# How can one express $\sqrt{2+\sqrt{2}}$ without using the square root of a square root?

I was trying to review some analysis, and came across problem 3 from page 78 of Walter Rudin's Principles of Mathematical Analysis. As part of the problem, I wanted to try to write $\sqrt{2+\sqrt{2}}$ without using the square root of a square root. In other words, I wanted to express the number in the form $q_1+q_2{q_3}^s$, where $q_1, q_2, q_3, s \in \mathbb{Q}$ (or perhaps something similar). I'm aware that this is probably a duplicate of another question, but I wasn't able to find it (I wasn't sure what to search for)... Many thanks in advance!

Edit (my work so far):

I tried expressing it as the solution to the quartic $x^4-4x^2+2=0$, but this seemed futile...

Edit 2 (the original problem):

The original problem from the text states:

If $s_1=\sqrt{2}$, and $$S_{n+1}=\sqrt{2+\sqrt{s_n}} (n=1,2,3,\ldots),$$ prove that $\{s_n\}$ converges, and that $s_n<2$ for $n=1,2,3,\ldots$.

The problem is from the 3rd chapter of the book, which talks about sequences and series. Rudin provides numerous theorems on this topic, such as the comparison, ratio, and root tests for convergence.

• For those of us without quick access to Rudin, what's the original problem itself? – Steven Stadnicki Aug 7 '12 at 16:13
• I'm certain that you can't express it using only rationals and $\sqrt 2$ (because the polynomial you mentioned is irreducible over rationals and of degree $4>2$). I don't believe the form you mentioned is possible, either. If you still want an answer to the question, you might want to add algebraic number theory tag. – tomasz Aug 7 '12 at 16:27
• @StevenStadnicki There's apparently a version available but I can't speak to whether or not one should use it. – rschwieb Aug 7 '12 at 16:28
• @Andrew: Not quite the same thing you asked about, but you might want to look at my comments about "square root of surds" (a standard topic in 19th century algebra texts) in this 3 March 2011 Math Forum archived post. – Dave L. Renfro Aug 7 '12 at 16:41
• One denesting is $\:\sqrt{2+\sqrt 2}\, =\, i^{1/4} + (-i)^{1/4}\$ – Bill Dubuque Aug 7 '12 at 17:13

## 3 Answers

Short answer: you can't. Call your element $\alpha$. You want $\alpha$ to live in a field extension of the form $\mathbb{Q}(\beta^{1/4})$ for some $\beta \in \mathbb{Q}$. It's clear that $L = \mathbb{Q}(\alpha)$ is a degree 4 field extension of $\mathbb{Q}$ (you've written down a minimal polynomial).

One can check easily that $L$ is Galois over $\mathbb{Q}$ (i.e. you can check that $\sqrt{2 - \sqrt{2}} \in L$). But $\mathbb{Q}(\beta^{1/4})$ (for $\beta$ squarefree) is not.

• OP wants $\alpha$ to be of the form $a + bc^{1/n}$. Since $\alpha$ has degree 4, $n$ would have to be 4; I'm proving that can't happen. (OP may be happy with a more general expression of $\alpha$ as a sum of a lot of roots; I think you will get a similar contradiction if you try that though.) – user29743 Aug 7 '12 at 16:34
• I think so, too, but it seems rather nontrivial to me (though I have only brief familiarity with algebraic number theory, so it wouldn't surprise me if I was wrong). These kinds of questions tend to have some methods fail quickly as you increase the number of summands. – tomasz Aug 7 '12 at 16:36
• I believe the galois group of $\mathbb Q(\sqrt{2-\sqrt{2}})$ is cyclic. In particular it can't be a bi-quadratic extension, because the galois group would have to be Klein 4. I think this generalizes fairly easily to showing that it can't be contained in $\mathbb Q(\sqrt{p_i})$ for some set of primes $p_i$. Because the galois group has no cyclic subgroup of order $4$. – JSchlather Aug 7 '12 at 16:57
• @JacobSchlather You mean "no cyclic quotient of order $4$", not "no cyclic subgroup of order $4$". Otherwise, good comment! – David E Speyer Aug 7 '12 at 18:35
• @DavidSpeyer Yes, I worked through a few of the details after posting that and realized that what I needed was quotient. But fortunately the galois group is $\mathbb Z_2^n$ so the argument is still fine. – JSchlather Aug 7 '12 at 18:48

For completeness's sake here is the solution to the original problem:

We first solve the equation $$\sqrt{2+\sqrt{x}}=x$$ and the only real root $y$ is about 1.812. Now we have $\sqrt{2+\sqrt{2}}\approx 1.847$. If we differentiate $\frac{\sqrt{2+\sqrt{x}}}{x}$ with respect to $x$, the derivative is always negative. So when $s_{n}\ge y$, $s_{n}$'s value would decrease under iteration until it is less than $y$, and when $s_{n}\le y$ it would increase, since on the interval $[0,y]$ the quotient is greater than $0$. This force the limit to be $y$.

Another way to see it is observing that $$s_{n}=\sqrt{2+\sqrt{s_{n-1}}}, |s_{n+1}-s_{n}|=|\sqrt{2+\sqrt{s_{n}}}-\sqrt{2+\sqrt{s_{n-1}}}|=|\frac{\sqrt{s_{n}}-\sqrt{s_{n-1}}}{s_{n}+s_{n+1}}|$$ therefore we have $$|\frac{s_{n+1}-s_{n}}{s_{n}-s_{n-1}}|=\frac{1}{|s_{n}+s_{n-1}||\sqrt{s_{n}}+\sqrt{s_{n-1}}|}\le \frac{1}{4}$$ and the rest follows from contraction mapping theorem.

• This inequality does not hold for all $x \le 2$. It fails at $x = 1.9$, for example: wolframalpha.com/input/… – user88319 Jun 5 '14 at 3:09
• Yes, there is an arithmetic mistake. The correct bound is $2^{-1/4}$, which is too small for it to be monotonely increasing. And judged from the graph it is also not monotonely increasing. I shall fix that. – Bombyx mori Jun 5 '14 at 17:51
• Hopefully fixed. – Bombyx mori Jun 5 '14 at 18:19

And then there is always trigonometry: $\sqrt{2+\sqrt{2}}=2\cos\left(\frac{\pi}{8}\right)$.