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Intro Abstract Math

by: Otilia Murray I

Intro Abstract Math MAT 108

Otilia Murray I
GPA 3.88


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This 48 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 108 at University of California - Davis taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/187460/mat-108-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15
Chapter 1 The Field of Reals and Beyond Our goal with this section is to develop review the basic structure that character izes the set of real numbers Much of the material in the rst section is a review of properties that were studied in MAT108 however there are a few slight dilTer ences in the de nitions for some of the terms Rather than prove that we can get from the presentation given by the author of our MAT127A textbook to the previous set of properties with one exception we will base our discussion and derivations on the new set As a general rule the de nitions olTered in this set of Compan ion Notes will be stated in symbolic form this is done to reinforce the language of mathematics and to give the statements in a form that clari es how one might prove satisfaction or lack of satisfaction of the properties YOUR GLOSSARIES ALWAYS SHOULD CONTAIN THE IN SYMBOLIC FORM DEFINITION AS GIVEN IN OUR NOTES because that is the form that will be required for suc cessful completion of literacy quizzes and exams where such statements may be requested 11 Fields Recall the following DEFINITIONS 0 The Cartesian product of two sets A and B denoted by A x B is abaeAbeB CHAPTER 1 THE FIELD OF REALS AND BEYOND o A function h from A into B is a subset of A x B such that i Va a e A 2 Elb b e B A a b e h ie domh A and ii Va Vb V0 61 b e h A a c e h gt b c ie his singlevalued o A binary operation on a set A is a function from A x A into A o A eld is an algebraic structure denoted by IF e f that includes a set of objects IF and two binary operations addition and multiplication that satisfy the Axioms of Addition Axioms of Multiplication and the Distributive Law as described in the following list A Axioms of Addition IF e is a commutative group under the binary operation of addition with the additive identity denoted by 6 Al IF gtlt IF gt IF A2 Vx Vy xy 6 IF 2 x y y x commutative with respect to addition A3 Vx Vy Vz xyz 6 IF 2 x y z x y z asso ciative with respect to addition A4 Ele e 6 IF A Vx x 6 IF gtx e e x x additive identity property A5 W x 6 IF gt 3 x x 6 IF A x x x 96 8H additive inverse property M Axioms of Multiplication IF f is a commutative group under the binary operation of multiplication with the multiplicative identity denoted by f Ml IF gtlt IF gt IF M2 Vx Vy xy 6 IF 2 x y y x commutative with respect to multiplication M3 Vx Vy Vz xyz 6 IF 2 x y z x y z associative with respect to multiplication M4 ElffeIFAf7 eVxx eIFgtxffxx mul tiplicative identity property M5 Vx x 6 IF e gt El x l x71 6 IF x x71 2 x71 x multiplicative inverse property 11 FIELDS 3 D The Distributive Law Vx Vy V2 Xayaz 6 IF gt x 02 x y x Z Remark 111 Properties A and M1 tell us that IF is closed under addition and closed under multiplication respectively Remark 112 The additive identity and multiplicative identity properties tell us that a eld has at least two elements namely two distinct identities To see that two elements is enough note that forIF 0 1 the algebraic structure IF 69 X 0 1 where 69 IF gtlt IF gt IF and X IF gtlt IF gt IF are de ned by thefollowing tables G9 0 1 0 1 0 0 1 0 0 0 1 1 0 1 0 1 is a eld Remark 113 The elds with which you are probably the most comfortable are the rationals Q 0 1 and the reals IR 0 1 A eld that we will discuss shortly is the complex numbers C 0 0 1 0 Since each ofthese distinctly different sets satis v the same list of eld properties we will expand our list of properties in search of ones that will give us distinguishing features When discussing elds we should distinguish that which can be claimed as a basic eld property AM and D from properties that can and must be proved from the basic eld properties For example given that IF is a eld we can claim that Vx Vy xy 6 IF 2 x y 6 IF as an alternative description of property A1 while we can not claim that additive inverses are unique The latter observation is important because it explains why we can t claim e w from IF e f being a eld and x w x e x we don t have anything that allows us to subtract from both sides of an equation The relatively small number of properties that are olfered in the de nition of a eld motivates our search for additional properties of elds that can be proved using only the basic eld properties and elementary logic In general we don t claim as axioms that which can be proved from the minimal set of axioms that comprise the de nition of a eld We will list some properties that require proof and olfer some proofs to illustrate an approach to doing such proofs A slightly dilferent listing of properties with proofs of the properties is offered in Rudin 4 CHAPTER 1 THE FIELD OF REALS AND BEYOND Proposition 114 Properties for the Additive Identity of a eld IF e f 1 Vxx eIFAxxxgtxe 2 VxerFgtxeexe 3 VxVyxyeIFAxyegt xeVye Proof of 1 Suppose that x 6 IF satis es x x x Since x 6 IF by the additive inverse property x 6 IF is such that x x x x e Now by substitution and the associativity of addition ex xxx xxx xxex of 3 Suppose that x y 6 IF are such that x y e and x 75 e Then by the multiplicative inverse property x71 e F satis es x x 1 x71 x f Then substitution the associativity of multiplication and 2 yields that yfy x 1 x yx 1xyx 1 ee Hence for xy 6 IF x y e A x e implies thaty e The claim now follows immediately upon noting that for any propositions P Q and M P 2 Q V M is logically equivalent to P A 39Q 2 M l Excursion 115 Use 1 to prove 2 The key here was to work from x e x e e Proposition 116 Uniqueness ofIdentities and Inverses for a eld IF e f 1 The additive identity ofa eld is unique 11 FIELDS 5 2 The multiplicative identity of a eld is unique 3 The additive inverse of any element in IF is unique 4 The multiplicative inverse ofany element in IF e is unique Proof of 1 Suppose that w 6 IF is such that Vxx eIFgtxwwxx In particular since e 6 IF we have that e e 11 Since e is given as an additive identity and w 6 IF e w 11 From the transitivity of equals we conclude that e 11 Therefore the additive identity of a eld is unique of 3 Suppose that a 6 IF is such that there exists w 6 IF and x 6 IF satisfying awwae anal axxae From the additive identity and associative properties wwe wax wax ex x Since a was arbitrary we conclude that the additive inverse of each element in a eld is unique I Excursion 117 Prove 4 Completing this excursion required only appropriate modi cation of the proof that was olTered for 3 You needed to remember to take you arbitrary element in F to not be the additive identity and then simply change the operation to multiplica tion Hopefully you remembered to start with one of the inverses of your arbitrary element and work to get it equal to the other one 6 CHAPTER 1 THE FIELD OF REALS AND BEYOND Proposition 118 Sums andProducts Involving Inversesfor a eld IF e f 1 Va Vb a b 6 IF gt a b a b 2 Va a e F gt a a 3 Va Vb a b 6 IF gt a b a b 4 Va Vb a b 6 IF gt a b a b 5 Va Vb a b 6 IF gt