AA1H Calculus Notes Math1115, Honours John Hutchinson

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1 AA1H Clculus Notes Mth1115, Honours John Hutchinson Author ddress: Deprtment of Mthemtics, School of Mthemticl Sciences, Austrlin Ntionl University E-mil ddress:

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3 Contents Chpter 1. Introduction v Chpter 2. The Rel Number System Introduction Importnt sets of rel numbers Algebric nd order properties Completeness property Properties of the rtionl nd irrtionls Functions Mthemticl induction Fields Deductions from the xioms Existence nd Uniqueness of the Rel Number System 20 Chpter 3. Limits Introduction Definition of limit for function Properties of limits of functions Functions of more thn one vrible 35 Chpter 4. Sequences Exmples of sequences Limit of sequences Monotone sequences Limits for functions vi limits for sequences 43 Chpter 5. Continuity Introduction Definition of continuity Properties of continuous functions Deeper properties of continuous functions Pthology nd continuity Uniform continuity Functions of two or more vribles 57 Chpter 6. Differentition Introduction The derivtive of function Computing derivtives Mximum nd minimum vlues Men Vlue Theorem Prtil derivtives 67 Chpter 7. Integrtion 69 iii

4 iv CONTENTS 7.1. Introduction The Riemnn integrl Riemnn sums Properties of the Riemnn integrl Fundmentl Theorem of Clculus 79 Chpter 8. Differentil Equtions Outline of proof of the Existence nd Uniqueness theorem Rigorous proof of the Existence nd Uniqueness theorem 85 Bibliogrphy 91 Index 5

5 CHAPTER 1 Introduction The im of the AA1H course is to give n introduction to modern mthemtics. In the process you will prove the mjor results used in the AA1 course nd thereby obtin more fundmentl understnding of tht mteril. Mthemtics is the study of pttern nd structure. It is studied both for its internl beuty nd for its universl pplicbility. In mthemtics we mke certin specific ssumptions (or xioms) bout the objects we study nd then develop the consequences of these ssumptions in precise nd creful mnner. The xioms re chosen becuse they re nturl in some sense; it usully hppens tht these xioms lso describe phenomen in other subjects, in which cse the mthemticl conclusions we drw will lso pply to these phenomen. Ares of mthemtics developed for mthemticl resons usully turn out to be pplicble to wide vriety of subjects; spectculr recent exmple being the pplictions of differentil geometry to understnding the fundmentl forces of nture studied in physics, nd nother being the ppliction of prtil differentil equtions nd geometric mesure theory to the study of visul perception in biology nd robotics. There re countless other exmples in engineering, economics, nd the physicl nd biologicl sciences. On the other hnd, the study of these disciplines cn usully only be done by pplying the techniques nd lnguge of mthemtics. Mthemtics is used s tool in such investigtions. But the study of these subjects cn lso led to the development of new fields of mthemtics nd insights into old fields. In this course we will study the rel number system, the concepts of limit nd continuity, differentibility nd integrbility, nd differentil equtions. While most of these terms will be fmilir from high school in more or less informl setting, we will study them in much more precise wy. This is necessry both for pplictions nd s bsis for generlising these concepts to other mthemticl settings. One very importnt question we investigte is: when do certin types of differentil equtions hve solution, when is there exctly one solution, nd when is there more thn one solution? The solution of this problem uses lmost ll the mteril tht is developed throughout the course. The study of differentil equtions is of tremendous importnce in mthemtics nd for its pplictions. Any phenomen tht chnges with position nd/or time is usully represented by one or more such equtions. The ides we develop re bsic to further developments in mthemtics. The concepts we study generlise in mny wys, such s to functions of more thn one vrible, nd to functions whose vribles re themselves functions (!); ll these generlistions re fundmentl to further pplictions. At the end of the first semester you should hve much better understnding of ll these ides. These Notes re intended so tht you cn concentrte on the lectures rther thn trying to write everything down. Occsionlly there my be lecture mteril not mentioned in the Notes, in which cse I will indicte this precisely, but generlly v

6 vi 1. INTRODUCTION you should not tke your own notes. So why come to the lectures? Becuse the Notes re frequently rther forml (this is consequence of the precision of mthemtics) nd it is often very difficult to see the underlying concepts. In the lectures I explin the mteril in less forml mnner, single out nd discuss the key nd underlying ides, nd generlly explin nd discuss the subject in mnner which it is not possible to do efficiently in print. It would be very big mistke to skip lectures. If you still do not believe me, sk students from my previous courses. Do not think tht you hve covered ny of this mteril in school; the topics my not pper new, but the mteril certinly will be. Do the ssigments, red the lecture notes before clss. Mthemtics is not body of isolted fcts; ech lecture will depend on certin previous mteril nd you will understnd the lectures much better if you keep up with the course. In the end this pproch will be more efficient s you will gin more from the lectures. Throughout the course I will mke vrious digressions nd dditionl remrks, mrked clerly by str. This is non-exminble nd generlly more chllenging mteril. But you should still red nd think bout it. It is included to put the subject in broder perspective, to provide n overview, to indicte further directions, nd to generlly round out the subject. In ddition, studying this more dvnced mteril will help your understnding of the exminble mteril. Moreover, you will need to know nd understnd the sttements of the results in the Sections 3.4, 5.6, 5.7 nd 6.6, for the proof of the Fundmentl Existence nd Uniqueness Theorem in Section 8.2, nd to lesser extent in Section 8.1. The other references for the course re the book [Adms] nd the notes [Wrd]. Both of these supplement the mteril here, but re t less theoreticl level. The book [Spivk] is excellent, but t slightly higher level thn the current course. The book [Stromberg] is for the extremely dedicted; it is very terse nd essentilly only pproprite for lter yer courses. For interesting discussions nd host of exmples on the beuty nd utility of mthemtics, see [Devlin], [Hildebrndt nd Tromb] nd [Dvis nd Hersh]. All these books re on two-dy lon through the reserve section of the Hncock librry. If you re hving difficulty with some of the concepts, sk your tutor or come nd see me during office hours. Do not let things slide!

