🚀 go-pugleaf

>	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).....  .

Presumably this operation can be made legitimate in some simple way ?
--------------------------------------------------------------------------
            Bill Taylor              wft@math.canterbury.ac.nz 
--------------------------------------------------------------------------
                  Kleeneness is next to Godelness.
--------------------------------------------------------------------------