a b a b 6 Va a 6 IF e gt a 1 as e a1 1 a a 1 a 1 7 Va Vb a b 6 IF e gt a b 1 a l b1 Proof of2 Suppose that a 6 IF By the additive inverse property a 6 IF and a 6 IF is the additive inverse of a ie a a e Since a is the additive inverse ofa a a a a e which alsojusti es that a is an additive inverse of a From the uniqueness of additive inverses Proposition 116 we conclude that a a l Excursion 119 Fill in what is missing in order to complete the following proof of 6 Proof Suppose that a 6 IF e From the multiplicative inverse property a 1 6 IF satis es Ifa 1 e then by Proposition 1 1142 a 1 a e Since multiplication is single valued this would imply that which contradicts part of the prop 2 3 erty Thus a 1 75 e Since a 1 e IF e by the property ailr1 e 4 IF and satis es a lr1 a 1 a 1 a lr1 f but this equation alsojusti es that a 171 is a multiplicative inversefor a l From Proposition 5 71 we conclude that a l a 11 FIELDS 7 From 5 a 1 a a 1 a ffrom which we conclude that a l is a for a Since a 1 is a mul 6 tiplicative inverse for a and multiplicative inverses are unique we have that a l a 1 as claimed l Acceptable responses are 1aa 1 f 2 e f 3 multiplicative identity 4 multiplicative inverse 5 1164 and 6 multiplicative inverse Proposition 1110 Solutions to Linear Equations Given a eld F 0 l 1 VaVb ab 6 F2 Elbe x e FAa x 2 b 2 VaVbab elFAa 750 Elxx elFAax b Proof of1 Suppose that a b 6 IF and a 75 0 Since a 6 IF 0 there exists a 1 6 IF such that a a 1 a 1 a 1 Because a 1 6 IF and b 6 F x dzf a 1 b e IFfrom lH Substitution and the associativity ofmultiplication e yieldthat axaailbaa 1blbb Hence x satis es a x b Now suppose that w 6 IF also satis es a w b Then 1112111ailawa 1awa 1bx Since a and b were arbitrary Va Vb ab 6 F2 Elbe x e FAa x 2 b l Remark 1111 As a consequence of Proposition 1110 we now can claim that if xwz e Fandx w x z then 11 z and ifwz 6 F x e IF O and w x z x then 11 z The justi cation is the uniqueness of solutions to linear equations in a eld In terms of your previous experience with elementary algebraic manipulations used to solve equations the proposition justi es what is commonly referred to as adding a real number to both sides of an equation and dividing both sides of an equation by a nonzero real number 8 CHAPTER 1 THE FIELD OF REALS AND BEYOND Proposition 1112 Addition and Multiplication Over Fields Containing Three or More Elements Suppose that IF is a eld and a b c d 6 IF Then 1 abcacbcba 2 abcacbcba 3 acbdabcd 4 acbdabcd Proposition 1113 Multiplicative Inverses in a eld IF 0 1 1 VaVbVcVdabcdeIFAb7E0d 0 gtbdi Oab 1cd 1acbd 1 2 Va Vb Vc a b c 6 IF A c as 0 gt a c l b 1 a b c l 3 Va Vb a b 6 IF A b as 0 gt a 13 a b 1 a 134 4 Va Vb Vc Va a 6 IF A b cd 6 IF 0 gt c ail as 0 A a b l c ai l 1 a a b c1 a b l a 1 5 Va Vb Vc Va a c 6 IF A bd 6 IF 0 gt b a as 0A ab 1cd 1adbcbd 1 Proof of3 Suppose a b 6 IF and b 75 0 Since b 75 0 the zero ofthe eld is its own additive inverse and additive inverses are unique we have that b 75 0 Since a 6 IF and b 6 IF 0 implies that a 6 IF and b 1 6 IF 0 by Proposition 1184 a b 1 a b l From Proposition 1186 we know that b l b 1 From the distributive law andProposition 1182 a b 1ab 1a b 1b 1a b 1b 1a0o from which we conclude that a b 1 is an additive inverse for a b l Since additive inverses are unique itfollows that a b 1 a b l Combining our results yields that a b 1 a 13 a b 1 as claimed l 11 FIELDS 9 Excursion 1114 Fill in what is missing in order to complete the following proof of4 Proof of4 Suppose that a 6 IF and b c d 6 IF 0 Since d 6 IF 0 by Proposition d 1 75 0 From the contrapositive ofProposition 1143 1 c 75 0 and d71 75 0 implies that In the following the justi cations 2 for the step taken is provided on the line segment to the right of the change that has been made a W ltcd1 1 ltab1c1ltd1 1 lt3 lta b1ltc1d lt4 altb1ltcld lt5 2 ab 1c 1d lt6 altdltb1 c1 lt7 adbc 1 lt8 adbc 1 lt9 From Proposition 1187 combined with the associative and commutative prop erties of addition we also have that adbc 1 adb 1c 1 adb 1c 1 adb 1c 1 lt 1 0 1 a b d c Eab 1dc 1 Consequently ab 1cd 171adbc 1ab 1dc 1as claimed l Acceptab1e responses are 1 1186 2 cd 1 75 0 3 Proposition 1187 4 Proposition 1186 5 associatiVity of multiplication 6 associatiVity of 10 CHAPTER 1 THE FIELD OF REALS AND BEYOND multiplication 7 commutativity of multiplication 8 Proposition ll87 9 associativity of multiplication 10 a b l d c The list of properties given in the propositions is by no means exhaustive The propositions illustrate the kinds of things that can be concluded proved from the core set of basic eld axioms Notation 1115 We have listed the properties without making use of some nota tional conventions that can make things look simpler The two that you might nd particularly helpful are that o the expression a b may be written as a b a b may be written as a b and a o the express1on a b 1 may be written as Note that applying this nota tional convention to the Properties of Multiplicative Inverses stated in the last proposition can make it easier for you to remember those properties Excursion 1116 On the line segments provided ll in appropriate justi cations for the steps given in the following outline of a proof that for a b c d in a eld ab c da cbd l Observation l l Justi cation l a b c d a b lt c d 103533quot a b lt ltc lt d a b ltlt c 61 a b ltlt c 61 a b cgt d I a b ltlt c d a b ltlt c d 2 a b ltlt c d a b c d 3 a b lt c d a ltltlt c b d 4 a c b d a c b d 5 a c b d a c b d a cbda cbd lt8 12 ORDERED FIELDS l l Acceptable responses are 1 Proposition ll8l 2 Proposition ll82 3 and 4 associativity of addition 5 commutativity of addition 6 and 7 as sociativity of addition and 8 notational convention 12 Ordered Fields Our basic eld properties and their consequences tell us how the binary operations function and interact The set of basic eld properties doesn t give us any means of comparison of elements more structure is needed in order to formalize ideas such as positive elements in a eld or listing elements in a eld in increasing order To do this we will introduce the concept of an ordered eld Recall that for any set S a relation on S is any subset of S x S De nition 121 An order denoted by lt M a set S is a relation on S that satis es the following two properties 1 The T richotomy Law If x e S and y e S then one and only one of x lty or xy oryltx is true 2 The TransitiveLaw Vx Vy V2 xyz e S A x lt y Ay lt z gt x lt 2 Remark 122 Satisfaction of the Trichotomy Law requires that VxVyxy 6 Sgt x yVx ltyVy ltx be true and that each of VxVyxy E Sgt 96 y gt cOC lty AcO lt x VxVyxy E Sgt 96 lty gt cOC y AcO lt x and VxVyxy E Sgt 0 ltxgt cOC ycx lty be true The rst statement Vx Vy xy e S gt x y V x lt y V y lt x is not equivalent to the Trichotomy Law because the disjunction is not mutually exclusive 12 CHAPTER 1 THE FIELD OF REALS AND BEYOND Example 123 For S a b c with a b andc distinct lt a b b c a c is an order on S The notational convention for a b e lt is a lt b The given or dering has the minimum and maximum number of ordered pairs that is needed to meet the de nition This is because given any two distinct elements of S x and y we must have one and only one of xy e lt or y x e lt After making free choices of two ordered pairs to go into an acceptable ordering for S the choice of the third