How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides

Question

I have been asked to prove:

$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$

Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.

However I know you can't square a side during proof because it generates an extraneous solution. So: how do you go about this proof without squaring both sides? Or can my method be made valid if I do this: $$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$ $$...=...$$ $$2=2$$ $$\lvert\sqrt2\rvert=\lvert\sqrt2\rvert$$ $$\sqrt2=\sqrt2\text{ hence proved.}$$ Cheers in advance :)

Answer 1

\begin{eqnarray}\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &=& \sqrt{4+2\sqrt3 \over 2}-\sqrt{4-2\sqrt3 \over 2}\\ &=&\sqrt{(\sqrt{3} +1)^2 \over 2}-\sqrt{(\sqrt{3} -1)^2\over 2}\\ &=&{\sqrt{3} +1 \over \sqrt{2}}-{\sqrt{3} -1\over \sqrt{2}}\\ &=&\sqrt2 \end{eqnarray}

Answer 2

Let $a=\sqrt{2+\sqrt3}\,$, $b = \sqrt{2-\sqrt3}\,$, then:

$$\require{cancel} a^2+b^2 = 2+\bcancel{\sqrt{3}}+2 - \bcancel{\sqrt{3}} = 4 \\ ab = \sqrt{(2+\sqrt3)(2-\sqrt3)} = \sqrt{2^2 - (\sqrt{3})^2} = \sqrt{4-3} = \sqrt{1} = 1 $$

It follows that: $$(a-b)^2 = a^2+b^2-2ab = 4 - 2 \cdot 1 = 2$$

Since $\sqrt{2+\sqrt3} \gt \sqrt{2-\sqrt3}\,$, $a-b \gt 0$ must be the positive root, so $a-b=\sqrt{2}\,$.

Answer 3

First of all we're not trying to find a solution of the equation here, what you are suggesting is to prove that $\mathrm{lhs} =\sqrt2 $ To do so we square the lhs (first read it fully) and we get $2$. So lhs would be $\sqrt2$ or $-\sqrt2$.

Now we observe the fact that lhs was positive initially ( as $ 2+\sqrt3 > 2-\sqrt3 $) hence lhs would take the positive value ie. $ +\sqrt2$, which is equal to rhs.

So I think it can be solved by observation and easy maths.

Answer 4

This one is a little bit round about.

$\sin \frac {\pi}{12} = \sin (\frac {\pi}{3} - \frac {\pi}{4})$ by angle addition rules

and

$\sin \frac {\pi}{12} = \sqrt {\frac {1-\cos \frac {\pi}{6}}{2}}$ by the half angle rules.

$\sin (\frac {\pi}{3} - \frac {\pi}{4}) = \frac {\sqrt 6 - \sqrt 2}{4}$

$\sqrt {\frac {1-\cos \frac {\pi}{6}}{2}} = \frac {\sqrt {2-\sqrt 3}}{2}$

$\frac {\sqrt 6 - \sqrt 2}{4} = \frac {\sqrt {2-\sqrt 3}}{2}$

similarly

$\cos \frac {\pi}{12} = \cos (\frac {\pi}{3} - \frac {\pi}{4}) = \sqrt {\frac {1+\cos \frac {\pi}{6}}{2}}\\ \frac {\sqrt 6 + \sqrt 2}{4} = \frac {\sqrt {2+\sqrt 3}}{2}$

$2\cos \frac{\pi}{12} - 2\sin \frac {\pi}{12} = \frac {\sqrt 6 + \sqrt 2}{2} - \frac {\sqrt 6 - \sqrt 2}{2} = \sqrt 2 = \sqrt {2+\sqrt 3} - \sqrt {2-\sqrt 3}$

Answer 5

$$a=\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\\a^2=2+\sqrt3 +2-\sqrt3 +2\sqrt{2+\sqrt3}\times\sqrt{2-\sqrt3}\\a^2=2+2-2\sqrt{4-3}\\a^2=2\\a>0\\a=\sqrt2$$

Answer 6

This is how I would write it

$$\begin{align} \left(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\right)^2 &= 2 + \sqrt3 + 2 - \sqrt3 - 2 \sqrt{2+\sqrt3}\sqrt{2-\sqrt3}\\ &= 4 - 2\sqrt{2^2-3}\\ &= 2 \end{align}$$ Hence, since $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}⩾0$ ($\sqrt⋅$ being increasing), it follows from the definition of the square root that $$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$

The key point here is to remember that the square root of $2$ is by definition¹ the only positive real number $x$ such that $x²=2$.

Also, please don't do the “write equivalent equalities and arrive at something trivially true” thing. Ever. It is never better than directly chaining $=$ and can backfire in interesting ways if one of your $\Leftrightarrow$s is really a $\Rightarrow$.

