Question

Write the rationalising factor of the denominator in $\frac{1}{\sqrt{2}+\sqrt{3}}$.

Answer 1

$\frac{1}{\sqrt{2}+\sqrt{3}}$
$$\begin{gathered} =\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{\sqrt{3}-\sqrt{2}}{3-2} \\ =\frac{\sqrt{3}-\sqrt{2}}{1} \end{gathered}$$
Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in $\frac{1}{\sqrt{2}+\sqrt{3}}$ is $\sqrt{3}-\sqrt{2}$.