Number Theory: An Introduction to Mathematics

2 Integers and Rational Numbers 11

However, two other natural numbersm′,n′may have the same difference asm,n,and
anyway what doesm−nmean ifm<n? To make things precise, we proceed in the
following way.
Consider the setN×Nof all ordered pairs of natural numbers. For any two such
ordered pairs,(m,n)and(m′,n′), we write

(m,n)∼(m′,n′) ifm+n′=m′+n.

We will show that this is anequivalence relation. It follows at once from the definition
that(m,n)∼(m,n)(reflexive law) and that(m,n)∼(m′,n′)implies(m′,n′)∼
(m,n)(symmetric law). It remains to prove the transitive law:

(m,n)∼(m′,n′)and(m′,n′)∼(m′′,n′′)imply(m,n)∼(m′′,n′′).

This follows from the commutative, associative and cancellation laws for addition
inN. For we have

m+n′=m′+n, m′+n′′=m′′+n′,

and hence

(m+n′)+n′′=(m′+n)+n′′=(m′+n′′)+n=(m′′+n′)+n.

Thus

(m+n′′)+n′=(m′′+n)+n′,

and som+n′′=m′′+n.
The equivalence class containing (1,1) evidently consists of all pairs(m,n)with
m=n.
We d e fi n e a nintegerto be an equivalence class of ordered pairs of natural numbers
and, as is now customary, we denote the set of all integers byZ.
Addition of integers is defined componentwise:

(m,n)+(p,q)=(m+p,n+q).

To justify this definition we must show that it does not depend on the choice of repre-
sentatives within an equivalence class, i.e. that

(m,n)∼(m′,n′)and(p,q)∼(p′,q′)imply(m+p,n+q)∼(m′+p′,n′+q′).

However, if

m+n′=m′+n, p+q′=p′+q,

then

(m+p)+(n′+q′)=(m+n′)+(p+q′)

=(m′+n)+(p′+q)=(m′+p′)+(n+q).