Peter is right that Q as a field contains order information absent from Q as either a linear or affine space. (−*x*² ≤ 0 ≤ *x*².)

For the order to be definable from the algebraic structure, you need to consider Q as a field

It certainly suffices, and indeed is traditional. But is it necessary for the construction of the affine real line? In particular do we have to define multiplication on the rationals only to throw it away at a later stage? Obviously the ordered dyadic rationals would suffice. Equally obviously they don't form a field. But can they be defined as an ordered affine space with the same degree of formality and economy as we routinely define the ordered linear space of rationals? Claim. The ordered algebra (D, xy, x+y, ≤) consisting of the set D of dyadic rationals under the two binary operations xy denoting 2y - x and x+y denoting (x + y)/2 can be defined order-algebraically up to isomorphism as the free ordered algebra on the ordinal 2 satisfying finitely many equations in the two operations along with the inference rules, from x ≤ y infer each of y ≤ xy yx ≤ x x ≤ x+y x+y ≤ y The Dedekind cuts being cuts in the rational line, the following claim depends on a distinct name for the corresponding notion of a cut in the ordered affine line of dyadic rationals, which I suggest calling a dyadic cut. (When cutting at a dyadic rational follow a consistent convention as to which side to associate that rational, as done with the Dedekind cuts, and always have both sides of the cut nonempty.) Claim. The dyadic cuts in the ordered affine space of dyadic rationals are in order-preserving bijection with the Dedekind cugts in the ordered linear space of rationals, with the cuts at dyadic rationals correctly matched to their Dedekind counterparts. Vaughan Pratt On Sun, Jun 10, 2018 at 6:54 AM, Peter Johnstone <ptj@dpmms.cam.ac.uk> wrote:

Sorry, what I wrote was a bit sloppy. Vaughan is right that the problem doesn't arise with the passage from considering Q as a linear space to considering it as an affine space, since it already has order- reversing linear automorphisms. For the order to be definable from the algebraic structure, you need to consider Q as a field, which is what the usual Dedekind-section construction does.

Peter Johnstone

On Thu, 7 Jun 2018, Vaughan Pratt wrote:

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Not quite: the affine rational line doesn't have a definable total order,

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since it has order-reversing automorphisms, so any definition using Dedekind sections is problematic.

Morphism-wise, since the affine transformations are just the composition of a linear transformation with a translation, and translation of the rational line preserves order, affinity can't be the problem here.

Structure-wise, one can equip the rational line with either its linear combinations or its linear order, or both. Using both eliminates the order-reversing linear transformations. "Affine" only makes sense in the context of having the linear combinations, as "affine" limits the linear combinations to those whose coefficients sum to one. If it is ok for the linear combinations and the linear order to coexist, it must be even more ok for the affine combinations and the linear order to coexist.

So whether one considers the morphisms or the structure they preserve, affinity (affineness?) must be a red herring here: any problem for the rational line as an affine space is surely also a problem for it as a vector space.

Vaughan Pratt

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