Question

If $\sqrt{2}=1.4142$ then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=?$

Answer 1

$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$

By rationalizing the denominator,
$=\sqrt{\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)}\times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}}$
$=\sqrt{\frac{(\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}}$
$=\sqrt{\frac{(\sqrt{2}-1{)}^2}{(\sqrt{2}{)}^2-1^2}}$
$=\sqrt{\frac{(\sqrt{2}-1{)}^2}{2-1}}$
$=\sqrt{(\sqrt{2}-1{)}^2}$

$=\sqrt{2}-1$
$=1.4142-1$ [$\because \sqrt{2}=1.4142$]

$=0.4142$