338 CONSTRUCTION OF NUMBER SYSTEMS Chapter 10
Proof For n, p E N, let Snp = {q E N (n(~ + q) = np + nq}. First, we have
1 E Snp, because
n(p + 1) = np + n [by (411
= np + n. 1 [by (3) and wd of addition]
Second, assume q E S,,, so that n(p + q) = np + nq. We must prove
q + 1 E Snp; to do this, we need to show n(p + (q + 1)) = np + n(q + 1).
Now:n(p + (q + 1)) = n((p + q) + 1)) [by (2) and wd of multiplication]
= n(p + q) + n [by (411
= (np + nq) + n (by induction hypothesis and wd
of addition)
= np + (nq + n) [by Theorem 5(a)]
= np + n(q + 1) [by (4) and wd of addition]We next consider properties of multiplication, some of which correspond
to additive properties in Theorem 5.
THEOREM 7
Let n, p, q E N. Then:(a) (np)q = n(pq) (multiplication is associative)
(b) n-1 =l-n=n (multiplicative identity)
(c) nP=Pn (multiplication is commutative)
(d) If p # 1, then pq # 1 for all q E N (positive integers #^1 do not
have reciprocals in N)
(e) If nq = pq, then n = p (multiplicative cancellation)Proof Parts (b), (c) and (e) are left as exercises (see Exercise 4). We now
consider the proof of (a). Given n, p E N; denote by Snp the set {q E
N l(np)q = n(pq)}. Note first that 1 E S.,, since (np). 1 = np = n(p I),
where the latter step follows from (3) and the well-definedness of the
multiplication operation. Next, assume q E S,,, so that (np)q = n(pq).
To prove q + 1 E Snp, we must show that (np)(q + 1) = n(p(q + 1)). Now:
n(p(q + 1)) = n(pq + p). [by (4) and wd of multiplication]
= n(pq) + np (by Theorem 6)
= (np)q + np (by induction hypothesis and wd of
addition)
= (np)(q + 1) [by (411
As for (d), assume p E N and p # 1, and let S, = (q E Nlpq # 1). We
prove S,, = N by induction. Namely, 1 E S,, since p. 1 = p # 1. Also,
if we assume q E S,, SO that pq # 1 then pq = a(x) for some x E N. We