Proof

Suppose aa is an element of A,A, and let a'<aa'<a be given. By way of contradiction suppose that a'a' does not belong to A.A. Then, by Condition (1) of the definition of a cut, it must be that a'∈B.a'∈B. But then, by Condition (2) of the definition of a cut, we must have that a≤a',a≤a', and this is a contradiction, because a'<a.a'<a. This proves part (1). Part (2) is proved in a similar manner.

To prove part (3), let the rational number ϵ>0ϵ>0 be given, and set r=ϵ/2.r=ϵ/2. Choose an element a0∈Aa0∈A and an element b0∈B.b0∈B. Such elements exist, because AA and BB are nonempty sets. Choose a natural number NN such that a0+Nr>b0.a0+Nr>b0. Such a natural number NN must exist. For instance, just choose NN to be larger than the rational number (b0-a0)/r.(b0-a0)/r. Now define a sequence {ak}{ak} of rational numbers by ak=a0+kr,ak=a0+kr, and let KK be the first natural number for which aK∈B.aK∈B. Obviously, such a number exists, and in fact KK must be less than or equal to N.N. Now, aK-1aK-1 is not in B,B, so it must be in A.A. Set a=AK-1a=AK-1 and b=AK.b=AK. Clearly, a∈A,a∈A,b∈B,b∈B, and

and this proves part (3).

Proof

Let A'=-B,A'=-B, i.e., the set of all the negatives of the elements of B,B, and let B'=-A,B'=-A, i.e., the set of all the negatives of the elements of A.A. It is immediate that the pair (A',B')(A',B') is a Dedekind cut. Let us show that (A,B)+(A',B')(A,B)+(A',B') is equivalent to the zero cut. Let (C,D)=(A,B)+(A',B').(C,D)=(A,B)+(A',B'). Then, by the definition of the sum of two cuts, we know that DD consists of all the elements of the form d=b+b'=b-a,d=b+b'=b-a, where b∈Bb∈B and a∈A.a∈A. Since a≤ba≤b for all a∈Aa∈A and b∈B,b∈B, we see then that the elements of DD are all greater than or equal to 0. To see that (C,D)(C,D) is equivalent to the 0 cut, it will suffice to show that DD contains all the positive rational numbers. (Why?) Hence, let ϵ>0ϵ>0 be given, and choose an a∈Aa∈A and a b∈Bb∈B such that b-a<ϵ.b-a<ϵ. This can be done by Condition (3) of Theorem 1. Then, the number b-a∈D,b-a∈D, and hence, by part (2) of Theorem 1, ϵ∈D.ϵ∈D. It follows then that the cut (C,D)(C,D) is equivalent to the zero cut (A0,B0),(A0,B0), as desired.

Proof

The first five axioms for a field, given in (Reference), have been established for FF in the preceding exercises, so that we need only verify axiom 6 to complete the proof that FF is a field. Thus, let x∈Fx∈F be a nonzero element. We must show the existence of an element yy of FF for which x×y=1.x×y=1. Suppose first that xx is a positive element of F.F. Then xx is the equivalence class of a positive cut (A,B),(A,B), and therefore AA contains some positive rational numbers. Let a0a0 be a positive number that is contained in A.A. It follows then that every element of BB is greater than or equal to a0a0 and hence is positive. Define B^B^ to be the set of all rational numbers rr for which r≥1/br≥1/b for every b∈B.b∈B. Then define A^A^ to be the set of all rationals rr for which r≤b^r≤b^ for every b^∈B^.b^∈B^. It follows directly that the pair (A^,B^)(A^,B^) is a Dedekind cut.

Let (C,D)=(A,B)×(A^,B^),(C,D)=(A,B)×(A^,B^), and note that every element d∈Dd∈D is of the form d=bb^,d=bb^, and hence is greater than or equal to 1. We claim that (C,D)(C,D) is equivalent to the cut (A1,B1)(A1,B1) that determines the element 1 of F.F. To see this we must verify that DD contains every rational number rr that is greater than 1. Thus, let r>1r>1 be given, and set ϵ=a0(r-1).ϵ=a0(r-1). From Condition (3) of Theorem 1, choose an a'∈Aa'∈A and a b'∈Bb'∈B such that b'-a'<ϵ.b'-a'<ϵ. Without loss of generality, we may assume that a'≥a0.a'≥a0. Finally, set b^=1/a'.b^=1/a'. Clearly b^≥1/bb^≥1/b for all b∈B,b∈B, so that b^∈B^.b^∈B^. Also d=b'b^∈D,d=b'b^∈D, and

implying that r∈D.r∈D. Therefore, (C,D)(C,D) is equivalent to the cut (A1,B1),(A1,B1), implying that (A,B)×(A^,B^)(A,B)×(A^,B^) is equivalent to the cut (A1,B1).(A1,B1). Therefore, if yy is the element of FF that is the equivalence class of the cut (A^,B^),(A^,B^), then x×y=1,x×y=1, as desired.

