$$\frac { \sqrt { 2 } \cdot \sqrt { 2 } } { \sqrt { 3 } + 1 }$$
$\sqrt{3}-1\approx 0.732050808$
$$\frac{2}{\sqrt{3}+1}$$
$$\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}$$
$$\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^{2}-1^{2}}$$
$$\frac{2\left(\sqrt{3}-1\right)}{3-1}$$
$$\frac{2\left(\sqrt{3}-1\right)}{2}$$
$$\sqrt{3}-1$$
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$\sqrt{3} - 1 = 0.732050808$
