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The set of constructible numbers

https://en.wikipedia.org/wiki/Constructible_number

is the smallest field extension of $\mathbb{Q}$ that is closed under square root and complex conjugation. I am looking for an algorithm that decides if two constructible numbers are equal (or, what is the same, if a constructible number is zero).

As equality of rational numbers is trivial, this algorithm probably needs to reduce the complexity of the involved number in several steps.

Does such an algorithm exist? If yes, what does it look like?

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    $\begingroup$ In what form would the two constructible numbers be presented to the algorithm? $\endgroup$ Commented Mar 16, 2017 at 19:34
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    $\begingroup$ A constructible number would be defined by a finite sequence of $+,-,*,/, \sqrt_\pm{}$ and complex conjugation applied to rational numbers, where $\sqrt_\pm{}$ would be one of the possible square roots of the given number. $\endgroup$ Commented Mar 16, 2017 at 19:37
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    $\begingroup$ There exist algorithms to deal with algebraic numbers in $\mathbb{C}$ (i.e., compute their sums, differences, products and inverses, solve algebraic equations on them, take their real and imaginary parts, and compare them for equality and, when they are real, for order). This solves your problem and much more. The basic idea is to represent an algebraic by a polynomial in $\mathbb{Q}[t]$ of which it is a root, and an interval (in real and imaginary parts if necessary) in which it is the only root. $\endgroup$
    – Gro-Tsen
    Commented Mar 16, 2017 at 19:39
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    $\begingroup$ Furthermore, I should add that such algorithms are not purely theoretical: they are implemented, e.g., in Sage, which is capable of handling exact algebraics (although, to be fair, complexity becomes quickly awful whenever you do something not too trivial). $\endgroup$
    – Gro-Tsen
    Commented Mar 16, 2017 at 19:41
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    $\begingroup$ See, for example, Henri Cohen, A Course in Computational Algebraic Number Theory (Springer 1993 / GTM 138), specifically §4.2. But this is merely one possible reference among many: the subject is really quite standard, and googling "algorithm exact algebraic arithmetic" returns various relevant results (which probably themselves point to other references). More abstractly, Rabin's theorem on the algebraic closure can also be relevant. $\endgroup$
    – Gro-Tsen
    Commented Mar 16, 2017 at 21:08

1 Answer 1

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Although this can be done using the complicated algorithms for general algebraic numbers, there’s a much simpler recursive algorithm for constructible numbers that I implemented in the Haskell constructible library.

A constructible field extension is either $\mathbb Q$ or $F{\left[\sqrt r\right]}$ for some simpler constructible field extension $F$ and $r ∈ F$ with $\sqrt r ∉ F$. We represent an element of $F{\left[\sqrt r\right]}$ as $a + b\sqrt r$ with $a, b ∈ F$. We have these obvious rules for $a, b, c, d ∈ F$:

$$\begin{gather*} (a + b\sqrt r) + (c + d\sqrt r) = (a + c) + (b + d)\sqrt r, \\ -(a + b\sqrt r) = -a + (-b)\sqrt r, \\ (a + b\sqrt r)(c + d\sqrt r) = (ac + bdr) + (ad + bc)\sqrt r, \\ \frac{a + b\sqrt r}{c + d\sqrt r} = \frac{ac - bdr}{c^2 - d^2 r} + \frac{bc - ad}{c^2 - d^2 r}\sqrt r, \\ a + b\sqrt r = c + d\sqrt r \iff a = c ∧ b = d. \end{gather*}$$

To compute the square root of $a + b\sqrt r ∈ F{\left[\sqrt r\right]}$:

  • If $\sqrt{a^2 - b^2 r} ∈ F$ and $\sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2}} ∈ F$, then $$\sqrt{a + b\sqrt r} = \sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2}} + \frac{b}{2\sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2}}}\sqrt r ∈ F{\left[\sqrt r\right]}.$$
  • If $\sqrt{a^2 - b^2 r} ∈ F$ and $\sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2r}} ∈ F$, then $$\sqrt{a + b\sqrt r} = \frac{b}{2\sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2r}}} + \sqrt{\frac{a + \sqrt{a^2 - b^2 r}}{2r}}\sqrt r ∈ F{\left[\sqrt r\right]}.$$
  • Otherwise, $\sqrt{a + b\sqrt r} ∉ F{\left[\sqrt r\right]}$, so we represent it as $$0 + 1\sqrt{a + b\sqrt r} ∈ F{\left[\sqrt r\right]}\left[\sqrt{a + b\sqrt r}\right].$$

In order to compute with numbers represented in different field extensions, we need to rewrite them in a common field extension first. To rewrite $a + b\sqrt r ∈ F{\left[\sqrt r\right]}$ and $c ∈ G$ in a common field extension, first rewrite $a, b, r ∈ F$ and $c ∈ G$ in a common field extension $H$. If $\sqrt r ∈ H$, then we have $a + b\sqrt r, c ∈ H$; otherwise we have $a + b\sqrt r, c + 0\sqrt r ∈ H{\left[\sqrt r\right]}$.

I implemented the constructible real numbers and built the constructible complex numbers generically on top of those, to enable ordering relations and to avoid having to think too hard about branch cuts.

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