Simplify $\frac{1}{\sqrt{-1+\sqrt{2}}}$ In an exercise I got as solution $\frac{1}{\sqrt{-1+\sqrt{2}}}$, it now holds that $\frac{1}{\sqrt{-1+\sqrt{2}}} = \sqrt{1+\sqrt{2}}$. But I really don't see how you can manipulate the left hand side to become the right hand side (quite shameful as a mathematician, I must admit). 
Does anyone know how to do it?
 A: Multiply the numerator and denominator of the left by $\sqrt{1+\sqrt{2}}$ to get the right.
A: $\frac{1}{\sqrt{\sqrt{2} - 1}} \cdot \frac{\sqrt{\sqrt{2} + 1}}{\sqrt{\sqrt{2} + 1}}$
= $\frac{\sqrt{\sqrt{2} + 1}}{\sqrt{(\sqrt{2})^2 - 1}}$
= $\frac{\sqrt{\sqrt{2} + 1}}{\sqrt{2 - 1}}$
= $\sqrt{\sqrt{2} + 1}$
= $\sqrt{1 + \sqrt{2}}$
A: Here is another take.
Let $\alpha=\sqrt{-1+\sqrt{2}}$.
Then $\alpha^2=-1+\sqrt2$ and $(\alpha^2+1)^2=2$, that is, $\alpha^4+2\alpha^2=1$.
Therefore $\alpha^2(\alpha^2+2)=1$ and so $\dfrac1{\alpha^2}=\alpha^2+2=1+\sqrt2$. This implies that $\dfrac1{\alpha}=\sqrt{1+\sqrt2}$.
A: You are given the fraction: $$\frac{1}{\sqrt{-1 + \sqrt{2}}} \tag1$$
Square it: $$\bigg(\frac{1}{\sqrt{-1 + \sqrt{2}}}\bigg)^2 = \frac{1^2}{\sqrt{-1 + \sqrt{2}}^2} = \frac{1}{-1 + \sqrt{2}} \tag2$$
Multiply by $1$: $$\frac{1}{-1 + \sqrt{2}}\times 1 = \frac{1}{-1 + \sqrt{2}}\times \frac{-1 - \sqrt{2}}{-1 - \sqrt{2}} = \frac{-1 - \sqrt{2}}{-1} \tag3$$
Now simplify (divide both terms in the numerator by $-1$): $$\frac{-1}{-1} - \frac{\sqrt{2}}{-1} = 1 - (-\sqrt{2}) = 1 + \sqrt{2} \tag4$$ And now finally, to cancel out the first square that we did in $(2)$, square root the $RHS$: $$\sqrt{1 + \sqrt{2}} \tag5$$
And there is you answer. You should end up with the following equation below: $$\frac{1}{\sqrt{-1 + \sqrt{2}}} = \sqrt{1 + \sqrt{2}} \tag{$\checkmark$}$$

EDIT:
The reason we make $1 = (-1 - \sqrt{2})/(-1 - \sqrt{2})$ is because $-1 - \sqrt{2}$ is the conjugate of $-1 + \sqrt{2}$. So pursuant to what @GeorgeN.Missailidis said, we did not need to square the fraction first, because all we could do was multiply it by $(\sqrt{-1 - \sqrt{2}})/(\sqrt{-1 - \sqrt{2}})$ since $\sqrt{-1 - \sqrt{2}}$ is the conjugate of $\sqrt{-1 + \sqrt{2}}$ anyway, but it is neater to cancel out the major roots.
