}}$?

Hint: In the above question, we asked to simplify $\dfrac{2}{\sqrt{3}}$. So, we will rationalize to solve this problem. In rationalization, we multiply the numerator and denominator with the denominator. Suppose we have $\dfrac{a}{\sqrt{b}}$, so for rationalizing we have to multiply the numerator and denominator with $\sqrt{b}$. So let’s see how we can solve our problem.

Step by step solution:
We have to simplify $\dfrac{2}{\sqrt{3}}$
We will rationalize the numerator and denominator by multiplying them with $\sqrt{3}$.
Also, $\dfrac{\sqrt{3}}{\sqrt{3}}$ is 1 so it would not change any value of the expression.
 $=\dfrac{2}{\sqrt{3}}.\dfrac{\sqrt{3}}{\sqrt{3}}$
 $=\dfrac{2\sqrt{3}}{\sqrt{9}}$
As we know the square root of 9 is 3, so we will get,
 $=\dfrac{2\sqrt{3}}{3}$

Therefore, $\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}$.

Additional Information:
The above problem was based on simple rationalization but there are different forms of rationalization as well. For example, $\dfrac{1}{3-\sqrt{2}}$ , we will solve this by multiplying the numerator and denominator with $\dfrac{3+\sqrt{2}}{3+\sqrt{2}}$. $3+\sqrt{2}$ is the conjugate of $3-\sqrt{2}$. There are more other types of rationalization so we will study them later.

Note:
In the above problem, we used rationalization. Rationalization is a process where a fractional part is rewritten so that the denominator of the fraction contains rational numbers only. Numbers like 4 and 5 are rational whereas numbers like $\sqrt{2}$ and $\sqrt{3}$ are irrational and we make this rational when it is in the denominator.