The real numbers can be axiomatically defined (up to isomorphism) as a Dedekind-complete ordered field.

What is a similar standard axiomatic definition of the integer numbers?

A commutative ordered ring with positive induction?

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    $\begingroup$ A free commutative ring with 1 of rank 0? An initial object in the category of rings with 1? $\endgroup$ – Arturo Magidin May 1 '10 at 19:55
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    $\begingroup$ I don't think that this question belongs to MO. your description is right and characterizes Z. also Z is characterized by a univeral property: it is initial in the category of commutative rings. $\endgroup$ – Martin Brandenburg May 1 '10 at 19:55
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    $\begingroup$ I cannot answer this question, but I believe that “being the initial object in the category of rings” characterises the integers up to isomorphism; I was wondering if this property can be used somehow in order to construct a (second order?) categorical theory of integers. $\endgroup$ – Antonio E. Porreca May 1 '10 at 20:05
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    $\begingroup$ Isn't it considerable overkill to talk about categories? $\endgroup$ – Gerald Edgar May 1 '10 at 20:12
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    $\begingroup$ Here is another characterization: $\mathbb{Z}$ is a ordered ring (commutative, nontrivial, with identity) with "double induction". Every subset $S$ such that $0 \in S$ and $x \in S \Rightarrow x \pm 1 \in S$ is already $\mathbb{Z}$. $\endgroup$ – Martin Brandenburg May 1 '10 at 20:34

It's the unique commutative ordered ring whose positive elements are well-ordered.

Edit: Oh, François basically already said this, didn't notice. Should I delete this?

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    $\begingroup$ It's an elegant formulation. Don't delete. $\endgroup$ – François G. Dorais May 1 '10 at 23:39
  • $\begingroup$ I would be very grateful for any references on proofs of the uniqueness theorem. $\endgroup$ – Victor Makarov May 2 '10 at 1:25
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    $\begingroup$ you should prove it by yourself. $\endgroup$ – Martin Brandenburg May 2 '10 at 9:27
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    $\begingroup$ @Martin: Victor didn't say that he doesn't know how to prove uniqueness (perhaps he did it already a long time ago). Victor only said, and very politely, that he would appreciate a reference. $\endgroup$ – Włodzimierz Holsztyński Jun 26 '13 at 2:36
  • $\begingroup$ @Martin: A proof (not a simple one) of uniqeness is in the book "Number Systems and the Foundations of Analysis" by Elliot Mendelson. $\endgroup$ – Victor Makarov Jan 29 '14 at 21:56

The ring $\mathbb{Z}$ is the unique ordered ring which satisfies full second-order induction: $$\forall X(0 \in X \land (\forall n \geq 0)(n \in X \to n+1 \in X) \to (\forall n \geq 0)(n \in X)),$$ where $X$ varies over all subsets of $\mathbb{Z}$ (or even all sets). In the comments, Martin Brandenburg has given yet another characterization of $\mathbb{Z}$ which does not assume the ordering.

A dual characterization is that every nonempty subset of $\mathbb{Z}$ which is bounded below has a minimal element. This is closer to the characterization of $\mathbb{R}$. Note that all of these characterizations only make sense in standard second-order logic, but the proposed characterization of $\mathbb{R}$ has the same problem.

The ring of integers also has categorical characterizations. For example, as proposed in the comments, $\mathbb{Z}$ is initial object in the category of (ordered) rings. See this question for related information.

  • $\begingroup$ Having read the above comments, let me clarify that by ring I mean a not necessarily commutative ring with identity. The fact that $\mathbb{Z}$ is commutative can be seen by applying induction to the subset $X = \{x : \forall y (xy = yx)\}$ of $\mathbb{Z}$ and realizing that $X$ is closed under negation. $\endgroup$ – François G. Dorais May 1 '10 at 21:03

Part 2: Cyclands and integers

(This text is a continuation of Part 1 from the same thread).

First I'll axiomatize here the cyclic groups--I'll call them cyclands (to avoid any confusion during the definition stage; later one can go back to the standard naming; the same goes for Abelian groups and minusops). Then there will come time for integers.


By definition, a cycland is an ordered triple   $(X\ -\ 1)$   such that   $X$   is an arbitrary set,   symbol   $-$   stands for a binary operation in   $X$,   $1\in X$, and   the following (algebraic induction-like) axiom holds:

$$ \left(\left(1\in A\subseteq X\right)\ \ \&\ \ \left(\forall_{x\ y\in A}\ x-y\in A\right)\right)\quad\Rightarrow\quad \left(A=X\right)$$


DEFINITION   Integers (i.e. the system of integers) is a cycland   $(\mathbb Z\ -\ 1)$   such that the following (additional) two axioms hold:

  1. $1-x\ne x$
  2. $1-x=x-1\quad\Rightarrow\quad x=1$

for arbitrary   $x\in\mathbb Z$.

Integers from scratch

To have an easy overview of the definition of integers let me list all the relevant axioms directly (without mentioning minusops and cyclands).

The system of integers is an ordered triple   $(\mathbb Z\ -\ 1)$   such that the following six axioms hold:

  1. $ x-(y-z)\ =\ z-(y-x)$
  2. $ x-x\ =\ y-y$
  3. $ x-(x-x) = x$
  4. $ 1-x\ne x$
  5. $ \left(1-x=x-1\right)\ \Rightarrow\ \left(x=1\right)$
  6. $ \left(\left(1\in A\subseteq X\right)\ \ \&\ \ \left(\forall_{x\ y\in A}\ x-y\in A\right)\right)\quad\Rightarrow\quad \left(A=X\right)$

for arbitrary   $x\ y\ z\ \in\ \mathbb Z$.

