How to rewrite $7-\sqrt 5$ in root form without a minus sign ? For clarity "root form " means an expression that only contains a finite amount of positive integers , additions , substractions , multiplications and root extractions (sqrt, cuberoot etc).

For example some quintic equations cannot be solved in root form.

A " root form without a minus sign " means an expression that only contains a finite amount of positive integers , additions , multiplications and root extractions (sqrt , cuberoot etc).

So the solution could look something like this :

$$ 7-\sqrt5 = \sqrt{...+1+(...)^{\frac{2}{3}}}+\sqrt{...+2(...)^{\frac{1}{3}}}$$

How to solve such problems ?


Warning: $\dfrac{44}{7+\sqrt 5}$ is not a solution, since no divisions are allowed!

I got that answer 3 times now so I put it in the OP as a warning , not just the comments.

  • 2
    $\begingroup$ Is $\frac{44}{7+\sqrt{5}} \,$ valid? $\endgroup$
    – K. Rmth
    Commented Jan 7, 2014 at 23:17
  • $\begingroup$ No it contains a division ! @K.Rmth $\endgroup$
    – mick
    Commented Jan 7, 2014 at 23:21
  • 3
    $\begingroup$ Do you have some reason to suspect that an answer exists? I am skeptical. $\endgroup$ Commented Jan 7, 2014 at 23:44
  • 4
    $\begingroup$ $\displaystyle{\large -\left(-7 + \sqrt{5}\right)}$. $\endgroup$ Commented Jan 8, 2014 at 0:23
  • 8
    $\begingroup$ I don't understand the downvotes on nor closure of this question. The idea of expressions formed recursively using positive integers, addition, multiplication and root extractions is perfectly sensible and clear. I can't think of Galois or algebraic number theory machinery that keeps track of negative signs "internal" to expressions in order to tackle this problem, so I find it interesting. Upvoted and voted to reopen. Also, as the question was clear from the very beginning, it's strange there are three incorrect answers (one now deleted by others) that assumed it is simpler than it is. $\endgroup$
    – anon
    Commented Jan 12, 2014 at 4:10

5 Answers 5


$7-\sqrt{5}$ cannot be written in root form without a minus sign. I'm not exactly sure why you want to do this, but Galois theory proves you can't. My original answer was very brief, just quickly quoting Galois theory, but my friend convinced me it would be a good idea to add more details. So let me explain the key concept from Galois theory that we need:

We will say a number $\alpha$ is "Galois-maximal" if $\alpha$ is a positive real and there is some polynomial $f(x)$, with rational coefficients, such that $\alpha$ is a root of $f(x)$, and all roots $\beta$ of $f(x)$ satisfy $|\beta| \leq \alpha$.

There are two facts that allow us to show $7- \sqrt{5}$ cannot be written in root form without a minus sign. The first is that any number written in root form without a minus sign is Galois-maximal. The second is that $7- \sqrt{5}$ is not Galois-maximal. From these two, the conclusion follows trivially.

To prove the first claim, we do induction on the number of positive integers, additions, multiplications, and root extractions. The base case is that positive integers are Galois-maximal. Indeed, $n$ is the unique root of the polynomial $x-n$. For the induction step, we need to check that if $\alpha$ is Galois-maximal, then $\alpha^{1/n}$ is, and that if $\alpha$ and $\beta$ are Galois-maximal, then $\alpha+\beta$ and $\alpha \beta$ are.

For $\alpha^{1/n}$, take a polynomial $f(x)$ with rational coefficients whose largest root is $\alpha$, and consider the polynomial $f(x^n)$. The roots of this polynomial are the $n$th roots of the roots of $f(x)$, so $\alpha^{1/n}$ is a root, and all other roots are the $n$th root of something with absolute value at least $\alpha$, hence have absolute value at most $\alpha$.

