We can uniquely define the set of real numbers as a complete ordered field. But can we do something similar with the set of rational numbers ? I think we need to change the completness axiom with some other one. I know that we can define integers in ordered field so maybe this axiom should state that every number is quotient of two integers ? Or maybe we can choose some simpler one. So what axiom/axioms should we choose instead of completness axiom to define rational numbers ? Thanks in advance.
Definition of rational numbers
axiomsdefinitionordered-fieldsrational numbers
Related Solutions
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:
Up to isomorphism, $ \mathbb{Q} $ is the only field with characteristic $ 0 $ that contains no proper subfield.
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} $.
I don't know how to write a formal proof in ZF and even if I did, I would still write an informal proof here because it's easier to understand and it will be easy for an expert to figure out how to write a complete formal proof that there is a complete ordered field which is unique up to isomorphism from reading this answer.
First, we can define a natural number as a finite ordinal number. Let's define 0 to be the empty set. For each natrual number $x$, let's define its the successor function $S$ to be a function from the set of all finite ordinals to itself such that for every finite ordinal $x$, $S(x) = x \bigcup {\text{{x}}}$. Now we can see that the function $S$ satisfies closure on the set of all natural numbers and every natrual number can be gotten by starting from 0 and applying the successor function. Now we can define addition recursively as
- $\forall x \in \mathbb{N}, x + 0 = x$
- $\forall x \in \mathbb{N}\forall y \in \mathbb{N}, x + S(y) = S(x + y)$
Now we can define multiplication in terms of addition as
- $\forall x \in \mathbb{N}, x \times 0 = 0$
- $\forall x \in \mathbb{N}\forall y \in \mathbb{N}, x \times S(y) = (x \times y) + x$
Now it's easy to prove the following statements using this definition but I won't bother writing the proof
- $\forall x \in \mathbb{N}\forall y \in \mathbb{N}\forall z \in \mathbb{N}, (x + y) + z = x + (y + z)$
- $\forall x \in \mathbb{N}\forall y \in \mathbb{N}, x + y = y + x$
- $\forall x \in \mathbb{N}\forall y \in \mathbb{N}, x \times y = y \times x$
- $\forall x \in \mathbb{N}\forall y \in \mathbb{N}\forall z \in \mathbb{N}, x \times (y + z) = (x \times y) + (x \times z)$
- $\forall x \in \mathbb{N}\forall y \in \mathbb{N}\forall z \in \mathbb{N}, (x \times y) \times z = x \times (y \times z)$
Now we can construct the set of all integers as follows. 0 is not a solution to $x + 1 = 0$ so we can invent the solution -1. We can again invent a solution to the equation $x + 1 = -1$, -2 and keep going for ever and call the result the set of all integers $\mathbb{Z}$. We can also invent an intuitive definition of + and \times and show that it still has those properties and agrees with those operations on $\mathbb{N}$.
Now we can construct the dyadic rationals as follows. For each odd number $x$, we can invent a solution to the equation $2 \times y = x$. Each newly invented solution $y$ is still not a solution to the equation $2 \times z = y$ so we can also invent a solution to that and keep going and call the result the set of all dyadic rationals. We can also create an intuitive definition for those operations on that set and show that it still satisfies those 5 laws and agrees with those operations on $\mathbb{Z}$. We can also create an intuitive definition of the relation $\leq$ on that set. Now that we invented that relation, take any subset of that set such that that set and its complement are nonempty and for any member of the subset, all smaller members of the set are in the subset. When the set has no maximal element and its complement has no minimal element, we will invent a number that's larger than all members of the subset and smaller than all members of the complement. Let's call those the real numbers. Now we can also invent an intuitive definition +, $\times$ and $\leq$ and show that $\exists x\exists y$ such that $(\mathbb{R}, x, y, +, \times, \leq)$ is a complete ordered field where some of the defining criteria are that $x$ is a multiplicative identity and $y$ is an additive identity and those 5 laws still hold. We can also show that +, $\times$, and $\leq$ on this set agree with the way they were defined on the previous set and that 0 is the additive identity and $S(0)$ is the multiplicative identity.
Now take any complete ordered field. Either it has infinitesimal numbers or it doesn't. If it does, then it's not complete. Therefore, no complete ordered field has infinitesimal numbers. Also, every complete ordered field without infinitesimal numbers is isomorphic to that complete ordered field. Therefore, all complete ordered fields are isomorphic to that one.
Best Answer
We can define the field of rational numbers as the smallest ordered field.
More precisely, let $F$ be an ordered field without subfields (except for $F$ itself). Then it is not hard to prove that