As promised, here lie the \emph{binomial denominators}. There are a couple of prerequisites to understanding this mathematics, the first is that you must totally grasp $(a+b)(a-b) = a^2 - b^2$. The significance of this result is that if we make $a$ and $b$ surds like so: $(\sqrt{a}+\sqrt{b})$, then multiplying by $(\sqrt{a}-\sqrt{b})$ yields a fantastic result, namely: \vspace{0.5cm} \begin{equation} (\source{\sqrt{a}}+\source{\sqrt{b}})(\target{\sqrt{a}}-\target{\sqrt{b}})= (\sqrt{a})^2 - (\sqrt{b})^2\mbox{\drawarrows}\\ = a - b \end{equation} \vspace{0.5cm} This act of starting with $\sqrt{a} + \sqrt{b}$, and multiplying it with the sign flipped: $\sqrt{a}-\sqrt{b}$ is called \emph{multiplying by the conjugate}. Let's practise this method: \begin{examplebox} \subsection{Examples:} \begin{questions} \Question[1] \[\frac{2 \sqrt{5}}{2 \sqrt{5}-2}=\] \begin{solutionordottedlines}[3cm] Multiply by the conjugate: $\frac{2 \sqrt{5}}{2 \sqrt{5}-2} \cdot \frac{2 \sqrt{5}+2}{2 \sqrt{5}+2} = \frac{4 \cdot + 4 \sqrt{5}}{20 - 4} = \frac{20 + 4 \sqrt{5}}{16} = \frac{5}{4} + \sqrt{5}$ \end{solutionordottedlines} \Question[1] \[\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}}\] \begin{solutionordottedlines}[3cm] Multiply by the conjugate: $\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}} \cdot \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}} = \frac{3 \cdot 2 - 2 \cdot 3}{6 - 4 \sqrt{6} + 4 \sqrt{6} - 12} = \frac{6 - 6}{-6} = 0$ \end{solutionordottedlines} \end{questions} \end{examplebox} \begin{examplebox} \subsection{Examples:} \begin{questions} \Question[1] \[\frac{2 \sqrt{5}}{2 \sqrt{5}-2}=\] \begin{solutionordottedlines}[3cm] \end{solutionordottedlines} \Question[1] \[\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}}\] \begin{solutionordottedlines}[3cm] \end{solutionordottedlines} \end{questions} \end{examplebox}