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So I have a (nested?) square root $ \sqrt{2-\sqrt{2}} $. I know that $ \sqrt{2-\sqrt{3}} = \frac{\sqrt{6}-\sqrt{2}}{2} $. I know how to turn the simplified version into the complex one, but not vice versa. What is $ \sqrt{2-\sqrt{2}} $ simplified in this fashion and what are the steps?

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Related. I particularly like this answer. –  Git Gud Mar 17 '13 at 22:09
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If you're looking for this radical as the sum of square roots, no such one exists. –  George V. Williams Mar 17 '13 at 22:14
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Can't do much better than this: en.wikipedia.org/wiki/Nested_radical#Denesting_nested_radicals –  Lepidopterist Mar 17 '13 at 22:14
    
I think $\sqrt{2-\sqrt{2}}$ does not have simplicated form, because let $\sqrt{2-\sqrt{2}}=\sqrt{a}-\sqrt{b}$ and calculate it, then $a$, $b$ have more complex form... –  tetori Mar 17 '13 at 22:15
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Perhaps someone can post a proof that $\sqrt{2-\sqrt{2}}$ is not a linear combination of rational roots (including more than two)? –  Martin Brandenburg Mar 17 '13 at 23:21
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4 Answers

this is a general way to do it:

$\sqrt{2-\sqrt{2}}=\sqrt x+\sqrt y $

then $2-\sqrt 2=2 \sqrt{xy}+x+y$ and then make the rational parts equal to the rational parts as such:

$2=x+y$

now we know $y=2-x$. substitute this into the irrational part:

$-\sqrt 2 =2\sqrt{(x)}\sqrt{(2-x)}$ square both sides to get

$2=4(x)(2-x) \rightarrow 2=8x-4x^2$

However the soulutions to this quadratic equation are $1-\frac{1}{\sqrt{2}}$ and $1+\frac{1}{\sqrt{2}}$

so $ \sqrt{ 2-\sqrt{2}}=\sqrt{1-\frac{1}{\sqrt{2}}} +\sqrt{1+\frac{1}{\sqrt{2}}}$

In other words there is no way to denest your radical.

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I think you need $\sqrt(xy)$ in line two. –  Chris Leary Mar 18 '13 at 15:11
    
yes, thanks for noticing. –  user4140 Mar 18 '13 at 15:32
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There is no simpler form because the answer is imaginary: $ \sqrt{1+i}-\sqrt{1-i} $. But at least now I how to find it

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want to explain why? –  nbubis Mar 17 '13 at 23:02
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There cannot be a formula like $$ \sqrt{2-\sqrt2}=q_1\sqrt{a_1}+q_2\sqrt{a_2}+\cdots+q_k\sqrt{a_k}, $$ where $a_1,a_2,\ldots,a_k$ are integers, and $q_1,q_2,\ldots,q_k$ are rational numbers. Here the number of terms, $k$, can be as large as we want. This follows from elementary Galois theory. If you haven't learned about Galois theory, then this answer is gonna be all Greek to you. It is probably also overkill, but if you know basic Galois theory you may appreciate this. Or may be not :-/

Consider the polynomial $$ p(x)=x^4-4x^2+2. $$ We easily see that its zeros are $$ x_1=\sqrt{2+\sqrt2},\quad x_2=\sqrt{2-\sqrt2},\quad x_3=-x_1\quad\text{and}\quad x_4=-x_2. $$ It is easy to see that none of these are rational (square them!) Similarly we can easily verify that none of the polynomial $(x-u)(x-v)$ where $u,v$ are any distinct zeros of $p(x)$ has rational coefficients. Therefore $p(x)$ is irreducible in the ring $\mathbb{Q}[x]$.

Claim 1. $F=\mathbb{Q}(x_1)$ is the splitting field of $p(x)$ over $\mathbb{Q}$.

Proof. We have $x_1^2=2+\sqrt2$, so $\sqrt2\in F$. But we also have $$ x_1x_2=\sqrt{(2+\sqrt2)(2-\sqrt2)}=\sqrt2, $$ so $x_2=\sqrt2/x_1\in F$, too. Obviously then also $x_3,x_4\in F$. As the splitting field also needs to contain $F$, we are done.

Claim 2. The Galois group $G=Gal(F/\mathbb{Q})$ is cyclic of order 4.

Proof. By general Galois theory there is an automorphism $\sigma\in G$ such that $\sigma(x_1)=x_2$. We shall see that $\sigma$ is of order four. As $|G|=[F:\mathbb{Q}]=4$ this will prove the claim. Here $$ \sigma(\sqrt2)=\sigma(x_1^2-2)=\sigma(x_1)^2-\sigma(2)=x_2^2-2=-\sqrt2. $$ Using that bit of information we can see that $$ \sigma(x_2)=\sigma(\frac{\sqrt2}{x_1})=\frac{\sigma(\sqrt2)}{\sigma(x_1)}=-\frac{\sqrt2}{x_2}=-\frac{\sqrt2}{\sqrt2/x_1}=-x_1=x_3. $$ More or less the same calculation will then show that $\sigma(x_3)=x_4$, and as $\sigma$ permutes the roots of $p(x)$ we get that $\sigma(x_4)=x_1$. Therefore $\sigma$ acts on the roots as the 4-cycle $x_1\mapsto x_2\mapsto x_3\mapsto x_4\mapsto x_1$ proving our claim.

Now we can prove our main claim. It is known that all the fields $K=\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_k})$, where $a_1,a_2,\ldots,a_k$ are (w.l.o.g. square-free) integers, have Galois groups isomorphic to $C_2^\ell$ for some integer $\ell\le k$. As $F=\mathbb{Q}(x_1)=\mathbb{Q}(x_2)$ no such field can contain the number $x_2$, for otherwise the Galois group $Gal(F/\mathbb{Q})$ would have to be a quotient of the elementary abelian 2-group $Gal(K/\mathbb{Q})$. This is obviously not the case. Q.E.D.

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A general way to solve such problems:

$$\sqrt{2-\sqrt{2}}=a$$

then $a^2=2-\sqrt{2}\rightarrow(2-a^2)^2=2\rightarrow 4-4a^2+a^4=2\rightarrow a^4-4a^2+2=0$

Let $b=a^2$

then

Solve $$b^2-4b+2=0$$

As @Martin suggested, this is a way but the conslusion is not right. The right way should be

$$\sqrt{a+b\sqrt{c}}=\sqrt{d}+\sqrt{e}$$ and the following arithmetic similar to above.

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How does this answer the question? $a$ is already "solved". –  Martin Brandenburg Mar 17 '13 at 23:20
    
@MartinBrandenburg $b$ seems not nested or? –  Seyhmus Güngören Mar 17 '13 at 23:25
    
Yes $a^2$ is not nested, but $a$ still is. –  Martin Brandenburg Mar 17 '13 at 23:26
    
Yes thats true and I conlude that I can not get a simplified version for this case. –  Seyhmus Güngören Mar 17 '13 at 23:28
    
This "conclusion" is not more than a guess and becomes wrong for other nested square roots (one already mentioned in the question). –  Martin Brandenburg Mar 17 '13 at 23:31
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