Bridge to Abstract Mathematics: Mathematical Proof and Structures

332 CONSTRUCTION OF NUMBER SYSTEMS Chapter 10

Axiom 1 the operations of addition and multiplication. The theorem by

which we will bring these operations into being is indeed the most impor-

tant and difficult in this article (see Theorems 3 and 4). Therefore, before

tackling this problem, let us first get our bearings by proving a few prelimi-

nary results. An important aspect of "getting our bearings" in the current

context is an early appreciation of the preponderance here of proof by in-

duction. Considering what there is to work with at this stage, namely, parts

(a) and (b) of Axiom 1, it should not be surprising that induction proofs are

so prominent; yet it will probably take some time to get used to this fact.

THEOREM 1

a(n) # n for all n E N; that is, the mapping a has no fixed points.

Proof Let S = (n E Nlo(n) # n). We will attempt to prove S = N by

verifying (I) and (11) of (b) of Axiom 1. (I) Clearly 1 E S. We cannot have

1 = a(1) since 1 4 im (a). (11) Assume m E S; we claim a(m) E S; that is,

o(m) # a(o(m)). For if the latter equation were valid, then by the one-

to-one property of the mapping a, we could conclude m = a(m), con-

tradicting the induction hypothesis.

THEOREM 2

im (a) = N - {I); that is, 1 is the only element of N that is not the successor of

some element of N.

Proof Our strategy is as follows. Let S = im (a) u (1). If we can prove

S = N, then by elementary set theory, we have N - (1) = S - (1) =

(im (a) u (1 )) n ( 1 )' = im (a), as desired. Note that the last equality

depends on the fact that 1 4 im (a). Hence we claim S = N; we turn to

(b) of Axiom 1 to support this claim. Clearly 1 E S, by our definition of

S. So suppose m E S; we claim a(m) E S. Now since m E S, then either

m = 1 or m E im (a). If m = 1, then a(m) = o(1) E im (a) G S, so that

a(m) E S, as desired, in this case. If m E im (a), then m = a(mf) for some

mf E N. Hence a(m) = o(o(mf)) so that a(m) E im (a), as desired. By (b)

of Axiom 1, we conclude S = N. The result now follows from the argu-

ment given at the outset.

The next result is the centerpiece of the development of N, for in it we
prove the existence and uniqueness of an operation on N corresponding
to our intuitive idea of addition. As we will see, the definition of multipli-
cation is formulated in an analogous fashion. Our problem is somehow
to "get at" an operation of addition. Let us first stop to realize what kind
of mathematical object addition is; addition on the positive integers is a
mapping s (for sum) of N x N into N, satisfying certain properties. For the
I sake of simplicity, let us fix a positive integer m and try to create a mapping
in one independent variable. Such a mapping might be denoted s,: N -, N

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