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Appendix: Existence and Uniqueness of a Complete Ordered Field

Module by: Lawrence Baggett. E-mail the author

Summary: This appendix is devoted to the proofs of Theorems 1 and 2 in "The Real and Complex Numbers: The Real Numbers," which together assert that there exists a unique complete ordered field. Our construction of this field will follow the ideas of Dedekind, which he presented in the late 1800's.

This appendix is devoted to the proofs of (Reference) and (Reference), which together assert that there exists a unique complete ordered field. Our construction of this field will follow the ideas of Dedekind, which he presented in the late 1800's.

Definition 1:

By a Dedekind cut, or simply a cut, we will mean a pair (A,B)(A,B) of nonempty (not necessarily disjoint) subsets of the set QQ of rational numbers for which the following two conditions hold.

  1.  AB=Q.AB=Q. That is, every rational number is in one or the other of these two sets.
  2. For every element aAaA and every element bB,bB,Ab.Ab. That is, every element of AA is less than or equal to every element of B.B.

Recall that when we define the rational numbers as quotients (ordered pairs) of integers, we faced the problem that two different quotients determine the same rational number, e.g., 2/36/9.2/36/9. There is a similar equivalence among Dedekind cuts.

Definition 2:

Two Dedekind cuts (A1,b1)(A1,b1) and (A2,B2)(A2,B2) are called equivalent if a1b2a1b2 for all a1A1a1A1 and all b2B2,b2B2, and a2b1a2b1 for all a2A2a2A2 and all b1B1.b1B1. In such a case, we write (A1,B1)(A2,B2).(A1,B1)(A2,B2).

Exercise 1

  1. Show that every rational number rr determines three distinct Dedekind cuts that are mutually equivalent.
  2. Let BB be the set of all positive rational numbers rr whose square is greater than 2, and let AA comprise all the rationals not in B.B. Prove that the pair (A,B)(A,B) is a Dedekind cut. Do you think this cut is not equivalent to any cut determined by a rational number rr as in part (a)? Can you prove this?
  3. Prove that the definition of equivalence given above satisfies the three conditions of an equivalence relation. Namely, show that
    1. (Reflexivity) (A,B)(A,B) is equivalent to itself.
    2. (Symmetry) If (A1,B1)(A2,B2),(A1,B1)(A2,B2), then (A2,B2)(A1,B1).(A2,B2)(A1,B1).
    3. (Transitivity) If (A1,B1)(A2,B2)(A1,B1)(A2,B2) and (A2,B2)(A3,B3),(A2,B2)(A3,B3), then (A1,B1)(A3,B3).(A1,B1)(A3,B3).

There are three relatively simple-sounding and believable properties of cuts, and we present them in the next theorem. It may be surprising that the proof seems to be more difficult than might have been expected.

Theorem 1

Let (A,B)(A,B) be a Dedekind cut. Then

  1. If aAaA and a'<a,a'<a, then a'A.a'A.
  2. If bBbB and b'>b,b'>b, then b'B.b'B.
  3. Let ϵϵ be a positive rational number. Then there exists an aAaA and a bBbB such that b-a<ϵ.b-a<ϵ.

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 aa',aa', 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 a0Aa0A and an element b0B.b0B. 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 aKB.aKB. 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, aA,aA,bB,bB, and

b - a = a K - a K - 1 = a 0 + K r - a 0 - ( K - 1 ) r = r = ϵ 2 < ϵ , b - a = a K - a K - 1 = a 0 + K r - a 0 - ( K - 1 ) r = r = ϵ 2 < ϵ ,
(1)

and this proves part (3).

We will make a complete ordered field FF whose elements are the set of equivalence classes of Dedekind cuts. We will call this field the Dedekind field. To make this construction, we must define addition and multiplication of equivalence classes of cuts, and verify the six required field axioms. Then, we must define the set PP that is to be the positive elements of the Dedekind field F,F, and then verify the required properties of an ordered field. Finally, we must prove that this field is a complete ordered field; i.e., that every nonempty set that is bounded above has a least upper bound. First things first.