For $\alpha+\beta$ and $\alpha \beta$, take polynomials $f$ and $g$ with rational coefficients such that $\alpha$ is a root of $f$, $\beta$ is a root of $g$, and they are each at least as large as the absolute value of all other roots. The key fact is that there are polynomials $h_1$ and $h_2$, each with rational coefficients, such that the roots of $h_1$ are exactly the sums of a root of $f$ and a root of $g$ and the roots of $h_2$ are the products of a root of $f$ and a root of $g$.

If we believe this fact, then we're done: certainly $\alpha+\beta$ is a root of $h_1$ and $\alpha \beta$ is a root of $h_1$. Moreover, any root of $h_1+h_2$ has the form $x+y$ with $x$ a root of $h_1$ so $|x| \leq \alpha$ and $y$ a root of $h_2$ so $|y| \leq \beta$ hence $|x+y| \leq |x|+|y| \leq \alpha+\beta$, and the same for $h_2$ except with multiplication.

It is easy to construct a polynomial with a given set of roots, but not completely obvious why, for this particular set of roots, it has rational coefficients. The slickest proof I know uses linear algebra, specifically, the companion matrix. This allows us to write matrices with rational entries $M_1$ and $M_2$ whose eigenvalues are the roots of $f$ and $g$ respectively. Then the eigenvalues of $M_1 \otimes M_2$ are the roots of $f$ times the roots of $g$, and the eigenvalues of $M_1 \otimes I + I \otimes M_2$ are the roots of $f$ plus the roots of $g$. Because $M_1$ and $M_2$ have rational entries, their characteristic polynomials have rational coefficients, and the eigenvalues are exactly the roots of the characteristic polynomial. This finishes the induction step.

Finally, we need to check $7- \sqrt{5}$ is not Galois-maximal. To do this, its sufficient to check every polynomial with rational coefficients that has $7- \sqrt{5}$ as a root also has $7 + \sqrt{5}$. By expanding $f(7 - \sqrt{5})$ out, we obtain a polynomial in $-\sqrt{5}$, which still has rational coefficients, so we have a polynomial with rational coefficients that has $-\sqrt{5}$ as a root, and we wish to show it has $\sqrt{5}$ as a root. Any time we see $x^2$ appearing in the polynomial, we can replace it with $5$ and the value of the polynomial at $-\sqrt{5}$ and $\sqrt{5}$ will not change. Once we have simplified in this way, we have a polynomial of degree at most one with rational coefficients. So it is of the form $a - b\sqrt{5}=0$ for $a,b$ rational. If $a$ or $b$ is nonzero, this implies $\sqrt{5}= a/b$, which is impossible, so then $a$ and $b$ are both zero, but then $a + b\sqrt{5}$ is zero as well, so $\sqrt{5}$ is indeed a root, as desired. QED.

Using the same argument, it is possible to produce many other examples of positive real numbers that can be written in root form with a minus sign but can't be written in root form without it.

  • $\begingroup$ Im not sure what you meant by Galois conjugate. I assume by conjugate you meant $7+\sqrt 5$. If $x>y$ then $x^{1/n}>y^{1/n}$ So I do not know how those powers change anything. What is clear to me is that $x>1$ is neccessary since we cannot expect a "root form without a MINUS" to be smaller than 1. ( notice there are no divisions ! ). Im not an expert in Galois theory. $\endgroup$
    – mick
    Commented Jan 10, 2014 at 20:55
  • 1
    $\begingroup$ This is an enticing answer, but it needs more justification than a single sentence. You need a rigorous proof for why the set A of numbers larger than all their Galois conjugates is (i) closed under addition, (ii) closed under multiplication, and (iii) closed under $n$th roots. (Of course, the positive integers are all contained in $A$, so that part is fine.) $\endgroup$ Commented Dec 30, 2014 at 6:16
  • 1
    $\begingroup$ It's closed under addition because the galois conjugates of a sum are the sum of the galois conjugates, and addition is monotonic. Same with multiplication. For $n$th roots, the galois conjugates of the $n$th roots are all of the $n$th roots of the galois conjugates. The largest is the principal root, and the principal root is a monotonic function, so again we are fine. $\endgroup$
    – Will Sawin
    Commented Dec 30, 2014 at 12:44
  • 1
    $\begingroup$ To be technical: you mean the real part of the principal root function is monotinc with respect to the real part of the argument, if the imaginary part of the argument is held fixed. One doesn't speak of complex numbers being "larger" than other complex numbers, except perhaps in terms of modulus (which is not how you're arguing, I don't think). $\endgroup$
    – anon
    Commented Jan 2, 2015 at 8:45
  • 1
    $\begingroup$ @Goos since products can be expanded I think this (ii) can be ignored. $\endgroup$
    – mick
    Commented Jan 10, 2015 at 23:18