ordered pair for inclusion will be determined by the need to have the Transitive Law satis ed Remark 124 The de nition of a particular order on a set S is to a point up to the de ner You can choose elements of S x S almost by preference until you start having enough elements to force the choice of additional ordered pairs in order to meet the required properties In practice orders are de ned by some kind of formula or equation Example 125 For Q the set ofrationals let ltC Q X Q be de ned by r s e ltltgt s r is apositive rational Then Q lt is an ordered set Remark 126 The treatment of ordered sets that you saw in MAT108 derived the Trichotomy Law from a set of properties that de ned a linear order on a set Given an order lt on a set we write x 5 yfor x lt y V x y With this notation the two linear ordering properties that could have been introduced and used to prove the Trichotomy Law are the Antisymmetric law VxVyxy E SAOCJ E S A0596 6 5 gt 6 y and the Comparability Law Vx VJWhy E Sgt my 6 5 V0 x E 5 Now because we have made satisfaction of the Trichotomy Law part of the def inition of an order on a set we can claim that the Antisymmetric Law and the Comparability Law are satis ed for an ordered set De nition 127 An ordered eld IF 0 1 lt is an ordered set that satis es the following two properties OF1VxVyVzxyz elFAx ltygtxz ltyz OFZ VxVyVzxyzeIFx lty0 ltzgtxz ltyz 12 ORDERED FIELDS 13 Remark 128 In the de nition of ordered eld o ered here we have deviated from one of the statements that is given in our text The second condition given in the text is that VxVyxyeIFAxgt0Aygt0gtxygt0 let s denote this proposition by altOF2 We will show that satisfaction ofOFl and alt OF2 is infact equivalent to satisfaction ofOFl and OF2 Suppose that OFl and OF2 are satis ed and letxy 6 IF be such that0 lt x and0 lt y From OF2 andProposition 1142 0 0 y lt x y Since x andy were arbi trary we conclude that Vx Vy xy 6 IF A x gt 0 Ay gt 0 gt x y gt 0 Hence OF2 2 altOF2 from which we have that OFl A OF2 2 OFl A altOF2 Suppose that OFl and altOF2 are satis ed and let xy z 6 IF be such that x lt y and 0 lt z From the additive inverse property x 6 IF is such that x x x x 0 From OFl we have that 0x xlty x From altOF2 the Distributive Law and Proposition 118 4 0 lt y x and0 lt 2 implies that o ltylt x zyzgtltlt xgtzgtltyzgtlt ltxz Because and are binary operations on IF x z 6 IF and y z x 2 6 IF It now follows from OF 1 and the associative property of addition that 0xz ltyz xzxzyz xzxzyz0 Hence x z lt y 2 Since x y and 2 were arbitrary we have shown that VxVyVzxyzeIFAx ltyA0 ltzgtxz ltyz which is OF2 Therefore OFl A altOF2 2 OFl A OF2 Combining the implications yields that OF1AOF2 ltgt OF1AaltOF2 as claimed To get from the requirements for a eld to the requirements for an ordered eld we added a binary relation a description of how the elements of the eld are or dered or comparable and four properties that describe how the order and the binary operations interac The following proposition offers a short list of other order properties that follow from the basic set 14 CHAPTER 1 THE FIELD OF REALS AND BEYOND Proposition 129 Comparison Properties Over Ordered Fields For an ordered eld IF 0 l lt we have each of the following 1 0ltl 2 VxVyxyeIFxgt0Aygt0gtxygt0 3 VxerFAxgt0gt xlt0 4 VxVyxyeIFx ltygt ylt x 5 VxVyVzxyzeIFxltyAzlt0gtxzgtyz 6 VxerFAx7E0gtxxx2gt0 7 VxVyxyeIF0ltxltygt0lty 1ltx 1 In the Remark 128 we proved the second claim We will prove two others Proofs for all but two of the statements are given in our text Proof or 1 By the Trichotomy Law one and only one of 0 lt 1 0 1 or 1 lt 0 is true in the eld From the multiplicative identity property 0 75 1 thus we have one and only one ofO lt l or 1 lt 0 Suppose that l lt 0 From OFl we have that 0 2 1 1 lt 0 1 1ie 0 lt 1 Hence OF2 implies that l 1 lt 0 1 which by Proposition 1183 is equivalent to 1 lt 0 But from the transitivity property 0 lt 1 A 1 lt 0 gt 0 lt 0 which is a contradiction l Excursion 1210 Fill in what is missing in order to complete the following proof ofProposition 12 94 Proof Suppose that x y 6 IF are such that x lt y In view of the additive inverse property x 6 IF and y 6 IF satis xxx x0 and 1 From 0 x x lt y x ie 2 3 and 0 y lt y Repeated use ofcommutativity and as 4 sociativity allows us to conclude that y x y x Hence y lt x as claimed l 12 ORDERED FIELDS 15 Acceptab1e responses are 1 yy y y 0 2 OF1 3 0 lt y x 4y x Remark 1211 From Proposition 12 91 we see that the two additional prop erties needed to get from an ordered set to an ordered eld led to the requirement that 0 1 be an element ofthe ordering binary relation From 0 lt 1 and OFl we also have that 1 lt 1 1 2 2 lt 2 1 3 etc Using the convention 1 1 1 1 2 n the general statement becomes 0 lt n lt n 1 n of them 121 Special Subsets of an Ordered Field There are three special subsets of any ordered eld that are isolated for special con sideration We oiTer their formal de nitions here for completeness and perspective De nition 1212 Let 1 0 1 5 be an ordered eld A subset S ofIF is said to be inductive ifand only if 116Sand 2 VxxeSgtxleS De nition 1213 For 1F 0 1 5 an ordered eld de ne N1 m S SEG where G S Q 1F S is inductive We will call NF the set ofnatural numbers of IF Note that T x 6 IF x 2 1 is inductive because 1 e T and closure ofIF un der addition yields that x 1 6 IF whenever x 6 IF Because Vu u lt 1 gt u T and T e G we immediately have that any n 6 NF satis es n 2 1 De nition 1214 Let 1F 0 1 5 be an ordered eld The set ofintegers of IF denoted ZF is ZFaelFaeNFV aeNFVa0 16 CHAPTER 1 THE FIELD OF REALS AND BEYOND It can be proved that both the natural numbers of a eld and the integers of a eld are closed under addition and multiplication That is VmVnneNFmENF2gtnmeNFAnmeNF and VmVnneZFmEZF2gtnmeZFAnmeZF This claim requires proof because the fact that addition and multiplication are bi nary operations on IF only places 71 m and n m in IF because NF C IF and ZF C IF Proofs of the closure of NF 2 N under addition and multiplication that you might have seen in MAT108 made use of the Principle of Mathematical Induction This is a useful tool for proving statements involving the natural numbers PRINCIPLE OF MATHEMATICAL INDUCTION PMI If S is an inductive set of natural numbers then S N In MAT108 you should have had lots of practice using the Principle of Mathe matical Induction to prove statements involving the natural numbers Recall that to do this you start the proof by de ning a set S to be the set of natural numbers for which a given statement is true Once we show that l e S and Vk k e S gt k l e S we observe that S is an inductive set of natural num bers Then we conclude by the Principle of Mathematical Induction that S N which yields that the given statement is true for all N Two other principles that are logically equivalent to the Principle of Mathemat ical Induction and still useful for some of the results that we will be proving in this course are the WellOrdering Principle and the Principle of Complete Induction WELL ORDERING PRINCIPLE W OP Any nonempty set S of natural num bers contains a smallest element PRINCIPLE OF COMPLETE INDUCTION PCI Suppose S is a nonempty set of natural numbers If VmmeNk NkltmCSgtmES then S N 12 ORDERED FIELDS 17 De nition 1215 Let F 0 1 5 be an ordered eld De ne QFr 