Definition 3:

If (A1,B1)(A1,B1) and (A2,B2)(A2,B2) are Dedekind cuts, define the sum of (A1,B1)(A1,B1) and (A2,B2)(A2,B2) to be the cut (A3,B3)(A3,B3) described as follows: B3B3 is the set of all rational numbers b3b3 that can be written as b1+b2b1+b2 for some b1B1b1B1 and b2B2,b2B2, and A3A3 is the set of all rational numbers rr such that r<b3r<b3 for all b3B3.b3B3.

Several things need to be checked. First of all, the pair (A3,B3)(A3,B3) is again a Dedekind cut. Indeed, it is clear from the definition that every element of A3A3 is less than or equal to every element of B3,B3, so that Condition (2) is satisfied. To see that Condition (1) holds, let rr be a rational number, and suppose that it is not in A3.A3. We must show that rr belongs to B3.B3. Now, since rA3,rA3, there must exist an element b3=b1+b2B3b3=b1+b2B3 for which r>b3.r>b3. Otherwise, rr would be in A3.A3. But this means that r-b2>b1,r-b2>b1, and so by part (2) of Theorem 1, we have that r-b2r-b2 is an element b1'b1' of B1.B1. Therefore, r=b1'+b2,r=b1'+b2, implying that rB3,rB3, as desired.

We define the 0 cut to be the pair A0={r:r0}A0={r:r0} and B0={r:r>0}.B0={r:r>0}. This cut is one of the three determined by the rational number 0.

Exercise 2

  1. Prove that addition of Dedekind cuts is commutative and associative.
  2. Prove that if (A1,B1)(C1,D1)(A1,B1)(C1,D1) and (A2,B2)(C2,D2),(A2,B2)(C2,D2), then (A1,B1)+(A2,B2)(C1,D1)+(C2,D2).(A1,B1)+(A2,B2)(C1,D1)+(C2,D2).
  3. Find an example of a cut (A,B)(A,B) such that (A,B)+0(A,B).(A,B)+0(A,B).
  4. Prove that (A,B)+0(A,B)(A,B)+0(A,B) for every cut (A,B).(A,B).

We define addition in the set FF of all equivalence classes of Dedekind cuts as follows:

Definition 4:

If xx is the equivalence class of a cut (A,b)(A,b) and yy is the equivalence class of a cut (C,D),(C,D), then x+yx+y is the equivalence class of the cut (A,B)+(C,D).(A,B)+(C,D).

It follows from the previous exercise, that addition in FF is well-defined, commutative, and associative. We are on our way.

We define the element 0 of FF to be the equivalence class of the 0 cut. The next theorem establishes one of the important field axioms for F,F, namely, the existence of an additive inverse for each element of F.F.

Theorem 2

If (A,B)(A,B) is a Dedekind cut, then there exists a cut (A',B')(A',B') such that (A,B)+(A',B')(A,B)+(A',B') is equivalent to the 0 cut. Therefore, if xx is an element of F,F, then there exists an element yy of FF such that x+y=0.x+y=0.

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 bBbB and aA.aA. Since abab for all aAaA and bB,bB, 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 aAaA and a bBbB such that b-a<ϵ.b-a<ϵ. This can be done by Condition (3) of Theorem 1. Then, the number b-aD,b-aD, and hence, by part (2) of Theorem 1, ϵD.ϵD. It follows then that the cut (C,D)(C,D) is equivalent to the zero cut (A0,B0),(A0,B0), as desired.

We will write -(A,B)-(A,B) for the cut (A',B')(A',B') of the preceding proof.

Exercise 3

  1. Suppose (A,B)(A,B) is a cut, and let (C,D)(C,D) be a cut for which (A,B)+(C,D)(A,B)+(C,D) is equivalent to the 0 cut. Show that (C,D)(A',B')=-(A,B).(C,D)(A',B')=-(A,B).
  2. Prove that the additive inverse of an element xx of the Dedekind field FF is unique.

The definition of multiplication of cuts, as well as multiplication in F,F, is a bit more tricky. In fact, we will first introduce the notion of positivity among Dedekind cuts.

