Number Theory: An Introduction to Mathematics

2 Integers and Rational Numbers 13

Proposition 10For every a∈Z,a· 0 = 0.

Proof We h av e

a· 0 =a·( 0 + 0 )=a· 0 +a· 0.

Adding−(a· 0 )to both sides, we obtain the result.

Proposition 10 could also have been derived directly from the definitions, but we
prefer to view it as a consequence of the properties which have been labelled.

Corollary 11Fo r a l l a,b∈Z,

a(−b)=−(ab),(−a)(−b)=ab.

Proof The first relation follows from

ab+a(−b)=a· 0 = 0 ,

and the second relation follows from the first, sincec=−(−c).

By the definitions of 0 and 1 we also have

(AM2) 1 =0.

(In fact 1=0wouldimplya=0foreverya,sincea· 1 =aanda· 0 =0.)
We will say that an integeraispositiveif it is represented by an ordered pair
(m,n) withn<m. This definition does not depend on the choice of representative.
For ifn<mandm+n′=m′+n,thenm+n′<m′+mand hencen′<m′.
We will denote byPthe set of all positive integers. The law of trichotomy(O2)
for natural numbers immediately implies

(P1)for every a, one and only one of the following alternatives holds:

a∈P, a= 0 , −a∈P.

Wesaythatanintegerisnegativeif it has the form−a,wherea ∈P,andwe
denote by−Pthe set of all negative integers. Sincea=−(−a),(P1)says thatZis
the disjoint union of the setsP,{ 0 }and−P.
From the property(O3)of natural numbers we immediately obtain

(P2)if a∈P and b∈P,then a+b∈P.

Furthermore, we have

(P3)if a∈P and b∈P,then a·b∈P.

To prove this we need only show that ifm,n,p,qare natural numbers such thatn<m
andq<p,then

mq+np<mp+nq.

Sinceq<p, there exists a natural numberq′such thatq+q′=p.Butthennq′<mq′,
sincen<m, and hence

mq+np=(m+n)q+nq′<(m+n)q+mq′=mp+nq.