# How do you rationalize (sqrt3-1) /(sqrt3+1)?

Multiply and divide by $\sqrt{3} - 1$ and get:
$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = {\left(\sqrt{3} - 1\right)}^{2} / \left(3 - 1\right) = {\left(\sqrt{3} - 1\right)}^{2} / 2 = \frac{3 - 2 \sqrt{3} + 1}{2} = \frac{4 - 2 \sqrt{3}}{2} = 2 \frac{2 - \sqrt{3}}{2} = 2 - \sqrt{3}$
$\frac{s q r 3 - 1}{s q r 3 + 1} . \frac{s q r 3 - 1}{s q r 3 - 1}$= ${\left(s q r 3 - 1\right)}^{2} / 2$