How to rewrite $7-\sqrt 5$ in root form without a minus sign?

Question

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 ?

EDIT

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.

Answer 1

$7-\sqrt{5}$ 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.

Answer 2

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

Answer 3

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 $7-\sqrt{5}=7+(\sqrt{-1})^2\sqrt{5}$.

Answer 4

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

Answer 5

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

$r_0 =r$

$r_i$

( 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.

QED.

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