I considered simplifying $\sqrt {1+\sqrt 5}$.

So I started $(a+b \sqrt 5)^2 = 1 + \sqrt 5$.

This gave me $a\color{blue}{^2} + 5b^2 =1 , 2ab = 1$ so the result was

$ \sqrt{1+\sqrt 5} = \sqrt{1/2 - i} + \dfrac{\sqrt 5}{\sqrt{2 - 4i}}.$

I continued in the same spirit : $(a_2+b_2 i)^2 = 1/2 - i. $

$a_2^2 - b_2^2 = 1/2, 2a_2b_2 = -1.$ This gave me :

$ \sqrt{1+\sqrt 5} = \dfrac{\sqrt{1+\sqrt 5}}{2} - \frac{1}{4} (\sqrt 5 - 1)(\sqrt{1+\sqrt 5})\color{blue}i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}\tag1$

I was a bit surprised by the self-similarity.

I got the feeling of " periodic " or " fractal " , like running in circles.

It this normal to get this kind of self-similarity when the initial radical cannot be rewritten in simpler shorter form ( without complex numbers or nesting ) ?

Are they ways to compute or theorems to know about just how many steps it takes to get the self-similarity ?

Does this imply that the number of nontrivial non-nested expressions for a real radical is Always finite ?


Perhaps this edit clarifies a bit what I meant.

Basicly the only identities and substitutions for identities we can have are :

$ \sqrt{1+\sqrt 5} = \sqrt{1/2 - i} + \sqrt{1/2 + i}$

$ \sqrt{1/2+ i} = \dfrac{\sqrt{1+\sqrt 5}}{2} - \frac{1}{4} (\sqrt 5 - 1)(\sqrt{1+\sqrt 5})i$

$ \sqrt{1/2- i} = \dfrac{\sqrt{1+\sqrt 5}}{2} + \frac{1}{4} (\sqrt 5 - 1)(\sqrt{1+\sqrt 5})i$

And no more than those 3. Or 2 if we do not count conjugates as seperate.

Now I suspect this mysterious number 2 is no coincidence ;


If a nested root form containing only sqrts , positive integers and additions (thus no divisions or substractions ) is given that cannot be simplified within the same ring then the amount of identities ( as in the example above , ignoring conjugates ) is at most the amount of sqrt's in the expression.

In this case 2 sqrt's => 2 identities.



A similar thing happens with $\sqrt{5+\sqrt 3}$. You end up with a $\sqrt{10+2\sqrt 22}$ term and when you go on you end up with $\sqrt{5+\sqrt 3}$ again.

Here $\sqrt{10+2\sqrt 22}$ takes on the " role " of $\sqrt{1/2 + i}$ but this examples demonstrates that the phenomenon can stay within the reals too.

These examples are fairly easy to check and in that sense they may appear trivial , but there must be more behind this. The same thing happens for cuberoots for example , and that gives us a mimimum polynomial of degree 9 which is much less well understood then degree 4.

Similar things happen if we use Bring radicals.

Although intuitive and easy to check any particular example by basic algebra , group and ring theory ... The basic framework is unclear to me. I do not think Galois theory alone explains it completely.

I gave a bounty but without succes.

  • 1
    $\begingroup$ How do you mean simplify the expression? (layman here ;) ) $\endgroup$ – Chinny84 Dec 25 '14 at 23:43
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    $\begingroup$ I assume the OP is referring to the fact that, sometimes, $\sqrt{a + b \sqrt{c}}$ can be rewritten as $d + e \sqrt{f}$, although the OP should clarify.... $\endgroup$ – user14972 Dec 25 '14 at 23:46
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    $\begingroup$ This just seems like an unholy mess to me, what do I know though? $\endgroup$ – Suzu Hirose Dec 25 '14 at 23:59
  • $\begingroup$ Also notice that the square root is a multi-valued function on $\mathbb{C}$ - what do you mean by $\sqrt{1/2-i}$? $\endgroup$ – Jack D'Aurizio Dec 26 '14 at 0:02
  • $\begingroup$ You might want to look at this: math.stackexchange.com/questions/720114/… $\endgroup$ – Pauly B Dec 26 '14 at 1:16

I can't quite follow your computations, or your notion of "self-similarity", but there is no simpler form of $\alpha=\sqrt{1+\sqrt 5}$.

