Proving Richardson's theorem for constants

Question

(I also asked this 18 hours ago on mathoverflow, but did not yet get any responses there.)

Richardson's theorem is given in this wikipedia article. $\:$ In this answer, Eric Towers states that

An algebraic expression is decidable. This is also the Tarski-Seidenberg theorem.

Algebraic expressions plus $\exp$ are decidable if Schanuel's conjecture holds.
Otherwise, this is unknown. See Tarski's exponential function problem.

Throw in sine and determining whether the constant expression vanishes is undecidable. (This can be linked to the undecidability of models of the integers (a la
Godel's Incompleteness Theorems) via the zeroes of $\sin (\pi k)$ for any integer $k$.
See Hilbert's Tenth Problem and Matiyasevich's Theorem.)
Sine isn't critical; any (real-)periodic function would do.

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How can the last bullet point of that quote be proven?

Eric Towers · Answer 1 · 2015-11-13T06:49:24.280

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Sun Zhi-Wei in "Reduction of unknowns in Diophantine representations", building on work by Matiyasevich and the MRDP theorem (Matiyasevich, Robinson, Davis, Putnam), has shown that solvability of Diophantine equations can be undecidable with as few as nine variables in $\mathbb{Z}$. Let $0 = D$ be any Diophantine equation with undecidable solvability and a single solution, but where the variables $x_1, x_2, ..., x_k$ of $D$ are taken to be real. Then for $0 = D + \sum_{i=1}^k \sin^2 \pi x_i$ to be satisfied, it is necessary that $(x_1, x_2, \dots, x_k) \in \mathbb{Z}^k$. I.e., this is an encoding of the problem "solve $D=0$ over the integers". By construction, it is undecidable whether this equation has a solution.

edited Nov 13 '15 at 06:49

answered Nov 13 '15 at 06:42

Eric Towers

67,037

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The wikipedia article's description does not allow $\pi$ in the expressions. $:$ (Although the question you answered did allow $\pi$, your answer enumerated operations but did not include $\pi$ as one of them.) $:$ Furthermore, even with $\pi$, the expressions you give are not constant expressions. $;;;;$ – Nov 13 '15 at 15:30
@RickyDemer : This equation has one solution, therefore, it represents a constant. The article you cite has "Specifically, the class of expressions for which the theorem holds is that generated by rational numbers, the number $\pi$,...". – Eric Towers Nov 13 '15 at 18:39
Do you mean "at least one solution" or "exactly one solution"? $:$ In either case, are you referring to the equation immediately after "for"? $:$ That equation does not necessarily have one solution, since for example, D could be everywhere strictly positive. $;;;;$ – Nov 13 '15 at 19:29
@RickyDemer : If $D$ is everywhere strictly positive, equality with zero is decidable. Undecidability requires being indefinite (having both positive and negative values). Additionally, if one starts with a $D$ having multiple solutions, it is a basic exercise to intersect ideals eliminating all but one solution, still over the integers. – Eric Towers Nov 14 '15 at 09:22
Okay, so based on your last sentence and the existence of a universal diophantine equation, one can get a decidable set of diophantine equations such that each of them has at most one solution and determining solvability is still as hard as the halting problem (and so undecidable). $:$ (continued …) $;;;;$ – Nov 14 '15 at 17:58
(… continued) $;;;$ How do you go from there to undecidability of promise problem whose yes instances are [uniquely-solvable equations whose solution satisfies $: x_{\hspace{.02 in}0} = 0 :$] and whose no instances are [uniquely-solvable equations whose solution satisfies $: x_{\hspace{.02 in}0} \neq 0 :$]? $;;;;;;;;$ – Nov 14 '15 at 17:58
@RickyDemer : I don't know why (1) you can't read plain English, and (2) you choose to continue arguing when you seem to be fundamentally unfamiliar with the large existing literature on this subject matter. You write "one can get a decidable set of diophantine equations". I have referenced the existence of undecidable Diophantine equations. Do you understand what is undecidable about these? – Eric Towers Nov 14 '15 at 22:38
Similarly, I also don't know why 1+1=3. $:$ On the other hand, I do understand that [the problem for which each instance is a diophantine equation and the outputs are [if the diophantine equation is solvable over the integers then YES else NO]] is undecidable. $:$ (By "a decidable set of diophantine equations", I meant "[a set of diophantine equations] for which membership in the set is decidable", not "[a set of diophantine equations] such that solvability becomes decidable when restricted to that set", which might have caused confusion.) $;;;;$ – Nov 15 '15 at 08:53
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@EricTowers You seem to misunderstand the intent of this question. The question asks for a proof of undecidability of the following problem: you are given an expression not involving any variables which uses these-and-these functions, e.g. $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} - 1$. Now the question is, does the constant expressed by this expression equal $0$? (in this example, it does. The question is about general procedure of determining this) – Wojowu Nov 15 '15 at 11:01
Yes, that was my intent. $:$ Additionally, I don't even see a way to show undecidability of the promise problem I mentioned in my comment about 25 hours ago, which is at least as hard, since one could just do x minus expression_with_no_variables = 0. $:$ ("I don't even see ..." is as long as the algorithm is allowed to run forever when the promise fails. $:$ I certainly see how to show that no total algorithm can solve that promise problem.) $;;;;$ – Nov 15 '15 at 19:10

Proving Richardson's theorem for constants

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