61F EImEn mn e ZFAn 7E OAr mn71 The set Q11 is called the set of rational numbers of IF Properties 1 and 5 from Proposition 1113 can be used to show the set of rationals of a eld is also closed under both addition and multiplication The set of real numbers R is the ordered eld with which you are most familiar Theorem 119 in our text asserts that R is an ordered eld the proof is given in an appendix to the rst chapter The notation and numerals for the corresponding special subsets of R are N J 1 2 34 5 the set ofnatural numbers Z m m e N V m 0 V m e N 3 2 10 123 Qpq 1 pq Zq0 Remark 1216 The set of natural numbers may also be referred to as the set of positive integers while the set of nonnegative integers is J U 0 Another common term for JU0 is the set of whole numbers which may be denoted by W InMAT 08 the letter N was used to denote the set of natural numbers while the author of our MAT127 text is using the letter J To make it clearer that we are referring to special sets of numbers we will use the blackboard bold form of the capital letter Feel free to use either the old N or the new J for the natural numbers in the eld of reals While N and Z are not elds both Q and R are ordered elds that have several distinguishing characteristics we will be discussing shortly Since Q C R and R Q 75 13 it is natural to want a notation for the set of elements of R that are not rational Towards that end we let llrr dzf 1R Q denote the set of irrationals It e was shown in MAT108 that 2 is irrational Because 2 0 llrr and J2 2 2 llrr we see that llrr is not closed under either addition or multiplication 122 Bounding Properties Because both Q and R are ordered elds we note that satisfaction of the set of ordered eld axioms is not enough to characterize the set of reals This natu rally prompts us to look for other properties that will distinguish the two algebraic 18 CHAPTER 1 THE FIELD OF REALS AND BEYOND systems The distinction that we will illustrate in this section is that the set of ra tionals has certain gaps During this motivational part of the discussion you might nd it intuitively helpful to visualize the old numberline representation for the reals Given two rationals r and s such that r lt s it can be shown that m r s2 1 e Qis suchthatr lt m lt s Thenrl r m2 1 e Qand s1 m s 2 1 e Q are such thatr lt r1 lt m and m lt s1 lt s Continuingthis process inde nitely and marking the new rationals on an imagined numberline might entice us into thinking that we can ll in most of the points on the number line between r and s A rigorous study of the situation will lead us to conclude that the thought is shockingly inaccurate We certainly know that not all the reals can be found this way because for example 5 could never be written in the form of r s 2 1 for r s e Q The following excursion will motivate the property that we want to isolate in our formal discussion of bounded sets Excursion 1217 LetA p e Q p gt 0 p2 lt 2 anal B p e Q p gt 0 A p2 gt 2 Now we will expand a bit on the approach used in our text to show that A has no largest element anal B has not smallest element For p a positive rational let p2 22p2 61 1 p2 p2 Then 2 q2 2 2p 2 11722 a For p e Ajusti that q gt p analq e A b Forp e B justify thatq lt p analq e B 12 ORDERED FIELDS 19 Hopefully you took a few moments to nd some elements of A and B in order to get a feel for the nature of the two sets Finding a q that corresponds to a p e A and a p e B would pretty much tell you why the claims are true For a you should have noted that q gt p because p2 2 p 2 1 lt 0 whenever p2 lt 2 then p2 2 p 2 1 gt 0impliesthatq p p2 2 p 2Y1 gt p0 p That q is rational follows from the fact that the rationals are closed under multi plication and addition Finally q2 2 2 p2 2 p 2 2 lt 0 yields that q e A as claimed For b the same reasons extend to the discussion needed here the only change is that for p e B p2 gt 2 implies that p2 2 p 2 1 gt 0 from which it follows that p2 2 p 2 1 lt 0 and q p p2 2 p 2 1 lt p 0 p Now we formalize the terminology that describes the property that our example is intended to illustrate Let S 5 be an ordered set ie lt is an order on the set S A subset A of S is said to be bounded above in S if Eluu eSAVaaeAgta 5u Any element u e S satisfying this property is called an upper bound of A in S De nition 1218 Let S 5 be an ordered set For A C S u is a least upper bound or supremum of A in S if and only if 1 ueSAVaaeAgta5uand 2 VbbeSVaaeAgta5bgtu5b Notation 1219 For S 5 an ordered set and A C S the least upper bound of A is denoted by lub A or sup A Since a given set can be a subset of several ordered sets it is often the case that we are simply asked to nd the least upper bound of a given set without specifying the parent ordered set When asked to do this simply nd if it exists the u that satis es VaaeAgta5u and VbVaaeAgta5bgtu5b The next few examples illustrate how we can use basic preadvanced calculus knowledge to nd some least upper bounds of subsets of the reals 20 CHAPTER 1 THE FIELD OF REALS AND BEYOND Example 1220 Find the lub x x e R 1 x2 From Proposition 1295 we know that for x e R 1 x2 2 0 this is equivalent to 1x222x 1 1 from which we conclude that Vx x e R 2L 5 Thus is an upper 1 x2 2 2 boundfor x 39 x e R Since 1 1 x2 itfollows that 1 112 x 1 lub xeR 2 1x2 2 The way that this example was done and presented is an excellent illustration of the difference between scratch work Phase II and presentation of an argument Phase III in the mathematical process From calculus MAT21A or its equivalent has a relative minimum at x 1 and a relative we can show that f x 2 maximum at x 1 we also know that y 0 is a horizontal asymptote for the 1 graph Armed with the information that 1 is a maximum for f we know x 1 2 5 In the scratch work phase we can work backwards from this inequality to try to nd something that we can claim from what we have done thus far simple algebra gets use from that all we need to do is use inequalities to show that x 1 m 5 E to 1 2x x2 2 0 Once we see that desire to claim 1 x2 2 0 x we are home free because that property is given in one of our propositions about ordered elds Excursion 1221 Find the lub A for each of the following Since your goal is simply to nd the least upper bound you can use any pre advanced calculus infor mation that is helpful 12 ORDERED FIELDS 21 3 1quot 1 AWIVIEJ 2 A sinx cosx x e R 3 l quot 1 For 1 let xquot then xgj 22171 is a sequence that is strictly l decreasing from E to 0 while x2 171 is also decreasing from E to 0 Consequently I l the terms in A are never greater than with the value of being achieved when n l and the terms get arbitrarily close to 0 as n approaches in nity Hence 1 lub A For 2 it is helpful to recall that sinx cos x 5 sin 2x The well 1 known behavior of the sine function immediately yields that lub A Example 1222 Find lub A where A x e R x2 x lt 3 What we are lookingfor here is sup A where A f 1 00 3for f x x2 x Because 2 Q 1 12 x x x y y 2 l l f is aparabola with vertex Hence 2 4 Af1 m3xeRlmltxltlm lvl3 2 from which we conclude that sup A 22 CHAPTER 1 THE FIELD OF REALS AND BEYOND Note that the set A p e Q p gt 0 A p2 lt 2 is a subset on and a subset of R We have that Q 5 and R 5 are ordered sets where lt is de ned by r lt s ltgt s r is positive Now lub A 2 2 Q hence there is no least