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\frac{\left(\sqrt{5}-1\right)\times 2}{2\left(\sqrt{5}+1\right)}
Divide \frac{\sqrt{5}-1}{2} by \frac{\sqrt{5}+1}{2} by multiplying \frac{\sqrt{5}-1}{2} by the reciprocal of \frac{\sqrt{5}+1}{2}.
\frac{\sqrt{5}-1}{\sqrt{5}+1}
Cancel out 2 in both numerator and denominator.
\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}
Rationalize the denominator of \frac{\sqrt{5}-1}{\sqrt{5}+1} by multiplying numerator and denominator by \sqrt{5}-1.
\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)}{\left(\sqrt{5}\right)^{2}-1^{2}}
Consider \left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)}{5-1}
Square \sqrt{5}. Square 1.
\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)}{4}
Subtract 1 from 5 to get 4.
\frac{\left(\sqrt{5}-1\right)^{2}}{4}
Multiply \sqrt{5}-1 and \sqrt{5}-1 to get \left(\sqrt{5}-1\right)^{2}.
\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}+1}{4}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-1\right)^{2}.
\frac{5-2\sqrt{5}+1}{4}
The square of \sqrt{5} is 5.
\frac{6-2\sqrt{5}}{4}
Add 5 and 1 to get 6.