$$7 - \sqrt{5} = (7 - \sqrt{5}) \bigg(\frac{7 + \sqrt{5}}{7 + \sqrt{5}} \bigg) = \frac{44}{7 + \sqrt{5}}$$

  • 7
    $\begingroup$ see the edit I made. $\endgroup$
    – mick
    Commented Jan 7, 2014 at 23:25

Note: the following argument appears to fail. For example, $\sqrt{1+\sqrt{5}}$ cannot be written in the form I suggest.

Assuming that we are taking principal roots, I don't believe that this is possible. Inductively, every such expression can be written in the form $\alpha+\beta\sqrt{5}$, where $\alpha$ and $\beta$ are nonnegative real algebraic numbers fixed by some fixed Galois action $\sigma:\sqrt{5}\mapsto -\sqrt{5}$ (as achille hui points out, it is necessary to choose $\sigma$ to fix $\mathbb{R}$, or at least, to fix the algebraic reals in some radical extension, which is a subtle point, possibly a major flaw in the argument).

First, note that such a representation (with $\alpha,\beta$ algebraic, not necessarily nonnegative reals) always exists and is unique: if $x=\alpha+\beta\sqrt{5}$ is algebraic, then $\sigma x = \alpha-\beta\sqrt{5}$, so $\alpha=\frac{x+\sigma x}{2}$ and $\beta=\frac{x-\sigma x}{2\sqrt{5}}$.

The base case for our induction is the positive integers. The inductive steps of addition and multiplication are easy, so it remains to look at $n$-th roots, say $(\alpha+\beta\sqrt{5})^{1/n} = \gamma + \delta\sqrt{5}$ with $\alpha$ and $\beta$ as above. Then $\gamma$ and $\delta$ are certainly real, by the above formulas and the fact that $(\alpha+\beta\sqrt{5})^{1/n}$ is real. By expanding out $(\gamma+\delta\sqrt{5})^n$ and comparing signs, we can see that $\gamma$ and $\delta$ are both nonnegative.

It follows that any expression involving only addition, multiplication, principal roots, and positive integers, cannot produce $7-\sqrt{5}$ or anything like it, since the $\sqrt{5}$ coefficient is negative and both coefficients are rational (hence fixed by any choice of $\sigma$).

On the other hand, if we are not taking principal roots, then the problem is almost trivial, because we can write $\sqrt{-1} = 1^{1/4}$ for a suitable fourth root of $1$, and then write $7-\sqrt{5}=7+(\sqrt{-1})^2\sqrt{5}$.

