# Can we alter Hilbert's axioms to have $\mathbb{Q}^3$ as a unique model?

Can we alter Hilbert's axioms to have $\mathbb{Q}^3$ as a unique model?

The critical axioms seem to be the congruence axioms IV.1 and IV.4, and presumably the line completeness axiom V.2.

But how are they to be modified?

IV.1 might be replaced by requiring that there are counter-examples (irrationality of $\sqrt{2}$) and appropriately relaxing "congruent" to "almost congruent" (= "arbitrarily close to congruence").

But what about line completeness then, since it might be possible to add irrational points to $\mathbb{Q}^3$ such that the modified axioms still hold?

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Why would you want to? Q^3 is not a very good model of geometry. – Qiaochu Yuan Dec 2 '10 at 16:30
Which geometry? And what is a "good" model? – Hans Stricker Dec 2 '10 at 16:53
Physicists are in search for discrete space(-time). So I wondered why not start with a "somehow-discrete" space? How far do we get? – Hans Stricker Dec 2 '10 at 16:54
Not far at all - you would end up embedding in R^3 if you wanted to do calculus (useful calculus). It is generally accepted the real numbers() are the "best we can do". For instance, preservation of compact sets doesn't hold, intermediate value theorem fails, unfirm continuity on compact sets fails, zero derivitve but non constant function exists - let alone what would happen in Q^3; say bye-bye to stokes theorem, etc. () perhaps complex. – Adam Jan 25 '12 at 23:14
There is a mathematician who prefers to work with rational numbers when it comes to geometry. He has written a book about it. In case you are interested: web.maths.unsw.edu.au/~norman – Dejan Govc Mar 16 '12 at 20:55