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_{1. According to my favourite teacher, 99% of maths just follows from definitions, and the other 1% takes your sanity away.}

Answer 7

$$\left(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\right)^2 \\\\ =\ (2+\sqrt3)+(2-\sqrt3)-2\sqrt{(2+\sqrt3)(2-\sqrt3)} \\\\ =2$$

Also

$$\sqrt{\sqrt2+\sqrt3}-\sqrt{\sqrt2-\sqrt3}>0$$

Answer 8

Since $\sqrt{2+\sqrt3}\cdot\sqrt{2-\sqrt3} = 1$ we may write the $\text{LHS}$ as $$\sqrt{2+\sqrt3}\cdot\sqrt{2-\sqrt3}\cdot\left(\sqrt{2+\sqrt 3}-\sqrt{2-\sqrt 3}\right)$$ $$= \left(2+\sqrt 3\right)\sqrt{2-\sqrt3} - \left(2-\sqrt3\right)\sqrt{2+\sqrt3}$$

$$=-2\left(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\right) + \sqrt3\left(\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}\right)$$

Multiplying and dividing the second term with the original LHS we have

$$-2x + \sqrt 3\cdot{2+ \sqrt 3 - 2 +\sqrt 3 \over x}$$

$$\implies\frac 6x - 2x = x$$ $$\implies\frac 6x = 3x$$ $$\implies x = \sqrt2$$

Answer 9

$$$$

\begin{align} \sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &= x \tag{A}\\ \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)\bigg(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\bigg) &= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\ (2+\sqrt 3)-(2-\sqrt 3) &= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\ \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x &= 2\sqrt 3 \\ \sqrt{2+\sqrt3} + \sqrt{2-\sqrt3} &= \dfrac{2\sqrt 3}{x} \tag{B}\\ 2\sqrt{2+\sqrt 3} &= \dfrac{2\sqrt 3}{x} + x \tag{B+A}\\ \sqrt{2+\sqrt 3} &= \dfrac{\sqrt 3}{x} + \dfrac x2 \tag{C}\\ 2\sqrt{2-\sqrt 3} &= \dfrac{2\sqrt 3}{x} - x \tag{B-A}\\ \sqrt{2-\sqrt 3} &= \dfrac{\sqrt 3}{x} - \dfrac x2 \tag{D}\\ 1 &= \dfrac{3}{x^2} - \dfrac{x^2}{4} \tag{CD} \\ 4x^2 &= 12 - x^4 \\ x^4 + 4x^2 - 12 &= 0 \\ (x^2 - 2)(x^2 + 6) &= 0 \\ x &= \sqrt 2 \\ \sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &= \sqrt 2 \end{align}

Answer 10

To be proven:

$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$

Denote: $a^2=2+\sqrt3,b^2=2-\sqrt3$. Then: $$a^2+b^2=4\\ a^2b^2=1 \Rightarrow ab=1\\ a^2+b^2+2ab=4+2=6 \Rightarrow a+b=\sqrt6\\ a^2-b^2=2\sqrt3 \Rightarrow (a-b)(a+b)=2\sqrt3 \Rightarrow a-b=\frac{2\sqrt3}{\sqrt6}=\sqrt2.$$

Answer 11

You can square both sides in a proof if you note the extraneous solutions are added.

Example $\sqrt{2 + \sqrt{3}} -\sqrt{2 - \sqrt{3}} = k$.

First $2 > \sqrt 3$ so $\sqrt{2 - \sqrt{3}} > 0$ so $k > 0$. !!!TAKE NOTE OF THAT!!!

$(\sqrt{2 + \sqrt{3}} -\sqrt{2 - \sqrt{3}})^2 = k^2$

$2 + \sqrt3 + 2-\sqrt 3 - 2(\sqrt{2 + \sqrt{3}}\sqrt{2 - \sqrt{3}}) = k^2$

$-2(\sqrt{2 + \sqrt{3}}\sqrt{2 - \sqrt{3}}) = k^2 - 4$

$\sqrt {4 -3} = \frac {4-k^2}{2}$

$4-3 = (\frac {4-k^2}{2})^2$ !!!NOTE!!! $\frac {4-k^2}{2} \ge 0$.

$1 = (\frac {4-k^2}{2})^2$ so

$\frac {4-k^2}{2} = \pm 1$. !!!BUT we took note that $\frac {4-k^2}{2}\ge 0$.!!!

So $\frac {4-k^2}{2} = 1$

So $4-k^2 = 2$

and $k^2 = 2$ so

$k = \pm \sqrt 2$ !!!But we took note that $k>0$ so $k = \sqrt 2$.

That is valid.

Answer 12

Actually, $(\sqrt6\pm\sqrt2)^2=4(2\pm\sqrt3)$ and so $2\sqrt{2\pm\sqrt3}=\sqrt6\pm\sqrt2$. Thus $$ 2(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}) = (\sqrt6+\sqrt2) - (\sqrt6-\sqrt2) = 2\sqrt2 $$ and so $$ \sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} = \sqrt2 $$