upper bound ofA in S Q but A C S R has a least upper bound in S R This tells us that the parent set is important gives us a distinction between Q and R as ordered elds and motivates us to name the important distinguishing property De nition 1223 An ordered set S lt has the least upper bound property if and only if VE ESAEy AGI M eSAVaaeEgta 53m 2 Elu u lub E Au 6 S Remark 1224 As noted above Q 5 does not satisfy the lub property while R 5 does satisfy this property The proof of the following lemma is left an exercise Lemma 1225 Let X 5 be an ordered set and A Q X IfA has a least upper bound in X it is unique We have analogous or companion de nitions for subsets of an ordered set that are bounded below Let S 5 be an ordered set ie lt is an order on the set S A subset A ofS is said to be bounded below in S if Eloo eSAVaa eAgto 5a Any element u e S satisfying this property is called a lower bound of A in S De nition 1226 Let S 5 be a linearly ordered set A subset A ofS is said to have a greatest lower bound or in mum in S if 1 EIggeSVaaeAgtg5aand 2 VcceSAVaaeAgtc agtc g l 2 Example 1227 Find the glb A where A 1quot Z n e N n l 2 1 Let xquot 1quot then for n odd xquot Z and for n even n l 2 n xn 4 12 ORDERED FIELDS N L 1 Suppose that n 2 4 By Proposition 1297 it follows that 5 n 1 anal 5 1 4 2 n 4 2 24 E and Proposition 1294 we have that gt n 2 2 1 2 1 1 Then OF2 anal OFl yield that 5 n 2 respectively From n N HAI HA H 1 2 Thus 2 n 1291 that n S 1 1 E Z from OFl Now itfollows from Proposition gt 0 for any n e N From Proposition 1297 anal OFl Alt 1 1 n gt 0 anal2 gt 0 implies that gt 0 and Z 2 Similarly from Proposition n n 2 2 1 2 1 1 1293 analOF1 gt 0 implies that lt 0 analZ lt Z 0 n n Combining our observations we have that VnneN 123A2ngt 5xquot lt and Vn nell 1232ngt All 5xquot 5 12 4 4 3 Comparing the values leads to the conclusion that glb A Z 7 3 5 1 1 Finally x1 x2 Z and x3 each ofwhich is outside of Excursion 1228 Final glb A for each of the following Since your goal is simply to nal the greatest lower bound you can use any pre aalvanceal calculus informa tion that is helpful 1AMne 2n1 24 CHAPTER 1 THE FIELD OF REALS AND BEYOND 1 1 2A nmeN 53m Our earlier discussion in Excursion 1221 the set given in 1 leads to the con 1 1 clusion that glb A 0 For 2 note that each of and are strictly de 2quot 3 quot creasing to 0 as n and m are increasing respectively This leads us to conclude that 5 glb A 0 although it was not requested we note that sup A We close this section with a theorem that relates least upper bounds and greatest lower bounds Theorem 1229 Suppose S lt is an ordered set with the least upper bound prop erty and that B is a nonempty subset of S that is bounded below Let LgeSVaaeBgtg5a Then a sup L exists in S anda infB Proof Suppose that S lt is an ordered set with the least upper bound property and that B is a nonempty subset of S that is bounded below Then LgeSVaaeBgtg5a is not empty Note that for each b e B we have that g 5 b for all g e L ie each element ofB is an upper bound for L Since L C S is bounded above and S satis es the least upper bound property the least upper bound of L exists and is in S Let a sup L 13 THERE4LFIELD 25 Now we want to show that a is the greatest lower bound for B De nition 1230 An ordered set S lt has the greatest lower bound property if and only if VE EcSAEy AGIy yeSAVaaeEgty5a gt ElwwglbEw e 3 39 Remark 1231 Theorem 1229 tells us that every ordered set that satis es the least upper bound property also satis es the greatest lower bound property 13 The Real Field The Appendix for Chapter 1 of our text offers a construction of the reals from the rationals In our earlier observation of special subsets of an ordered eld we olTered formal de nitions of the natural numbers of a eld the integers of a eld and the rationals of a eld Notice that the de nitions were not tied to the objects symbols that we already accept as numbers It is not the form of the objects in the ordered eld that is important it is the set of properties that must be satis ed Once we accept the existence of an ordered eld all ordered elds are alike While this identi cation of ordered elds and their corresponding special subsets can be made more formal we will not seek that formalization It is interesting that our mathematics education actually builds up to the formu lation of the real number eld Of course the presentation is more handson and intuitive At this point we accept our knowledge of sums and products involving real numbers I want to highlight parts of the building process simply to put the properties in perspective and to relate the least upper bound property to something 26 CHAPTER 1 THE FIELD OF REALS AND BEYOND tangible None of this part of the discussion is rigorous First de ne the sym bols 0 and l by dzf 0 and dzf l and suppose that we have an ordered eld e e R 0 l 5 Furthermore picture a representation of a straight horizontal line e gt on which we will place elements of this eld in a way that attaches some geometric meaning to their location The natural numbers of this eld NR is the smallest inductive subset it is closed under addition and multiplication It can be proved Some of you saw the proofs in your MAT108 course that VxxeNRgtx21 and VwUJ NRgt EIUU NRAUJltUltUJ1 This motivates our rst set of markings on the representative line Let s indicate the rst mark as a place for 1 Then the next natural number of the eld is 1 1 while the one after that is l l 1 followed by l l l 1 etc This naturally leads us to choose a xed length to represent 1 or 1 unit and place a mark for each successive natural number 1 away from and to the right of the previous natural number It doesn t take too long to see that our collections of added 1 s is not a pretty or easy to read labelling system this motivates our desire for neater representations The symbols that we have come to accept are l 2 3 4 5 6 7 8 and 9 In the space provided draw a picture that indicates what we have thus far The fact that in an ordered eld 0 lt l tells us to place 0 to the left of l on our representative line then 0 l U 13 ljusti es placing 0 l uni away from the 1 Now the de nition of the integers of a eld ZR adjoins the additive inverses of the natural numbers of a eld our current list of natural numbers leads to acceptance of l 2 3 4 5 6 7 8 and 9 as labels ofthe markings of the new special elements and their relationship to the natural numbers mandates their relative locations Use the space provided to draw a picture that indicates what 13 THE REAL FIELD 27 we have thus far Your picture should show several points with each neighboring pair having the same distance between them and lots of space with no labels or markings but we still have the third special subset of the ordered eld namely the rationals of the eld QR We are about to prove an important result concerning the density of the rationals in an ordered eld But for this intuitive discussion our grade school knowledge of fractions will suf ce Picture or use the last picture that you drew to illustrate the following process Mark the midpoint of the line segment from 0 to 1 and label it 2 1 or 5 then mark the midpoint of each of the smaller 1 1 line segments the one from 0 to E and the one from E to 1 and label the two new points 1 and 3 respectively repeat the process with the four smaller line segments to get i g g 2 3 g as the marked rationals between 0 and 1 It doesn t take too many iterations of this process to have it look like you have lled the interval Of course we know that we haven t because any rational in the from p q 1 where 0 lt p lt q and q 75 2quot for any n has been omitted What turned out to be a surprise at the time of discovery is that all the rationals r such the 0 5 r 5 1 will not be enough to ll the interval 0 1 At this point we have the set of elements of the eld that are not in any of the special subsets R QR and the set of vacancies on our model line We don t know that there is a onetoone correspondence between them That there is a correspondence follows from the what is proved in the Appendix to Chapter 1 of our text Henceforth we use 1R 0 1 lt to denote the ordered eld of reals that satis es the least upper bound property and may make free use of the fact that for any x 6 1R we have that x is either rational or the least upper bound of a set of rationals Note that the sub eld Q 0 1 lt is an ordered eld that does not satisfy the least upper bound property 131 Density Properties of the Reals In this section we prove some useful density properties for the reals 28 CHAPTER 1 THE FIELD OF REALS AND BEYOND Lemma 131 IfS Q R has L as a least upper bound L then V88eR8 gt 0gt Elss eSAL 8 lts 5L Proof Suppose S is a nonempty subset of R such that L sup S and let 8 e R be such that 8 gt 0 By Proposition 1293 and OFl 8 lt 0 and L 8 lt L From the de nition of least upper bound each upper bound of S is greater than or equal to L Hence L 8 is not an upper bound for S from which we conclude that 39 Vs s e S gt s 5 L 8 is satis ed ie Elss eSAL 8 lts Combining this with L sup S yields that Elss eSAL 8 lts 5L Since 8 was arbitrary V8 8 e RA8 gt 0 2 Els s e SAL 8 lt s 5 L as claimed l Theorem 132 The Archimedean Principle for Real Numbers If a and B are positive real numbers then there is some positive integer n such that not gt 3 Proof The proof will be by contradiction Suppose that there exist positive real numbers a and B such that na 5 B for every natural number n Since a gt 0 a lt 205 lt 305 lt lt na lt is an increasing sequence of real numbers that is bounded above by 3 Since R 5 satis es the least upper bound property nor n e N has a least upper bound in R say L Choose 6 7a which is positive because a gt 0 Since L sup n05 n e N from Lemma 131 there exists s e not n e N such thatL 8 lt s 5 L Ifs Na then for all natural numbers m gt Nwe alsohavethatL 8 lt ma 5 L Henceform gt N0 5 L m0i lt 8 In particular 1 05L Nlalt8 a and l OSL N2OClt8EOL Thus L a lt N 1a and N 20i lt L lt L a But adding a to both sides of the rst inequality yields L a lt N 20i which contradicts N 20i lt L a Hence contrary to our original assumption there exists a natural number n such that na gt B l 13 THE REAL FIELD 29 Corollary 133 Density of the Rational Numbers If a and B are real numbers with a lt B then there is a rational number r such that a lt r lt 3 Proof Since 1 and a are positive real numbers by the Archimedean Prin ciple there exists a positive integer m such that 1 lt m a or equivalently ma 1 5 m Let n be the largest integer such that n 5 ma It follows that n15ma15m Since n is the largest integer such that n lt ma we know that ma lt n 1 Consequently ma lt n 1 lt m which is equivalent to having n1 m alt lt3 Therefore we have constructed a rational number that is between a and B l Corollary 134 Density of the Irrational Numbers If a and B are real num bers with a lt B then there is an irrational number 1 such that a lt y lt 3 Proof Suppose that a and B are real numbers with a lt B By Corollary 133 there is a rational r that is between and E Since 5 is irrational we conclude that y r is an irrational that is between a and B l 132 Existence of nth Roots The primary result in this connection that is offered by the author of our text is the following Theorem 135 For R x e R x gt 0 we have that VxVnxERAneJgtEIyyeRyquotx 30 CHAPTER 1 THE FIELD OF REALS AND BEYOND Before we start the proof we note the following fact that will be used in the presentation Fact 136 Vy Vz Vn yz e R An 6 J A0 lt y lt z gt yquot lt 2quot To see this for y z e R satis ving are 0 lt y lt 2 let SneJyquot ltzquot Our set up automaticallyplaces l e S Suppose thatk e S ie k e J andyk lt 2quot Since 0 lt y by OF2 ykJr1 y yk lt y 2quot From 0 lt z and repeated use ofProposition 12 92 we canjusti v that 0 lt 2quot Then OF2 with 0 lt 2k and y lt zyields thaty 2quot lt z 2k 2 2quot As a consequence ofthe transitive law yk1 lt yZk AyZk lt Zk1gtyk1 lt Zk1 that is k l e S Since k was arbitrary we conclude that Vkke Sgt kl e S From 1 e S A Vk k e S gt k l e S S is an inductive subset ofthe natural numbers By the Principle ofMathematical Induction PMI S J Since y and 2 were arbitrary this completes the justi cation of the claim Fact 137 Vw Vn w 6 Rm 6 J 1A0 lt w lt 1 gt 11quot lt 11 Since n 2 2 n l 2 land byFact 136 wquot 1 lt 1 1 1 From OF2 0 lt wAw 1 lt limplies that wquot wquot 1 w lt 111 2 w ie wquot lt w as claimed Fact 138 Va Vb Vn a b 6 RA n e J 1 A0 lt a lt b gt 1quot aquot lt b a nbn l From Fact 136 n 2 2 A0 lt a lt b gt aquot 1 lt bnil while OF2 yields that a bj lt b bl bj1for j l 2 n 2 It can be shown by repeated application of Exercise 6a that bnil bn72a ban72 anil lt bnil bnil bnil nbnil this with OF2 implies that bquot aquot b a 1 bHa 1m a lt b a nbnil as claimed Proof of the theorem Let RJr u e R u gt 0 When n 1 there is nothing to prove so we assume that n 2 2 For xed x e RJr and n e J 1 set EteRtquotltx 13 THERE4LFIELD 31 x Excursion 139 Use 11 1 tojustz that E 75 x Now let u 1 x and suppose thatt gt u gt 0 Fact 136 yields that tquot gt uquot From Proposition 1297 u gt 1 gt 0 lt lt 1 It follows from u 1 1 Fact 137 and Proposition 1297 that 0 lt n 5 and uquot 2 u By transitiVity u u tquot gt uquot A uquot 2 u implies that tquot gt u Finally since u gt x transitiVity leads to the conclusion that tquot gt x Hence I E Since I was arbitrary Vt t gt u gt t E which is equivalent to Vt t e E gt t 5 u Therefore E C R is bounded above From the least upper bound property lub E exists Let y lubE Since E C R we have that y 2 0 By the Trichotomy Law one and only one of yquot x yquot lt x or yquot gt x In what follows we will that neither of the possibilities yquot lt x or yquot gt x can hold n Case 1 Ifyquot lt xthenx yquot gt 0Sincey1gt Oandn 21 x yl gt n y 1quot 0 and we can choose h such that 0 lt h lt 1 and x yquot n y 1quot Taking a y and b y h in Fact 138 yields that yhquot yquotlthnyhquot 1ltx yquot 32 CHAPTER 1 THE FIELD OF REALS AND BEYOND Excursion 1310 Use this to obtain contradict that y sup E Case 2 If0 lt x lt yquot then 0 lt yquot x lt nyquot Hence kynx nynil is such that 0 lt k lt y Fort 2 y k Fact 136 yields that tquot 2 y kquot From Fact 138 with b y and a y k we have that yntn5ynyknltknynilZynx Excursion 1311 Use this to obtain another contradiction From case 1 and case 2 we conclude that yquot x this concludes the proof that there exists a solution to the given equation The uniqueness of the solution follows from Fact 136 To see this note that ifyquot x and w is such that 0 lt w 75 y then 11 lt y implies that wquot lt yquot x while y lt 11 implies thatx yquot lt 11quot In either