# How can the surd $\sqrt{2-\sqrt{3}}$ be expressed?

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

• 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 $$\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{2}}$$ Oct 5, 2016 at 11:26
• You mean $\frac{\sqrt{3}-1}{\sqrt{2}}>0$? Oct 5, 2016 at 11:34
• 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. Oct 5, 2016 at 12:32
• math.stackexchange.com/questions/196155/… Oct 5, 2016 at 17:02
• Here is the MathJax guide. Oct 7, 2016 at 1:26

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.

• 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? Nov 12, 2016 at 1:54
• @frog1944 Oh wait, there's a sign change that I added wrong... Anyways, it's now fixed. Nov 12, 2016 at 2:23
• ok, thank you very much. It makes more sense now Nov 12, 2016 at 2:50