Question:
If $\sqrt{2}=1.4142 \ldots$, then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ is equal to
(a) $2.4142 \ldots$
(b) $5.8282 \ldots$
(c) $0.4142 \ldots$
(d) $0.1718 \ldots$
Solution:
$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} \cdot \frac{\sqrt{2}-1}{\sqrt{2}-1}}$ [inside the root, multiplying numerator and denominator by $(\sqrt{2}-1)]$
$=\sqrt{\frac{(\sqrt{2}-1)^{2}}{(\sqrt{2})^{2}-(1)^{2}}}$ [using identity $(a-b)(a+b)=a^{2}-b^{2}$ ]
$=\sqrt{\frac{(\sqrt{2}-1)^{2}}{2-1}}=\sqrt{2}-1=(1.4142 \ldots)-1=0.4142 \ldots$