case 11quot 75 x I For Excursion 139 you want to justify that the given 11 is in E Because 0 lt x x lt1x0 lt 11 21 lt1Inviewoffact137wquot lt wforn 2 Zorwquot 5 w x 1 x forn21Butxgt01xgt1impliesthat lt1 ltx1x 1 x 1 x From trans1t1V1ty wquot lt w A w lt x gt wquot lt x Le 11 e E To obtain the desired contradiction for completion of Excursion 1310 hope fully you notices that the given inequality implied that y hquot lt x which would 13 THERE4LFIELD 33 place y h in E since y h gt y this would contradict that y sup E from which we conclude that yquot lt x is not true The work needed to complete Excursion 1311 was a little more involved In this case the given inequality led to tquot lt x or tquot gt x which justi es that t E hence t gt y k implies that t E which is logically equivalent to t e E implies that t lt y k This would make y k an upper bound for E which is a contradiction Obtaining the contradiction yields that x lt yquot is also not true Remark 1312 For x a positive real number anal n a natural number the number y that satis es the equation yquot x is written as x anal is reaal as the nth root of x Repeated application of the associativity and commutativity of multiplication can be used to justify that for positive real numbers a and and n a natural number quot3quot 063quot From this identity and the theorem we have the following identity involving nth roots of positive real numbers Corollary 1313 If a anal b are positive real numbers anal n is a positive integer then abln alnbln Proof For a a1quot and buquot we have that ab 2 MWquot 2 MBquot Hence a is the unique solution to yquot ab from which we conclude that ab139l a as needed I 133 The Extended Real Number System The extended real number system is RU 00 00 where 1R 0 1 lt is the ordered eld that satis es the least upper bound property as discussed above and the symbols 00 and 00 are de ned to satisfy 00 lt x lt 00 for all x 6 1R With this convention any nonempty subset S of the extended real number system is bounded above by 00 and below by 00 if S has no nite upper bound we write lub S 00 and when S has no nite lower limit we write glb S 00 The 00 and 00 are useful symbols they are not numbers In spite of their appearance 00 is not an additive inverse for 00 This means that there is no 34 CHAPTER 1 THE FIELD OF REALS AND BEYOND oo oo mean1ng attached to any of the eXpress1ons 00 00 or or 1n fact these expressions should never appear in things that you write Because the symbols 00 and 00 do not have additive or multiplicative inverses R U 00 00 is not a eld On the other hand we do have some conventions concerning interaction of the special symbols with elements of the eld R namely 0 IfxeRthenxoooox oo ooand i0 oo oo o Ifx gt 0thenx 00 2 00 andx oo 2 oo o Ifx lt 0thenx 00 2 oo andx oo 00 Notice that nothing is said about the product of zero with either of the special sym bols 14 The Complex Field For C R x R de ne addition and multiplication by x1y1x2y2x1xzy1y2 and x1y139x2y2 961962 y1y2x1y2 y1X2 respectively That addition and multiplication are binary operations on C is a con sequence of the closure of R under addition and multiplication It follows immedi ately that xay070x7y and Hence 0 0 and l 0 satisfy the additive identity property and the multiplica tive identity eld property respectively Since the binary operations are de ned as combinations of sums and products involving reals direct substitution and appro priate manipulation leads to the conclusion that addition and multiplication over C are commutative and associative under addition and multiplication The actual manipulations are shown in our text on pages 1213 To see that the additive inverse property is satis ed note that x y e C implies that x e R A y e R The additive inverse property in the eld R yields that x e R 14 THE COMPLEXFIELD 35 and y e R It follows that x y e C and xy x y 00 and needed Suppose xy e C is such that xy 75 0 0 Then x 75 0 Vy 75 0 from which we conclude that x2 y2 75 0 and a b if m x2 y2 is well de ned Now x y xay 39 Gab xay39 m x y y x x x2y2 y x2y2 x x2y2 y x2y2 xxyy xyyx x2y2 x2y2 x2y2 xyyx x2y2 x2y2 10 Hence the multiplicative inverse property is satis ed for C Checking that the distributive law is satis ed is a matter of manipulating the appropriate combinations over the reals This is shown in our text on page 13 Combining our observations justi es that C 0 0 l 0 is a eld It is known as the complex eld or the eld of complex numbers Remark 141 Identifying each element ofltC in theform x 0 with x e R leads to the corresponding identi cation of the sums and products x a 2 x 0 a 0 x a 0 andx a x 0 a 0 x a 0 Hence the real eld is a sub eld of the complex eld The following de nition will get us to an alternative formulation for the complex numbers that can make some of their properties easier to remember De nition 142 The complex number 0 l is de ned to be i With this de nition it can be shown directly that 12 10 1 and o ifa and b are real numbers then a b a bi 36 CHAPTER 1 THE FIELD OF REALS AND BEYOND With these observations we can write C abi ab gunk 1 with addition and multiplication being carried out using the distributive law com mutativity and associativity We have two useful forms for complex numbers the rectangular and trigono metric forms for the complex numbers are freely interchangeable and olTer different geometric advantages From Rectangular Coordinates Complex numbers can be represented geometrically as points in the plane We plot them on a rectangular coordinate system that is called an Argand Graph In 2 x iy x is the real part of z denoted by Re 2 and y is the imaginary part of z denoted by Imz When we think of the complex number x i y as a vector 573 joining the origin 0 0 0 to the point P x y we grasp the natural geometric interpretation of addition in C De nition 143 The modulus of a complex number 2 is the magnitude of the vector representation and is denoted by z Ifz x iy then z 1x2 y2 De nition 144 The argument of a nonzero complex number 2 denoted by arg z is a measurement of the angle that the vector representation makes with the positive real axis De nition 145 For 2 x i y the conjugate of z denoted by E is x i y Most of the properties that are listed in the following theorems can be shown fairly directly from the rectangular form 14 THE COMPLEXFIELD 37 Theorem 146 For 2 and w complex numbers 1 z 2 0 with equality only ifz 0 2 IEI IZI 3 lzwl lzl lwl 4 lRezl 5 z andllmzl 5 lzl 5 lz ml2 2 z2 2RezE lwl2 The proofs are left as exercises Theorem 147 The Triangular Inequalities For complex numbers 21 and 22 2122S2122 and 3921 22 2 21 Izzll Proof To see the rst one note that I2122I2 2122Re2122222 s 212 2 l21 Izzl 222 l21l 222 The proof of the second triangular inequality is left as an exercise I Theorem 148 If 2 and w are complex numbers then 4 22 is a nonnegative real number From Polar Coordinates For nonzero z x iy e C let r xx2 y2 and 6 arctan argz x Then the trigonometric form is z r cos6 i sin6 In engineering it is customary to use cis 6 for cos 6 i sin6 in which case we write 2 r cis 6 NOTE While r 6 uniquely determines a complex number the converse is not true 38 CHAPTER 1 THE FIELD OF REALS AND BEYOND Excursion 149 Use the polar form for complex numbers to develop a geometric interpretation of the product of two complex