7 CHAPTER 2 The Rel Number System The reference for this chpter is [Adms, Chpter P], minly P1, P2 to pge 14, P4 to pge 29, nd P5. But you should red ll of this chpter; it gives slightly different nd somewht more elementry pproch to the mteril covered here Introduction Rel numbers hve deciml expnsions. They cn be represented s points on n infinite line. The deciml expnsions nd represent the sme rel number. Rel numbers hve deciml expnsions, for exmple: 2= =1.5= π = , lso written The... indicte the expnsions go on forever, nd the 14 6 indicte tht the pttern 146 is repeted forever. In the first two cse the expnsion continues with zeros nd in the third cse one cn compute the expnsion to ny required degree of ccurcy. Rel numbers cn lso be represented geometriclly s points on n infinite line. 0 1/2 π 2 In this chpter we will give creful nlysis of wht is ment by rel number. (Sometimes we sy rel number, nd sometimes we just sy number. You will lso lter meet the complex numbers (if you hve not lredy done so), these include the rel numbers nd llow one to give mening to 1.) There is one point tht sometimes cuses confusion. Is it the cse tht 1=. 9?, or is it tht. 9 is little less thn one? By. 9 we men, s usul, , with the 9 s repeted forever. Any of the pproximtions to. 9,.9 = ,.99 =,.999 =,.9999 = ,... is certinly strictly less thn one. On the other hnd,. 9 is defined to be the limit of the bove infinite sequence (we discuss limits of sequences in lter chpter). Any mthemticlly useful wy in which we define the limit of this sequence will in fct imply tht. 9 = 1. To see this, let =. 9 = Then, for ny resonble definition of infinite sequence nd limit, we would wnt tht Subtrcting, gives 9 = 9, nd hence =1. 10 =

8 2 2. THE REAL NUMBER SYSTEM 2.2. Importnt sets of rel numbers Beginning from the set R of rel numbers we define the sets N of nturl numbers, Z of integers, Q of rtionls, nd the set of irrtionls. We discuss intervls. We introduce some generl nottion for describing sets. Finlly we discuss n-tuples of numbers nd n-dimensionl spce Nottion for sets. By set (or clss, orfmily) we men collection, often infinite, of objects of some type. 1 Members of set re often clled elements of the set. If is member (i.e. element) of the set S, we write If is not member of S we write S. S. If set is finite, we my describe it by listing its members. For exmple, A = { 1, 2, 3 }. Note tht {1, 2, 3}, {2, 3, 1}, {1, 1, 2, 2, 2, 3} re different descriptions of exctly the sme set. Some infinite sets cn lso be described by listing their members, provided the pttern is cler. For exmple, the set of even positive integers is We often use the nottion E = { 2, 4, 6, 8,...}. S = { x : P (x) }, where P (x) is some sttement, or proposition, involving x. We red this s S is the set of ll (rel numbers) x such tht P (x) is true. It is usully understood from the context of the discussion tht x is restricted to the rel numbers. But if there is ny possible mbiguity, then we write Note tht this is exctly the sme set s or S = { x R : P (x) }. { y : P (y) }, S = { y R : P (y) }. The vribles x nd y re sometimes clled dummy vribles, they re ment to represent ny rel number with the specified properties. One lso sometimes uses insted of : when describing sets. The union of two or more sets is the set of numbers belonging to t lest one of the sets. The intersection of two or more sets is the set of numbers belonging to ll of the sets. We use for union nd for intersection. For exmple { x :0<x<1or2<x 3}=(0,1) (2, 3] (0, 2)=(0,1) [1, 2)=(0,1] [1, 2)=(0,1) ( 1 2, 2) { x :0<x<2 nd 1 x 3 } =[1,2) 1 There is mthemticl theory of sets, nd in fct ll of mthemtics cn be formulted within the theory of sets. However, this is normlly only useful or prcticl when considering fundmentl questions bout the foundtions of mthemtics.

9 2.2. IMPORTANT SETS OF REAL NUMBERS Different types of rel numbers. The set of rel numbers is denoted by The set N of nturl numbers is defined 2 to by The set Z of integers is defined by The set Q of rtionl numbers is defined by R. N := {1, 2, 3,...}. Z := {..., 3, 2, 1,0,1,2,3,...}. Q := {m/n : m, n Z, n 0}. Rtionl numbers re those whose deciml expnsion either termintes fter finite number of plces, s for 2 nd 1.5, or re recurring, s for or 3/7. See [Adms, Ex. 1, pge 4]. A rel number is irrtionl if it is not rtionl. It cn be proved tht π nd e re irrtionl, see [Spivk] Intervls. An intervl is set of rel numbers with the property tht it contins t lest two numbers, nd moreover it contins ll rel numbers between ny two of its members. (It follows tht n intervl must contin n infinite number of members. 3 ) Bounded intervls re intervls of the following type: [, b] :={x: x b}, (, b) :={x:<x<b}, (, b] :={x:<x b}, [, b) :={x: x<b}. An intervl my lso be unbounded in either or both directions: [, ) :={x: x}, (, ) :={x:<x}, (,b]:={x:x b}, (,b):={x:x<b}. Finlly, R is lso n intervl, which we could write s (, ). Note tht is not number, nd by itself does not hve ny mening here, just s { or : does not hve ny mening by itself. An intervl is open if it does not contin ny of its endpoints, nd is closed if it contins ll of its endpoints. Thus open intervls re those of the form (, b), (, ), (,b) nd R, while closed intervls re those of the form [, b], [, ), (,b] nd R. (If this seems confusing, remember tht ± re not numbers, nd in prticulr cnnot be endpoints of intervls.) In prticulr, R is both open nd closed. The end-points or boundry points for the previous exmples re the points nd b (note tht they my or my not belong to the respective intervl). The interior points re ll other points in the intervl. Thus in (1, 2] the endpoints re 1 nd 2, of which only 2 belongs to the intervl, nd the interior points re ll points in the open intervl (1, 2). 2 We often use the nottion := to men by definition is equl to. 3 If, b re two numbers in the intervl, then, for exmple. + b 2 b b, +, + 3 4,... is n infinite set of numbers between nd b which re distinct from one nother.

