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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 :)

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    $\begingroup$ "However I know you can't square a side during proof..." That's not true. It's true that it may happen that you can generate more solutions. If that were the case, you simply have to check that these are not really solutions. The crucial part is to pay attention to always make "iff" moves. $\endgroup$ – Kezer Dec 17 '17 at 16:58
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    $\begingroup$ Since $x^2$ is injective from $[0,\infty)$ to $[0,\infty)$ and $\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt3}>0,\ \sqrt{2}>0$ this means that squaring preserves the equality, i.e if $a,b>0$ then $a=b\iff a^2=b^2$. $\endgroup$ – kingW3 Dec 17 '17 at 17:00
  • $\begingroup$ Simplifying Square Roots of Square Roots by Denesting. $\endgroup$ – BPP Dec 18 '17 at 17:00
  • $\begingroup$ Out of curiosity, what is the "index form" you are mentioning to? $\endgroup$ – Taladris Dec 18 '17 at 17:16
  • $\begingroup$ "you can't square a side during proof". That's not true. You can square during a proof, as soon as you know the sign (positive or negative) of the things you squared. $\endgroup$ – Taladris Dec 18 '17 at 17:17
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\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}

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    $\begingroup$ Is this approach generalizable? That is, it is always possible to turn $a\pm \sqrt b,\,\,\,a \in \mathbb{Z}, b \in \mathbb{Z^+}$ into a useful expression of the form $(x+y)^2$? $\endgroup$ – Tiwa Aina Dec 17 '17 at 21:33
  • $\begingroup$ I spy an exponent in there. $\endgroup$ – DonielF Dec 18 '17 at 0:19
  • $\begingroup$ @DonielF Not one that generated any additional solutions. Demonstrating $a^2 = b^2$ does not show that $a=b$, but showing the converse $a=b$ indeed implies $a^2=b^2$ $\endgroup$ – Anti Earth Dec 18 '17 at 16:56
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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 $a-b=\sqrt{2}\,$.

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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.

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  • $\begingroup$ So is there a way to write "taking the positive value of √2" using symbols or something? Or would you literally jsut write that? $\endgroup$ – Adam Bromiley Dec 17 '17 at 17:01
  • $\begingroup$ We see that in the previous answer, sir has (without talking about it) written the positive answers ie. √3+1 & √3-1 which I mentioned that I am taking the positive value. $\endgroup$ – Sri Krishna Sahoo Dec 17 '17 at 17:04
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    $\begingroup$ @Khrishn Not quite. Since $\;\sqrt3\pm1>0\;$ what John wrote $\;\sqrt{(\sqrt3\pm1)^2}=\sqrt3\pm1\;$ is correct. $\endgroup$ – DonAntonio Dec 17 '17 at 18:15
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    $\begingroup$ You do not have to put everything into symbols! There is nothing wrong with writing "taking the positive value of √2" in words, exactly as you said in your previous comment. $\endgroup$ – David Dec 17 '17 at 23:02
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    $\begingroup$ $\sqrt2$ is the positive value of $\pm\sqrt2.$ $\endgroup$ – David K Dec 18 '17 at 3:26
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$$a=\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\\a^2=2+\sqrt3 +2-\sqrt3 +2\sqrt{2+\sqrt3}\times\sqrt{2-\sqrt3}\\a^2=2+2-2\sqrt{4-3}\\a^2=2\\a>0\\a=\sqrt2$$

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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 $\Rightarrow$.


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

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

Also

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

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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

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

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

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  • $\begingroup$ Nice one +1..... $\endgroup$ – Maria Mazur Feb 5 at 19:15
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$$$$

\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}

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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.

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