numbers The following identity can be useful when working with complex numbers in polar form Proposition 1410 DeMoivre s Law For 6 real and n e Z cis 6quot cis n6 Example 1411 Find all the complex numbers that when cubed give the value one We are lookingfor all r e C such that 3 l DeMoivre s Law o ers us a nice toolfor solving this equation Let r r cis 6 Then3 2 l ltgt r3 cis 36 1 Since lr3 cis 36 r3 we immediately conclude that we must have r 1 Hence we need only solve the equation cis 36 1 Due to the periodicity of the sine and cosine we know that the last equation is equivalent to nding all 6 such that cis 36 2k cis 2k7rfor k e Z whichyields that 36 2k7r fork e Z But k e Z 2 2 g 0 Thus we have three distinct complex numbers whose cubes are 27f 271 one namely 18 01 0 1 and 018 Y In rectangular form the three 3 complex numbers whose cubes are one are i 0 and 2 2 2 2 Theorem 1412 Schwarz s Inequality If a1 an and b1 bn are complex numbers then 2 n 2 n 2 j1 j1 n 26 ij 11 14 THE COMPLEXFIELD 39 Proof First the statement is certainly true if bk 2 0 for all k l 5 k 5 n Thus we assume that not all the bk are zero Now for any xi 6 C note that n 2 jl Excursion 1413 Make use of this inequality anal the choice of to complete the proof Remark 1414 A special case of Schwarz s Lemma contains information relating the modulus of two vectors with the absolute value of their alot product For ex ample a1 a2 anal a 2 b1 b2 are vectors in R x R then Schwarz s Lemma merely reasserts that o a1b1 a2b2 5 141 Thinking Complex Complex variables provide a very convenient way of describing shapes and curves It is important to gain a facility at representing sets in terms of expressions involving 40 CHAPTER 1 THE FIELD OF REALS AND BEYOND complex numbers because we will use them for mappings and for applications to various phenomena happening within shapes Towards this end let s do some work on describing sets of complex numbers given by equations involving complex variables One way to obtain a description is to translate the expressions to equations in volving two real variables by substituting z x i y Example 1415 Find all complex numbers 2 that satisfy 2 z 2Imz l Letz x iy Then 2z2Imz l 2 x2y22y l 1 4x2y2 4y2 4y1y 2 5 1 4x2 4yly25 x2 1A gt1 1 l l The last equation implies thaty 5 Since y 5 Ay 2 is never satis ed we ltgt ltgt ltgt ltgt conclude that the set of solutions for the given equation is empty Excursion 1416 Find all z e C such that z z l 2i 3 Your work should have given the E 2i as the only solution 14 THE COMPLEX FIELD 41 Another way which can be quite a time saver is to reason by TRANSLAT ING TO THE GEOMETRIC DESCRIPTION In order to do this there are some geometric descriptions that are useful for us to recall ZIIZZor Z I IZle IZZzl Z I IZZ1Z22 p for a constant p gt IZ122I is the locus of all points z equidistant from the xed point 20 with the distance being r gt 0 a circle is the locus of all points z equidistant from two xed points 21 and 22 the perpendicular bisec tor ofthe line segmentjoining 21 and 22 is the locus of all points for which the sum of the distances from 2 xed points 21 and 22 is a con stant greater than 21 22 an ellipse Excursion 1417 For each of the following without substituting xi y for z sketch the set of points 2 that satisfy the given equations Provide labels names andor important points for each object z 2i z32i 2 z 4iz7i212 42 CHAPTER 1 THE FIELD OF REALS AND BEYOND 3 4Z3 i53 The equations described a straight line an ellipse and a disk respectively In set notation you should have obtained x 139 y e C y Zx g 3 2 EC39 y 2 1 d x 1y 23 62 an e c 3 Hf 5 32 Remark 1418 In general z39fk is a positive real number and a b e C then Z CI bzhk ZECI 2 describes a circle Excursion 1419 Use the space below to justify this remark 15 PROBLEM SETA 43 Z a k leads to b Simplifying Z 1 k2 z2 2Re a2 2k2 Re b2 W k2 W from which the remark follows 15 Problem Set A 1 For IF p q r let the binary operations of addition 69 and multiplication X be de ned by the following tables 8 r a Is there an additive identity for the algebraic structure IF 69 8 Brie y justify your position b Is the multiplicative inverse property satis ed If yes specify a multi plicative inverse for each element of IF that has one 44 CHAPTER 1 THE FIELD OF REALS AND BEYOND c Assuming the notation from our eld properties nd r em p 69 fl d IS 10 P 7 P r 7 1 1 7 P 1 r r a eld orderng 011 IF Brie y justify your claim 2 For a eld IF e f prove each of the following parts of Proposition a The multiplicative identity of a eld is unique b The multiplicative inverse of any element in IF e is unique LA For a eld IF e f prove each of the following parts of Proposition 118 a V0 Vb a b 6 IF gt 0 b a b b Va Vb a b 6 IF 2 a b a b c Va Vb a b 6 IF 2 a b a b d Va Vb a b 6 IF 2 a b a b e Va Vb ab e IF e gt a brl a l 134 4 For a eld IF 01 prove Proposition ll10l VaVb ab e IFgt Elbe x e IFAa x 2 b 5 For a eld IF 01 show that for a b c 6 IF a bca b c and a b ca bc Give reasons for each step of your demonstration 6 For an ordered eld IF 0 l lt prove that a Va Vb Vc Vdabcd e IFa lt b Ac lt 61 a c lt b d b VaVbVcVaIabcaI eIFAO lt a lt b0 lt c lt 61 ac lt bd 7 For an ordered eld IF 0 l lt prove each of the following 15 PROBLEMSETA 45 a Va Vb Va 61 b c 6 IF c 75 0 gt ac 1bc 1a b 0 1 b Va Vb Vc Vd ac 6 IF 1351 6 IF 0 gt 1 d 7A 0A 1cd 1adbcbd 1 Find the least upper bound and the greatest lower bound for each of the fol lowing 9 xeR 0 gtlt 1 g x5ltxlt2 Let X 5 be an ordered set and A Q X Prove that if A has a least upper bound in X it is unique 50 10 Suppose that S Q R is such that inf S M Prove that VsselRsgt 0gt ElggeSM5g ltMs 2 1 11 For fx 2 nd x x a SUP f 100 3 b inff 1 3 00 46 N 4 UI N O N For any integers k and 71 show that 139quot CHAPTER 1 THE FIELD OF REALS AND BEYOND Suppose that P C Q C R and P 75 IfP and Q are bounded above show that sup P S SUP Q LetA x e R x 2 x 3 1 lt 2 Findthe sup A andthe infA Use the Principle of Mathematical Induction to prove that for a 2 0 and n a natural number 1 aquot 2 l na Find all the values of d 1i4 e 1 1n 1 z39quot a 234 1 b l213932i 236i c 11 Show that the following expressions are both equal to one 3 3 1 M 1 zxi T b T 39quot4k How many distinct values can be assumed by 139quot Use the Principle of Mathematical Induction to prove DeMoivre s Law If 21 3 4139 and 22 2 3139 obtain graphically and analytically a 2Z1422 d Z122 b 32125 e In 22 c 21 5 4 t mama 1 Prove that there is no ordering on the compleX eld that will make it an or dered eld Carefully justify the following parts of Theorem 146 For 2 and 1 complex numbers a z 2 0 with equality only if z 0 b IEI IZI c lzwl lzl lwl 15 PROBLEM SETA 22 N L 47 d lRezl S lzl and IImzl S lzl e lz ml2 2 z2 2RezE lwlz Prove the other triangular inequality For complex numbers 21 and 22 2122IZ Z122 Carefully justify the following parts of Theorem 148 If 2 and w are complex numbers then azw2u b 2u 22 c Rez ImzZ z 2139 d 23 is a nonnegative real number 24 Find the set of all z e C that satisfy a l ltz53 d z lZ12 g z 2Z25 b 23 1 e Im22gt0 h zlRez z2 39 22 cRe22gt0 f z 1i2239 25 When does az bi c 0 represent a line 26 Prove that the vector 21 is parallel to the vector 22 if and only if 1m 2122 0 48 CHAPTER 1 THE FIELD OF REALS AND BEYOND


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