10 4 2. THE REAL NUMBER SYSTEM n-tuples of rel numbers. We will lter work with pirs (x, y) of rel numbers, triples of rel numbers (x, y, z), nd more generlly n-tuples (x 1,...,x n). Just s rel numbers cn be represented geometriclly by points on line, so cn pirs be represented s points in the plne nd triples s points in spce (we need to mke choice of origin, coordinte xes, nd scles on the xes). We could lso define n-dimensionl spce to be the set of n-tuples of rel numbers! y.(x,y) x (x,y,z) 2 1. z y.. x Algebric nd order properties We introduce the lgebric nd order xioms for the rel number system nd indicte very briefly how ll the usul lgebric nd order properties follow from these. The rtionl numbers lso stisfy these xioms, but this is not the cse for the nturl numbers, the integers, or the irrtionls. The bsolute vlue of rel number is defined nd the bsic properties re proved. The rel number system consists of the rel numbers, together with the opertions ddition (denoted by +) nd multipliction (denoted by ) nd the less thn reltion (denoted by <). One lso singles out two prticulr rel numbers, zero or 0 nd one or 1. If nd b re rel numbers, then so re + b nd b; nd the reltion <bmust be either true or flse. We will usully write b for b. We will soon see tht one cn define subtrction nd division in terms of + nd ; nd, > etc. cn be defined from <. There re three ctegories of properties of the rel number system: the lgebric properties, the order properties nd the completeness properties Algebric properties. These re the properties of ddition, multipliction, subtrction nd division. It turns out tht there re certin bsic properties, usully clled xioms, from which we cn prove ll the other lgebric properties. These xioms re: Axioms (Addition). If, b nd c re rel numbers then: A1: + b = b + ; A2: ( + b)+c=+(b+c); A3: +0=0+=; A4: there is exctly one rel number, denoted by, such tht +( )=( )+=0. Axioms (Multipliction). If, b nd c re rel numbers then: A5: b = b ; A6: ( b) c = (b c);

11 2.3. ALGEBRAIC AND ORDER PROPERTIES 5 A7: 1=1 =, 1 0; A8: if 0 there is exctly one rel number, denoted by 1, such tht 1 = 1 =1. Axioms (The Distributive Property). If, b nd c re rel numbers then: A9: (b + c) = b+ c. Note We re not relly using subtrction in xiom A4; we could just s well hve written for, but it is more consistent with stndrd nottion to write. We re merely sserting tht unique rel number, with certin property, exists. Similrly, in A8, we could just s well hve written #,sy,for By the symbol = for equlity we men denotes the sme thing s, or equivlently, is the sme rel number s. We tke = to be logicl notion nd do not write down xioms for it. 4 Insted, we use ny properties of = which follow from its logicl mening. For exmple: = ; if = bthen b = ; if = bnd b = c then = c; if=bnd something is true of then it is lso true of b (since nd b denote the sme rel number!). When we write b, we men tht is not the sme rel number s b. 3. Some of the xioms re redundnt. For exmple, from A1 nd the property +0 = it follows tht 0 + =, (why?). Similr comments pply to A4; nd becuse of A5 to A7 nd A8. 4. One cn show tht prt from these cses the xioms re not redundnt; in other words tht no one xiom follows from the others. More precisely, one cn construct exmples where, sy, A8 is flse but ll the other xioms re true (see the following section on Fields); nd similrly for ny of the other xioms More lgebric properties. All the usul lgebric properties of the rel numbers follow from A1 A9, in prticulr, one cn solve simultneous systems of liner equtions. We will not spend much time on indicting how one deduces other lgebric properties from these xioms, but will continue to use ll the usul properties of ddition, multipliction, subtrction nd division tht you hve used in the pst. None-the-less, it is useful to hve some ide of the methods involved in mking deductions from A1 A9. Lter in this course, when we discuss vector spces, you will hve more prctice t mking deductions from lgebric sets of xioms somewht like those bove. The first thing is to define subtrction nd division. For this, suppose nd b re ny 5 two rel numbers (except tht b 0 in the definition of division). Then we define b = +( b) b= b 1 for b 0 This my look like circulr definition; it my pper tht we re defining subtrction in terms of subtrction. But this is not the cse. Given b, from A4 there is certin rel number, which we denoted by b, with certin properties. We then define b to be the sum of nd this rel number b. Similr comments pply to the definition of division. We lso write /b or for b. b We cn lso now define other numbers nd opertions. For exmple, we define 2 = 1 + 1, 3 = 2 + 1, etc. Also, we define x 2 = x x, x 3 = x x x, x 2 = ( x 1) 2, etc. etc. We define b, for b 0, to be tht number 0 such tht 2 = b. Similrly, if n is nturl number, then n b is tht number 0 such tht n = b. To prove there is lwys such number requires the completeness xiom (see lter). 4 One cn write down bsic properties, i.e. xioms, for = nd the logic we use. See lter courses on the foundtions of mthemtics. 5 We do not even ssume b.

12 6 2. THE REAL NUMBER SYSTEM As n exmple of the wy in which one cn use A1 A9 to derive other lgebric properties, we prove the cncelltion property of ddition: Theorem 2.2. If, b nd c re rel numbers nd + c = b + c, then = b. Proof. Assume + c = b + c. Since + c nd b + c denote the sme rel number, we obtin the sme result if we dd c to both; i.e. ( + c)+( c)=(b+c)+( c). (This used the existence of the number c from A4.) Hence +(c+( c)) = b +(c+( c)) from A2 pplied twice, once to ech side of the eqution. Hence +0=b+0 from A4 gin pplied twice. Finlly, = b from A3. We will not pursue this ide of mking deductions from the xioms, but see Section 2.9 if you re interested in knowing more. In future we will forget bout the xioms nd just use ll the usul properties of ddition, multipliction, subtrction nd division. Here we list few: Theorem 2.3. If, b, c, d re rel numbers nd c 0,d 0then 1. if c = bc then = b =0 3. ( ) = 4. (c 1 ) 1 = c 5. ( 1) = 6. ( b) = (b) =( )b 7. ( )+( b)= (+b) 8. ( )( b) =b 9. (/c)(b/d) =(b)/(cd) 10. (/c)+(b/d) =(d + bc)/cd For those who re interested, I indicte the proofs of these fcts from the xioms in Section Order properties. The rel numbers hve nturl ordering, denoted by < which we red s is less thn. Bsic properties re: Axioms (Less thn). If, b nd c re rel numbers then: A10: one nd only one of the following hold <bor = b or b<; A11: if <bnd b<c, then <c; A12: if <bthen + c<b+c; A13: if <bnd 0 <c, then c<bc; If 0 <we sy is positive nd if <0wesyis negtive.

