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Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.

I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.

Any help would be greatly appreciated.

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    $\begingroup$ It is $1$, so it is rational. $\endgroup$ Dec 30 '16 at 0:36
  • $\begingroup$ The body of your Question (apart from just the title) should be used to give a self-contained presentation of the problem you want help with, including an indication of what you tried and/or where you found difficulty with any such approach. Please review How to Ask. $\endgroup$
    – hardmath
    Dec 30 '16 at 2:56
  • $\begingroup$ @i707107 : how do we get 1? if you knew how should have posted it as answer not a comment. $\endgroup$
    – jimjim
    Dec 31 '16 at 14:21
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    $\begingroup$ @Arjang I thought that giving the answer 1 provides an enough hint, and wanted the poster to figure out. $\endgroup$ Dec 31 '16 at 17:28
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Note that $4+2\sqrt{3}=(1+\sqrt{3})^2$

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  • $\begingroup$ You are a very quick magician there young sir. $\endgroup$ Dec 30 '16 at 0:39
  • $\begingroup$ Argh... beat me! $\endgroup$
    – Frank
    Dec 30 '16 at 0:40
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Method 1: Consider this denesting algorithm:


Denested square roots: Given a radical of the form $\sqrt{X\pm {Y}}$ with $X,Y\in\mathbb{R}$ and $X>Y$, we have a possible simplification as$$\sqrt{X\pm Y}=\sqrt{\dfrac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\dfrac {X-\sqrt{X^2-Y^2}}2}\tag1$$


Using $(1)$ on $\sqrt{4+2\sqrt3}$, we have$$\sqrt{4+2\sqrt3}=\sqrt{\dfrac {4+2}2}+\sqrt{\dfrac {4-2}2}=\sqrt3+1\tag2$$ So the original expression becomes$$\color{brown}{\sqrt{4+2\sqrt3}}-\sqrt3=\color{brown}{\sqrt3+1}-\sqrt3=1$$ Which is rational.


Method 2: If you don't like $(1)$ and think it's too complicated, I present you an alternative method. Simply set $\sqrt{4+2\sqrt{3}}-\sqrt3$ equal to a variable and simplify!

Here, we have$$\begin{align*} & \sqrt{4+2\sqrt3}-\sqrt3=\alpha\\ & \sqrt{4+2\sqrt3}=\alpha+\sqrt3\\ & 4+2\sqrt3=(\alpha+\sqrt3)^2\\ & 4+2\sqrt3=\alpha^2+3+2\alpha\sqrt3\end{align*}$$ To solve for $\alpha$, we have $2\alpha\sqrt3=2\sqrt3\implies\alpha=1$. Checking with the other half, $\alpha^2+3=1+3=4$ holds. Hence,$$\sqrt{4+2\sqrt3}-\sqrt3=1$$

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Many questions with sum or difference of square roots can be solved with conjugating.

So, if $s = \sqrt{4 + 2\sqrt{3}} - \sqrt{3}$, and $t = \sqrt{4 + 2\sqrt{3}} + \sqrt{3}$,

$\begin{array}\\ st &=(\sqrt{4 + 2\sqrt{3}} - \sqrt{3})(\sqrt{4 + 2\sqrt{3}} + \sqrt{3})\\ &=4 + 2\sqrt{3}-3\\ &=1 + 2\sqrt{3}\\ \end{array} $

Since $t-s = 2\sqrt{3}$, $st = 1+t-s$ or $s(t+1) = 1+t$, and, by magic, we get $s = 1$ (unless $t+1 = 0$ which it does not since $t > 0$).

I am surprised that this worked so well.

I will suppress my need to generalize and submit this as is.

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  • $\begingroup$ (+1) I like this approach! And Happy Holidays Marty! -Mark $\endgroup$
    – Mark Viola
    Dec 30 '16 at 1:08
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    $\begingroup$ Thanks. And may 2017 not be as bad as I fear. $\endgroup$ Dec 30 '16 at 1:10
  • $\begingroup$ Yes, I believe I know the source of the fear. I am still in shock. $\endgroup$
    – Mark Viola
    Dec 30 '16 at 1:13
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Let $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$. Then:

$$(x+\sqrt3)^2=4+2\sqrt{3}\implies x^2-1=2(1-x)\sqrt3\implies(x-1)(x+1+2\sqrt3)=0$$

So, certainly $x=1$ or $x=-1-2\sqrt3$. But a moment's thought (e.g. considering that $x>0$) convinces us that the first of these must be the case - i.e., $x=1$ (a known rational number).

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Personally I like this method which is the same thing as the other answers.$\sqrt{4+2\sqrt{3}}=\sqrt{a}+\sqrt b$ squaring both sides $4+2\sqrt 3=a+2\sqrt{ab}+b$ for this to be true we must equate the parts with and without the radicals so $a+b=4$ and $2\sqrt{ab}=2\sqrt{3}$ so $ab=3$ now we have a system of equations that can be solved through substitution $a(4-a)=3\to a^2-4a+3=0$ which has roots 3, 1. So if a=1 then b=3 and vice versa if a=3. So our answer is $\sqrt{4+2\sqrt{3}}-\sqrt 3=1+\sqrt{3}-\sqrt3$

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  • $\begingroup$ refreshingly different, very nice $\endgroup$
    – jimjim
    Jan 18 '17 at 5:09
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Such square roots can be computed by a very Simple Denesting Rule:

Here $\, 4+2\sqrt 3\ $ has norm $= 4.\:$ $\rm\ \color{blue}{Subtracting\ out}\,\ \sqrt{norm}\ = 2\,\ $ yields $\,\ 2+2\sqrt 3\:$

which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{4}\, =\, 2.\,\ \ \rm \color{brown}{Dividing\ it\ out}\,\ $ of the above yields $\ \ 1+\sqrt 3$

Remark $\ $ Many more worked examples are in prior posts on this denesting rule.

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