Definition 5:

A Dedekind cut x=(A,B)x=(A,B) is called positive if AA contains at least one positive rational number.

Exercise 4

  1. Suppose (A,B)(A,B) and (C,D)(C,D) are equivalent cuts, and assume that (A,B)(A,B) is positive. Prove that (C,D)(C,D) also is positive. Make the obvious definition of positivity in the set F.F.
  2. Show that the sum of two positive cuts is positive. Conclude that the sum of two positive elements of F,F, i.e., the sum of two equivalence classes of positive cuts, is positive.
  3. Let (A,B)(A,B) be a Dedekind cut. Show that one and only one of the following three properties holds for (A,B).(A,B). (i) (A,B)(A,B) is a positive cut, (ii) -(A,B)-(A,B) is a positive cut, or (iii) (A,B)(A,B) is equivalent to the 0 cut.
  4. Establish the law of tricotomy for F:F: That is, show that one and only one of the following three properties holds for an element xF.xF. (i) xx is positive, (ii) -x-x is positive, or (iii) x=0.x=0.

We first define multiplication of cuts when one of them is positive.

Definition 6:

Let (A1,B1)(A1,B1) and (A2,B2)(A2,B2) be two Dedekind cuts, and suppose that one of these cuts is a positive cut. We define the product(A3,B3)(A3,B3) of (A1,B1)(A1,B1) and (A2,B2)(A2,B2) as follows: Set B3B3 equal to the set of all b3b3 that can be written as b1b2b1b2 for some b1B1b1B1 and b2B2.b2B2. Then set A3A3 to be all the rational numbers rr for which r<b3r<b3 for all b3B3.b3B3.

Again, things need to be checked.

Exercise 5

  1. Show that the pair (A3,B3)(A3,B3) of the preceding definition for the product of positive cuts is in fact a Dedekind cut.
  2. Prove that multiplication of Dedekind cuts, when one of them is positive, is commutative.
  3. Suppose (A1,B1)(A1,B1) is a positive cut. Prove that
    (A1,B1)((A2,B2)+(A3,B3))=(A1,B1)(A2,B2)+(A1,B1)(A3,B3)(A1,B1)((A2,B2)+(A3,B3))=(A1,B1)(A2,B2)+(A1,B1)(A3,B3)
    (2)
    for any cuts (A2,B2)(A2,B2) and (A3,B3).(A3,B3).
  4. Show that, if (A1,B1)(A2,B2)(A1,B1)(A2,B2) and (C1,D1)(C2,D2)(C1,D1)(C2,D2) and (a1,B1)(a1,B1) and (A2,B2)(A2,B2) are positive cuts, then (A1,B1)(C1,D1)(A2,B2)(C2,D2).(A1,B1)(C1,D1)(A2,B2)(C2,D2).
  5. Show that the product of two positive cuts is again a positive cut.

We are ready to define multiplication in F.F.

Definition 7:

Let xx and yy be elements of F.F.

If either xx or yy is positive, define the product x×yx×y to be the equivalence class of the cut (A,B)(C,D),(A,B)(C,D), where xx is the equivalence class of (A,B)(A,B) and yy is the equivalence class of (C,D).(C,D).

If either xx or yy is 0, define x×yx×y to be 0.

If both xx and yy are negative, i.e., both -x-x and -y-y are positive, define x×y=(-x)×(-y).x×y=(-x)×(-y).

The next exercise is tedious. It amounts to checking a bunch of cases.

Exercise 6

  1. Prove that multiplication in FF is commutative.
  2. Prove that multiplication in FF is associative.
  3. Prove that multiplication in FF is distributive over addition.
  4. Prove that the product of two positive elements of FF is again positive.

We define the element 1 of FF to be the equivalence class of the cut (A1,B1),(A1,B1), where A1={r:r1}A1={r:r1} and B1={r:r>1}.B1={r:r>1}.

Exercise 7

  1. Prove that the elements 0 and 1 of FF are not equal.
  2. Prove that x×1=xx×1=x for every element xF.xF.
  3. Use the associative law and part (b) to prove that if xy=1xy=1 and xz=1,xz=1, then y=z.y=z.

