Christina Pospisil
Proseminar Numbers
The construction of the integers from the natural numbers
1
Basic Definitions
We begin with the relation
(x1 , y1 ) ~(x2 , y2 ) ⇔ x1 + y2 = y1 + x2
(1)
on N x N.
Proposition
The formula (1) defines an equivalence relation on N x N.
Proof:
i) Reflexivity:
Let (x1 , y1 ) ∈ N x N. Then x1 + y1 = y1 + x1 and therefore (x1 , y1 ) ~(y1 , x1 ) holds for all (x1 , y1 ) ∈ N x N.
ii) Symmetry:
Let (x1 , y1 ), (x2 , y2 ) ∈ N x N and (x1 , y1 ) ~(x2 , y2 ). Then x1 + y2 = y1 + x2 . Hence x2 + y1 = y2 + x1 and finally
(x2 , y2 ) ~(x1 , y1 ) holds for all (x1 , y1 ), (x2 , y2 ) ∈ N x N.
iii) Transitivity
Let (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) ∈ N x N with (x1 , y1 ) ~(x2 , y2 ) and (x2 , y2 ) ~(x3 , y3 ). Then x1 + y2 = y1 + x2 and
x2 + y3 = y2 + x3 . With combination follows x1 + y2 + x2 + y3 = y1 + x2 + y2 + x3 . Subtracting y2 + x2 from both
sides leads to x1 + y3 = y1 + x3 and as result (x1 , y1 ) ~(x3 , y3 ) holds for all (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) ∈ N x N.
Consequently all three characteristics of an equivalence relation hold. We denote the equivalence class including (x1 , y1 ) by [(x1 , y1 )], so that
[(x1 , y1 )] = {(x2 , y2 ) ∈ N x N| x1 + y2 = y1 + x2 }
(see Figures 1 and 2).
(2)
Figure 1: The construction of the integers from the natural numbers
1
Figure 2: The equivalence classes of N x N under ~are the points on the diagonal lines whose
coordinates are natural numbers. The ’usual’ notation for the corresponding integer is given by
the intersection of the line with the horizontal axis.
Definition
The set of integers is defined by
Z := {[(x1 , y1 )] | x1 , y1 ∈ N}.
(3)
The arithmetic of the integers is defined as follows.
Definition
For [(x1 , y1 )], [(x2 , y2 )] ∈ Z we define
[(x1 , y1 )] ⊕ [(x2 , y2 )] := [(x1 + x2 , y1 + y2 )] ∈ Z (Addition),
[(x1 , y1 )] [(x2 , y2 )] := [(x1 x2 + y1 y2 , x2 y1 + x1 y2 )] ∈ Z (Multiplication).
(4)
(5)
Proposition
The operations ⊕ and are well defined.
Proof:
Let [(x1 , y1 )], [(x2 , y2 )], [(x3 , y3 )], [(x4 , y4 )] ∈ N x N with [(x1 , y1 )] = [(x3 , y3 )] and [(x2 , y2 )] = [(x4 , y4 )]
For the operation ⊕ we have that [(x1 , y1 )] ⊕ [(x2 , y2 )] = [(x1 + x2 , y1 + y2 )] and [(x3 , y3 )] ⊕ [(x4 , y4 )] = [(x3 +
x4 , y3 + y4 )].
Therefore x1 + x2 + y3 + y4 = x1 + y3 + x2 + y4 = y1 + x3 + y2 + x4 and from this we find that [(x1 + x2 , y1 + y2 )] =
[(x3 + x4 , y3 + y4 )]. 2
For the operation we have that [(x1 , y1 )] [(x2 , y2 )] = [(x1 x2 + y1 y2 , x2 y1 + x1 y2 )] and [(x3 , y3 )] [(x4 , y4 )] =
[(x3 x4 + y3 y4 , x4 y3 + x3 y4 )].
Therefore we can transform
x1 x2 + y1 y2 + x3 y4 + x4 y3 + x2 y3 + x3 y2
= x2 (x1 + y3 ) + y2 (y1 + x3 ) + x3 y4 + x4 y3
= x2 (y1 + x3 ) + y2 (x1 + y3 ) + x3 y4 + x4 y3
= x3 (x2 + y4 ) + y3 (y2 + x4 ) + x2 y1 + x1 y2
= x3 (y2 + x4 ) + y3 (x2 + y4 ) + x2 y1 + x1 y2
= x1 y2 + x2 y1 + x3 x4 + y3 y4 + y3 x2 + x3 y2
and this leads to [(x1 x2 + y1 y2 , x2 y1 + x1 y2 )] = [(x3 x4 + y3 y4 , x4 y3 + x3 y4 )]. Next we turn to the basic properties of the arithmetic operations.
Lemma
The operations ⊕ and are associative and commutative. Furthermore is distributive over .
Proof of associativity
Let [(x1 , y1 )], [(x2 , y2 )], [(x3 , y3 )] ∈ N x N. For addition we find that
([(x1 , y1 )] ⊕ [(x2 , y2 )]) ⊕ [(x3 , y3 )] = [(x1 + x2 , y1 + y2 )] ⊕ [(x3 , y3 )]
= [((x1 + x2 ) + x3 , (y1 + y2 ) + y3 )]
= [(x1 + x2 + x3 , y1 + y2 + y3 )]
= [(x1 + (x2 + x3 ), y1 + (y2 + y3 ))]
= [(x1 , y1 )] ⊕ ([(x2 + x3 , y2 + y3 )])
= [(x1 , y1 )] ⊕ ([(x2 , y2 )] ⊕ [(x3 , y3 )]).
For multiplication we find that
([(x1 , y1 )] [(x2 , y2 )]) [(x3 , y3 )] = [(x1 x2 + y1 y2 , x2 y1 + x1 y2 )] [(x3 , y3 )]
= [((x1 x2 + y1 y2 )x3 + (x2 y1 + x1 y2 )y3 , (x1 x2 + y1 y2 )y3 + (x2 y1 + x1 y2 )x3 )]
= [(x1 x2 x3 + y1 y2 x3 + x2 y1 y3 + x1 y2 y3 , x2 y1 x3 + x1 y2 x3 + x1 x2 y3 + y1 y2 y3 )]
= [(x1 x2 x3 + x1 y2 y3 + y1 y2 x3 + x2 y1 y3 , x2 y1 x3 + y1 y2 y3 + x1 y2 x3 + x1 x2 y3 )]
= [(x1 (x2 x3 + y2 y3 ) + y1 (y2 x3 + x2 y3 ), y1 (x2 x3 + y2 y3 ) + x1 (y2 x3 + x2 y3 ))]
= [(x1 , y1 )] [(x2 x3 + y2 y3 , x2 y3 + x3 y2 )]
= [(x1 , y1 )] ([(x2 , y2 )] [(x3 , y3 )]).
Proof of commutativity:
Again let [(x1 , y1 )], [(x2 , y2 )] ∈ N x N.
For addition we find that
[(x1 , y1 )] ⊕ [(x2 , y2 )] = [(x1 + x2 , y1 + y2 )] = [(x2 + x1 , y2 + y1 )] = [(x2 , y2 )] ⊕ [(x1 , y1 )]
and for multiplication we have
[(x1 , y1 )] [(x2 , y2 )] = [(x1 x2 + y1 y2 , x2 y1 + x1 y2 )] = [(x2 x1 + y2 y1 , x1 y2 + x2 y1 )] = [(x2 , y2 )] [(x1 , y1 )].
Proof of distributivity:
Again let [(x1 , y1 )], [(x2 , y2 )], [(x3 , y3 )] ∈ N x N, so that
([(x1 , y1 )] ⊕ [(x2 , y2 )]) [(x3 , y3 )] = [(x1 + x2 , y1 + y2 )] [(x3 , y3 )]
= [((x1 + x2 )x3 + (y1 + y2 )y3 , (x1 + x2 )y3 + (y1 + y2 )x3 )]
= [(x1 x3 + x2 x3 + y1 y3 + y2 y3 , x1 y3 + x2 y3 + y1 x3 + y2 x3 )]
= [(x1 x3 + y1 y3 + x2 x3 + y2 y3 , x1 y3 + y1 x3 + x2 y3 + y2 x3 )]
= [(x1 x3 + y1 y3 , x1 y3 + y1 x3 )] ⊕ [(x2 x3 + y2 y3 , x2 y3 + y2 x3 )]
= ([(x1 , y1 )] [(x3 , y3 )]) ⊕ ([(x2 , y2 )] [(x3 , y3 )]). 3
Next we consider the neutral elements of ⊕ and .
Lemma
The neutral element for ⊕ is [(1, 1)] ∈ Z.
Proof:
[(x1 , y1 )] ⊕ [(x, x)] = [(x1 + x, y1 + x)] = [(x1 , y1 )] for all [(x1 , y1 )] ∈ Z. Lemma
The neutral element for is [(1,0)] ∈ Z.
