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Hardy and Wright mention ( though don't give a proof ) that any finite combination of real quadratic surds is an algebraic number. For example $\sqrt{11+2\sqrt{7}}$. Are all finite combinations of cube root, fourth root ... $n^{th}$ root also algebraic ? such as $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$.

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    $\begingroup$ Yes, they are all algebraic. $\endgroup$ Feb 2, 2017 at 17:06
  • $\begingroup$ Yep. You are simply using elementary operators and functions, and the field of elementary functions is closed under (finitely many of) those operations. $\endgroup$ Feb 2, 2017 at 17:07
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    $\begingroup$ They are algebraic, but there are algebraic numbers which can't be expressed only with nested radicals. $\endgroup$
    – Watson
    Feb 2, 2017 at 17:08
  • $\begingroup$ @Watson throw up a concrete example and that would be an answer in and of itself. $\endgroup$ Feb 2, 2017 at 17:10
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    $\begingroup$ Note that changing your example can invalidate answer. In particular, I was working on mine and had almost posted it when you changed the target. $\endgroup$ Feb 2, 2017 at 17:29

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The usual ways to prove this is to use some basic linear algebra.

Assume $\alpha$ and $\beta$ are algebraic numbers.

The field $\mathbb Q[\alpha][\beta]$ is a finite-dimensional vector space over $\mathbb Q$. Let $d$ be the dimension of the vector space.

Then $1,\alpha+\beta,(\alpha+\beta)^2,\cdots,(\alpha+\beta)^d$ are $d+1$ elements of that vector space, and thus must be linearly dependent. But that means that $\alpha+\beta$ is the root of a rational polynomial of degree at most $d$.

The same for $\alpha\beta$.

More specifically, if $\alpha$ is the root of a rational polynomial of degree $d_1$ and $\beta$ is the root of a rational polynomial of degree $d_2$ then $\alpha+\beta$ and $\alpha\beta$ are both roots of a polynomial of degree at most $d_1d_2$.

Finally, if $\alpha$ is the root of the rational polynomial $p(x)$ then $\sqrt[n]\alpha$ is a root of $p(x^n)$.

So this means that $5+3\sqrt6$ is the root of a polynomial of degree $2$, and $3\sqrt[7]{5+3\sqrt6}$ is a root of a polynomial of degree $14$ and $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt6}}$ is the root of a polynomial of degree $42$ and finally $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$ is a root of a polynomial of degree $42\cdot 9=378$. There might be a polynomial of smaller degree, but we get this upper bound.

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    $\begingroup$ Shouldn't the lowest degree that you have shown be $42 \cdot 9=378?$ As you say, it could be lower. $\endgroup$ Feb 2, 2017 at 17:28
  • $\begingroup$ Yep, brain freeze. @RossMillikan $\endgroup$ Feb 2, 2017 at 17:44
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    $\begingroup$ But you stated without proof the key part of the demonstration: that $\Bbb Q[\alpha][\beta]$ is finite dimensional over $\Bbb Q$. That is the important part. The rest is just interpreting what this means. $\endgroup$ Feb 3, 2017 at 1:12
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    $\begingroup$ Yeah, I didn't prove everything. @PaulSinclair $\endgroup$ Feb 3, 2017 at 2:02
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Yes, you can unpack them into a polynomial that they satisfy. For your example, we can write $$x=\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}\\x^3=2+3\sqrt[7]{5+3\sqrt{6}}\\\frac 13(x^3-2)=\sqrt[7]{5+3\sqrt{6}}\\\left(\frac 13(x^3-2)\right)^7=5+3\sqrt 6\\\frac 19\left(\left(\frac 13(x^3-2)\right)^7-5\right)^2-6=0$$

Note: this uses a previous example in the question. The approach is the same. Note that the algebraics are closed under the field operations, so the sum of algebraics is again algebraic.

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    $\begingroup$ The OP edited the example, for $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$ this is more difficult. $\endgroup$
    – Watson
    Feb 2, 2017 at 17:12
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    $\begingroup$ @Watson: That is why I appealed to closure under addition. $\endgroup$ Feb 2, 2017 at 17:26
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Hint: Let $x = \sqrt{11+2\sqrt{7}}$, square this expression $x^2=11+2\sqrt{7} \implies x^2-11=2\sqrt{7}$. Now square again to get a polynomial with rational coefficients. You can use a similar approach for the second expression.

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    $\begingroup$ true, yes, but this isn't super easy to generalize. A bit more care must be taken in proving this for all nested, $n$th degree surds $\endgroup$ Feb 2, 2017 at 17:09
  • $\begingroup$ @MrYouMath I updated my example. $\endgroup$ Feb 2, 2017 at 17:10
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I think you're asking if a sum of two algebraic numbers is algebraic, where "algebraic" is defined as "the root of a polynomial with rational coefficients".

While there are linear algebra reasons for this to be true, there is actually a nice direct calculation in terms of resultants.

If $\alpha_i$ are the roots of a polynomial $f$ and $\beta_j$ are the roots of a polynomial $g$, both with rational coefficients, then

$$ \prod_{i} \prod_j (x - (\alpha_i + \beta_j)) $$

when expanded, is a polynomial with rational coefficients. This polynomial can be computed by a resultant:

$$ \operatorname{Res}_t(f(t), g(x-t)) $$

and other formulations of the resultant make it obvious the result is a polynomial with rational coefficients.

Alternatively, Galois theory nearly immediately proves the product polynomial defined above has rational coefficients, because the polynomial is preserved by any permutation of the $\alpha$'s and of the $\beta$'s.

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All the numbers you list are algebraic.

Lemma. If $\alpha$ and $\beta$ are algebraic, then $\alpha+\beta$ and $\alpha\beta$ are algebraic.