If xx is negative, then -x-x is positive. If we write zz for the multiplicative inverse of the positive element -x,-x, then -z-z is the multiplicative inverse of the element x.x. Indeed, by the definition of the product of two negative elements of F,F,x×(-z)=(-x)×z=1.x×(-z)=(-x)×z=1.

The properties that guarantee that FF is an ordered field also have been established in the preceding exercises, so that the proof of this theorem is complete.

Proof

We have that x≤yx≤y if and only if the element y-x=y+-xy-x=y+-x is positive or 0. Writing, as before, (A',B')(A',B') for the cut -(A,B),-(A,B), we have that y-xy-x is the equivalence class of the cut (C,d)-(A,B)=(C,D)+(A',B'),(C,d)-(A,B)=(C,D)+(A',B'), so we need to determine when the cut (G,H)=(C,D)+(A',B')(G,H)=(C,D)+(A',B') is a positive cut or the 0 cut; which is the case when the set HH only contains nonnegative numbers. By definition of addition, the set HH contains all numbers of the form h=d+b'h=d+b' for some d∈Dd∈D and some b'∈B'.b'∈B'. Since B'=-A,B'=-A, this means that HH consists of all elements of the form h=d-ah=d-a for some d∈Dd∈D and a∈A.a∈A. Now these numbers hh are all greater than or equal to 0 if and only if each a∈Aa∈A is less than or equal to each d∈D,d∈D, i.e., if and only if (A,B)≤(C,D).(A,B)≤(C,D). This proves the theorem.

Proof

Let SS be a nonempty subset of F,F, and suppose that there exists an upper bound for S;S; i.e., an element MM of FF such that x≤Mx≤M for all x∈S.x∈S. Write (A,B)(A,B) for a cut such that MM is the equivalence class of (A,B).(A,B). We must show that there exists a least upper bound for S.S.

For each x∈S,x∈S, let (Ax,Bx)(Ax,Bx) be a Dedekind cut for which xx is the equivalence class of (Ax,Bx),(Ax,Bx), and note that ax≤bax≤b for all ax∈Axax∈Ax and all b∈B.b∈B. Let A0A0 be the union of all the sets AxAx for x∈S.x∈S. Let B0B0 be the set of all rational numbers rr for which r≥a0r≥a0 for every a0∈A0.a0∈A0. we claim first that the pair (A0,B0)(A0,B0) is a Dedekind cut. Both sets are nonempty; A0A0 because it is the union of nonempty sets, and B0B0 because it contains all the elements of the nonempty set B.B. Clearly Condition (2) for a cut holds from the very definition of this pair. To see Condition (1), let rr be a rational number that is not in B0.B0. We must show that it is in A0.A0. Now, since rr is not in B0,B0, there must exist some a0∈A0a0∈A0 for which r<a0.r<a0. But a0∈∪x∈SAx,a0∈∪x∈SAx, so that there must exist an x∈Sx∈S such that a0∈Ax,a0∈Ax, and hence rr is also in Ax.Ax. But then r∈A0,r∈A0, and this proves that (A0,B0)(A0,B0) is a Dedekind cut.

Let M0M0 be the equivalence class determined by the cut (A0,B0).(A0,B0). Since each Ax⊆A0,Ax⊆A0, we see that ax≤b0ax≤b0 for every ax∈Axax∈Ax and every b0∈B0.b0∈B0. Hence, (Ax,Bx)≤(A0,B0)(Ax,Bx)≤(A0,B0) for every x∈S,x∈S, and therefore, by Theorem A.4, x≤M0x≤M0 for all x∈S.x∈S. This shows that M0M0 is an upper bound for S.S.

Finally, suppose M'M' is another upper bound for S,S, and let (A',B')(A',B') be a cut for which M'M' is the equivalence class of (A',B').(A',B'). Then ax≤b'ax≤b' for every ax∈Axax∈Ax and every b'∈B',b'∈B', implying that a0≤b'a0≤b' for every a0∈A0a0∈A0 and every b'∈B'.b'∈B'. Therefore, (A0,B0)≤(A',B'),(A0,B0)≤(A',B'), implying that M0≤M'.M0≤M'. This shows that M0M0 is the least upper bound for S,S, and the theorem is proved.