5
$\begingroup$

I posted this question at math.stackexchange.com but didn't get an answer.

Motivation

Physicists are in search for a model of discrete space(-time) for a long time. So I wondered why not start with a "somehow discrete" space? How far do we get?

Question

Can we alter the axioms of Euclidean space, e.g. Hilbert's axioms, to have $\mathbb{Q}^3$ as a unique model?

The crucial 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?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ The reasons that physicists tend to give for considering discrete models of spacetime also eliminate metric spaces like $\mathbb{Q}^3$ from consideration. In particular, path integration seems to be significantly harder to axiomatize in your model than in the traditional picture. $\endgroup$
    – S. Carnahan ♦
    Commented Dec 10, 2010 at 7:10
  • $\begingroup$ That's a good point, thank you! (Nevertheless, in principle it might be the case, that physical space is locally isomorphic to $\mathbb{Q}^3$, isn't it? But we just could not - by finitary means - distinguish it from $\mathbb{R}^3$?) $\endgroup$
    – Hans-Peter Stricker
    Commented Dec 10, 2010 at 7:53
  • $\begingroup$ For me a more natural candidate would be the $A^3$, where $A$ is the set of real rational integers. Otherwise, you would end up with certain standard curves (which physisist are interested) only having finitely many points. The question is then: what is arclenght or area and how do you calculate it? $\endgroup$
    – Nick S
    Commented Dec 10, 2010 at 19:04
  • $\begingroup$ What are the real rational integers? Googling (google.com/search?q="real+rational+integers") didn't reveal a lot. $\endgroup$
    – Hans-Peter Stricker
    Commented Dec 12, 2010 at 15:13

2 Answers 2

Reset to default
6
$\begingroup$

Every consistent first-order axiomatic system has models of every cardinality. Given this fact, if your modified axioms has ${\mathbb{Q}}^3$ as a model, then ${\mathbb{Q}}^3$ will not be a unique model. However it still makes sense to ask if ${\mathbb{Q}}^3$ is a unique model.

So long as your modification uses only countably many symbols, then there will always be a countable models. Countable models are "discrete" in a sense.

In Hilbert's original writings, he came up with a concrete countable model to his axiomatization of Euclidean geometry. This model is the smallest Pythaogrean field containing ${\mathbb{Q}}$, i.e. $\sqrt{a^2+b^2}$ is in the field if $a$ and $b$ are. If you allow adding irrational points, this countable model is an instance of a discrete model.

Improve this answer
$\endgroup$
2
  • 3
    $\begingroup$ From the wikipedia article: "Hilbert's axioms, unlike Tarski's axioms, do not constitute a first-order theory because the axioms V.1–2 cannot be expressed in first-order logic". On the other hand, if you use first-order axioms, no infinite field (and no theory of a projective plane that can define a field, i.e. a Desarguesian one: see <a>en.wikipedia.org/wiki/Projective_plane</a>) can be omega-categorical. $\endgroup$
    – Antongiulio
    Commented Dec 9, 2010 at 13:10
  • 3
    $\begingroup$ So, of course, answering this question requires axioms that (like Hilbert's) are not first-order. $\endgroup$
    – Gerald Edgar
    Commented Dec 9, 2010 at 15:17
1
$\begingroup$

As Colin Tan said, "[using] only countably many symbols, then there will always be a countable models." Whereas the field $\mathbb{Q}$ of rational numbers uses a finite-number of symbols for an uncountable number of rational numbers. I misunderstood and misapplied a concept.

The field generated by rational numbers is quite different from the "approximation space" rendered by using a finite number of bits interpreted as a floating-point number low-precision approximation to real numbers. I'm editing my answer to point out my misunderstanding. @Hans-Stricker, I've fixed my error by pointing it out, but leaving it up (below the ruled line) so that some other bit-flipper like me will see why {0,1}$^n \times${0,1}$^n$ is not equivalent to $\mathbb{Q}$

below this is my original (erroneous) answer

Similarly, every numerical simulation in physics (or chemistry, biology, physiology, or medicine) always has to use finite precision representation of values, such that there is a limit to the largest and smallest integer represented by a fixed number of bits, and such that there is a limited amount of "floating-point-precision" available in dividing the bits of a floating-point representation of a real number into a set number of bits for the mantissa and a set number of bits for the exponent.

For example, assuming that $d=64$-bits are used to represent "real numbers" as floating point numbers in computations, $m=48$ bits may be allocated to the mantissa, allowing the numerator to be $2^{48}$ yielding approximately $14$ digits of base-ten specificity to the numerator; this leaves $d-m=16$ bits to the exponent which may be signed (+/-) yielding a range of -32768 to +32767.

In this case, the floating point number is in the range $n\times 2^{d-m}$, where $-(2^{47} \le n \le +(2^{47}-1)$, and ${-32768} \le d \le {32768}$.

If the total number of bits is $d$, the number of bits allocated to the exponent, $m$, may be decreased while simultaneously increasing the number of bits, $d-m$, allocated to the mantissa, increasing the "precision" of the numerator while decreasing the range over $\mathbb{R}$ spanned by this particular approximating set of {0,1}$^m \times ${0,1}$^{d-m}$ (which is not equivalent to $\mathbb{Q}$, as I erroneously stated originally)

Thus every numerical simulation is already, in a way, based on $\mathbb{Q}^d$ when models of $d$-dimensional systems are created and iterated using Euler or Runge-Kutta of whatever order.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Hi, criticism noted; however, an explanation of criticism would be more useful than just a down-vote with no words behind it. $\endgroup$
    – sleepless in beantown
    Commented Dec 10, 2010 at 14:30
  • $\begingroup$ It wasn't me who down-voted, but I cannot see that this an answer to my question. $\endgroup$
    – Hans-Peter Stricker
    Commented Dec 10, 2010 at 15:09
  • 1
    $\begingroup$ @sleepless: the OP is asking for a particular system of axioms which has a certain structure as a unique model. Although this is motivated by physics, it seems to be a pure math question. Your answer speaks of numerical approximations and applications in the sciences, so seems to be getting at the applied math notion of "model". In other words, it might possible be based on a misinterpretation, but in any event I don't see how it addresses the question. $\endgroup$
    – Pete L. Clark
    Commented Dec 10, 2010 at 17:34
  • $\begingroup$ Note also that what Colin Tan says is true for first-order theories (by Lowenheim-Skolem) but does not apply to Hilbert's axioms. By way of analogy, the standard axioms for the real numbers -- including completeness, which is not first-order -- use only countably (indeed finitely) many symbols but have only an uncountable model. $\endgroup$
    – Pete L. Clark
    Commented Dec 10, 2010 at 17:36
  • $\begingroup$ @Pete-Clark, you are correct. I went on a left-turn after "countably many" symbols, as you are correct in that the field generated by rational numbers is quite different from the "approximation space" rendered by using a finite number of bits interpreted as a floating-point number low-precision approximation to real numbers. I'm editing my answer to point out my misunderstanding. @Hans-Stricker, I'll fix my error, but leave it up so that some other bit-flipper like me will see why {0,1}$^n \times${0,1}$^n$ is not equivalent to $\mathbb{Q}$. $\endgroup$
    – sleepless in beantown
    Commented Dec 10, 2010 at 18:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .