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We have the well-known Peano axioms for the natural numbers and the real numbers can be characterized by demanding them to be a Dedekind-complete, totally ordered field (or some variation of this).

But I never saw any axiomatic characterization of the rational numbers! Either they are constructed out of the natural numbers or are found as a subset of the reals.

I know that the rational numbers are unique in the sense that they are the smallest totally ordered field. But it is somewhat unsatisfactory for me to define them to be a totally ordered field which has an order-preserving embedding into any other totally ordered field. This would be like defining the real numbers to be an Archimedean totally ordered field such that every other Archimedean totally ordered field can be order-preservingly embedded into it - this is some ugly definition (for me) and I find the usual one much better.

So what is a nice axiomatic characterization of the rational numbers?

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Smallest field containing the natural numbers. –  i. m. soloveichik Dec 19 '12 at 23:36
    
@AlexE: I have posted a solution that you might be looking for. –  Haskell Curry Dec 20 '12 at 0:42
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I'm a bit confused - the real numbers are the unique totally ordered field satisfying the least upper bound property. The natural numbers are not the unique thing satisfying the Peano axioms. In what sense do you mean "characterization"? –  Jason DeVito Dec 20 '12 at 1:29
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Using second-order logic, both the Peano axioms and the axioms for real numbers characterize uniquely. Going down to first-order, it is not possible for both to characterize them uniquely, if I recall correctly. –  AlexE Dec 20 '12 at 8:48
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6 Answers

up vote 14 down vote accepted

Preamble

I believe that the OP is seeking a characterization of $ \mathbb{Q} $ using only the first-order language of fields, $ \mathcal{L}_{\text{Field}} $. Restricting ourselves to this language, we can try to uncover new axioms, in addition to the usual field axioms (i.e., those that relate to the associativity and commutativity of addition and multiplication, the distributivity of multiplication over addition, the behavior of the zero and identity elements, and the existence of a multiplicative inverse for each non-zero element), that describe $ \mathbb{Q} $ uniquely.

Any attempt to describe the smallest field satisfying a given property must prescribe a method of comparing one field with another (namely using field homomorphisms, which are injective if not trivial), but such a method clearly cannot be formalized using $ \mathcal{L}_{\text{Field}} $.


1. There Exists No First-Order Characterization of $ \mathbb{Q} $

The answer is ‘no’, if one is seeking a first-order characterization of $ \mathbb{Q} $. This follows from the Upward Löwenheim-Skolem Theorem, which is a classical tool in logic and model theory.

Observe that $ \mathbb{Q} $ is an infinite $ \mathcal{L}_{\text{Field}} $-structure of cardinality $ \aleph_{0} $. The Upward Löwenheim-Skolem Theorem then says that there exists an $ \mathcal{L}_{\text{Field}} $-structure (i.e., a field) $ \mathbb{F} $ of cardinality $ \aleph_{1} $ that is an elementary extension of $ \mathbb{Q} $. By definition, this means that $ \mathbb{Q} $ and $ \mathbb{F} $ satisfy the same set of $ \mathcal{L}_{\text{Field}} $-sentences, so we cannot use first-order logic to distinguish $ \mathbb{Q} $ and $ \mathbb{F} $. In other words, as far as first-order logic can tell, these two fields are identical (an analogy may be found in point-set topology, where two distinct points of a non-$ T_{0} $ topological space can be topologically indistinguishable). However, $ \mathbb{Q} $ and $ \mathbb{F} $ have different cardinalities, so they are not isomorphic. This phenomenon is ultimately due to the fact that the notion of cardinality cannot be formalized using $ \mathcal{L}_{\text{Field}} $. Therefore, any difference between the two fields can only be seen externally, outside of first-order logic.


2. Finding a Second-Order Characterization of $ \mathbb{Q} $

This part is inspired by lhf's answer below, which I believe deserves more credit. We start by formalizing the notion of proper subfield using second-order logic.

Let $ P $ be a variable for unary predicates. Consider the following six formulas: \begin{align} \Phi^{P}_{1} &\stackrel{\text{def}}{\equiv} (\exists x) \neg P(x); \\ \Phi^{P}_{2} &\stackrel{\text{def}}{\equiv} P(0); \\ \Phi^{P}_{3} &\stackrel{\text{def}}{\equiv} P(1); \\ \Phi^{P}_{4} &\stackrel{\text{def}}{\equiv} (\forall x)(\forall y)((P(x) \land P(y)) \rightarrow P(x + y)); \\ \Phi^{P}_{5} &\stackrel{\text{def}}{\equiv} (\forall x)(\forall y)((P(x) \land P(y)) \rightarrow P(x \cdot y)); \\ \Phi^{P}_{6} &\stackrel{\text{def}}{\equiv} (\forall x)((P(x) \land \neg (x = 0)) \rightarrow (\exists y)(P(y) \land (x \cdot y = 1))). \end{align} What $ \Phi^{P}_{1},\ldots,\Phi^{P}_{6} $ are saying is that the set of all elements of the domain of discourse that satisfy the predicate $ P $ forms a proper subfield of the domain. The domain itself will be a field if we impose upon it the first-order field axioms. Hence, $$ \{ \text{First-order field axioms} \} \cup \{ \text{First-order axioms defining characteristic $ 0 $} \} \cup \{ \neg (\exists P)(\Phi^{P}_{1} ~ \land ~ \Phi^{P}_{2} ~ \land ~ \Phi^{P}_{3} ~ \land ~ \Phi^{P}_{4} ~ \land ~ \Phi^{P}_{5} ~ \land ~ \Phi^{P}_{6}) \} $$ is a set of first- and second-order axioms that characterizes $ \mathbb{Q} $ uniquely because of the following two reasons:

