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I was wondering how $\sqrt{2-\sqrt{3}}$ could be expressed in terms of $\frac{\sqrt{3}-1}{\sqrt{2}}$. I did try to solve both the expressions separately but none of them seemed to match. I would appreciate it if someone could also mention the procedure

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    $\begingroup$ Since $$\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)^2=\frac{3-2\sqrt{3}+1}{2}=2-\sqrt{3}\qquad\text{and}\qquad 2-\sqrt{3}>0$$ It follows that $$\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{2}}$$ $\endgroup$ – Ángel Mario Gallegos Oct 5 '16 at 11:26
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    $\begingroup$ You mean $\frac{\sqrt{3}-1}{\sqrt{2}}>0$? $\endgroup$ – Element118 Oct 5 '16 at 11:34
  • $\begingroup$ i was wondering whether we could do it the other way round i.e $\sqrt{2-\sqrt{3}}$ = $\frac{\sqrt{3}-1}{\sqrt{2}}$. Could you mention the procedure aswell. $\endgroup$ – Anamika Ghosh Oct 5 '16 at 12:32
  • $\begingroup$ math.stackexchange.com/questions/196155/… $\endgroup$ – Hans Lundmark Oct 5 '16 at 17:02
  • $\begingroup$ Here is the MathJax guide. $\endgroup$ – Simply Beautiful Art Oct 7 '16 at 1:26
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Theorem: Given a nested radical of the form $\sqrt{X\pm Y}$, it can be rewritten into the form $$\sqrt{\frac {X+\sqrt{X^2-Y^2}}{2}}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}{2}}\tag{1}$$ Where $X>Y$.


Therefore, we have $X=2,Y=\sqrt{3}$ because $2>\sqrt{3}$. So plugging that into $(1)$ gives us $$\sqrt{\frac {2+\sqrt{4-3}}{2}}-\sqrt{\frac {2-\sqrt{4-3}}{2}}\tag{2}$$ Simplifying $(2)$ gives us $$\sqrt{\frac {2+1}{2}}-\sqrt{\frac {2-1}{2}}\implies \sqrt{\frac 32}-\sqrt{\frac 12}$$

$$\therefore\sqrt{2-\sqrt{3}}=\frac {\sqrt{3}-1}{\sqrt{2}}$$


Alternatively, one can rewrite it as a sum of two surds, and simplify from there. Specifically, let $\sqrt{2-\sqrt3}$ equal $\sqrt d-\sqrt e$. Squaring, we get\begin{align*} & 2-\sqrt3=d+e-2\sqrt{de}\\ & \therefore\begin{cases}d+e=2\\de=\frac 34\end{cases}\end{align*} With solving for $d$ and $e$ gives the simplification.

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  • $\begingroup$ when you solve for d and e though, you get 1.5 and 0.5. When you sub this into the equation aren't you then saying that $\sqrt{2-\sqrt{3}}=\sqrt{1.5} + \sqrt{0.5}$? How do you get around this? $\endgroup$ – frog1944 Nov 12 '16 at 1:54
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    $\begingroup$ @frog1944 Oh wait, there's a sign change that I added wrong... Anyways, it's now fixed. $\endgroup$ – Frank Nov 12 '16 at 2:23
  • $\begingroup$ ok, thank you very much. It makes more sense now $\endgroup$ – frog1944 Nov 12 '16 at 2:50

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