3
$\begingroup$

Euclidean numbers are those real number that can be constructed from the natural numbers by a finite chain of +,-,*,/ and $\sqrt{}$. They are also called Constructible Numbers.

I am now interested in whether there is an (easy) algorithm to determine if two Euclidean Numbers are equal (as real numbers) or, equivalently, if a given Euclidean Number is zero.

$\endgroup$
1
  • 4
    $\begingroup$ Given that the complexity of the square-root-sum problem is open, no efficient algorithm is known which could tell if some such presentation is well-formed in the sense of only including square roots of nonnegative numbers. $\endgroup$
    – Noah Stein
    Commented Jul 14, 2015 at 16:03

1 Answer 1

6
$\begingroup$

If $\alpha$ is constructible, you can construct a minimal polynomial of $\alpha$ recursively as follows: If $\alpha=\sqrt{\beta}$, and $\beta$ has minimal polynomial $P(x)$, then $P(x^2)$ is a polynomial with $P(\alpha)=0$. If $\alpha=\beta+\gamma$, and $\beta, \gamma$ have minimal polynomials $P, Q$ with roots $x_1, \ldots, x_n$ and $y_1, \ldots, y_m$, respectively, then $R(x) = \prod_{i,j}(x-x_i-y_j)$ is a polynomial with $R(\alpha)=0$. Note that using elementary symmetric functions you can compute $R$ without knowing the $x_i, y_j$'s. Factorize the resulting polynomial over $\mathbb{Q}$, and check numerically which factor contains $\alpha$.

The last step uses that roots of integral polynomial cannot be too close together. An efficient version of this statement is Liouville's theorem, using this you know that you have to compute $\alpha$ only to a certain predictable precision.

Note that you cannot do without some numerical computation, because the usual convention that the square root is a map $\sqrt{\phantom{x}}:[0, \infty)\rightarrow[0, \infty)$ is quite artificial from an algebraic point of view, so distinguishing $2+\sqrt{2}$ from $2-\sqrt{2}$ requires some non-field theoretic input, like positivity or order.

This algorithm is not polynomial, as the degree of the minimal polynomial grows too fast, however, in practice it can be used for rather difficult looking expressions.

$\endgroup$
4
  • $\begingroup$ This may be a dumb question, but what are the details on "using elementary symmetric functions you can compute R without knowing the $x_i$,$y_j$'s"? Is this always true? $\endgroup$ Commented Jul 23, 2019 at 3:31
  • $\begingroup$ In fact, I'm still not convinced that $R(x)$ will be a polynomial with rational coefficients. Why is that true? $\endgroup$ Commented Jul 23, 2019 at 3:37
  • $\begingroup$ @AndréPorto: A symmetric polynomial $P$ in $x_1,\ldots,x_n$ can be written as $Q(\omega_1,\ldots,\omega_n)$, where $\omega_1,\ldots,\omega_n$ are the elementary symmetric polynomials in $x_1,\ldots,x_n$, and the coefficients of $Q$ are contained in the same field as those of $P$. $Q$ is computed by a greedy algorithm, which is fast. For $R$ you have to repeat this: View $R$ as a polynomial in $x, y_1\ldots, y_i$, compute the coefficients of this polynomial in terms of the coefficients of $P$, then view this polynomial as a polynomial in $x$, and compute $R$ in terms of the coefficients of $Q$. $\endgroup$ Commented Jul 24, 2019 at 11:14
  • $\begingroup$ @AndréPorto: Exampe: If $P, Q$ are quadratic, the coefficient of $x^2$ in $R$ (the most difficult one) is $x_1^2+x_2^2+4x_1x_2+y_1^2+y_2^2+4y_1y_2+x_1y_1+x_2y_1+x_2y_1+x_2y_2$, which is $\omega_1(x_1, x_2)^2+2\omega_2(x_1, x_2) +\omega_1(y_1, y_2)^2+2\omega_2(y_1, y_2) + \omega_1(x_1, x_2)\omega_1(y_1, y_2)$. $\endgroup$ Commented Jul 24, 2019 at 11:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .