Lecture Notes in Calculus. Raz Kupferman Institute of Mathematics The Hebrew University

Transcription

1 Lecture Notes in Clculus Rz Kupfermn Institute of Mthemtics The Hebrew University July 10, 2013

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3 Contents 1 Rel numbers Axioms of field Axioms of order (s tught in 2009) Axioms of order (s tught in 2010, 2011) Absolute vlues Specil sets of numbers The Archimeden property Axiom of completeness Rtionl powers Rel-vlued powers Addendum Functions Bsic definitions Grphs Limits Limits nd order Continuity Theorems bout continuous functions Infinite limits nd limits t infinity Inverse functions Uniform continuity

4 ii CONTENTS 3 Derivtives Definition Rules of differentition Another look t derivtives The derivtive nd extrem Derivtives of inverse functions Complements Tylor s theorem Integrtion theory Definition of the integrl Integrtion theorems The fundmentl theorem of clculus Riemnn sums The trigonometric functions The logrithm nd the exponentil Integrtion methods Sequences Bsic definitions Limits of sequences Infinite series

5 CONTENTS iii Foreword 1. Texbooks: Spivk, Meizler, Hochmn.

6 iv CONTENTS

7 Chpter 1 Rel numbers In this course we will cover the clculus of rel univrite functions, which ws developed during more thn two centuries. The pioneers were Isc Newton ( ) nd Gottfried Wilelm Leibniz ( ). Some of their followers who will be mentioned long this course re Jkob Bernoulli ( ), Johnn Bernoulli ( ) nd Leonhrd Euler ( ). These first two genertions of mthemticins developed most of the prctice of clculus s we know it; they could integrte mny functions, solve mny differentil equtions, nd sum up lrge number of infinite series using welth of sophisticted nlyticl techniques. Yet, they were sometimes very vgue bout definitions nd their theory often lid on shky grounds. The sound theory of clculus s we know it tody, nd s we re going to lern it in this course ws mostly developed throughout the 19th century, notbly by Joseph-Louis Lgrnge ( ), Augustin Louis Cuchy ( ), Georg Friedrich Bernhrd Riemnn ( ), Peter Gustv Lejeune-Dirichlet ( ), Joseph Liouville ( ), Jen- Gston Drboux ( ), nd Krl Weierstrss ( ). As clculus ws being estblished on firmer grounds, the theory of functions needed thorough revision of the concept of rel numbers. In this context, we should mention Georg Cntor ( ) nd Richrd Dedekind ( ). 1.1 Axioms of field Even though we hve been using numbers s n elementry notion since first grde, rigorous course of clculus should strt by putting even such bsic con-

8 2 Chpter 1 cept on xiomtic grounds. The set of rel numbers will be defined s n instnce of complete, ordered.(שדה סדור שלם ) field Definition 1.1 A field F (ש ד ה ) is non-empty set on which two binry opertions re defined 1 : n opertion which we cll ddition, nd denote by +, nd n opertion which we cll multipliction nd denote by (or by nothing, s in b = b). The opertions on elements of field stisfy nine defining properties, which we list now. 1. Addition is ssocitive :(קיבוצי ) for ll, b, c F, ( + b) + c = + (b + c). 2 It should be noted tht ddition ws defined s binry opertion. As such, there is no -priori mening to + b + c. In fct, the mening is mbiguous, s we could first dd + b nd then dd c to the sum, or conversely, dd b + c, nd then dd the sum to. The first xiom sttes tht in either cse, the result is the sme. Wht bout the ddition of four elements, + b + c + d? Does it require seprte xiom of equivlence? We would like the following dditions (( + b) + c) + d ( + (b + c)) + d + ((b + c) + d) + (b + (c + d)) ( + b) + (c + d) to be equivlent. It is esy to see tht the equivlence follows from the xiom of ssocitivity. Likewise (lthough it requires some non-trivil work), we 1 A binry opertion in F is function tht tkes n ordered pir of elements in F nd returns n element in F. Tht is, to every, b F corresponds one nd only one + b F nd b F. In forml nottion, (, b F)(!c F) : (c = + b). 2 A few words bout the equl sign; we will tke it literlly to mens tht the expressions on both sides re the sme. More precisely, equlity is n equivlence reltion, concept which will be explined further below.

9 Rel numbers 3 cn show tht the n-fold ddition is defined unmbiguously n 2. Existence of n dditive-neutrl element נייטרלי לחיבור ) :(איבר there exists n element 0 F such tht for ll F + 0 = 0 + =. 3. Existence of n dditive inverse נגדי ) :(איבר for ll F there exists n element b F, such tht + b = b + = 0. The would like to denote the dditive inverse of, s customry, ( ). There is however one problem. The xiom ssumes the existence of n dditive inverse, but it does not ssume its uniqueness. Suppose there were two dditive inverses: which one would be denote ( )? With the first three xioms there re few things we cn show. For exmple, tht + b = + c implies b = c. The proof requires ll three xioms. Denote by d n dditive inverse of. Then, + b = + c d + ( + b) = d + ( + c) (existence of inverse) (d + ) + b = (d + ) + c (ssocitivity) 0 + b = 0 + c (property of inverse) b = c. (0 is the neutrl element) It follows t once tht every element F hs unique dditive inverse, for if b nd c were both dditive inverses of, then + b = 0 = + c, from which follows tht b = c. Hence we cn refer to the dditive inverse of, which justifies the nottion ( ). We cn lso show tht there is unique element in F stisfying the neutrl property (it is not n priori fct). Suppose there were, b F such tht + b =.

10 4 Chpter 1 Then dding ( ) (on the left!) nd using the lw of ssocitivity we get tht b = 0. Comment: We only postulte the opertions of ddition nd multipliction. Subtrction is short-hnd nottion for the ddition of the dditive inverse, b def = + ( b)..(חבורה ) Comment: A set stisfying the first three xioms is clled group By themselves, those three xioms hve mny implictions, which cn fill n entire course. 4. Addition is commuttive :(חילופי ) for ll, b F + b = b +. With this lw we finlly obtin tht ny (finite!) summtion of elements of F cn be re-rrnged in ny order. Exmple: Our experience with numbers tells us tht b = b implies tht = b. There is no wy we cn prove it using only the first four xioms! Indeed, ll we obtin from it tht + ( b) = b + ( ) (given) ( + ( b)) + (b + ) = (b + ( )) + ( + b) (commuttivity) + (( b) + b) + = b + (( ) + ) + b = b b + = b + b, but how cn we deduce from tht tht = b? (ssocitivity) (property of dditive inverse) (property of neutrl element) With very little comments, we cn stte now the corresponding lws for multipliction: 5. Multipliction is ssocitive: for ll, b, c F, (b c) = ( b) c.

11 Rel numbers 5 6. Existence of multiplictive-neutrl element: there exists n element 1 F such tht for ll F, 1 = 1 =. 7. Existence of multiplictive inverse הופכי ) :(איבר for every 0 there exists n element 1 F such tht 3 1 = 1 = 1. The condition tht 0 hs strong implictions. For exmple, from the fct tht b = c we cn deduce tht b = c only if 0. Comment: Division is defined s multipliction by the inverse, /b def = b Multipliction is commuttive: for every, b F, b = b. The lst xiom reltes the ddition nd multipliction opertions: 9. Distributive lw הקיבוץ ) :(חוק for ll, b, c F, (b + c) = b + c. We cn revisit now the b = b exmple. We proceed, + = b + b = 1 b + 1 b (property of neutrl element) (1 + 1) = (1 + 1) b (distributive lw). We would be done if we knew tht , for we would multiply both sides by (1 + 1) 1. However, this does not follow from the xioms of field! 3 Here gin, the uniqueness of the multiplictive inverse hs to be proved, without which there is no justifiction to the nottion 1.

12 6 Chpter 1 Exmple: Consider the following set F = {e, f } with the following properties, + e f e e f f f e e f e e e f e f It tkes some explicit verifiction to check tht this is indeed field (in fct the smllest possible field), with e being the dditive neutrl nd f being the multiplictive neutrl (do you recognize this field?). Note tht e f = e + ( f ) = e + f = f + e = f + ( e) = f e, nd yet e f, which is then no wonder tht we cn t prove, bsed only on the field xioms, tht f e = e f implies tht e = f. With 9 xioms t hnd, we cn strt proving theorems tht re stisfied by ny field. Proposition 1.1 For every F, 0 = 0. Proof : Using sequentilly the property of the neutrl element nd the distributive lw: 0 = (0 + 0) = Adding to both sides ( 0) we obtin 0 = 0. Comment: Suppose it were the cse tht 1 = 0, i.e., tht the sme element of F is both the dditive neutrl nd the multiplictive neutrl. It would follows tht for every F, = 1 = 0 = 0, which mens tht 0 is the only element of F. This is indeed field ccording to the xioms, however very boring one. Thus, we rule it out nd require the field to stisfy 0 1.

