Does there exist an ordered field $(F;+,-,*,0,1,<)$, which is not $\mathbb{Q}$, such that its natural part and/or its integer part and/or its rational part is definable without parameters? By natural part, I mean the set $\{0,1,1+1,1+1+1,1+1+1+1,...\}$. By integer part, I mean the union of the natural part with the negatives of the natural part. By rational part, I mean the quotient of the integer part with itself.
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What does "definable without parameters" mean? What about the real numbers? – Henry Nov 27 '22 at 17:43
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1@Henry These are technical terms in mathematical logic; see e.g. here. – Noah Schweber Nov 28 '22 at 16:12
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Yes, as Atticus mentioned, in number fields you can always define the natural numbers. Check out theorem 3 page 208 of Rumely, Undecidability and Definability for the Theory of Global Fields . The idea is that you can quantify over finite sets in number fields, and then you can encode the natural numbers via saying that $n$ is a natural number iff there is a finite set that includes $0$ and whenever any $t$ is in the finite set, then either $t=n$ or $t+1$ is also in the finite set. Originally this is a theorem by Julia Robinson. Then just take any real number field and you are done.
Florian Felix
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hi Florian; nice to see you on here! :) I have erased my comment now that you've given a reference – Atticus Stonestrom Nov 29 '22 at 18:04