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4) after developing the factson the implicit

function theorem needed for the proof. In the twelfth, and last, chapter we discuss vector functions inE N .We prove the theoremsof Green and Stokesand the divergence theorem, not in full generality but of sufficient scope for most applications. The ambitiousreader can get a moreinsight either by referring to the book or the text by Walter Rudin.

Murray H. Protter

Berkeley, CA

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AV nδ xContents Chapter 4Elementary Theory of DifferentiationSSSSSSSSSSSSS67 zmS θtb PbN>δx/>δb >.R S ηi zmr E.δbNIb UD.u/>A.I >.R S ii Chapter 5 Elementary Theory of IntegrationSSSSSSSSSSSSSSSSS81 emS θtb PxNβADR E./baNx) sAN UD.u/>A.IA.R S vS emr θtb α>bkx.. E./baNx) πS emG θtb lAaxN>/tk x.d ,RnA.b./>x) UD.u/>A.Iπη Chapter 6 Elementary Theory of Metric SpacesSSSSSSSSSSSSSS101 ηmS θtb futHxNX x.d θN>x.a)b E.bτDx)>/>bI? qb/N>u fnxubISpS

ηmr θAnA)AaT >. qb/N>u fnxubI

Spη

ηmG LAD./xβ)b x.d ε.uAD./xβ)b fb/I

SSz ηmz LAknxu/ fb/Ix.d /tb .bVgANb) θtbANbkSSv

ηme UD.u/>A.IA. LAknxu/ fb/I

Srr

Chapter 7 Differentiation and Integration inR

p

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imz θtb PxNβADR E./baNx) >.R SzS ime θtb α>bkx.. E./baNx) >.R Sze Chapter 8 Infinite SeriesSSSSSSSSSSSSSSSSSScSSSSSSSSSSSSSSSSSScSSS150 vmS θbI/I sAN LA.δbNab.ub x.d P>δbNab.ub Sep vmr fbN>bI As BAI>/>δb x.d Cbax/>δb θbNkI? BAHbN fbN>bISee vmG ε.>sANk LA.δbNab.ub

Sηr

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Chapter 9 The Derivative of an Integral.

Improper IntegralsSSSSSSSSSSSSSSSSSScSSSSSSSSSSSSSSSS178 πmS θtb PbN>δx/>δb As x UD.u/>A. Pb?.bd βT x. E./baNx)m

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Chapter 11 The Implicit Function Theorem.

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Contentsxi

Chapter 12 Vector Functions onS

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21. The Real Number System

Axioms of Addition

A-1. Closure property

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A-2. Commutative law

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A-3. Associative law

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A-4. Existence of a zero

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A-5. Existence of a negative

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Theorem 1.1

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axb 46 0 8 C > N"

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Ja(a)Kxxx.

1.1. Axioms for a Field3

We conclude thatxb(a), and the uniqueness of the solution is established. %#The numberb(a)isdenoted byba. The next theorem establishes familiar properties of negative numbers.

Theorem 1.2

(i)"a ,(a)a# (ii)"aNb , (ab))(a)(b). Proof (i) From the definition of negative, we have (a)[(a)]0,(a)aa(a)0. Axiom A-5 states that the negative of(a)is!#Therefore,a (a). To establish (ii), we know from the definition of negative that (ab)[(ab)]0.

Furthermore, using the axioms, we have

(ab)[(a)(b)][a(a)][b(b)]000. The result follows from the "only one" part of Axiom A-5.

Axioms of Multiplication

M-1. Closure property

"aNb , N ,NN ab( abaRb ),N urasdTn#

M-2. Commutative law

$ aNb baab N#

M-3. Associative law

a ,b,Nc, ! (ab)ca(bc) N#

41. The Real Number System

M-4. Existence of a unit

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M-5. Existence of a reciprocal

lng teE: aoyutgaprHHtgtai Hgny ,tgn i:tgt rP nat eap na-0 nat aoyutgx

PoE: i:eiaxC d:rP aoyutgxrP Ee--tp i:t

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Special Axiom onMurayuH.auPuao

D. Distributive law

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Theorem 1.3

vHarP ea0 aoyutg &i:taaRC

1.1. Axioms for a Field5

Proof Letbbe any number. Thenb0b, and thereforea(b0)ab. From the distributive law (Axiom D), we find that (ab)(aR0)(ab), so thataR00 by Axiom A-4.

Theorem 1.4

"aNb Na0 , N x aRxb# x xba 1 The proof of Theorem 1.4 isjust like the proof of Theorem 1.1 with addition replaced by multiplication, 0 by 1, andabya 1 . The details are left to the reader. %#The expression "if and only if," a technical one used fre- quently in mathematics, requires some explanation. SupposeAandB stand for propositions that may be true or false. To say thatAistrueB istrue meansthat the truth ofBimpliesthe truth ofA. The statementAis true Bistrue meansthat the truth ofAimpliesthe truth ofB. Thus the shorthand statement "Aistrue if and only ifBistrue" isequivalent to theN that the truth ofAimpliesand isimplied by the truth ofB. As a further shorthand we use the symbolto represent "if and only if," and we write AB for the two implicationsstated above. The term N (is used as a synonym for "if and only if." We now establish the familiar principle that is the basis for the solution of quadratic and other algebraic equationsby factoring.

Theorem 1.5

(i) ab0 N a0b0 # (ii) a0Nb0 N ab0# Proof We must prove two statements in each of parts (i) and (ii). To prove (i), observe that ifa0orb0 or both, then it followsfrom Theorem 1.3 thatab0. Going the other way in (i), suppose thatab0. Then there are two cases: eithera0ora0. Ifa0, the result follows. Ifa0, then we see thatquotesdbs_dbs12.pdfusesText_18