13 2.3. ALGEBRAIC AND ORDER PROPERTIES More order properties. One cn define >, nd in terms of < s follows: >bif b<, bif ( <bor = b), b if ( >bor = b). (Note tht the sttement 1 2, lthough it is not one we re likely to mke, is indeed true, why?) All the usul properties of inequlities cn in fct now be proved from A10 A13, (together with A1 A9). For exmple, Theorem 2.4. If, b nd c re rel numbers then 1. <bnd c<0implies c > bc 2. 0 < 1 nd 1 < 0 3. >0implies 1/ > <<bimplies 0 < 1/b < 1/ Henceforth we will forget bout the xioms nd use ll the properties of inequlities nd ll the lgebric properties tht you hve used before. Remrk 2.5. It is simple consequence of the stndrd properties of < tht there is no smllest positive number, becuse if s is ny positive number then s/2 (for exmple) is smller positive number. Remrk 2.6. The set Q of rtionl numbers, together with 0 nd 1, the opertions of ddition nd multipliction, nd the < reltion, is lso model of the corresponding versions of xioms A1 A13, with rel replced by rtionl. The min points to note re tht 0 nd 1 re of course rtionl, if is rtionl then so re nd 1 (ssuming 0), nd the sum nd product of rtionl numbers is rtionl. Aprt from this, the xioms A1 A13 re stisfied for rtionl numbers becuse they re stisfied for rel numbers (every rtionl number is certinly rel). The set of irrtionl numbers does not stisfy the corresponding versions of A1 A13. For exmple, 0 nd 1 re not irrtionl, nd the sum nd product of irrtionl numbers need not be irrtionl (exmples?). Which of the xioms A1 A13 re stisfied by N? ByZ? Definition 2.7. The bsolute vlue of rel number is defined by { if 0 = if <0 (If we wnted the definition to look more symmetric we could hve considered the cses >0 nd = 0 seprtely. But otherwise there is no rel point to doing this.) It follows by considering the cses 0 nd <0 seprtely tht 6 (2.1) = 2,. Note lso tht ( ) (2.2) <b if nd only if <bnd <b. The following properties lso follow from the properties of inequlities. Theorem 2.8. If nd b re rel numbers then: 1. b = b 2. ± b + b (tringle inequlity) 3. b b Proof. (We cn use ny of the usul propeties of inequlities, together with the definition of.) 6 When we write p q r we men p q nd q r. Similrly for p<q retc.

14 8 2. THE REAL NUMBER SYSTEM 1. We consider the vrious possible cses. If either or b (or both) re zero, then both sides re zero. If >0 nd b>0 then lso b > 0 nd both sides equl b. If >0 nd b<0 then b < 0 nd both sides equl b; similrly if <0 nd b>0. Finlly, if <0 nd b<0 then both sides equl b. Hence one hs equlity in ll cses b is either + b or ( + b), while b is either b or ( b). Thus ± b is one of + b, b, b or + b. The result now follows from (2.1), since ech of these four quntities is + b 3. From (2.2) it is sufficient to prove b b nd ( b ) b. From the tringle inequlity we hve = ( b)+b b + b, nd so (2.3) b b. Since this is true for ny rel numbers nd b, we cn switch nd b to get b b (= b ), i.e. (2.4) ( b ) b. It follows from (2.3) nd (2.4) tht b b. By repeted pplictions of the tringle inequlity it follows tht (2.5) n n, for ny nturl number n. See [Adms, pges 8 11] for more on inequlities Completeness property The Completeness Axiom is introduced. It is true for the rel numbers, but the nlogous result is not true for the rtionls. We define the notion of upper bound (lower bound) nd lest upper bound (gretest lower bound) of set of rel numbers. This is the finl xiom for the rel number system, nd is probbly not one you hve met before. It is more difficult to understnd thn the other properties, but it is essentil in proving mny of the importnt results in clculus. Axioms (Completeness). A14: If A is ny non-empty set of rel numbers with the property tht there is some rel number x such tht x for every A, then there is smllest (or lest) rel number x with this sme property. A is non-empty mens tht A contins t lest one number. Note tht the number x in the xiom need not belong to A. For exmple, if A is the intervl [0, 1) then the smllest (or lest ) number x s bove is 1, but 1 A. On the other hnd, if A =[0,1] then the smllest number x s bove is gin 1, but now 1 A. There is some useful nottion ssocited with the Completeness xiom. Definition 2.9. If A is set of rel numbers nd x is rel number such tht x for every A, then x is clled n upper bound for A. If x is the smllest upper bound then x is clled the lest upper bound or supremum of A. In this cse one write x =l.u.b.a or x = sup A.

15 2.4. COMPLETENESS PROPERTY 9 If x for every A, then x is clled n lower bound for A. If x is the lrgest lower bound then x is clled the gretest lower bound or infimum of A. In this cse one write x =g.l.b.a or x = inf A. Thus the Completeness Axiom sy tht if non-empty set A hs n upper bound then it lso hs lest upper bound. (Remember tht when we sy A hs lest upper bound x we do not require tht x A.) There is n equivlent form of the xiom, which sys: If A is ny non-empty set of rel numbers with the property tht there is some rel number x such tht x for every A, then there is lrgest rel number x with this sme property. In other words if non-empty set A hs lower bound then it lso hs gretest lower bound. It is not too hrd to see tht this form does indeed follow from the Completeness xiom. The trick is to consider, insted of A, the set A := { x:x A}, which is obtined by reflecting A bout 0. A 0 A* Lowerbounds for A correspond under reflection to upperbounds for A,ndg.l.b. corresponds to l. u. b.. IfAis bounded below then A is bounded bove, nd so by the completeness xiom hs l. u. b.. After reflection, this l. u. b. for A gives g. l. b. for A. (To ctully write this out crefully needs some cre you need to just show check the relevnt definitions nd the properties of inequlities tht the first two sentences in this prgrph re indeed correct.) Similrly, the completeness xiom does follow from this equivlent form. Unlike in the cse of A1 A13, we will lwys indicte when we use the completeness xiom (or property ). The completeness xiom implies there re no gps in the rel numbers. For exmple, the rtionl numbers re not model of the corresponding version of A14. This is becuse there re sets of rtionl numbers A which hve the property tht there is some rtionl number x such tht x for every A, but there is no smllest rtionl number x with this sme property. For exmple, let A = { Q :0 nd 2 < 2 } = { Q :0 < 2}. (The first definition for A hs the dvntge tht A is defined without ctully referring to the existence of 2, even s rel number.) There re certinly rtionl numbers x such tht x for every A, just tke x = 23. But we clim there is no smllest such rtionl number. Proof. This clim bsiclly follows from the fct tht 2 is not rtionl, see Theorem 2.11, nd so cnnot be the required rtionl number. On the other hnd, the required rtionl number x cnnot be < 2, since there is lwys rtionl number between ny such x nd 2 (see Theorem 2.16), contrdicting the fct x for every A. Finlly, the required x cnnot be > 2, since there is lwys rtionl number between 2 nd ny such x (see Theorem 2.16), nd this rtionl number mens we hve contrdicted the fct x is the smllest rtionl number such tht x for every A. Remrk We defined n intervl to be set of rel numbers with the property tht it contins t lest two numbers, nd moreover it contins ll rel numbers between ny two of its members. It follows from the completeness xiom tht ny intervl is indeed one of the 9 types described in Section While this my seem obvious, we do need the completeness xiom to prove it, essentilly since the uppper bound of n intervl my not be rtionl.

16 10 2. THE REAL NUMBER SYSTEM For exmple, suppose the intervl I is bounded. Then it hs g. l. b. ndl.u.b. b,sy. If Ind b I then I =[, b] since I contins ll numbers between nd b, nd no others by definition of g. l. b. nd l. u. b.. If I nd b I then I =[, b). The min point is to prove tht if x<bthen x I. But if x I then neither cn ny number greter thn x be in I (by definition of intervl ) nd hence x is n upper bound for I, contrdicting the fct tht b is the lest upper bound. In similr wy, we estblish tht ny intervl is of one of the 9 given forms Properties of the rtionl nd irrtionls We prove tht 2 is irrtionl. We show the Completeness Axiom implies tht there is indeed rel number whose squre is 2; tht there is no rel number greter thn every integer; nd tht for ny positive number (no mtter how smll) there is smller number of the form 1/n for some nturl number n. We prove tht between ny two rel numbers there re n infinite number of rtionls nd n infinite number of irrtionls. The ncient Greeks, in the school of Pythgors, first discovered tht not ll rel numbers re rtionl. This ws considered to be very serious problem, since the Greeks thought in terms of the nturl numbers nd rtios! They proved tht 2 is irrtionl, even though it rose in the very nturl mnner s the length of the hypotenuse of right-ngled tringle whose other two sides were ech of length Theorem is not rtionl. Proof. We rgue by contrdiction. Tht is, we ssume 2=m/n where m nd n re integers. Multiplying numertor nd denomintor by 1 if necessry, we cn tke m nd n to be positive. By cnceling if necessry, we cn reduce to the sitution where m nd n hve no common fctors. Squring both sides of the eqution, we hve for these new m nd n tht 2=m 2 /n 2 nd hence m 2 =2n 2. It follows tht m is even, since the squre of n odd number is odd. (More precisely, if m were odd we could write m =2r+ 1 for some integer r; but then m 2 =(2r+1) 2 = 4r 2 +4r+ 1 = 2(2r 2 +2r) + 1, which is odd, not even.) But since m is even, we cn write m =2p for some integer p, nd hence m 2 =4p 2. Substituting this into m 2 =2n 2 gives 4p 2 =2n 2, nd hence 2p 2 = n 2.

17 2.5. PROPERTIES OF THE RATIONAL AND IRRATIONALS 11 But now we cn rgue s we did before for m, nd deduce tht n is lso even. Thus m nd n both hve the common fctor 2, which contrdicts the fct they hve no common fctors. This contrdiction implies tht our originl ssumption ws wrong, nd so 2isnot rtionl. The fct tht there is rel number 2, i.e. positive rel number x such tht x 2 = 2, is hrdly surprising, but it does not ctully follow from A1 A13. This is becuse the rtionl numbers themselves stisfy A1 A13, but the previous proof shows there is no rtionl number x with this property. Although we give the proof of the following theorem t this stge, it follows much more esily lter on from the Intermedite Vlue Theorem, see Theorem Theorem There exists positive number x with the property tht x 2 =2.We write x = 2. Proof. Let A = { R :0 nd 2 < 2 }. Let x = sup A, which exists by the Completeness xiom, since A is certinly bounded bove, by 23, sy. The existence of such n x requires the completeness xiom. But once we know such number exists, it hs the usul lgebric nd order properties of ny rel number. It is now just mtter of some messy mnipulting of inequlities to rule out x 2 < 2 nd x 2 > 2, thereby showing tht in fct x 2 =2. ( There re three possibilities: x 2 < 2, x 2 =2, x 2 > 2. If x 2 < 2, then by tking y to be slightly bigger number thn x, we still hve y 2 < 2. This is not surprising, but to write it out crefully is little tricky. To do it, let y = x + ε where ε>0 is yet to be chosen. Then y 2 = x 2 + ε(2x + ε). Now choose ε>0so { } ε<min 1, 2 x2. 2x +1 Becuse ε<1, it follows tht 2x + ε<2x+1. Becuse lso ε< 2 x2 it then follows tht 2x+1 y 2 = x 2 + ε(2x + ε) <x 2 +ε(2x +1)<x 2 +(2 x 2 )=2, nd so ) y 2 < 2. It follows tht y A from the definition of A, but then we hve contrdiction to y>x since x ny member of A. If x 2 > 2, then by tking y to be slightly smller number thn x, we lso hve y 2 > 2 (the proof is similr to the bove, exercise). But y 2 > 2 implies y>for every A (since if y nd y 0, then y 2 2, which in turn implies y 2 < 2, contrdiction). Hence y is n upper bound for A, contrdicting the fct tht x is the smllest upper bound for A. Hence x 2 = 2, since we hve ruled out x 2 < 2 nd x 2 > 2. One very useful fct is tht between ny two distinct rel numbers there is rtionl nd n irrtionl number (in fct n infinite number of ech type). We sy the set of rtionls nd the set of irrtionls re both dense in R.

