358 CONSTRUCTION OF NUMBER SYSTEMS Chapter 10
what should be true about the componentwise difference of the two se-
quences? Those of you whose immediate reaction is "null sequence" are
"right on the money"! We now have a way of characterizing, in a manner
that operates entirely within Q, our intuitive, but vague, idea of two se-
quences of rationals seeming to want to converge to the same number (pos-
sibly not rational). We formalize this in Definition 4.
DEFINITION 4
Define a relation -- on the set 6 of all Cauchy sequences of rational numbers,
by the rule {a,) -- (6,) if and only if the sequence {a, - 6,) of componentwise
differences is a null sequence.THEOREM 3
The relation -- is an equivalence relation on 6.As indicated earlier, we denote by R the collection of all equivalence
classes generated on & by -. Also, we denote by [a,] the equivalence class
determined by the sequence {a,}. The proof of Theorem 3 is an easy exer-
cise [see Exercise 2(a)] for anyone with any experience whatsoever in work-
ing with sequences. Recalling the process used in Article 10.2 to construct
Z from N and Q from Z, the next step should be evident; it is time to
define operations of addition and multiplication on R.
DEFINITION 5
We define operations of "addition" and "multiplication," denoted + and respec-
tively, on R, as follows. Let [a,] and [b,] be elements of R. Then we define:
(a) Thesum[a,]+[b,]tobe[a,+b,],and
(6) The product [a,][b,] to be [a,b,].If we have done our job well thus far in this chapter, you should be at
least one move ahead of the discussion at this stage and thinking about
what must clearly be the next step. We must prove that + and -, as defined
in Definition 5, are well-defined operations. That is, we must show that if
[a,] = [a;] and [b,] = [b:], then [a, + b,] = [a; + K] and [a&,] = [aLK].
More specifically, we need the following result.
THEOREM 4
Let [a,], [b,], [a;], and [b;] be elements of R. Then if {a, - a;) and {b, - 6;)
are null sequences, we have:
(a) {(a, + 6,) - (a; + g)) is a null sequence, and
(6) {a,b, - a;%) is a null sequence.The proof of Theorem 4, which we again do not include in the text,
should be a reasonable exercise [see Exercise 2(b)] for those who are at-