  1. Up to isomorphism, $ \mathbb{Q} $ is the only field with characteristic $ 0 $ that contains no proper subfield.

  2. If $ \mathbb{F} \ncong \mathbb{Q} $ is a field with characteristic $ 0 $, then $ \mathbb{F} $ does not model this set of axioms. Otherwise, interpreting “$ P(x) $” as “$ x \in \mathbb{Q}_{\mathbb{F}} $” yields a contradiction, where $ \mathbb{Q}_{\mathbb{F}} $ is the copy of $ \mathbb{Q} $ sitting inside $ \mathbb{F} $.

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I think you have made my question much more precise than I could have ever done. Thanks! –  AlexE Dec 19 '12 at 23:45
    
You're welcome! It is only after rephrasing the problem precisely that I recognize its non-triviality. –  Haskell Curry Dec 20 '12 at 0:00
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The Peano axioms for arithmetic and the axiomatization of the reals (which OP seem to like) also fail to be a first order characterization of any system (by the same Loewenheim-Skolem argument). –  Ittay Weiss Dec 20 '12 at 1:02
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The first two paragraphs of the question give the impression that the OP wants a characterization of $\mathbb Q$ along the same lines as the well-known characterizations of $\mathbb N$ and $\mathbb R$. Then what's wanted would be a second-order characterization, not first-order. –  Andreas Blass Dec 20 '12 at 14:49
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@AndreasBlass: Well, at first, this was what I wanted. But the reformulation of the question (i.e., using only first-order logic) has a much more interesting answer. Also, I'm quite satisfied with lhf's answer (i.e., defining the rational numbers to be a totally ordered field, which does not have any proper subfield). –  AlexE Dec 20 '12 at 15:19
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Among all fields the rationals can be characterized as follows. The field of rational numbers is, up to isomorphism, the smallest field of characteristic $0$.

As you say, the rationals can be built up from the integers via a construction which is a special case of the construction known as the field of fractions of an integral domain. In this way the rationals can be characterized, among all integral domains rings, as follows: The field of rational numbers is, up to isomorphism, the field of fractions of the ring of integers.

Whether or not these characterizations fit your expectations of being nice is a matter of taste. In any case, the second characterization is a special case of a the very common phenomenon of a universal property.

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I would like to emphasise the charaterisation by universal property which you mentioned. The OP might like to consider looking at this from a category theoretic point of view in which case this definition of the rationals is very natural. –  Daniel Rust Dec 19 '12 at 22:59
    
This definition requires us to know what the integers are. And if we know them, we can just construct the rationals - "so no need to define them". (well ... defining them to be the field of fractions of the integers is in some sense a construction out of the integers). I'm looking for a axiomatic characterization which does not require us to know the integers (or the naturals, or the reals). Just like the axioms for the reals. –  AlexE Dec 19 '12 at 23:04
    
Using the universal property (that they are the smallest such field) is some way to define them without using the integers, but I'm not quite satisfied with this. I'm looking for something like the axioms for the reals, which do not mention the universal property (that the reals are the biggest Archimedean ordered field). –  AlexE Dec 19 '12 at 23:11
    
Surely the dedekind complete definition of the reals is, in itself, a universal property definition. Most 'completeness' constructions can be characterised by a universal property as they simply infer that they are the unique smallest object with a particular property. –  Daniel Rust Dec 19 '12 at 23:17
    
Under universal property I understand an axiom which gives us a map under some conditions (e.g., defining the reals to be the biggest Archimedean ordered field means, that for every other such field there is an embedding into them). But the Dedekind-completeness axiom does not give us some map - it gives us the existence of special elements (just like the field axioms give us the existence of special elements called neutral elements and inverse elements). So a priori the Dedekind-completeness axiom is not a universal property. –  AlexE Dec 19 '12 at 23:29
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One simple axiomatic characterization of the rational numbers is that it is an ordered field generated by $1$. This implies that it is the prime field of characteristic $0$.

