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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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    $\begingroup$ Related. I particularly like this answer. $\endgroup$
    – Git Gud
    Commented Mar 17, 2013 at 22:09
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    $\begingroup$ If you're looking for this radical as the sum of square roots, no such one exists. $\endgroup$ Commented Mar 17, 2013 at 22:14
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    $\begingroup$ Can't do much better than this: en.wikipedia.org/wiki/Nested_radical#Denesting_nested_radicals $\endgroup$ Commented Mar 17, 2013 at 22:14
  • $\begingroup$ 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... $\endgroup$
    – Hanul Jeon
    Commented Mar 17, 2013 at 22:15
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    $\begingroup$ Perhaps someone can post a proof that $\sqrt{2-\sqrt{2}}$ is not a linear combination of rational roots (including more than two)? $\endgroup$ Commented Mar 17, 2013 at 23:21

4 Answers 4

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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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  • $\begingroup$ I think you need $\sqrt{xy}$ in line two. $\endgroup$ Commented Mar 18, 2013 at 15:11
  • $\begingroup$ yes, thanks for noticing. $\endgroup$
    – Asinomás
    Commented Mar 18, 2013 at 15:32
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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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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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    $\begingroup$ want to explain why? $\endgroup$ Commented Mar 17, 2013 at 23:02
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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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  • $\begingroup$ How does this answer the question? $a$ is already "solved". $\endgroup$ Commented Mar 17, 2013 at 23:20
  • $\begingroup$ @MartinBrandenburg $b$ seems not nested or? $\endgroup$ Commented Mar 17, 2013 at 23:25
  • $\begingroup$ Yes $a^2$ is not nested, but $a$ still is. $\endgroup$ Commented Mar 17, 2013 at 23:26
  • $\begingroup$ Yes thats true and I conlude that I can not get a simplified version for this case. $\endgroup$ Commented Mar 17, 2013 at 23:28
  • $\begingroup$ This "conclusion" is not more than a guess and becomes wrong for other nested square roots (one already mentioned in the question). $\endgroup$ Commented Mar 17, 2013 at 23:31

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