Having done a lot of hard work putting the natural numbers
on a sound footing the next step, to build , is
rather more straightforward. The most direct method is
perhaps to define as a disjoint union of two copies
of , one copy being 0,1,2,3,
and the other copy being 1,2,3,
.
A slightly more sophisticated method is to use equivalence relations.
1. Getting the integers from the natural numbers
Definition.
Let be the following relation defined on
: we define (
_{1,
1)(
2,
2)
}
to mean
_{1+
2=
2+
1
}
.
Proposition.
in an equivalence relation on
.
Definition.
We write
,
for the equivalence class of
(,)
. The set is
the set
/
of equivalence classes.
Proposition.
The function defined by
()=
,0
is a onetoone function mapping
into .
Proof.
Let ,
with ()=()
.
Then
,0=
,0
hence
(,0)(,0)
hence +0=+0
hence
=
.
Definition.
We identify each
with its image ()
under the map .
In particular, 0 is the element 0,0
,
and 1 is 1,0
.
Definition.
We define addition, multiplication and order relations
on by

_{1,
1
}
+
_{2,
2
}
=
_{1+
2,
1+
2
}

_{1,
1
}
_{2,
2
}
=
_{1
2+
1
2
}
,
_{1
2+
1
2
}

_{1,
1
}
_{2,
2
}
_{1+
2
}
_{2+
1
}
Proposition.
The operations +, and relation on
are welldefined, i.e., the definitions above do not depend on the particular choice
of representatives (
_{1,
1),(
2,
2)
}
.
Proposition.
The embedding
is a homomorphism
respecting +, and :

(+)=()+()
;

(
)=()
()
; and

()
()
;
2. The ring structure
Having constructed the set of integers and its main operations,
we need to prove that it satisfies the key axioms we expect.
We list here the key axioms and in a few cases sketch the proof
that the set satisfies the axioms.
Firstly, , with the addition and multiplication
operations forms a nice algebraic structure. Division is not available
but almost everything else you'd hope works doe so. In particular
the integers forms a commutative ring with 1.
Theorem.
The set , with the addition and multiplication
operations, is a commutative ring with 1. I.e.,
 (,+) is an abelian group
with identity element 0.
 (
0,)
is an abelian semigroup with
identity element 1. (This means that the commutative and associative
laws hold, and also
(1
==
1)
, but that
multiplicative inverses don't necessarily exist.)
 The distibutivity law holds.
The ring of integers has a further nice property concerning multiplication.
Theorem.
The ring of integers is an integral domain. That is, it has no zero
divisors, i.e, satisfies the additional axiom
(
=0(=0
=0))
3. The order structure
So far we haven't yet mentioned the order structure on the integers.
This is easily remedied, and the axioms look very similar to those for Archimedean
ordered fields.
As for , etc., we can work with
or with the relation
defined to mean
=
.
We shall work with here.
Theorem.
defined by
is a linear
order on , i.e., satisfies
 Transitivity:
((
)
)
 Irreflexivity:
 Linearity:
(
=
)
We also need to add axioms stating the relationship between the order and
the arithmetic operations.
Theorem.
The integers with forms an ordered ring, i.e., also satisfies
the following axioms:
 One is positive: 01.
 Addition respects :
(
+
+)
 Multiplication respects :
((0
)
)
In addition, this order is discrete, that is, there are no elements
between and +1:

(
+1)
We have been presenting all the properties as they most naturally arise.
However, our axiomatisation is not optimal in the sense that some axioms
could be omitted since they follow from others. A case in point is where
the new axioms for the order automatically implies the nozerodivisor
property.
Exercise.
Show that any ordered ring is an integral domain.
4. The Archimedean property and induction
The se of integers forms an Archimedean structure in the same sense as in
Archimedean ordered fields. The direct translation
of this axiom from the earlier web page is perfectly correct, though looks
somewhat silly when one recalls the correct picture. It says that no single
integer
is greater than all natural numbers. Phrased in a
slightly more meaningful form for the integers, this becomes
Theorem.
No integer
is greater than all natural numbers.
In other words, the set of nonnegative integers
0
is exactly .
By the least number principle in , this can be
rephrased in an equivalent but rather more powerful form.
Theorem.
Let
()
hold for some integer
and
suppose that there is
such that
(
())
. Then there is a least
_{0}
in
such that
(
_{0)}
.
Proof.
Use the least number principle on the property ()
of natural numbers where ()
says that
(+)
holds.
For an example where the conditions of the last result are not met
and where the least number principle fails, consider the property
()
that says that
0
is prime
.
We now have all the information we need about the integers. The next task
is to combine these to make an Archimedean field, the field of rationals,
containing the integers.