10.3 OUTLINE OF THE CONSTRUCTION OF THE REALS 357tempting to fill in the proofs; (b) is by far the less routine and we include
hints for the proof in the statement of the exercise.
At this point we have constructed the object we've been looking for, but
we are far from finished. In fact, there are four major steps remaining. You
should stop for a moment and think through what those four steps are.
Let us now resume and begin to present, with most details omitted, the
four steps.
THEOREM 5
The system (R, +, -) is a field.Remarks on the proof Most parts of the verification of Theorem^5 (i.e., all
field axioms except the multiplicative inverse axiom) are routine and are
left to you in Exercise 3(a). We do note, however, in reference to Axiom
4, that the additive identity for R is the equivalence class consisting of all
the null sequences of rational numbers [you should verify in Exercise 3(b)
that the collection of all null sequences does indeed constitute an equiv-
alence class]. This fact is important in understanding why the multi-
plicative inverse axiom is satisfied in R. Suppose [a,], for instance, is a
nonzero equivalence class in R. We claim that [a,] has a multiplicative
inverse [b,] in R. That is, there exists a Cauchy sequence (b,) such that
[anbn] = [I], in other words, such that (1 - anbn) is a null sequence.
The fact that [a,] is nonzero means that (a,) is not a null sequence.
This means that there exists a positive rational E such that infinitely many
terms of the sequence are 2 E in absolute value. In symbols, 3~ > 0
(E rational) such that, for any N E N, there exists n E N such that n > N
and lanlz E. Combining this with the fact that {a,) is Cauchy, we can
prove that there exists a positive integer No such that la,l > ~/2 for all
m E N such that m > No. An immediate consequence of this fact is that
a, # 0 for all but (at most) finitely many terms of the sequence, specifi-
cally, a, # 0 whenever rn > No. We are now ready to define the sequence
(b,). Let b, = 0 if n I No and let b, = l/a, if n 2 No. Clearly the prod-
uct sequence (anbn) has values 0 for n 5 No and 1 for n > No, so that
the sequence (1 - a,b,), having all terms beyond the Noth equal to zero,
is clearly a null sequence, as desired. One final remark about this verifi-
cation: We must, of course, prove specifically that the sequence (b,), as
we've defined it, determines an equivalence class in R. For this we need
that (b,) is Cauchy. You should be able to fill in this detail [see Exercise
3(c)], if you wish to do so. 0
So (R, +, a) is a field, modulo, of course, verification of all the details of
Theorem 5. Before addressing the matters of ordering and completeness,
let us consider in exactly what sense Q can be regarded as a subset of R.
We say that a rational number q is associated with the real number [q,] if
and only if the sequence {q, - q) is null. Note that if q is associated with
the equivalence class containing the sequence {q,), then q is associated with