13 Rel numbers 7 Comment: In principle, every lgebric identity should be proved from the xioms of field. In prctice, we will ssume henceforth tht ll known lgebric mnipultions re vlid. For exmple, we will not bother to prove tht b = (b ) (even though the proof is very esy). The only exception is (of course) if you get to prove n lgebric identity s n ssignment. Exercise 1.1 Let F be field. Prove, bsed on the field xioms, tht ➀ If b = for some field element 0, then b = 1. ➁ 3 b 3 = ( b)( 2 + b + b 2 ). ➂ If c 0 then /b = (c)/(bc). ➃ If b, d 0 then Justify every step in your proof! b + c d d + bc =. bd 1.2 Axioms of order (s tught in 2009) The rel numbers contin much more structure thn just being field. They lso form n ordered set. The property of being ordered cn be formlized by three xioms: Definition 1.2 ס דוּר ) (שׁ ד ה A field F is sid to be ordered if it hs distinguished subset P F, which we cll the positive elements, such tht the following three properties re stisfied: 1. Trichotomy: every element F stisfies one nd only one of the following properties. Either (i) P, or (ii) ( ) P, or (iii) = Closure under ddition: if, b P then + b P. 3. Closure under multipliction: if, b P then b P.

14 8 Chpter 1 We supplement these xioms with the following definitions: < b mens b P > b mens b < b mens < b or = b b mens > b or = b. Comment: > 0 mens, by definition, tht 0 = P. Similrly, < 0 mens, by definition, tht 0 = ( ) P. We cn then show number of well-known properties of rel numbers: Proposition 1.2 For every, b F either (i) < b, or (ii) > b, or (iii) = b. Proof : This is n immedite consequence of the trichotomy lw, long with our definitions. We re sked to show tht either (i) b P, or (ii) b = ( b) P, or (iii) tht b = 0. Tht is, we re sked to show tht b stisfies the trichotomy, which is n xiom. Proposition 1.3 If < b then + c < b + c. Proof : (b + c) ( + c) = b P. Proposition 1.4 (Trnsitivity) If < b nd b < c then < c. Proof : c = (c b) + (b ) } {{ }} {{ }. in P in P } {{ } in P by closure under ddition

15 Rel numbers 9 Proposition 1.5 If < 0 nd b < 0 then b > 0. Proof : We hve seen bove tht, b < 0 is synonymous to ( ), ( b) P. By the closure under multipliction, it follows tht ( )( b) > 0. It remins to check tht the xioms of field imply tht b = ( )( b). Corollry 1.1 If 0 then 2 > 0. Proof : If 0, then by the trichotomy either > 0, in which cse 2 > 0 follows from the closure under multipliction, or < 0 in which cse 2 > 0 follows from the previous proposition, with = b. Corollry > 0. Proof : This follows from the fct tht 1 = 1 2. Another corollry is tht the field of complex numbers (yes, this is not within the scope of the present course) cnnot be ordered, since i 2 = ( 1), which implies tht ( 1) hs to be positive, hence 1 hs to be negtive, which violtes the previous corollry. Proposition 1.6 If < b nd c > 0 then c < bc. Proof : It is given tht b P nd c P. Then, bc c = (b ) } {{ } in P c }{{} in P } {{ } in P by closure under multipliction Exercise 1.2 Let F be n ordered field. Prove bsed on the xioms tht.

16 10 Chpter 1 ➀ < b if nd only if ( b) < ( ). ➁ If < b nd c > d, then c < b d. ➂ If > 1 then 2 >. ➃ If 0 < < 1 then 2 <. ➄ If 0 < b nd 0 c < d then c < bd. ➅ If 0 < b then 2 < b 2. Exercise 1.3 Prove using the xioms of ordered fields nd using induction on n tht ➀ If 0 x < y then x n < y n. ➁ If x < y nd n is odd then x n < y n. ➂ If x n = y n nd n is odd then x = y. ➃ If x n = y n nd n is even then either x = y of x = y. ➄ Conclude tht x n = y n if nd only if x = y or x = y. Exercise 1.4 Let F be n ordered field, with multiplictive neutrl term 1; we lso denote = 2. Prove tht ( + b)/2 is well defined for ll, b F, nd tht if < b, then < + b 2 < b. Exercise 1.5 Show tht there cn be no ordered field tht hs finite number of elements. 1.3 Axioms of order (s tught in 2010, 2011) The rel numbers contin much more structure thn just being field. They lso form n ordered set. The property of being ordered cn be formlized by four xioms: Definition 1.3 ס דוּר ) (שׁ ד ה A field F is sid to be ordered if there exists reltion < ( reltion is property tht every ordered pir of elements either stisfy or not), such tht the following four properties re stisfied:

17 Rel numbers 11 O1 Trichotomy: every pir of elements, b F stisfies one nd only one of the following properties. Either (i) < b, or (ii) b <, or (iii) = b. O2 Trnsitivity: if < b nd b < c then < c. O3 Invrince under ddition: if < b then + c < b + c for ll c F. O4 Invrince under multipliction by positive numbers: if < b then c < bc for ll 0 < c. Comment: The set of numbers for which 0 < re clled the positive numbers. The set of numbers for which < 0 re clled the negtive numbers. We supplement these xioms with the following definitions: (2 hrs, 2013) > b mens b < b mens < b or = b b mens > b or = b. Proposition 1.7 The set of positive numbers is closed under ddition nd multipliction, nmely If 0 <, b then 0 < + b nd 0 < b. Proof : Closure under ddition follows from the fct tht 0 < implies tht b = 0 + b (neutrl element) < + b, (O3) nd b < + b follows from trnsitivity. Closure under multipliction follows from the fct tht 0 = 0 (proved) < b. (O4) (2 hrs, 2010) (2 hrs, 2011)

18 12 Chpter 1 Proposition < if nd only if ( ) < 0, i.e., number is positive if nd only if its dditive inverse is negtive. Proof : Suppose 0 <, then ( ) = 0 + ( ) (neutrl element) < + ( ) (O3) = 0 (dditive inverse) i.e., ( ) < 0. Conversely, if ( ) < 0 then 0 = ( ) + (dditive inverse) < 0 + (O3) = (neutrl element) Corollry 1.3 The set of positive numbers is not empty. Proof : Since by ssumption 1 0, it follows by O1 tht either 0 < 1 or 1 < 0. In the ltter cse 0 < ( 1), hence either 1 or ( 1) is positive. Proposition < 1. Proof : By trichotomy, either 0 < 1, or 1 < 0, or 0 = 1. The third possibility is ruled out by ssumption. Suppose then, by contrdiction, tht 1 < 0. Then for every 0 < (nd we know tht t lest one such exists), = 1 < 0 = 0, i.e., < 0 which violtes the trichotomy xiom. Hence 0 < 1.

19 Rel numbers 13 Proposition < if nd only if 0 < 1, i.e., number is positive if nd only if its multiplictive inverse is lso positive. Proof : Let 0 < nd suppose by contrdiction tht 1 becuse 1 = 1). Then, < 0 (it cn t be zero 1 = 1 < 0 = 0, which is contrdiction. Proposition 1.11 If < 0 nd b < 0 then b > 0. Proof : We hve seen bove tht, b < 0 implies tht 0 < ( ), ( b). By the closure under multipliction, it follows tht ( )( b) > 0. It remins to check tht the xioms of field imply tht b = ( )( b). Corollry 1.4 If 0 then 2 > 0. Proof : If 0, then by the trichotomy either > 0, in which cse 2 > 0 follows from the closure under multipliction, or < 0 in which cse 2 > 0 follows from the previous proposition, with = b. Another corollry is tht the field of complex numbers (which is not within the scope of the present course) cnnot be ordered, since i 2 = ( 1), which implies tht ( 1) hs to be positive, hence 1 hs to be negtive, which is violtion. 1.4 Absolute vlues Definition 1.4 For every element in n ordered field F we define the bsolute,(ערך מוחלט ) vlue 0 = ( ) < 0.

20 14 Chpter 1 We see right wy tht = 0 if nd only if = 0, nd otherwise > 0. Proposition 1.12 (Tringle inequlity שיוויון המשולש ) ((אי For every, b F, + b + b. Proof : We need to exmine four cses (by trichotomy): 1. 0 nd b nd b nd b nd b 0. In the first cse, = nd b = b, hence + b = + b = + b. (Hey, does this men tht + b + b?) In the fourth cse, = ( ), b = ( b), nd 0 < ( + b), hence + b = ( + b) = ( ) + ( b) = + b. It remins to show the second cse, s the third cse follows by interchnging the roles of nd b. Let 0 nd b 0, i.e., = ( ) nd b = b. We need to show tht + b ( ) + b. We cn divide this cse into two sub-cses. Either + b 0, in which cse If + b < 0, then + b = + b ( ) + b = + b. + b = ( + b) = ( ) + ( b) ( ) + b = + b, which completes the proof 4. (3 hrs, 2013) 4 We hve used the fct tht 0 implies ( ) nd 0 implies ( ). This is very esy to show. For exmple, nd ( ) < follows from trnsitivity. > 0 = + ( ) > 0 + ( ) = 0 > ( ),

21 Rel numbers 15 Comment: By replcing b by ( b) we obtin tht b + b. The following version of the tringle inequlity is lso useful: Proposition 1.13 (Reverse tringle inequlity) For every, b F, b + b. Proof : By the tringle inequlity,, c F, c + c. Set c = b +, in which cse b + b +, or b b +. By interchnging the roles of nd b, b + b. It follows tht 5 b + b, Proposition 1.14 For every, b, c F, + b + c + b + c. 5 Here we used the fct tht ny property stisfied by nd ( ) is lso stisfied by.

22 16 Chpter 1 Proof : Use the tringle inequlity twice, + b + c = + (b + c) + b + c + b + c. (2 hrs, 2009) Exercise 1.6 Let min(x, y) nd mx(x, y) denote the smllest nd the lrgest of the two rguments, respectively. Show tht mx(x, y) = 1 2 (x + y + x y ) nd min(x, y) = 1 (x + y x y ) Specil sets of numbers Wht kind of sets cn hve the properties of n ordered field? We strt by introducing the nturl numbers הטבעיים ),(המספרים by tking the element 1 (which exists by the xioms), nd nming the numbers 1 + 1, , , etc. Tht these numbers re ll distinct follows from the xioms of order (ech number is greter thn its predecessor becuse 0 < 1). Tht we give them nmes, such s 2, 3, 4, nd my represent them using deciml system is immteril. We denote the set of nturl numbers by N. Clerly the nturl numbers do not form field (for exmple, there re no dditive inverses becuse ll the numbers re positive). We cn then ugment the set of nturl numbers by dding the numbers 0, 1, (1 + 1), ( ), etc. This forms the set of integers,(השלמים ) which we denote by Z. The integers form commuttive group with respect to ddition, but re not field since multiplictive inverses do not exist 6. 6 How do we know tht (1 + 1) 1, for exmple, is not n integer? We know tht it is positive nd non-zero. Suppose tht it were true tht 1 < (1 + 1) 1, then which is violtion of the xioms of order = (1 + 1) 1 < (1 + 1) (1 + 1) 1 = 1,

23 Rel numbers 17 The set of numbers cn be further ugmented by dding ll integer quotients, m/n, n 0. This forms the set of rtionl numbers, which we denote by Q. It should be noted tht the rtionl numbers re the set of integer quotients modulo n equivlence reltion 7. A rtionl number hs infinitely mny representtions s the quotient of two integers. Proposition 1.15 Let b, d 0. Then b = c d if nd only if d = bc. Proof : Multiply/divide both sides by bd (which cnnot be zero if b, d 0). It cn be checked tht the rtionl numbers form n ordered field. (4 hrs, 2010) (4 hrs, 2011) In this course we will not tech (xiomticlly) mthemticl induction, nor inductive definitions (lso clled recursive definitions); we will ssume these concepts to be understood. As n exmple of n inductive definition, we define for ll Q nd n N, the n-th power, by setting, 1 = nd k+1 = k. As n exmple of n inductive proof, consider the following proposition: Proposition 1.16 Let, b > 0 nd n N. Then > b if nd only if n > b n. 7 A few words bout equivlence reltions שקילות ).(יחס An equivlence reltion on set S is property tht ny two elements either hve or don t. If two elements, b hve this property, we sy tht is equivlent to b, nd denote it by b. An equivlence reltion hs to be symmetric ( b implies b ), reflexive ( for ll ) nd trnsitive ( b nd b c implies c). Thus, with every element S we cn ssocite n equivlence clss שקילות ),(מחלקת which is {b : b }. Every element belongs to one nd only one equivlence clss. An equivlence reltion prtitions S into collection of disjoint equivlence clsses; we cll this set S modulo the equivlence reltion, nd denote it S/.

24 18 Chpter 1 Proof : We proceed by induction. This is certinly true for n = 1. Suppose this were true for n = k. Then > b implies k > b k implies k+1 = k > b k > b b k = b k+1. Conversely, if n > b n then b would imply tht n b n, hence > b. Another inequlity tht we will need occsionlly is: Proposition 1.17 (Bernoulli inequlity) For ll x > ( 1) nd n N, (1 + x) n 1 + nx. Proof : The proof is by induction. For n = 1 both sides re equl. Suppose this were true for n = k. Then, (1 + x) k+1 = (1 + x)(1 + x) k (1 + x)(1 + kx) = 1 + (k + 1)x + kx (k + 1)x, where in the left-most inequlity we hve used explicitly the fct tht 1 + x > 0. Exercise 1.7 The following exercise del with logicl ssertions, nd is in preprtion to the type of ssertions we will be deling with ll long this course. For ech of the following sttements write its negtion in hebrew, without using the word no. Then write both the sttement nd its negtion, using logicl quntifiers like nd. ➀ For every integer n, n 2 n holds. ➁ There exists nturl number M such tht n < M for ll nturl numbers n. ➂ For every integer n, m, either n m or n m. ➃ For every nturl number n there exist nturl numbers, b,, such tht b < n, 1 < nd n = b. (3 hrs, 2009)

25 1.6 The Archimeden property Rel numbers 19 We re iming t constructing n entity tht mtches our notions nd experiences ssocited with numbers. Thus fr, we defined n ordered field, nd showed tht it must contin set, which we clled the integers, long with ll numbers expressible t rtios of integers the rtionl numbers Q. Whether it contins dditionl elements is left for the moment unnswered. The point is tht Q lredy hs the property of being n ordered field. Before proceeding, we will need the following definitions: חסום) Definition 1.5 A set of elements A F is sid to be bounded from bove (מלעיל if 8 there exists n element M F, such tht M for ll A, i.e., A is bounded from bove ( M F) : ( A)( M). Such n element M is clled n upper bound מלעיל ) (חסם for A. A is sid to be bounded from below מלרע ) (חסום if ( m F) : ( A)( m). Such n element m is clled lower bound מלרע ) (חסם for A. A set is sid to be bounded (חסום ) if it is bounded both from bove nd below. Note tht if set is upper bounded, then the upper bound is not unique, for if M is n upper bound, so re M + 1, M + 2, nd so on. Proposition 1.18 A set A is bounded if nd only if ( M F) : ( A)( M). Proof : 9 Suppose A is bounded, then ( M 1 F) : ( A)( M 1 ) ( M 2 F) : ( A)( M 2 ). 8 In definitions if lwys stnds for if nd only if. 9 Once nd for ll: P if Q mens tht if Q holds then P holds, or Q implies P. P only if Q mens tht if Q does not hold, then P does not hold either, which implies tht P implies Q. Thus, if nd only if mens tht ech one implies the other.

26 20 Chpter 1 Tht is, ( M 1 F) : ( A)( M 1 ) ( M 2 F) : ( A)(( ) ( M 2 )). By tking M = mx(m 1, M 2 ) we obtin ( A)( M ( ) M), i.e., ( A)( M). The other direction isn t much different. A few nottionl conventions: for < b we use the following nottions for n open פתוח ),(קטע closed סגור ),(קטע nd semi-open חצי סגור ) (קטע segment, (, b) = {x F : < x < b} [, b] = {x F : x b} (, b] = {x F : < x b} [, b) = {x F : x < b}. Ech of these sets is bounded (bove nd below). We lso use the following nottions for sets tht re bounded on one side, (, ) = {x F : x > } [, ) = {x F : x } (, ) = {x F : x < } (, ] = {x F : x }. The first two sets re bounded from below, wheres the lst two re bounded from bove. It should be emphsized tht ± re not members of F! The bove is purely nottion. Thus, being less thn infinity does not men being less thn field element clled infinity. (5 hrs, 2011) Consider now the set of nturl numbers, N. Does it hve n upper bound? Our geometricl picture of the number xis clerly indictes tht this set is unbounded, but cn we prove it? We cn esily prove the following:

27 Rel numbers 21 Proposition 1.19 The nturl numbers re not upper bounded by nturl number. Proof : Suppose n N were n upper bound for N, i.e., ( k N) : (k n). However, m = n + 1 N nd m > n, i.e., ( m N) : (n < m), which is contrdiction. Similrly, Corollry 1.5 The nturl numbers re not bounded from bove by ny rtionl number Proof : Suppose tht p/q Q ws n upper bound for N. Since, by the xioms of order, p/q p, it would imply the existence of nturl number p tht is upper bound for N. 10 (5 hrs, 2010) It my however be possible tht n irrtionl element in F be n upper bound for N. Cn we rule out this possibility? It turns out tht this cnnot be proved from the xioms of n ordered field. Since, however, the boundedness of the nturls is so bsic to our intuition, we my impose it s n dditionl xiom, known s the xiom of the Archimeden field: the set of nturl numbers is unbounded from bove 11. (5 hrs, 2013) This xiom hs number of immedite consequences: 10 Convince yourself tht p/q p for ll p, q N. 11 This dditionl ssumption is temporry. The Archimeden property will be provble once we dd our lst xiom.

28 22 Chpter 1 Proposition 1.20 In n Archimeden ordered field F: ( F)( n N) : ( < n). Proof : If this weren t the cse, then N would be bounded. Indeed, the negtion of the proposition is ( F) : ( n N)(n ), i.e., is n upper bound for N. Corollry 1.6 In n Archimeden ordered field F: ( ɛ > 0)( n N) : (1/n < ɛ). Proof : By the previous proposition, ( ɛ > 0)( n N) : (0 < 1/ɛ < n). } {{ } (0<1/n<ɛ) Corollry 1.7 In n Archimeden ordered field F: ( x, y > 0)( n N) : (y < nx). Comment: This is relly wht is ment by the Archimeden property. For every x, y > 0, segment of length y cn be covered by finite number of segments of length x. y x x x x x x

29 Rel numbers 23 Proof : By the previous corollry with ɛ = y/x, ( x, y > 0)( n N) : (y/x < n). } {{ } (y<nx) 1.7 Axiom of completeness The Greeks knew lredy tht the field of rtionl numbers is incomplete, in the sense tht there is no rtionl number whose squre equls 2 (wheres they knew by Pythgors theorem tht this should be the length of the digonl of unit squre). In fct, let s prove it: Proposition 1.21 There is no r Q such tht r 2 = 2. Proof : Suppose, by contrdiction, tht r = n/m, where n, m N (we cn ssume tht r is positive) stisfies r 2 = 2. Although it requires some number theoreticl knowledge, we will ssume tht it is known tht ny rtionl number cn be brought into form where m nd n hve no common divisor. By ssumption, m 2 /n 2 = 2, i.e., m 2 = 2n 2. This mens tht m 2 is even, from which follows tht m is even, hence m = 2k for some k N. Hence, 4k 2 = 2n 2, or n 2 = 2k 2, from which follows tht n is even, contrdicting the ssumption tht m nd n hve no common divisor. Proof : Another proof: suppose gin tht r = n/m is irreducible nd r 2 = 2. Since, n 2 = 2m 2, then m 2 < n 2 < 4m 2, it follows tht nd Consider now the rtio m < n < 2m < 2n, 0 < n m < m nd 0 < 2m n < n. q = 2m n n m,

30 24 Chpter 1 whose numertor nd denomintor re both smller thn the respective numertor nd denomintor of r. By elementry rithmetic q 2 = 4m2 4mn + n 2 n 2 2nm + m 2 = 4 4n/m + n2 /m 2 n 2 /m 2 2n/m + 1 = 6 4n/m 3 2n/m = 2, which is contrdiction. Exercise 1.8 Prove tht there is no r Q such tht r 2 = 3. Exercise 1.9 Wht fils if we try to pply the sme rguments to prove tht there is no r Q such tht r 2 = 4 (n ssertion tht hppens to be flse)? A wy to cope with this missing number would be to dd 2 by hnd to the set of rtionl numbers, long with ll the numbers obtined by field opertions involving 2 nd rtionl numbers (this is clled in lgebr field extension) (שדה הרחבה ) 12. But then, wht bout 3? We could dd ll the squre roots of 3 ll positive rtionl numbers. And then, wht bout 2? Shll we dd ll n-th roots? But then, wht bout solution to the eqution x 5 + x + 1 = 0 (it cnnot be expressed in terms of roots s result of Glois theory)? It turns out tht single dditionl xiom, known s the xiom of completeness השלמות ),(אכסיומת completes the set of rtionl numbers in one fell swoop, such to provide solutions to ll those questions. Let us try to look in more detil in wht sense is the field of rtionl numbers incomplete. Consider the following two sets, A = {x Q : 0 < x, x 2 2} B = {y Q : 0 < y, 2 y 2 }. Every element in B is greter or equl thn every element in A (by trnsitivity nd by the fct tht 0 < x y implies x 2 y 2 ). Formlly, ( A b B)( b). Does there exist rtionl number tht seprtes the two sets, i.e., does there exist n element c Q such tht ( A b B)( c b)? 12 It cn be shown tht this field extension of Q consists of ll elements of the form { + b 2 :, b Q}.

