20 I The Expanding Universe of Numbers
Proposition 16For any cuts A,B, there exists a cut C such that A+C=B if and
only if A<B.
Proof We prove the necessity of the condition by showing thatA<A+Cfor any
cutsA,C.Ifa∈Aandc∈C,thena<a+c.SinceA+Cis a cut, it follows that
a∈A+C. ConsequentlyA≤A+C, and Lemma 15 implies thatA=A+C.
Suppose now thatAandBare cuts such thatA<B,andletCbe the set of all
c∈Psuch thatc+d∈Bfor somed∈P\A. We are going to show thatCis a cut
and thatA+C=B.
The setCis not empty. For chooseb∈B\Aand thenb′∈Bwithb<b′.Then
b′=b+c′for somec′∈P, which impliesc′∈C. On the other hand,C≤B,since
c+d∈Bandd∈Pimplyc∈B. ThusCis a proper subset ofP.
Supposec∈C,p∈Pandp<c.Wehavec+d∈Bfor somed∈P\Aand
c=p+efor somee∈P.Sinced+e∈P\Aandp+(d+e)=c+d∈B,it
follows thatp∈C.
Suppose now thatc∈C,sothatc+d∈Bfor somed∈P\A. Chooseb∈Bso
thatc+d<b.Thenb=c+d+efor somee∈P. If we putc′=c+e,thenc<c′.
Moreoverc′∈C,sincec′+d=b. This completes the proof thatCis a cut.
Supposea∈Aandc∈C.Thenc+d∈Bfor somed∈P\A. Hencea<d.It
follows thata+c<c+d,andsoa+c∈B. ThusA+C≤B.
It remains to show thatB≤A+C.Pickanyb∈B.Ifb∈A,thenalsob∈A+C,
sinceA < A+C. Thus we now assumeb ∈/ A. Chooseb′∈ Bwithb <b′.
Thenb′ =b+dfor somed ∈ P. By Lemma 15, there existsa ∈ Asuch that
a+d∈/A. Moreovera<b,sinceb∈/A, and henceb=a+cfor somec∈P.Since
c+(a+d)=b+d=b′, it follows thatc∈C. Thusb∈A+CandB≤A+C.
We can now show that addition of cuts satisfies the order relation(O3). Suppose
first thatA<B. Then, by Proposition 16, there exists a cutDsuch thatA+D=B.
Hence, for any cutC,
A+C<(A+C)+D=B+C.Suppose next thatA+C<B+C.ThenA=B.SinceB<Awould implyB+C<
A+C, by what we have just proved, it follows from the law of trichotomy thatA<B.
From(O3)and the law of trichotomy, it follows that addition of cuts satisfies the
cancellation law(A1).
We next define multiplication of cuts. For anyA,B∈P,letABdenote the set
of all rational numbersab, witha∈Aandb∈B. In the same way as forA+B,it
may be shown thatAB∈P. We note only that ifa∈A,b∈Bandc<ab,then
b−^1 c<a. Henceb−^1 c∈Aandc=(b−^1 c)b∈AB.
It follows from the corresponding properties of rational numbers that multiplication
of cuts satisfies the commutative law(M2)and the associative law(M3). Moreover
(M4)holds, the identity element for multiplication being the cutIconsisting of all
positive rational numbers less than 1.
We now show that the distributive law(AM1)also holds. The distributive law for
rational numbers shows at once that
A(B+C)≤AB+AC.