By MDPR, c.e. sets are the same as Diophantine sets. An existential definition of $\Bbb Z$ in $\Bbb Q$ would solve Hilbert's 10th problem over $\Bbb Q$.

The idea is not so crazy, you just input a pair $(a,b)$ of integers (I guess a pair of pairs of naturals, but that's a detail) and if $b \vert a$ you halt and if it doesn't you run forever. So the pairs $(a,b)$ for which the algorithm halts are rational numbers that are actually integers and if it doesn't they're rational numbers that are not integers.

N.B. This is so simple I know I cannot be correct. Many people much more experienced than me have tried and failed with many more complicated methods. This is more of a request to explain why this is wrong.

What I am guessing could be my incorrect assumption:

  1. You can just give an algorithm that distinguishes between integers and rationals and by MDPR the c.e. set this algorithm produces is a Diophantine set and hence a Diophantine definition of $\Bbb Z$ in $\Bbb Q$.
  2. You can sneak the integers into my algorithm by defining rationals as pairs of integers and not worry about circularity.

If I understand your idea correctly, you're not actually defining $\mathbb{Z}$ in $\mathbb{Q}$. You're defining $\mathbb{Z}$ in a structure $\tilde{\mathbb{Q}}$, which intuitively consists of $\mathbb{Q}$ together with the details of its construction via ordered pairs of integers. This is unavoidable when you try to check whether $a\vert b$ in a rational $q=(a,b)$: you need to be able to first "unpack" the rational into an ordered pair, and second check whether there is an integer (not a rational) which when multiplied by the left coordinate yields the right coordinate.

Now this absolutely works, but it doesn't help define $\mathbb{Z}$ in $\mathbb{Q}$ at all, let alone in a $\Sigma_1$ way. The structure $\tilde{\mathbb{Q}}$ is, when we dive into the details, $\mathbb{Z}$ itself (or perhaps more accurately, $\mathbb{Z}$ together with some inessentialy "syntactic sugar")! And an existential definition of $\mathbb{Z}$ in $\mathbb{Z}$ isn't very surprising.

It may help at this point to look at a context where we know such an idea must fail - for example, the field $\overline{\mathbb{Q}}\cap\mathbb{R}$ of algebraic real numbers. As a real closed field this is decidable, and so cannot define $\mathbb{Z}$ at all, but it can be implemented in $\mathbb{Z}$ in a manner similar to the implementation of rationals as (equivalence classes of) ordered pairs of integers.

I think the source of this confusion is a reflexive identification of computations with $\Sigma_1$-definitions. The fact that we can do this over $\mathbb{Z}$ (or similar) is a highly nontrivial property of $\mathbb{Z}$, and we're not guaranteed anything similar about $\mathbb{Q}$; in fact, this is essentially what you're trying to prove in the first place!

For this reason it's important to grapple with $\Sigma_1$ definitions themselves, not intuitive analogues. A $\Sigma_1$ definition of $\mathbb{Z}$ in $\mathbb{Q}$ can't "unpack" an element of $\mathbb{Q}$ into some explicit implementation in a different context, nor can it in any way "search" over natural numbers or similar objects.

  • $\begingroup$ So, is it reasonable to say that the method is flawed because of both of my guesses? Both that, by "packing" the rational into the ordered pair $(a,b)$, I am already sneaking $\Bbb Z$ into $\tilde{\Bbb Q}$, and that I am conflating computation with $\Sigma_1$ definition? On the second point, should I then believe that there is no sense in trying to go in this direction, [computation] $\to$ [$\Sigma_1$ definition], unless somehow I truly quantify over "sugar-free" rationals? $\endgroup$
    – BigSocks
    Feb 10 at 20:04
  • $\begingroup$ @BigSocks Yes, basically. $\endgroup$ Feb 10 at 20:04
  • $\begingroup$ Awesome, thank you very much for your input and careful analysis of my mistake! $\endgroup$
    – BigSocks
    Feb 10 at 20:06

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