10.1 AN AXlOMATlZATlON FOR THE SYSTEM OF POSITIVE INTEGERS 335(i)' and (ii)'. Thus the inductive step (11) is proved and we
conclude, by induction, that Y = N; that is, s, is a function.
With this, existence is established.(Uniqueness) Suppose r, is any mapping from N to N satisfying prop-
erties (i) and (ii). Since r, is a relation on N satisfying (i) and (ii) and
since s, is the intersection of all such relations, then necessarily s, is a
subset of r,. We claim that, in fact, r, is a subset of s,, so that s, equals
r,. Let (x, y) E r,. Since x E N and dom (s,) = N, there exists z E N such
that (x, z) E s,. Since s, c r,, then (x, z) E r,. Since (x, y) E r,, (x, Z) E r,,
and r, is a function, we may conclude y = z, so that (x, y) = (x, z) and
(x, y) E s,, as required.With a mapping s,: N -, N determined uniquely by each positive integer
m, we are now in a position to define the desired "sum function" s.
DEFINITION 1
If m, n E N, we define the sum s(m, n) of m and n by the rule s(m, n) = s,(n).
Following normal usage, we frequently denote s(m, n) by the symbol m + n.By Theorem 3, each function s, is uniquely determined by a given m E N
and properties (i) and (ii) of that theorem. For a given m, the value s,(n)
is uniquely determined by n E N, since each s, is a function. Hence each
pair (m, n) of positive integers determines uniquely a corresponding number
s(m, n). Furthermore, it is clear that each such value s(m, n) is contained in
N. The following statement summarizes these two properties.
COROLLARY(a) + is a function on N x N and so constitutes a well-defined operation on
N; that is, if a, a', b, b' E N with a = a' and b = b', then a + b = a' + b'.
(b) N is closed under the operation +; that is, if a, b E N, then a + b E N.Part (a) of the corollary is the assertion that "equals added to equals
yield equals," a particular case of which is the result "y = z implies
x + y = x + z for any x, y, z E N." The property in (b) is called closure
under addition. -
Before dealing with other properties of addition, let us first define the
multiplication operation. Our approach will be to look closer at what we
actually did in Theorem 3, when we formulated addition. It is not hard to
see that Theorem 3 is a special case of the much more abstract-appearing
result we are about to state. What is only slightly more difficult is the
verification [left as an exercise in Exercise 2(b)] that a proof essentially
identical to the proof of Theorem 3 works in this more general case as
well.