10.2 DEVELOPMENT OF THE INTEGERS AND RATIONAL NUMBERS 345answer. Suppose you come along and carry out the same procedure, select-
ing, however, representatives different from the ones I used (which you are
perfectly free to do!). Is the answer you obtain necessarily the same as
mine, assuming, of course, that we both calculate correctly? Note, by the
way, what it means for our answers to be "the same." We need not arrive
at identical ordered pairs as the sum; rather, we need only arrive at ordered
pairs that come from the same equivalence class, that is, equivalent ordered
pairs. If the operation of addition on Z, as defined previously, is to "make
sense," if it is to be well defined, the answer to the preceding question had
better be "yes." Fortunately, such is the case, with the analogous fact for
multiplication also holding true. We formalize these results in our next
theorem.
THEOREM 2
Let a, b, c, d, a', b', c', and d' be elements of N with (a, b) - (a', b') and (c, d) -
(c', d'). Then:
(a) (a + c, b + d) - (a' + c', b' + d')
(b) (ac + bd, ad + bc) - (a'c' + b'd', a'd' + b'c')Proof (a) To show that (a + c, b + d) - (a' + c', b' + d'), we need to
prove that (a + c) + (b' + d') = (b + d) + (a' + c'). Our assumptions
that (a, b) - (a', b') and (c, d) - (c', d') yield the equations a + b' = b + a'
and c + d' =d + c'. Adding these equations and regrouping, by using
additive associativity and commutativity, we obtain the desired result.
(b) Left to the reader (see Exercise 1).With the proof of Theorem 2, we now know that + and. are "honest"
operations on the set Z. The main questions now about the system (Z, +, 0)
are: (1) What are its algebraic properties? and (2) How is it related to the
system (N, +, a), from which it was constructed? We will address question
(2) first. Experience tells us we should expect Z to contain N as a proper
subset. Let us examine the precise sense in which this turns out to be true.
Let n E N and [(p, q)] E Z. Let us agree to say that n is associated with
[(p, q)] if and only if n + q = p (automatically implying that p > q). Sup-
pose n is associated with [(p, q)] and let (r, s) be an arbitrary element of
N x N. Then it is easy to verify that n is associated with the integer [(r, s)]
if and only if (r, s)-E [(p, q)]; that is, if and only if (r, s) - (p, q). (Proof: Sup-
pose n + q = p and n + s = r. We claim that (p, q) - (r, s); that is, p + s =
q + r. But p + n + s = n + q + r so, by additive cancellation, the desired
result holds. Conversely, suppose that n + q = p and (p, q) - (r, s), so that
p + s = q + r. We claim n + s = r. But (n + q) + (p + s) = p + (q + r), so
that, regrouping, we get (n + s) + (q + p) = r + (q + p), so that n + s = r,
as desired.) Noting that any positive integer n is surely associated with the
integer [(n + 1, I)], we conclude that each n E N is associated with a unique
equivalence class in Z, a class we will denote by [n], so that the mapping
n -, [n] of N into Z is well defined. Furthermore, this mapping is one to