3 Real Numbers 21It remains to show thata 1 b+a 2 c∈A(B+C)ifa 1 ,a 2 ∈A,b∈Bandc∈C.But
a 1 b+a 2 c≤a 2 (b+c) ifa 1 ≤a 2 ,and
a 1 b+a 2 c≤a 1 (b+c) ifa 2 ≤a 1.In either event it follows thata 1 b+a 2 c∈A(B+C).
We can now show that multiplication of cuts satisfies the order relation(O4).If
A<B, then there exists a cutDsuch thatA+D=Band henceAC<AC+DC=
BC. Conversely, supposeAC < BC.ThenA = B.SinceB < Awould imply
BC<AC, it follows thatA<B.
From(O4)and the law of trichotomy(O2)it follows that multiplication of cuts
satisfies the cancellation law(M1).
We next prove the existence of multiplicative inverses. The proof will use the fol-
lowing multiplicative analogue of Lemma 15:
Lemma 17For any cut A and any c∈P with c> 1 , there exists a∈A such that
ac∈/A.
Proof Choose anyb∈ A. We may supposebc∈ A, since otherwise we can take
a=b.Sinceb<bc,wehavebc=b+dfor somed∈P. By Lemma 15 we can
choosea∈Aso thata+d∈/A.Sinceb+d∈A, it follows thatb+d<a+d,and
sob<a. Henceab−^1 >1and
a+d<a+(ab−^1 )d=ab−^1 (b+d)=ac.Sincea+d∈/A, it follows thatac∈/A.
Proposition 18For any A∈P, there exists A−^1 ∈Psuch that A A−^1 =I.
Proof LetA−^1 be the set of allb∈ Psuch thatb<c−^1 for somec∈P\A.Itis
easily verified thatA−^1 is a cut. We note only thata−^1 ∈/A−^1 ifa∈Aand that, if
b<c−^1 ,thenalsob<d−^1 for somed>c.
We now show thatAA−^1 =I.Ifa∈Aandb∈A−^1 thenab<1, sincea≥b−^1
would implya >cfor somec ∈ P\A. ThusAA−^1 ≤ I. On the other hand, if
0 <d<1 then, by Lemma 17, there existsa∈Asuch thatad−^1 ∈/A. Choosea′∈A
so thata<a′, and putb=(a′)−^1 d.Thenb<a−^1 d.Sincea−^1 d=(ad−^1 )−^1 ,it
follows thatb∈A−^1 and consequentlyd=a′b∈AA−^1. ThusI≤AA−^1.
For any positive rational numbera,thesetAaconsisting of all positive rational
numberscsuch thatc<ais a cut. The mapa→AaofPintoPis injective and
preserves sums and products:
Aa+b=Aa+Ab,Aab=AaAb.Moreover,Aa<Abif and only ifa<b.
By identifyingawithAawe may regardPas a subset ofP. It is a proper subset,
since(P4)does not hold inP.