Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$

Question

Prove the following:
$\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$

How do you go about proving this?

$a^2=b^2$ implies $a=\pm b$, but in this case, both numbers are positive. — Callus - Reinstate Monica, Feb 28 '17 at 20:54
As a general rule, note that $2^2-3=1$ is a perfect square, then see this about denesting radicals. — dxiv, Feb 28 '17 at 20:56
Perhaps not an exact duplicate, but this will definitely help: http://math.stackexchange.com/a/1586834/272831 — Simply Beautiful Art, Feb 28 '17 at 21:11
They are approximately equal and they both are roots of $x^4-4x^2+1$, hence they are truely equal. — Jack D'Aurizio, Feb 28 '17 at 21:12
I'm voting to close this question as off-topic because no effort by the user to solve before asking the question. — P Vanchinathan, Mar 01 '17 at 02:28
For a general view of square root denesting problem, this very accessible paper is interesting: http://www.cybertester.com/data/denest.pdf — zwim, Jan 02 '18 at 07:03

score 4 · Answer 1 · answered Feb 28 '17 at 21:22

Hint:

In general a double radical $\sqrt{a\pm \sqrt{b}}$ can be denested if $a^2-b$ is a perfect square and in this case we have: $$ \sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}} $$

In Your case $a^2-b=1$ so....

score 3 · Accepted Answer · edited Mar 02 '17 at 13:58

3

\begin{align}\sqrt{2-\sqrt{3}} &=\frac{\sqrt{2}\sqrt{2-\sqrt{3}}}{\sqrt{2}}\\ &=\frac {\sqrt { 4-2\sqrt{3}}}{\sqrt{2}} \\ &=\frac {\sqrt{1-2\sqrt{3} +{ \left( \sqrt{3}\right)}^{2} } }{ \sqrt {2} } \\ &=\frac{\sqrt { { \left( 1-\sqrt{3}\right)}^{2}}}{ \sqrt{2}}\\ &=\frac{\sqrt {3} -1 }{ \sqrt {2} }\\ & = \frac{\sqrt{6}-\sqrt{2}}{2} \end{align}

edited Mar 02 '17 at 13:58

lioness99a

4,943

answered Feb 28 '17 at 20:54

haqnatural

21,578

Adren · Answer 3 · 2017-03-01T03:50:33.130

Previous comments and other answers show you the path to the solution.

I would like to propose another formula of the same kind, thats illustrates the fact that cancellation of nested square roots may sometimes occur ... but not necessarily in any obvious way !

Question Find a simpler form for $x=\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}$

Answer We have :

$$x\sqrt{2}=\sqrt{4+2\sqrt{3}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(1+\sqrt{3}\right)^{2}}+\sqrt{\left(1-\sqrt{7}\right)^{2}}=1+\sqrt{3}+\sqrt{7}-1=\sqrt{3}+\sqrt{7}$$

Hence : $2x^{2}=10+2\sqrt{21}$ and finally (since $x\ge0$) :

$$\boxed{x=\sqrt{5+\sqrt{21}}}$$

A slightly more general formula could be :

For all $(p,q)\in\mathbb{N}^2$ :

$$\sqrt{p+1+\sqrt{2p+1}}+\sqrt{q+1-\sqrt{2q+1}}=\sqrt{p+q+1+\sqrt{\left(2p+1\right)\left(2q+1\right)}}$$

Note also that $\sqrt{5+\sqrt{21}} = \frac{1}{2}(\sqrt{14}+\sqrt{6}),$. In general, radicals of the form $,\sqrt{a \pm \sqrt{b}},$ "denest" nicely if $a^2-b$ is a perfect square, see Emilio Novati's answer for example. — dxiv, Feb 28 '17 at 21:21

score 1 · Answer 4 · answered Feb 28 '17 at 20:54

1

$\sqrt{2-\sqrt3}=\sqrt{\frac{4-2\sqrt3}{2}}=\frac{\sqrt3-1}{\sqrt2}$

answered Feb 28 '17 at 20:54

Michael Rozenberg

194,933

Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$

4 Answers4