Proof:
[(x1 , y1 )] [(1, 0)] = [(x1 ∗ 1 + y1 ∗ 0, x1 ∗ 0 + y1 ∗ 1)] = [(x1 , y1 )] for all [(x1 , y1 )] ∈ Z It remains to consider the inverse element for ⊕.
Lemma
The inverse element of [(x1 , y1 )] with respect to ⊕ is [(y1 , x1 )] for all [(x1 , y1 )] ∈ Z.
Proof:
[(x1 , y1 )] ⊕ [(y1 , x1 )] = [(x1 + y1 , y1 + x1 )] = [(1, 1)] Remark
The above results show that (Z, ⊕, ) is a commutative ring with unity.
2
Definition of positive, negative numbers and the zero element
Definition
We define the set of positive and negative numbers and the zero element by
Z+ := { Px | x ∈ N and Px := (p + x, p)| p ∈ N }
Z− := { Nx | x ∈ N and Nx := (n, n + x)| n ∈ N}
0 = [(x, x)]
Proposition
The mapping ϕ: N → Z+ , x 7→ [(x + k, k)] is an isomorphism (N, +, ∗) → (Z+ , ⊕, ).
Proof:
The function φ is obviously bijective. Furthermore
φ(x + y) = [(k + x + y, k)]
= [(k + x + y + k, k + k)]
= [(k + x + k + y, k + k)]
= [(k + x, k)] + [(k + y, k)]
= φ(x) ⊕ φ(y)
and
φ(x ∗ y) = [(xy + 1, 1)]
= [(xy + x + y + 1, x + y + 1)]
= [(xy + x + y + 2, x + y + 2)]
= [(1 + xy + x + y + 1, 1 + x + y + 1)]
= [((1 + x) ∗ (1 + y) + 1), (1 + x) ∗ 1 + (1 + y) ∗ 1)]
= [(1 + x, 1)] [(1 + y, 1)]
= φ(x) φ(y) Lemma
The sets are Z+ , Z− and {0} mutually disjoint.
Proof
Clearly [(x + k, k)] , [(k, y + k)] for any x,y ∈ N so that Z+ , Z− are disjoint. Furthermore [(x + k, k)] , [(k, k)] and
[(k, y + k)] , [(k, k)] for any x,y ∈ N, so that Z+ , {0} and Z− , {0} are disjoint. 4
Lemma
The set Z is countably infinite.
Proof
We can write Z as the union of the three countable sets Z+ , Z− and 0. 3
The ordering of the integers
Definition
We define the relation
[(x1 , y1 )] ≥ [(x2 , y2 )] ⇔ x1 + y2 ≥ y1 + x2
(6)
on Z.
Proposition
The formula (6) defines an partial order on Z.
Proof of reflexivity:
Clearly x1 + y1 ≥ y1 + x1 , so that [(x1 , y1 )] ≥ [(x1 , y1 )].
Proof of transitivity:
Let [(x1 , y1 )] ≥ [(x2 , y2 )] and [(x2 , y2 )] ≥ [(x3 , y3 )], so that x1 + y2 ≥ y1 + x2 and x2 + y3 ≥ y2 + x3 . Adding y3 we find
that x1 + y2 + y3 ≥ y1 + x2 + y3 ≥ y1 + y2 + x3 . Subtracting y2 leads to x1 + y3 ≥ y1 + x3 and as result [(x1 , y1 )] ≥
[(x3 , y3 )].
Proof of antisymmetry:
Let [(x1 , y1 )] ≥ [(x2 , y2 )] and [(x2 , y2 )] ≥ [(x1 , y1 )], so that x1 + y2 ≥ y1 + x2 and x2 + y1 ≥ y2 + x1 .
Therefore x1 + y2 ≥ y1 + x2 ≥ y2 + x1 and [(x1 , y1 )] = [(x2 , y2 )].
Consequently all three characteristics of an partial order hold. Remark
‘<‘ is a total ordering on Z, which has the monotony properties
[(x1 , y1 )] < [(x2 , y2 )] ⇒ [(x1 , y1 )] ⊕ [(x3 , y3 )] < [(x2 , y2 )] ⊕ [(x3 , y3 )]
and
[(x1 , y1 )] < [(x2 , y2 )], [(x3 , y3 )] > 0 ⇒ [(x1 , y1 )] [(x3 , y3 )] < [(x2 , y2 )] [(x3 , y3 )]
5