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Jul 1, 2010 at 10:54 comment added Joel David Hamkins Yes, that's the right idea. Residue mod n is definable in PA. The m here is standard finite, sine it is less than n+1, and hence definable.
Jul 1, 2010 at 10:41 comment added Kate Hodesdon Sorry, thats formatted horribly. I meant the lines to have linebreaks between - I was just checking that the residues are definable.
Jul 1, 2010 at 10:27 comment added Kate Hodesdon Thanks Joel, that's great. So, will we have something like the following: Let a, b be the nonstandard numbers differing by finite n. a=k(n+1) + m b=k(n+1) + m+n \Exists y[yn+1 + m = x] in tp(a) \Exists y[yn+1 + m-1 = x] in tp(a) n is finite, so definible. And so are the m, m-1, right? Thanks again
Jul 1, 2010 at 9:36 vote accept Kate Hodesdon
Jun 30, 2010 at 17:44 history edited Joel David Hamkins CC BY-SA 2.5
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Jun 30, 2010 at 17:03 history answered Joel David Hamkins CC BY-SA 2.5