I think it boils down to what it means (in your system) to be an "integer". 1. ceil(x) always returns an integer, so "integer" is the range of ceil(). 2. integers are totally ordered. 3. between any two integers, there are only a finite number (including zero) of other integers. But this is basically well-ordering. There are a number of interesting models which are less powerful than the real numbers. There are "real closed fields", which don't support integers (proof: real closed fields are decidable, whereas full integers get you undecidability). There are decidable models which include a "successor" function, and can support addition, and multiplication by a constant, but not multiplication of two arbitrary numbers. At 09:40 PM 9/9/2008, James Propp wrote:

Using the well-orderedness property of the non-negative integers, plus the usual axioms of R as an ordered field and Z as an ordered ring, one can show that the the Archimedean property ("For all real x there exists an integer n with n > x") implies the existence of the ceiling function ("For all real x there exists a smallest integer n with n greater than or equal to x"). But does one need well-orderedness (or equivalently the axiom of induction) to prove this? That is, from the assumption that the ceiling function exists, can one derive the axiom of induction?

It seems to me that the proposition that the ceiling function exists has an intermediate character: it cannot be derived from the ordered field and ordered ring axioms, but neither is it strong enough to imply induction. Are there a couple of nice non-standard models of (R,Z,0,1,+,x,>) that make this completely patent?

Jim