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Apr 20, 2010 at 11:56 comment added David E Speyer Could you spell out this reduction? I don't see it. (Even assuming that you meant $\mathbb{Q}$, not $\mathbb{Z}$.)
Apr 19, 2010 at 20:59 comment added Donu Arapura Hilbert's 10th is undecidable over $\mathbb{Z}$, but probably you meant $\mathbb{Q}$. Yes, I thought about that, but the reduction isn't clear (to me).
Apr 19, 2010 at 20:14 history answered captain obvious CC BY-SA 2.5