Proof. Since $\beta$ is algebraic (over $\mathbb{Q}$), it is also algebraic over $\mathbb{Q}(\alpha)$. Hence $\mathbb{Q}(\alpha,\beta)$ is a finite extension of $\mathbb{Q}(\alpha)$, which in turn is finite over $\mathbb{Q}$. Therefore $\mathbb{Q}(\alpha,\beta)$ is finite by the dimension formula: $$ [\mathbb{Q}(\alpha,\beta):\mathbb{Q}]= [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)] [\mathbb{Q}(\alpha):\mathbb{Q}] $$ Any finite extension of $\mathbb{Q}$, say $n$-dimensional, consists of algebraic elements, because if $\gamma$ is an element, $\{1,\gamma,\gamma^2,\dots,\gamma^n\}$ is linearly dependent, so we find a polynomial with coefficients in $\mathbb{Q}$ that has $\gamma$ as root. In particular $\alpha+\beta$ and $\alpha\beta$ belong to $\mathbb{Q}(\alpha,\beta)$, so they're algebraic. QED

Note that this doesn't say that an algebraic extension (that is, all of whose elements are algebraic) is finite. Just that a finite extension is algebraic.

Lemma. If $\alpha$ is algebraic, then $\sqrt[n]{\alpha}$ is algebraic.

Proof. $\sqrt[n]{\alpha}$ is the root of $X^n-\alpha$, so it is algebraic over $\mathbb{Q}(\alpha)$, hence algebraic over $\mathbb{Q}$ by the same argument on dimensions as before. QED

A nested radical can be defined inductively as

  1. A rational number is a nested radical;
  2. If $\alpha_1,\dots,\alpha_k$ are nested radical, $q_1,\dots,q_k$ are rational numbers and $n$ is an integer, then $$\sqrt[n]{q_1\alpha_1+\dots+q_k\alpha_k}$$ is a nested radical;
  3. nothing else is a nested radical.

Proposition. Any rational linear combination of nested radicals is algebraic.

Proof. Dismantle the number using the two lemma above. QED

However, there are algebraic numbers that cannot be expressed as sums of nested radicals. If this were true, then every equation with rational coefficients would be solvable by radicals; this is false as proved by Galois: radical extensions have solvable Galois group, but not all Galois groups of polynomials are solvable.

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The complex algebraic numbers are what is called an algebraically closed field:

this implies that if $a_0,a_1,a_2,\dots, a_n$ are real algebraic numbers then the real roots of the polynomial $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ are all algebraic numbers. This is way stronger than what you need.

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Yes, this holds in general. Recall that an extension $F/\Bbb Q$ is algebraic iff for each $\alpha\in F$ we have

$$[\Bbb Q(\alpha):\Bbb Q]=\dim_{\Bbb Q} \Bbb Q(\alpha)< \infty$$

that is, if every element in it generates a finite extension. Similarly any finite extension is algebraic since if $\beta\in F$ then $\Bbb Q(\beta)\subseteq F\implies [\Bbb Q(\beta):\Bbb Q]\le [F:\Bbb Q]$. For a nested radical expression, this is easy then, take the most nested radical, adjoin the needed roots and proceed by induction.

In your first example $\alpha=\sqrt{11+2\sqrt 7}$, then we see that the field $\Bbb Q(\sqrt 7)$ has a polynomial for this which is ${1\over 2}(x^2-11)-\sqrt 7 = 0$ so the splitting field of this polynomial, which has degree at most $2$ over $\Bbb Q(\sqrt 7)$, and therefore degree at most $4$ over $\Bbb Q$, is algebraic, hence $\alpha$ is algebraic.

For your other example and more generally any example, you do similarly, since there are only finitely many nested radicals each time, you get splitting fields of finite degree, implying all such expressions are algebraic.

Just to work out the more complicated example, $\alpha = \sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$ first adjoin $\sqrt 6$, and let $F_1=\Bbb Q(\sqrt 6)$. Then look at the splitting field, $F_2$, of $(x/3)^7-5-3\sqrt 6$ over $\Bbb Q(\sqrt 6)$. Then the splitting field, $F_3$ of $x^3-2-3\sqrt[7]{5+3\sqrt 6}$ over $F_2$, finally let $F_4=F_3(\sqrt[9]{2})$. Because each step had finite degree, $[F_4:\Bbb Q]<\infty$ showing $\alpha$ is algebraic.

One more since we're on a roll, look at $\sqrt{\sqrt{23}+\sqrt 5}$ this gives you a slight problem at first it seems since $\sqrt{23}$ and $\sqrt{5}$ are both the most nested radical, but as usual $[F(\alpha,\beta): E]\le [F(\alpha):E][F(\beta):E]$ and we can reduce to the simpler case by the classical result that sums and products of algebraic numbers are algebraic (proved in the same way with finite degrees).

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The question in the title itself:

What real numbers do algebraic numbers cover?

has a rather amusing answer:

Practically none.

There are only countably many algebraic numbers, since each one is a root of some polynomial over $\mathbb{Q}$, and there are countably many such polynomials each of which has finitely many roots in $\mathbb{C}$. But there are uncountably many real numbers.

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  • $\begingroup$ Yes, but how real are all those uncountable reals actually? $\endgroup$ Feb 3, 2017 at 8:28
  • $\begingroup$ @CountIblis: We can be agnostic about it, if you don't believe in full semantics of power-sets or higher-order logic. A concrete analogous statement would be that given every computable enumeration of computable reals (represented by a program that always halts on an input $n$ and outputs the $n$ digit), there is a computable real that is computably distinguishable from every member of that enumeration. I view uncountability as indicating complexity, not necessarily size. $\endgroup$
    – user21820
    Feb 3, 2017 at 8:41

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