Timeline for Intersection of integers and rationals defined by logic
Current License: CC BY-SA 3.0
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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 |