  • $\begingroup$ +1 even though I'm still not sure about the part "$\gamma$ and $\delta$ are certainty real". $\endgroup$ Commented Jan 8, 2014 at 1:02
  • $\begingroup$ @achillehui $z=\gamma+\delta\sqrt{5}$ is real (since I am always taking the principal root, and we started with a nonnegative real), and $\sigma$ fixes $\mathbb{R}$. And $\gamma$ and $\delta$ can be written in terms of $z$ and $\sigma z$. $\endgroup$ Commented Jan 8, 2014 at 1:07
  • $\begingroup$ Okay, if $\sigma$ can be chosen to fix $\mathbb{R}$. But what happens if $\alpha - \beta\sigma < 0$ and $n$ is even? Shouldn't we have $(\alpha - \beta\sqrt{5})^{1/n} = \gamma - \delta\sqrt{5}$? that's what bothering me. $\endgroup$ Commented Jan 8, 2014 at 1:20
  • $\begingroup$ @achillehui You're right, I do need to pick $\sigma$ as such. But what you're saying can't happen by inductive hypothesis, if I understand correctly. The thing we're taking the root of already has both coefficients nonnegative by assumption. $\endgroup$ Commented Jan 8, 2014 at 1:23
  • $\begingroup$ @achillehui It may be that $\sigma$ cannot be chosen to fix $\mathbb{R}$; I have to think about it for a bit. But we only need to work in a finite extension, and it's true there. $\endgroup$ Commented Jan 8, 2014 at 1:31

Does multiplying by $\frac{7 +\sqrt 5}{7+ \sqrt 5}$ do what you want?

  • 3
    $\begingroup$ see the edit I made. Sorry to confuse. $\endgroup$
    – mick
    Commented Jan 7, 2014 at 23:26

Let $n$ be some positive integer.

$(7-\sqrt 5)^n = r$ where $r$ is in the desired form but there is no root symbol over the entire expression on the RHS.

Now every root symbol in $r$ denotes a principal root.

The number of ways that we can write the same identity $r$ but replace one or more of its roots by a nonprincipal is given by the product $a_1 a_2 ... a_m = k$ where the amount of roots symbols are $m$ and the roots are $*^{1/a_i}$. We denote the Original $r$ and the nonprincipal ones as

$r_0 =r$


( i is just an index )

Therefore (by very basic galois theory) there exists an integer polynomial $P$ of degree $k$ such that $P(x) = (x-r_0)(x-r_1)(x-r_2)...(x-r_i)...(x-r_k).$ ( notice that $P$ is completely factored and hence we have unique factorization )

Now clearly $r = r_0$ is the max in absolute value of all the $r_i$. Call this PROPERTY 1.

It is easy to show with some elementary ring theory or number theory that if

$(7-\sqrt 5)^n$ is a zero of an integer polynomial then so is $(7+\sqrt 5)^n$.

Therefore some $r_i$ must be equal to $(7+\sqrt 5)^n$. This is PROPERTY 2.

Now notice that |$(7+\sqrt 5)^n$| > |$(7-\sqrt 5)^n$|. This is PROPERTY 3.

Combining property 1,2 and 3 we see that $r$ must be both the largest in absolute value of all the $r_i$ and NOT the largest in absolute value of all the $r_i$.

Proof by contradiction.


Sorry if I did not use much of Galois theory , Im not so good with it.

  • $\begingroup$ This is a good idea, but poorly expressed. You'll want to clean up the exposition big time (you seem to have a big problem with communication). The key idea is what you call proposition 1; since it's so fundamental to the argument I recommend justifying it, even if it is intuitive. Here's a hint how: use the fact that root functions are monotonic increasing on positive reals, and show that if $r,s$ vary over the roots of two polynomials and $|r|$ and $|s|$ are both maximized when $r$ and $s$ are positive reals, then $|r+s|$ is maximized at those same values of $r$ and $s$. Induct. $\endgroup$
    – anon
    Commented Jan 2, 2015 at 8:41
  • $\begingroup$ @anon I think the main issue is that $(x-r_0)(x-r_1)...$ must indeed be an integer polynomial. For a fixed small degree polynomial this is easy to prove. Is there an easy way to prove that regardless of the structure ( amount of additions , roots etc ) of the roots $r_n$ for any degree ? $\endgroup$
    – mick
    Commented Jan 10, 2015 at 23:16

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