Other standard constructions or notions

We define in   $\mathbb Z$   the usual, like:

  • $0\ :=\ 1-1$
  • $\mathit{Neg}(x)\ :=\ 0-x$
  • $x+y\ :=\ x -\mathit{Neg}(y)$

Furthermore, in every cycland, in particular in   $\mathbb Z$,   there is exactly one binary operation   $\cdot$   which has the following two properties:

  • $x\cdot 1\ =\ x$
  • $x\cdot(y-z)\ =\ x\cdot y - x\cdot z$

for arbitrary   $x\ y\ z\ \in\ \mathbb Z$.


Part 1: Abelian group introduction

(See Part 2 for the final answer).

I feel that for a fully elegant definition of integers one needs to readdress the definition of an Abelian group. For the sake of communication I will even introduce a synonym minusop for abelian groups to stress the independence of their definition from the general groups. The general groups have an elegant definition (perhaps more than one). Then adding the commutativity axiom one gets Abelian groups. This seems to me unnecessarily complex. Let me define Abelian groups directly.

Symmetric operations

Symmetric operations should not be confused with much more special commutative operations.

By definition, an operation   $\#:X^2\rightarrow X$   is called symmetric   $\Leftarrow:\Rightarrow$

$$ \forall_{x\ y\ z\in X}\qquad x\ \#\ (y\ \#\ z)\ \ =\ \ z\ \#\ (y\ \#\ x) $$

This property will appear in the definition of minusop as an axiom. Both addition and subtraction are symmetric operations in any Abelian group. Every commutative operation is symmetric.


By definition, a minusop is an ordered pair   $(X\ -)$,   where   $X$   is an arbitrary set, and   $-$   is a binary operation in   $X$,   such that the following three axioms hold:

  1. $x-(y-z)\ =\ z-(y-x)$
  2. $x-x\ =\ y-y$
  3. $x-(x-x)\ =\ x$

for arbitrary   $x\ y\ z\in X$.

Every Abelian group admits a standard interpretation as a minusop; and every non-empty minusop admits its standard interpretation as an Abelian group, so that Abelian groups and non-empty minusops are essentially the same objects.

(I am not writing the last obvious statement in any detail to keep this post sensibly short).


A class of operations even more general than symmetric is still useful--I call operation   $\#:X^2\rightarrow X$   insider trading (in mathematics it's legal)   $\Leftarrow:\Rightarrow$

$$\forall_{u\ w\ x\ y\ \in\ X}\qquad (u\ \#\ w)\ \#\ (x\ \#\ y)\ \ =\ \ (u\ \#\ x)\ \#\ (w\ \#\ y)$$

Every symmetric operation is an insider trading.

A simplification by Emil Jeřábek

The last two of the three minusop axioms above (in the previous section before REMARK) can be replaced by one, as pointed out Emil Jeřábek, as now presented below. A minusop can be axiomatized by the following two conditions:

  1. Symmetry:     $x-(y-z)\ =\ z-(y-x)$
  2. Zero:     $x-(y-y)\ =\ x$

for arbitrary   $x\ y\ z\ \in\ X$.

It can be seen instantly that the earlier three axioms imply the two above (the new first axiom is a repetition of the old first axiom).

In the other direction, a substitution of $y$ by $x$ in axiom zero gives the old axiom 3. Furthermore, assuming the above two axioms we obtain:

$$ x-x\ =\ (x-x)-((y-y)-(y-y))\ =\ (y-y)-((y-y)-(x-x)) $$ $$ (y-y) - (y-y)\ =\ y-y $$

which proves the old axiom 2. The new 2-axiom system is equivalent to the old 3-axiom system.

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    $\begingroup$ You can simplify 2 and 3 to the single axiom $x-(y-y)=x$. $\endgroup$ – Emil Jeřábek Jun 26 '13 at 11:00
  • $\begingroup$ @Emil, nice! Thank you. I edited my "Answer" above, adding a section titled "A simplification by Emil Jeřábek" but my wireless Internet connection failed etc., and once again I lost my edition. I'll redo it in a moment. You'll let me know, please, if the new version is fine with you. $\endgroup$ – Włodzimierz Holsztyński Jun 27 '13 at 1:25
  • $\begingroup$ Ooooooooooph, done--the Internet gods were kind to me this time. $\endgroup$ – Włodzimierz Holsztyński Jun 27 '13 at 1:46
  • $\begingroup$ Are you sure that every commutative operation is also symmetric? I think this is unprovable without some further assumptions. $\endgroup$ – goblin Jan 7 '14 at 11:56
  • $\begingroup$ Bump. I've recently become interested in your answer again. Any thoughts about my comment? It seems to me that if $\#$ is associative, then commutativity implies symmetry. But I don't see how to go from commutativity to symmetry without associativity. $\endgroup$ – goblin Feb 28 '18 at 7:15

The integers can also be characterized as a well-ordered abelian group (Theorem 20.14 in "Modern Algebra" by Seth Warner).

  • $\begingroup$ This is already stated in the top answer to this question. $\endgroup$ – Emil Jeřábek Jan 29 '14 at 13:02
  • $\begingroup$ References may be useful (even though the result is an easy exercise), but the fact that your formulation includes the redundant condition on zero divisors makes it a weaker result. Also, note that a non-trivial ordered ring is never well-ordered. Only the positive part of the ring should be well-ordered in the characterization. $\endgroup$ – Emil Jeřábek Jan 29 '14 at 17:53
  • $\begingroup$ Mendelson defines "a well-ordered integral domain" as an ordered integral domain where the positive part is well ordered. $\endgroup$ – Victor Makarov Jan 29 '14 at 18:10

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