A cambridge interview problem

$\begin{aligned} \sqrt{3 - 2 \sqrt{2}} &= \sqrt{2 + 1 - 2 \sqrt{2}} \\ &= \sqrt{\sqrt{2}^{2} + 1^{2} - 2(\sqrt{2})(1)} \\ &= \sqrt{(\sqrt{2} - 1)^{2}} \\ &= \boxed{\sqrt{2} - 1} \end{aligned}$

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

Note that of the three possible answers, only $(\sqrt{2} - 1)$ is in the ring $\mathbb{Z}[\sqrt{2}]$ so it's the only possible answer.

See: ring theory .

Generally let's take

$\sqrt{x-\sqrt{y}}$ = $\sqrt{a}-\sqrt{b}$

$x-\sqrt{y}$ = $a +b-2\sqrt{ab}$ ( Squaring both sides )

So, $x= a+b$

And $\sqrt{y}$ = $2\sqrt{ab}\cdots$ Eq (a)

So, $y = 4ab$

$(a-b)$ = $±\sqrt{(a+b)^2 -4ab}$

$(a-b)$ = $±\sqrt{x^2-y} \cdots$ *Eq (b) *

Solving two equations we will end up with

$a= \frac{x+\sqrt{x^2-y}}{2}$

And $b=\frac{x-\sqrt{x^2-y}}{2}$

So $\sqrt{x-\sqrt{y}}$ = $\sqrt{a}-\sqrt{b}$

$\sqrt{x-\sqrt{y}}$ = $\sqrt{\frac{x+\sqrt{x^2-y}}{2}}-\sqrt{\frac{x-\sqrt{x^2-y}}{2}}$

As per question $x= 3$ and $y= 8$

So $\sqrt{3-\sqrt{8}}$ = $\sqrt{2}-1$