18 12 2. THE REAL NUMBER SYSTEM But first we need Theorem 2.13, which is logiclly equivlent to the sttement for every rel number x there exists nturl number n (which will depend on x) such tht 7 n>x. 1. Why does the theorem imply this fct? 2. Why does this fct imply the theorem? Note tht the ssertion in Theorem 2.13 is not just n lgebric or inequlity property. It is n ssertion bout the non-existence of rel number with certin property (or equivlently, by the previous prgrph, to the existence for ech rel number of (nturl) number with certin property.) For this reson, it is not surprising tht we need the Completeness xiom to prove it. The proof is little unusul, but it is logiclly correct. Theorem 2.13 (Archimiden Property). There is no rel number x with the property tht x n for every nturl number n. Proof. (This is nother proof by contrdiction.) Assume tht there is some number, which we denote by x, such tht x n for every n N. By the Completeness Axiom there is smllest such x; consider this prticulr x. Since n + 1 is nturl number if n is, we must lso hve tht x n + 1 for every nturl number n. But this is the sme s sying x 1 n for every nturl number n. In other words, x hs the property tht x 1 n for every n N!! This contrdicts the fct x ws the lest number with the property x n for every n N. Thus the ssumption t the beginning of the proof ws wrong, nd so there is no rel number x greter thn or equl to every nturl number. The next result is lso importnt. We could just s well hve written x insted of ε, but it is trditionl to write ε or δ for smll positive number. Corollry For ny rel number ε>0there is nturl number n such tht 1/n<ε. Proof. Suppose ε > 0. From the previous remrk, there is nturl number n such tht n>1/ε. This implies 1/n<ε. Remrk A theorem is n importnt result, nd corollry is something which is firly strightforwrd consequence of previous theorem. Theorem Assume <bre rel numbers. Then there is rtionl number x nd n irrtionl number y such tht <x<bnd <y<b. Proof. We hve just seen tht there is nturl number n such tht 1 n <b. First ssume 0. Consider the sequence 1 n, 2 n, 3 n, 4 n, /n 2/n 3/n m/n b > 1/n Since 1/n is less thn the difference between nd b, it follows tht t lest one member m/n of this sequence must lie between nd b. This is the required rtionl number x. 8 7 In fct, there exists n infinite number of such n. 8 To be more precise, use the Completeness Property to first choose x s the lest rel number with the property tht x p/n whenever p/n pnd n nturl numbers. One then shows tht x must itself be of the form p/n for some p, nd then deduces tht <(p+1)/n<b for this prticulr p.

19 2.6. FUNCTIONS 13 If <0, then similr proof works with the sequence... x 3 x 2 1 n, 2 n, 3 n, 4 n,... To find n irrtionl number y between nd b, first choose nturl number n such tht 2/n<b (why is this possible?). Then choose nturl number m s before so tht now <m( 2/n) <b. Since m 2/n is irrtionl (if it were rtionl, nd equled p/q, sy, then it would follow tht 2=np/mq nd so 2 would be rtionl, contrdicting Theorem 2.11) we cn tke this s the required irrtionl number y. The proof is gin similr if <0. Remrk We cn now choose nother rtionl number x 1 between nd the rtionl number x of the theorem, nd then nother rtionl number between nd x 2, etc. This gives n infinite set of rtionl numbers between nd b. x 1 x Similrly there is n infinite number of irrtionl numbers between ny two rel numbers Functions We define the notion of function, its domin nd its rnge. We discuss the ide of dependent nd of n independent vrible. We give number of exmples of functions. We show how functions cn be combined lgebriclly nd by composition to give new functions. One of the most importnt ides in mthemtics is tht of function. Definition A function f from set A into set B is something which ssigns to every number x A unique (i.e. exctly one) number f(x) B. We write f : A B, nd sy f mps (the set) A into (the set) B. b b c d A.. p=f() q r=f(b) s=f(c)=f(d) t B Note tht ech number in A is mpped to exctly one number in B. Different numbers in A my be mpped to the sme number in B, butnumberinacnnot be mpped to more thn one number in B. There my be numbers in B which re not of the form f() for ny A. In this course, A nd B will usully (but not lwys) be sets of numbers, but in lter work it is very importnt to tke more generl functions where A nd B my be much more generl sets of objects. The domin of f, written D(f), is the set A (the set of input vlues of f). The rnge of f, written R(f), is the set of ll numbers of the form f(x) for some x A (the set of output vlues of f). Thus D(f) =A, R(f) B. Here, R(f) ={p, r, s}. (By S T, where S nd T re two sets, we men tht every member of S is lso member of T. Notice tht if S nd T re the sme sets, which mens they hve the sme members, then it is lso true tht S T.) Mny texts sy function from A to B is rule which ssigns to ech member of A member of B. But it is necessry to interpret the mening of the word rule in very brod sense it is not necessry tht function be described by some sort of English or

20 14 2. THE REAL NUMBER SYSTEM mthemticl expression. All tht is ment is tht to ech number x(sy) in the domin there corresponds exctly one number, denoted by f(x). Often function will only be described in n indirect mnner; for exmple it my be the limit of sequence of other functions (see the proof of Theorem 8.2). In other cses we my prove function exists by using rgument by contrdiction to show tht f(x) hs vlue for every x in some given domin. We dopt the convention tht, unless indicted otherwise, the domin of function is the lrgest set of rel numbers for which the definition of the function mkes sense. Thus the domin of the function f described by f(x) = 1/(x 2) is (unless stted otherwise) the set of ll rel numbers other thn 2. We write this set s { x : x 2}. See lso Exmples 2,3,5 below. In order to describe function completely, we need to give both the domin nd the rule. Two functions f 1 nd f 2 re sid to be the sme function, or to be equl, if they hve the sme domin, nd if f 1(x) =f 2(x) for ll x in the domin. The grph of f is the set of ll points (x, f(x)) such tht x D(f). For exmple, the grph of the function f(x) =x 2 is Nottion If we denote n rbitrry input vlue of function f by x nd the corresponding output vlue f(x) byy,wesyxis the dependent vrible nd y is the independent vrible. We write y = f(x) nd sy y equls f of x. Besides letting y denotes n output vlue s bove, it is lso often convenient in computtions nd pplictions to let y denote the ctul function f itself. In this cse we sometimes write y = y(x) to indicte tht the function y hs output vlue y(x) for input vlue x. However, when we re looking t more theoreticl questions, this cn led to confusion nd mbiguity, nd so in those circumstnces we will usully void this prctice. Exmple Consider the function f defined by f(x) =x 2. Unless we sy otherwise the domin here is R nd the rnge is then [0, ). We cn write f : R R, f : R [0, ), or even f : R [ 3, ). Note tht the function f defined by f(x) =x 2 is exctly the sme s the function f defined by f(y) = y 2 or by f() = 2. We sy x, y, re dummy vribles. We cll f the squring function. It is lso the sme s the function g defined by g(x) =x 2.

