Question

If both a and b are rational numbers, find the value of a and b in the following equality :

$\frac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}$

Answer 1

$\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{(\sqrt{3}-1{)}^2}{2}=2-\sqrt{3}$

Therefore,

$2-\sqrt{3}=a+b\sqrt{3}$

$a=2$ and $b=-1$