Theorem 3

With respect to the operations of addition and multiplication defined above, together with the definition of positive elements, FF is an ordered field.

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 xFxF 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 r1/br1/b for every bB.bB. Then define A^A^ to be the set of all rationals rr for which rb^rb^ 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 dDdD 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 bB,bB, so that b^B^.b^B^. Also d=b'b^D,d=b'b^D, and

d = b ' b ^ = b ' a ' = a ' + b ' - a ' a ' < 1 + ϵ a ' 1 + ϵ a 0 = r , d = b ' b ^ = b ' a ' = a ' + b ' - a ' a ' < 1 + ϵ a ' 1 + ϵ a 0 = r ,
(3)

implying that rD.rD. 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.

So, the Dedekind field is an ordered field, but we have left to prove that it is complete. This means we must examine upper bounds of sets, and that requires us to understand when one cut is less than another one. We say that a cut (A,B)(A,B) is less than or equal to a cut C,D)C,D) if adad for every aAaA and dD.dD. We say that an element xx in the ordered field FF is less than or equal to an element yy if y-xy-x is either positive or 0.

Theorem 4

Let xx and yy be elements of F,F, and suppose xx is the equivalence class of the cut (A,B()(A,B() and yy is the equivalence class of the cut (C,D).(C,D). Then xyxy if and only if (A,B)(C,D).(A,B)(C,D).

Proof

We have that xyxy 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 dDdD 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 dDdD and aA.aA. Now these numbers hh are all greater than or equal to 0 if and only if each aAaA is less than or equal to each dD,dD, i.e., if and only if (A,B)(C,D).(A,B)(C,D). This proves the theorem.

We are now ready to present the first of the two main theorems of this appendix, that is (Reference) in (Reference).

Theorem 5

There exists a complete ordered field. Indeed, the Dedekind field FF is a complete ordered field.

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 xMxM for all xS.xS. 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 xS,xS, let (Ax,Bx)(Ax,Bx) be a Dedekind cut for which xx is the equivalence class of (Ax,Bx),(Ax,Bx), and note that axbaxb for all axAxaxAx and all bB.bB. Let A0A0 be the union of all the sets AxAx for xS.xS. Let B0B0 be the set of all rational numbers rr for which ra0ra0 for every a0A0.a0A0. 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 a0A0a0A0 for which r<a0.r<a0. But a0xSAx,a0xSAx, so that there must exist an xSxS such that a0Ax,a0Ax, and hence rr is also in Ax.Ax. But then rA0,rA0, 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 AxA0,AxA0, we see that axb0axb0 for every axAxaxAx and every b0B0.b0B0. Hence, (Ax,Bx)(A0,B0)(Ax,Bx)(A0,B0) for every xS,xS, and therefore, by Theorem A.4, xM0xM0 for all xS.xS. 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 axb'axb' for every axAxaxAx and every b'B',b'B', implying that a0b'a0b' for every a0A0a0A0 and every b'B'.b'B'. Therefore, (A0,B0)(A',B'),(A0,B0)(A',B'), implying that M0M'.M0M'. This shows that M0M0 is the least upper bound for S,S, and the theorem is proved.

We come now to the second major theorem of this appendix, i.e., (Reference) of (Reference). This one asserts the uniqueness, up to isomorphism, of complete ordered fields.

Theorem 6

Let F^F^ be a complete ordered field. Then there exists an isomorphism of F^F^ onto the Dedekind field F.F. That is, there exists a one-to-one function J:F^FJ:F^F that is onto all of F,F, and that satisfies

  1.  J(x+y)=J(x)+J(y).J(x+y)=J(x)+J(y).
  2.  J(xy)=J(x)J(y).J(xy)=J(x)J(y).
  3. If x>0,x>0, then J(x)>0.J(x)>0.

Proof

We know from (Reference) that, inside any ordered field, there is a subset that is isomorphic to the field QQ of rational numbers. We will therefore identify this special subset of F^F^ with Q.Q.

