If $$\sqrt 2 = 1.414{\text{,}}$$   the square root of $$\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}$$   is nearest to = ?

Solution (By Examveda Team)

$$\eqalign{ & = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \cr & = \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \times \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}} \cr & = {\left( {\sqrt 2 - 1} \right)^2} \cr & \therefore \sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \cr & = \left( {\sqrt 2 - 1} \right) \cr & = \left( {1.414 - 1} \right) \cr & = 0.414 \cr} $$