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Oct 26, 2012 at 14:57 comment added Emil Jeřábek Sure. If $Y\subseteq\mathbb Q^d$ is definable in $(\mathbb Q,+,\le)$, then $Y\cap\mathbb N^d$ is definable in $(\mathbb N,+)$ by quantifier elimination of $(\mathbb Q,+,\le)$.
Oct 26, 2012 at 14:43 history asked Steven CC BY-SA 3.0