$\alpha$ is a root of the polynomial $x^4-2x^2-4$, which is irreducible over the integers and therefore also over $\mathbb Q$. Thus, $x^4-2x^2-4$ is the unique monic minimal polynomial of $\alpha$ over $\mathbb Q$. Since it has degree 4, there is no way to write $\alpha$ with just one square root sign, because all numbers that arise in that way have minimal polynomials of degree 2 or 1.

We might still hope to write $\alpha$ with two square roots as $\sqrt p+\sqrt q+r$ for rational $p$, $q$ and $r$ -- that would arguably be simpler than nesting the two square roots. Unfortunately, that is not possible.

The minimal polynomial of $\sqrt p+\sqrt q$ (assuming neither of $p$, $q$, or $p/q$ are rational squares) is $$\tag{*} x^4-2(p+q)x^2+(p-q)^2$$ If $\alpha$ is of this form, we would have $2(p+q)=2$ and $(p-q)^2=-4$, but the latter equation clearly has no real solutions for $p$ and $q$. We do get $$ \sqrt{1+\sqrt5} = \sqrt{1/2+i}+\sqrt{1/2-i} $$ where in each case we select the square root with positive real part -- but that doesn't really look simpler to me (though your mileage may vary).

And we can't hope for $\alpha$ to have the form $\sqrt p+\sqrt q+r$ either, because then $\beta=\alpha-r$ would be $\sqrt p+\sqrt q$, but if we substitute $\alpha=\beta+r$ in $\alpha^4-2\alpha^2-4=0$ we get something with a nonzero third-degree term (unless $r=0$), which can't possibly have the form $(\text{*})$.

  • $\begingroup$ I'm not sure, but I think what he means by the "self-similarity" and "fractal" part is that he sees similar expressions in the LHS and RHS. Kindly look at my comment/answer below. $\endgroup$ – Tito Piezas III Dec 26 '14 at 3:49

(Too long for a comment.)

I think what he means is that since from $(1)$,

$$ \sqrt{1+\sqrt 5} = \dfrac{\sqrt{1+\sqrt 5}}{2} - \frac{1}{4} (\sqrt 5 - 1)(\sqrt{1+\sqrt 5}) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}\tag1$$


$$ \dfrac{\sqrt{1+\sqrt 5}}{2} = -\frac{\sqrt 5-1}{4}\Bigr(\sqrt{1+\sqrt 5}\Bigl) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}$$

So when he "... was surprised by the self-similarity", I assume it was because he saw that part of the RHS can be expressed in terms of the LHS,

$$ A = \color{blue}{\frac{1-\sqrt 5}{2} A i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}}$$

and we can iterate the "rule" for $A$ as,

$$ A = \frac{1-\sqrt 5}{2} \left(\color{blue}{\frac{1-\sqrt 5}{2} A i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}}\right) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}$$

and again,

$$ A = \frac{1-\sqrt 5}{2} \left({\frac{1-\sqrt 5}{2} \left(\color{blue}{\frac{1-\sqrt 5}{2} A i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}}\right) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}}\right) i +\dfrac{\sqrt 5}{\sqrt{2 - 4i}}$$

and so on to as many levels as one wishes, with $A = \frac{\sqrt{1+\sqrt 5}}{2}$, hence his "fractal".

  • $\begingroup$ Yes I noticed this. Thanks ! But I edited and tried to explain more clearly. $\endgroup$ – mick Dec 28 '14 at 19:21

I don't really understand what you are doing regarding self-similarity and the introduction of $i$. But as far as writing $\sqrt{1+\sqrt{5}}$ as $a+b\sqrt{5}$, since you are using that form in the first place, I would assume that you mean $a,b$ rational.

But as you have found (with a typo?), $a^2+5b^2=1$ and $2ab=1$, so

$$\begin{aligned} a^2+5b^2&=1\\ 4a^2b^2+20b^4&=4b^2\\ 1+20b^4&=4b^2\\ 20b^4-4b^2+1&=0\\ b^4-\frac15b^2+\frac1{20}&=0\\ 0<\left(b^2-\frac1{10}\right)^2+\frac1{25}&=0\end{aligned}$$

So there is no rational $b$ that works. So writing in the form $a+b\sqrt{5}$ is not going to make anything simpler, since you'd have to use irrational, (non real even) $b$.


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