7
Hence, we have |a
n
· b
n
− c
n
· d
n
| ≤ |b
n
| · |a
n
− c
n
| + |c
n
| · |b
n
− d
n
|. Now, from Theorem
5, there are numbers M and L such that |b
n
| ≤ M and |c
n
| ≤ L for all n. Taking some
number R (for example = M + L) which is bigger than both, we have
|a
n
· b
n
− c
n
· d
n
| ≤ |b
n
| · |a
n
− c
n
| + |c
n
| · |b
n
− d
n
| ≤ R(|a
n
− c
n
| + |b
n
− d
n
|).
Now, noting that both a
n
− c
n
and b
n
− d
n
tend to 0 and using the ω/2 trick (actually, this
time we’ll want to use ω/2R, I suggest you work out the details), we see that a
n
· b
n
− c
n
·
d
n
→ 0. Whew!
So, now that we have well-defined operations, it behooves us to prove that R, equipped
with them, is a field. This is a long and laborious task, and we shall not really work
through it here; the interested reader should be in a position to prove the field axioms
him- or herself! We will work out one slightly trickier example here, just to give a flavour.
Theorem 15. Given any real number s 6= 0, there is a real number t such that s · t = 1.
Proof. First we must properly understand what the theorem says. The premise is that s
is nonzero, which means that s is not in the equivalence class of (0, 0, 0, 0, . . .). In other
words, s = [( a
n
)] where a
n
− 0 does not converge to 0. From this, we are to deduce the
existence of a real number t = [(b
n
)] such that s · t = [(a
n
· b
n
)] is the same equivalence
class as [(1, 1, 1, 1, . . .)]. Doing so is actually an easy consequence of the fact that nonzero
rational numbers have multiplicative inverses, but there is a subtle difficulty. Just because
s is nonzero (i.e. (a
n
) does not tend to 0), there’s no reason any number of the terms in
(a
n
) can’t equal 0. However, it turns out that eventually, a
n
6= 0. That is,
Lemma 16. If (a
n
) is a Cauchy sequence which does not tend to 0, then there is an N such that,
for n > N, a
n
6= 0.
The proof of Lemma 16 is left to you. We will now use it to complete the proof of Theorem
15.
Let N be such that a
n
6= 0 for n > N. Define a sequence b
n
of rational numbers as fol-
lows: for n ≤ N, b
n
= 0, and for n > N, b
n
= 1/a
n
; (b
n
) = (0, 0, . . . , 0, 1/a
N +1
, 1/a
N +2
, . . .).
(This makes sense since, for n > N, a
n
is a nonzero rational number, so 1/a
n
exists.) Then
a
n
· b
n
is equal to a
n
· 0 = 0 for n ≤ N, and equals a
n
· b
n
= a
n
· 1/a
n
= 1 for n > N.
Well, then, if we look at the sequence (1, 1, 1, 1, . . .), we have (1, 1, 1, 1, . . .) − (a
n
· b
n
) is the
sequence which is 1 − 0 = 0 for n ≤ N and equals 1 − 1 = 0 for n > N. Since this sequence
is eventually equal to 0, it converges to 0, and so [(a
n
· b
n
)] = [(1, 1, 1, 1, . . .)] = 1 ∈ R. This
shows that t = [(b
n
)] is a multiplicative inverse to s = [(a
n
)].
Our choice of (b
n
) to start with 0s was arbitrary; we could have chosen it to equal 17
until after the Nth stage, or chosen each term randomly from Q; the only important choice
is that b
n
= 1/a
n
for n > N.
Following proofs that are generally easier than this one, we can show that R, +, · is a
field. Now, we want R to be an ordered field, which means there is an order relation < on
R which respects the field operations, as spelled out in §1.5 and 1.17 of [1]. Like before,
having newly constructed these ”real” numbers, we must define what it means for one to
be less than another. We will take our intuition from rational approximating sequences
again. How do we know that π > 3.141592? Well, 3 isn’t greater than 3.141592, nor is 3.14.