3 Real Numbers 23Proof LetSbe the set of all positive real numbersxsuch thatx^2 ≤a.ThesetSis
not empty, since it containsaifa≤1and1ifa>1. Ify>0andy^2 >a,then
yis an upper bound forS. In particular, 1+ais an upper bound forS.Letbbe the
least upper bound forS.Thenb^2 =a,sinceb^2 <awould imply(b+ 1 /n)^2 <afor
sufficiently largen>0andb^2 >awould imply(b− 1 /n)^2 >afor sufficiently large
n>0. Finally, ifc^2 =aandc>0, thenc=b,since
(c−b)(c+b)=c^2 −b^2 = 0. The unique positive real numberbin the statement of Proposition 21 is said to be a
square rootofaand is denoted by
√
aora^1 /^2. In the same way it may be shown that,
for any positive real numberaand any positive integern, there exists a unique positive
real numberbsuch thatbn=a,wherebn=b···b(ntimes). We say thatbis ann-th
rootofaand writeb=n
√
aora^1 /n.
A set is said to be afieldif two binary operations, addition and multiplication, are
defined on it with the properties(A2)–(A5),(M2)–(M5)and(AM1)–(AM2).Afield
is said to beorderedif it contains a subsetPof ‘positive’ elements with the properties
(P1)–(P3). An ordered field is said to becompleteif, with the order induced byP,it
has the property(P4).
Propositions 19–21 hold in any complete ordered field, since only the above prop-
erties were used in their proofs. By construction, the setRof all real numbers is a
complete ordered field. In fact, any complete ordered fieldFis isomorphic toR,i.e.
there exists a bijective mapφ:F→Rsuch that, for alla,b∈F,
φ(a+b)=φ(a)+φ(b),
φ(ab)=φ(a)φ(b),andφ(a)>0 if and only ifa∈P. We sketch the proof.
Letebe the identity element for multiplication inFand, for any positive integer
n,letne=e+···+e(nsummands). SinceFis ordered,neis positive and so has a
multiplicative inverse. For any rational numberm/n,wherem,n∈Zandn>0, write
(m/n)e=m(ne)−^1 ifm> 0 ,=−(−m)(ne)−^1 ifm<0, and=0ifm=0. The
elements(m/n)eform a subfield ofFisomorphic toQand we defineφ((m/n)e)=
m/n.Foranya∈F,wedefineφ(a)to be the least upper bound of all rational numbers
m/nsuch that(m/n)e≤a. One verifies first that the mapφ:F→Ris bijective and
thatφ(a)<φ(b)if and only ifa<b. One then deduces thatφpreserves sums and
products.
Actually, any bijective mapφ:F→Rwhich preserves sums and products is also
order-preserving. For, by Proposition 21,b>aif and only ifb−a=c^2 for some
c=0, and then
φ(b)−φ(a)=φ(b−a)=φ(c^2 )=φ(c)^2 > 0.Those whose primary interest lies in real analysis maydefineRto be a complete
ordered field and omit the tour throughN,Z,QandP. That is, one takes as axioms
the 14 properties above which define a complete ordered field and simply assumes that
they are consistent.