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

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 3

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

Answer 4

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

Also

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

Answer 5

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 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

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.