The notion of being generated by $1$ can be formulated as there are no proper subfields.

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Somehow, it seems more elegant to define it as having characteristic 0 and deriving that it is ordered rather than the other way around. –  asmeurer Dec 19 '12 at 23:37
    
@asmeurer, I was following the OP line, of removing things from the axioms for real numbers. –  lhf Dec 19 '12 at 23:38
    
This is a second-order characterisation, though. –  Zhen Lin Dec 20 '12 at 4:18
    
I like this axiom (no proper subfields) even if it is not first-order. If Haskell Curry had not written his answer I would have accepted this. –  AlexE Dec 20 '12 at 8:54
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There is no recursively enumerable axiomatization of the rational numbers in the first order logic. In particular, we cannot write a finite set of axioms or axiom schemata that completely define the theory of the rational numbers.

This follows from the following result of Julia Robinson [R].

Theorem (see [R]). The notion of an integer and that of a positive integer is arithmetically definable in terms of the notion of a rational and the operations of addition and multiplication on rationals.

Accordingly, if the theory of rational numbers had a recursively enumerable axiomatization, it would be decidable, and thus the arithmetic would be decidable. But it is not, so the theory of rational numbers doesn't have a recursively enumerable axiomatization.

On the other hand, we can write a non-complete axiomatization of rationals as follows. We define predicates “$x$ is an integer” and “$x$ is a positive integer” using Robinson's theorem. Then write

  • $\mathbb Q$ is a field.

  • Every element of $\mathbb Q$ is a ratio of two integers.

  • Positive integers satisfy Peano axioms.

[R] Julia Robinson. Definability and Decision Problems in Arithmetic. The Journal of Symbolic Logic, Vol. 14, No. 2 (Jun., 1949), pp. 98—114.

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With non-complete You mean not uniquely, right? –  AlexE Dec 20 '12 at 8:50
    
@AlexE: It's a stronger statement than the statement that the set of axioms doesn't uniquely characterize $\mathbb Q$. If we write a r.e. set of axioms $A$ then there will be a formula $\phi$ which is true in $\mathbb Q$ but which doesn't follow from $A$. In particular, there is another model ${\mathbb Q}'$ for $A$ in which $\phi$ is false. –  Yury Dec 20 '12 at 13:23
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http://en.wikipedia.org/wiki/Real_number#Axiomatic_approach

  • Let R denote the set of all real numbers. Then:

  • The set R is a field, meaning that addition and multiplication are defined and have the usual properties.

  • The field R is ordered, meaning that there is a total order ≥ such that, for all real numbers x, y and z:

    • if x ≥ y then x + z ≥ y + z;

    • if x ≥ 0 and y ≥ 0 then xy ≥ 0.

  • The order is Dedekind-complete; that is, every non-empty subset S of R with an upper bound in R has a least upper bound (also called supremum) in R.

The last property is what differentiates the reals from the rationals. For example, the set of rationals with square less than 2 has a rational upper bound (e.g., 1.5) but no rational least upper bound, because the square root of 2 is not rational.

So wouldn't simply removing that last property work?

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No, what is left are the axioms of ordered fields and there many different such fields. –  lhf Dec 19 '12 at 23:31
    
Saying that a field is not complete is not sufficient. There are lots of fields that are not complete, whose completion is $\mathbb{R}$. $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$ are two trivial examples. –  asmeurer Dec 19 '12 at 23:32
    
You need to remove completeness and add that the field is generated by 1. The fact that it is ordered implies that the characteristic is 0. –  lhf Dec 19 '12 at 23:32
    
How do you axiomatize is generated by 1? Since such an axiom might answer my question. –  AlexE Dec 19 '12 at 23:36
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Let $F$ be your totally ordered field. For all $x\in F$ there exists $n,m\in\mathbb{Z}$ such that $x=n(1)(m(1))^{-1}$. ($k(1)$ here is short hand for the sum $1+1+\ldots +1$ $k$ times.) –  Daniel Rust Dec 19 '12 at 23:48
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I have found it useful to construct the set of positive rational numbers $Q^+$ from the set of natural numbers $N=\{1,2,3,...\}$ using Cartesian products, power sets and subsets where, for example:

$1/2=\{(1,2), (2,4), (3,6),...\}$

$2/3=\{(2,3), (4,6), (6,9),...\}$

etc.

$(a,b)\longleftrightarrow a/b$

For the entire set of rational numbers, you might consider sets of ordered triples of natural numbers:

$-1/2=\{(1,2,2), (2,3,2), (2,4,4), ...\}$

$(a,b,c)\longleftrightarrow (a-b)/c$

The advantage of this approach is that you start with only the natural numbers and use the only the axioms of set theory to construct the rationals. No field axioms, etc. There is no need to redefine equality for the rationals -- it is just the usual equality of sets.

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