21 2.6. FUNCTIONS 15 Consider the function h defined by h(x) =x 2, x 1. This is the function whose domin is [1, ) nd which ssigns to ech x in the domin the number x 2. Thus the two functions f nd h re not equl since their domins re not equl. 2. The function f defined by f(x) = x2 +1 x 1 hs domin {x : x 1}. This is the sme set s (, 1) (1, ), the union of (, 1) nd (1, ). It is lso exctly the sme set s {y : y 1} 3. The function f defined by f(x) = xhs domin {x : x 0} =[0, ), unless otherwise indicted. 4. The function defined by { x x rtionl f(x) = x x irrtionl Its grph looks something like Of course this is somewht misleding, s both the rtionls nd the irrtionls re dense in R. 5. The function defined by f(x) = x2 1 x 1 hs domin {x : x 1}ccording to our conventions. For ech x in the domin we see tht f(x) =x+ 1. However, f is not the sme s the function g defined by g(x) =x+1, 6. since the domin of g, unless we specify otherwise, is ll of R. Of course we could extend the domin of f by defining f(1) = 2, nd the extended function would then be the sme s g.

22 16 2. THE REAL NUMBER SYSTEM The function g with domin [0, 1] defined by 1/2 if x =1/2 1/4 if x =1/4,3/4 1/8 if x =1/8,3/8,5/8,7/8 g(x) =. 1/2 k if x =1/2 k,3/2 k,5/2 k,...,(2 k 1)/2 k. g(x) = 0 otherwise. In other words, { 1/2 k if x = p/2 k where k N, p odd nd 1 p<2 k g(x) = 0 otherwise, for 0 x 1. We cn dd, subtrct, divide nd multiply functions to get new functions. We cn lso multiply function by rel number to get new function. Definition Let f nd g be functions nd c be rel number. Then we define functions f + g, f g, fg, f/g nd cf by (f + g)(x) =f(x)+g(x) (f g)(x) =f(x) g(x) (fg)(x) =f(x)g(x) (f/g)(x) =f(x)/g(x) (cf)(x) =cf(x). We restrict x in ll cses to be member of the domins of both f nd g, nd in the fourth cse we lso require tht g(x) 0. We cn lso combine two functions by tking the output of one to be the input of the other. Definition If f nd g re two functions, then the composition of f nd g is defined by (f g)(x) =f(g(x)) for ll x D(f) such tht f(x) D(g). For exmple, if h(x) = g(x) (for ll x D(f)) then h is the composition of the bsolute vlue function f, given by f(y) = y, with g. See [Adms, pges 33 36] for further discussion nd exmples.

23 2.7. MATHEMATICAL INDUCTION Mthemticl induction Suppose we wnt to prove some sttment is true for every integer n greter thn or equl to some fixed integer n 0. We often do this with the following. Principle of Mthemticl Induction. Suppose tht for some sttement P (n) bout integers n we know The sttement P (n 0) is true; if the sttement P (n) is true for some integer n n 0 then the sttement P (n +1) is lso true. In this cse, the sttement P (n) is true for ll integers n n 0. For n exmple, see the proof of Theorem 6.6. The Principle of Mthemticl Induction is esy to justify informlly. By the first ssumption, P (n 0) is true. By the second ssumption pplied with n replced by n 0 it follows P (n 0 + 1) is true. By the second ssumption pplied with n replced by n 0 +1it then follows P (n 0 + 2) is true. eetc. Remrk If one wnts to give more creful proof of the Principle of Mthemticl Induction, then one first needs to give more creful definition of the set N. One cn do this by defining N to be the set of ll rel numbers which belong to every inductive set S, where set S is defined to be inductive if it hs the property tht 1 S nd tht if x S for some rel number x, then lso x +1 S. One then shows tht N itself is inductive nd so is the smllest inductive set. In order now to prove the Principle of Mthemticl Induction in cse n 0 = 1, suppose the two ssumptions of the Principle re true nd let T be the set of ll integers n for which P (n) is true. Then T is inductive (why?), nd so every member of N is lso in T s N is the smllest inductive set (conversely, every member of T is in N s T ws lredy ssumed to be set of integers). The proof in cse n 0 > 1 is now esy. Just tke T to be the set of integers {1,...,n 0 1} together with the set of integers n n 0 for which P (n) is true. There is lso stronger version clled the Principle of Complete Mthemticl Induction, in which we my ssume not only tht P (n) is true, but lso tht P (n 0),...,P(n 1) re true. Principle of Complete Mthemticl Induction. Suppose tht for some sttement P (n) bout integers n we know The sttement P (n 0) is true; if the sttements P (n 0),...,P(n) re true for some integer n n 0 then the sttement P (n +1) is lso true. In this cse, the sttement P (n) is true for ll integers n n 0. Once gin, it is esy to justify informlly. By the first ssumption, P (n 0) is true. By the second ssumption pplied with n replced by n 0 it follows P (n 0 + 1) is true. By the second ssumption pplied with n replced by n it then follows P (n 0 + 2) is true. etc. The rigorous proof is similr to tht in the cse of ordinry induction. Exercise The Fiboncci sequence (see Section 4.1) is defined by 1 =1, 2=1, n= n 1+ n 2if n 2. Prove by complete induction tht n = ( ) 1+ n ( ) n Fiboncci first cme up with this sequence s model for rbbit popultion growth. Let n be the number of pirs born in the nth month. He ssumed there ws one pir born the first month nd one in the second. After tht he ssumed tht in the nth month there is one pir born for ech pir born in the previous month nd for ech pir born two months go.

