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

How do you go about proving this?

  • 2
    Have you tried squaring both sides? – 2017-02-28
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    $a^2=b^2$ implies $a=\pm b$, but in this case, both numbers are positive. – 2017-02-28
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    As a general rule, note that $2^2-3=1$ is a perfect square, then see [this](http://math.stackexchange.com/a/2136920/291201) about denesting radicals. – 2017-02-28
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    Perhaps not an exact duplicate, but this will definitely help: http://math.stackexchange.com/a/1586834/272831 – 2017-02-28
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    They are approximately equal and they both are roots of $x^4-4x^2+1$, hence they are truely equal. – 2017-02-28
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    I'm voting to close this question as off-topic because no effort by the user to solve before asking the question. – 2017-03-01
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    For a general view of square root denesting problem, this very accessible paper is interesting: http://www.cybertester.com/data/denest.pdf – 2018-01-02

4 Answers 4

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\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}

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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....

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

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

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    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. – 2017-02-28
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$\sqrt{2-\sqrt3}=\sqrt{\frac{4-2\sqrt3}{2}}=\frac{\sqrt3-1}{\sqrt2}$