home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.logic
- Path: sparky!uunet!caen!zaphod.mps.ohio-state.edu!magnus.acs.ohio-state.edu!wjcastre
- From: wjcastre@magnus.acs.ohio-state.edu (W.Jose Castrellon G.)
- Subject: Re: Non-standard integers.
- Message-ID: <1992Aug19.151419.8835@magnus.acs.ohio-state.edu>
- Sender: news@magnus.acs.ohio-state.edu
- Nntp-Posting-Host: bottom.magnus.acs.ohio-state.edu
- Organization: The Ohio State University,Math.Dept.(studnt)
- References: <1992Aug18.133344.385@csc.canterbury.ac.nz> <16rh2nINN6og@agate.berkeley.edu> <1992Aug19.172922.410@csc.canterbury.ac.nz>
- Date: Wed, 19 Aug 1992 15:14:19 GMT
- Lines: 37
-
- In article <1992Aug19.172922.410@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
- >> Kemeny's conjecture is false: If M is a non-standard model of
- >>PA, then there is a non-standard integer n such that no integer in the
- >>"standard row" of n is divisible by all standard integers.
- >>
- >> To see this, take n which is odd but is divisible by all
- >>standard odd integers. If m differs from n by a standard integer, then
- >>there must be a standard odd prime which does not divide the
- >>difference. Hence m is not divisible by that prime.
-
- >
- >Looking at this bit...
- >
-
- >>take n which is odd but is divisible by all standard odd integers.
-
- >
- >...at first it wasn't clear to me that every nonstandard model would have
- >such an n . But I suppose this can be seen by taking any odd nonstandard k,
- >and forming n = k(k-2)(k-4)(k-6)..... .
-
- If what is meant here is that n equals to the product of the odd predecessors
- in the _row_ of k, then there is no model with such n, just as there is no
- number equal to the product of all standard integers; the reason for that is
- overspill, or else that by being able to do that you could pin down the
- standard integers.(Boolos&Jeffry surely discusses that).
- To see that there is an odd non-standard divisible by all odd standard naturals
- just take any non-standard that is divisible by all standard numbers and get
- its greatest odd divisor. You can prove that there is always such a divisor in
- PA for all numbers, therefore it is true in all models of PA.
-
- >
- >Presumably this operation can be made legitimate in some simple way ?
-
- Yes, by Godel coding, but n will necessarily be divisible by some non-standard
- naturals.
- WJCG
-