Mathematical Analysis I, 2004a

§7. Integers and Rationals 35
In an ordered field,
N = {x ∈ J | x>0}. (Why?)
Theorem 1. If a and b are integers (or rationals) in F , so are a + b and ab.
Proof. For integers, this follows from Examples (a) and (d) in §§5–6; one only
has to distinguish three cases:
(i) a, b ∈ N;
(ii) −a ∈ N, b ∈ N;
(iii) a ∈ N, −b ∈ N.
The details are left to the reader (see Basic Concepts of Mathematics, Chapter
2, §7, Theorem 1).
Now let a and b be rationals, say,
a = p q and b = r s ,
where p, q, r, s ∈ J and q, s ≠ 0. Then, as is easily seen,
a ± b =
ps ± qr
qs
and ab = pr
qs ,
where qs ≠0;andqs and pr are integers by the first part of the proof (since
p, q, r, s ∈ J).
Thus a ± b and ab are fractions with integral numerators and denominators.
Hence, by definition, a ± b ∈ R and ab ∈ R. □
Theorem 2. In any field F , the set R of all rationals is a field itself , under
the operations defined in F , with the same neutral elements 0 and 1. Moreover,
R is an ordered field if F is. (We call R the rational subfield of F .)
Proof. We have to check that R satisfies the field axioms.
The closure law I follows from Theorem 1.
Axioms II, III, and VI hold for rationals because they hold for all elements
of F ; similarly for Axioms VII to IX if F is ordered.
Axiom IV holds in R because the neutral elements 0 and 1 belong to R;
indeed, they are integers, hence certainly rationals.
To verify Axiom V, we must show that −x and x −1 belong to R if x does.
If, however,
x = p (p, q ∈ J, q ≠0),
q
then
−x = −p
q ,
where again −p ∈ J by the definition of J; thus−x ∈ R.