If xx is an element of F^,F^, let Ax={rQ:rx}Ax={rQ:rx} and let Bx={rQ:r>x}.Bx={rQ:r>x}. We claim first that the pair (Ax,Bx)(Ax,Bx) is a Dedekind cut. Indeed, from the definition of AxAx and Bx,Bx, we see that Condition (2), i.e., that each axAxaxAx is less than or equal to each bxBx,bxBx, holds. To see that Condition (1) also holds, let rr be a rational number in F^.F^. Then, because F^F^ is an ordered field, either rxrx or r>x,r>x, i.e., rAxrAx or rBx.rBx. Hence, (Ax,Bx)(Ax,Bx) is a Dedekind cut.

We define a function JJ from F^F^ into FF by setting J(x)J(x) equal to the equivalence class determined by the cut (Ax,Bx).(Ax,Bx). We must check several things.

First of all, JJ is one-to-one. Indeed, let xx and yy be elements of F^F^ that are not equal. Assume without loss of generality that x<y.x<y. Then, according to (Reference), which is a theorem about complete ordered fields and hence applicable to F^,F^,, there exist two rational numbers r1r1 and r2r2 such that x<r1<r2<y,x<r1<r2<y, which implies that r1Bxr1Bx and r2Ay.r2Ay. Since r2>r1,r2>r1, the cut (Ay,By)(Ay,By) is not equivalent to the cut (Ax,Bx),(Ax,Bx), and therefore J(x)J(y).J(x)J(y).

Next, we claim that the function JJ is onto all of the Dedekind field F.F. Indeed, let zz be an element of F,F, and let (A,B)(A,B) be a Dedekind cut for which zz is the equivalence class determined by (A,B).(A,B). Think of AA as a subset of the complete ordered field F^.F^. Then AA is nonempty and is bounded above. In fact, every element of BB is an upper bound of A.A. Let x=supA.x=supA. (Here is another place where we are using the completeness of the field F^.F^.) We claim that the cut (A,B)(A,B) is equivalent to the cut (Ax,Bx),(Ax,Bx), which will imply that J(x)=z.J(x)=z. Thus, if axAx,axAx, then axx,axx, and xbxb for every bB,bB, because xx is the least upper bound of A.A. Similarly, if aA,aA, then ax,ax, and x<bxx<bx for every bxBx.bxBx. This proves that the cuts (A,B)(A,B) and (Ax,Bx)(Ax,Bx) are equivalent, as desired.

If xx and yy are elements of F^,F^, and bxBxbxBx and byBy,byBy, then bx>xbx>x and by>y,by>y, so that bx+by>x+y,bx+by>x+y, and therefore bx+byBx+ybx+byBx+y for every bxBxbxBx and byBy.byBy. On the other hand, if rBx+y,rBx+y, then r>x+y.r>x+y. Therefore, r-x>y,r-x>y, implying, again by (Reference), that there exists an element byBybyBy such that y<by<r-x.y<by<r-x. But then r-by>x,r-by>x, which means that r-by=bxr-by=bx for some bxBx.bxBx. So, r=bx+by,r=bx+by, and this shows that Bx+y=bx+By.Bx+y=bx+By. It follows from this that the cuts (Ax+y,Bx+y)(Ax+y,Bx+y) and (Ax,Bx)+(Ay,By)(Ax,Bx)+(Ay,By) are equal, and therefore J(x+y)=J(x)+J(y).J(x+y)=J(x)+J(y). A consequence of this is that J(-x)=-J(x)J(-x)=-J(x) for all xF^.xF^.

If xx and yy are two positive elements of F^,F^, then an argument just like the one in the preceding paragraph shows that J(xy)=J(x)J(y).J(xy)=J(x)J(y). Then, since J(-x)=-J(x),J(-x)=-J(x), the fact that J(xy)=J(x)J(y)J(xy)=J(x)J(y) for all x,yF^x,yF^ follows.

Finally, if xx is a positive element of F^,F^, then the set AxAx must contain some positive rationals, and hence the cut (Ax,Bx)(Ax,Bx) is a positive cut, implying that J(x)>0.J(x)>0.

We have verified all the requirements for an isomorphism between the two fields F^F^ and F,F, and the theorem is proved.

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