Algebra trick clarification question I have the equation $x = \sqrt{3} + \sqrt{2} \,\,$ and the trick is that $\frac{1}{x} = \sqrt{3} - \sqrt{2}$. I don't see how the minus sign appears just by taking 1 over x?

If we are told that $x = \sqrt{3}+\sqrt{2}$ then it follows that $\dfrac{1}{x} = \dfrac{1}{\sqrt{3}+\sqrt{2}}$
$=\dfrac{1}{\sqrt{3}+\sqrt{2}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} = \dfrac{\sqrt{3}-\sqrt{2}}{3-2}$
$=\sqrt{3}-\sqrt{2}$
This same pattern can be extended to if $x=\sqrt{n+1}+\sqrt{n}$ then $\dfrac{1}{x}=\sqrt{n+1}-\sqrt{n}$