24 18 2. THE REAL NUMBER SYSTEM The number of seeds in ny ring (lyer) of pine cone is member of the Fiboncci sequence. There is n entire journl devoted to Fiboncci sequences, The Fiboncci Qurterly Fields The rel numbers nd the irrtionls, s well s the integers modulo fixed prime number, form field. Any set S, together with two opertions nd nd two members 0 nd 1 of S, which stisfies the corresponding versions of A1 A9, is clled field. Thus R (together with the opertions of ddition nd multipliction nd the two rel numbers 0 nd 1) is field. The sme is true for Q, but not for Z since the nlogue of xiom A8 does not hold, why?. An interesting exmple of field is the set F p = {0, 1,...,p 1} for ny fixed prime p, together with ddition nd multipliction defined modulo p ; i.e. one performs the usul opertions of ddition nd multipliction, but then tkes the reminder fter dividing by p. Thus for p = 5 one hs: It is not too hrd to convince yourself tht the nlogues of xioms A1 A9 hold for ny prime p. The xiom which fils if p is not prime is A8, why?. Note tht since F p is field, we cn solve simultneous liner equtions in F p. The fields F p re very importnt in coding theory nd cryptogrphy Deductions from the xioms We sketch how the usul lgebric nd order properties of the rel numbers follow rigorously from the xioms A1 A13. Ech line in the following proofs will 1. be n exmple of one (or occsionlly more) of xioms A1 A9; 2. or be previously proved result; 3. or follow from previously proved results by rules of logic 9 including the properties of equlity. Proof of Theorem 2.3. Fill in the missing steps, nd go through the proofs line by line nd indicte wht is used in ech step. 1. Write out your own proof, following the ides of the proof of the similr result for ddition. 2. The trick here is to use the fct 0+0 = 0 (from A3), together with the distributive xiom A9. The proof is s follows: Proof. One hs (0 + 0)= 0 But the left side equls 0+0 nd the right side equls Hence = Hence 0 = 0. 9 For exmple, if we prove tht some sttement P implies nother sttement Q, nd if we lso prove tht P is true, then it follows from rules of logic tht Q is true.

25 2.9. DEDUCTIONS FROM THE AXIOMS Proof. We wnt to show ( ) =. By ( ) we men the negtive of, nd hence by A4 we know tht ( ) is the unique number which when dded to gives In other words, ( )+( ( )) = 0 nd if ( )+x=0 then we must hve x = ( ). Thus if we cn lso show tht (2.6) ( )+=0 then it will follow tht = ( )!! But (2.6) is just A3, nd so we re done. The proof cn be written more precisely s follows: Proof. From the second equlity in A4 one hs (2.7) ( )+=0. From the first equlity in A4 with there replced by ( ) one hs ( ) + ( ( )) = 0, nd moreover, if ( )+x= 0 then x = ( ). Hence, from (2.7), = ( ). 4. Write out your own proof, long similr lines to the preceding proof. 5. (As in the proof of 2) it is sufficient to show ( 1) + =0(why?) The proof is s follows: Proof. ( 1) + =( 1) +1 = (( 1)+ 1)(two xioms were used for this step) = 0 =0 Hence =( 1) from the uniqueness prt of A4. 6. Proof. ( b) =(( 1)b) =(( 1))b =(( 1))b =( 1)(b) = (b) Prove the second equlity yourself. 7. Prove this yourself using, in prticulr, 4 nd A9 8. Proof. ( )( b) =(( 1))( b) =( 1)(( b)) = (( b)) = ( (b)) = b 9. Proof. First note tht (/c)(b/d) =(c 1 )(bd 1 ) nd (b)/(cd) =(b)(cd) 1. But (c 1 )(bd 1 )=(b)(c 1 d 1 ) (fill in the steps to prove this equlity; which involve number of pplictions of A5 nd A6). If we cn show tht c 1 d 1 =(cd) 1 then we re done. Since, by A8, (cd) 1 is the unique rel number such tht (cd)(cd) 1 =1,itis 10 Since cn represent ny number in A4, we cn replce in A4 by. This might seem strnge t first, but it is quite legitimte.

26 20 2. THE REAL NUMBER SYSTEM sufficient to show 11 tht (cd)(c 1 d 1 )=1. Do this; use A5 A8 Importnt Remrk: There is tricky point in the preceding tht is esy to overlook; but will introduce some importnt ides bout logicl resoning. We used the number (cd) 1. To do this we need to know tht cd 0. We know tht c 0nd d 0nd we wnt to prove tht cd 0. This is equivlent to proving tht if cd 0 is flse, i.e. if cd = 0, then t lest one of c 0 nd d 0 is flse, i.e. t lest one of c =0ord= 0 is true. In other words, we wnt to show tht if cd =0then either c =0or d =0(possibly both). The rgument is written out s follows: Clim: Ifc 0 nd d 0 then cd 0 Proof. We will estblish the clim by proving tht if cd = 0 then c =0or d=0. 12 There re two possibilities concerning c; either c = 0, in which cse we re done or c 0. But in this cse, since cd = 0, it follows c 1 (cd) =c 1 0 nd so d =0 why?; fill in the steps Thus we hve shown tht if cd = 0 then c =0ord= 0. Equivlently, if c 0 nd d 0, then cd Exercise HINT: We wnt to prove c 1 + bd 1 =(d + bc)(cd) 1. First prove tht (c 1 + bd 1 )(cd) =d + bc. Then deduce the required result. And now the proofs of Theorem Existence nd Uniqueness of the Rel Number System We begn by ssuming tht the rel number system stisfies Axioms A1 A14. But it is possible to go bck even further nd begin with xioms for set theory, nd then prove the existence of set of objects stisfying Axioms A1 A14. This is done by first constructing the nturl numbers, then the integers, then the rtionls, nd finlly the rels. The nturl numbers re constructed s certin types of sets, the negtive integers re constructed from the nturl numbers, the rtionls re constructed s sets of ordered pirs s in [Birkhoff nd McLne, Chpter II-2]. Finlly, the rels re then constructed by the method of Dedekind Cuts s in in [Birkhoff nd McLne, Chpter IV-5] or the method of Cuchy Sequences s in [Spivk, Chpter 28]. The rel number system is uniquely chrcterised by Axioms A1 A14, in the sense tht ny two structures stisfying the xioms re essentilly the sme up to renming of 11 When we sy it is sufficient to show... we men tht if we cn show... then the result we wnt will follow. 12 Note; in mthemtics, if we sy *** or ### (is true) then we lwys include the possibility tht both *** nd ### re true.

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