31 Rel numbers 25 It cn be shown tht if there existed such c it would stisfy c 2 = 2, hence such c does not exist. This observtion motivtes the following definition: Definition 1.6 An ordered field F is sid to be complete if for every two nonempty sets, A, B F stisfying there exists n element c F such tht ( A b B)( b). ( A b B)( c b). We will soon see how this xiom completes the field Q. We next introduce more definitions. Let s strt with motivting exmple: Exmple: For ny set [, b) = {x : x < b}, b is n upper bound, but so is ny lrger element, e.g., b + 1. In fct, it is cler tht b is the lest upper bound. Definition 1.7 Let A F be set. An element M F is clled lest upper bound עליון ) (חסם for A if (i) it is n upper bound for A, nd (ii) if M is lso n upper bound for A then M M. Tht is, M is lest upper bound for A ( A)( M) ( M F)( if ( A)( M ) then (M M )). We cn see right wy tht lest upper bound, if it exists, is unique: Proposition 1.22 Let A F be set nd let M be lest upper bound for A. If M is lso lest upper bound for A then M = M. Proof : It follows immeditely from the definition of the lest upper bound. If M nd M re both lest upper bounds, then both re in prticulr upper bounds, hence M M nd M M, which implies tht M = M.

32 26 Chpter 1 We cll the lest upper bound of set (if it exists!) supremum, nd denote it by sup A. Similrly, the gretest lower bound תחתון ) (חסם of set is clled the infimum, nd it is denoted by inf A. 13 (7 hrs, 2011) We cn provide n equivlent definition of the lest upper bound: Proposition 1.23 Let A be set. A number M is lest upper bound if nd only if (i) it is n upper bound, nd (ii) ( ɛ > 0)( A) : ( > M ɛ). A M!! Proof : There re two directions to be proved. 1. Suppose first tht M were lest upper bound for A, i.e., ( A)( M) nd ( M F)(if ( A)( M ) then (M M )). Suppose, by contrdiction, tht there exists n ɛ > 0 such tht M ɛ for ll A, i.e., tht ( ɛ > 0) : ( A)( < M ɛ), then M ɛ is n upper bound for A, smller thn the lest upper bound, which is contrdiction. 13 In some books the nottions lub (lest upper bound) nd glb (gretest lower bound) re used insted of sup nd inf.

33 Rel numbers Conversely, suppose tht M is n upper bound nd ( ɛ > 0)( A) : ( > M ɛ). Suppose tht M ws not lest upper bound. Then there exists smller upper bound M < M. Tke ɛ = M M. Then, M = M ɛ for ll A, or in forml nottion, which is contrdiction. ( ɛ > 0) : ( A)( < M ɛ), Exmples: Wht re the lest upper bounds (if they exist) in the following exmples: ➀ [ 5, 15] (nswer: 15). ➁ [ 5, 15) (nswer: 15). ➂ [ 5, 15] {20} (nswer: 20). ➃ [ 5, 15] (17, 18) (nswer: 18). ➄ [ 5, ) (nswer: none). ➅ {1 1/n : n N} (nswer: 1). Exercise 1.10 Let F be n ordered field, nd A F. Prove tht M = inf A if nd only if M is lower bound for A, nd ( ɛ > 0)( A) : ( < M + ɛ). Exercise 1.11 Show tht the xiom of completeness is equivlent to the following sttement: for every two non-empty sets A, B F, stisfying tht A nd b B implies b, there exists n element c F, such tht c b for ll A nd b B. Exercise 1.12 For ech of the following sets, determine whether it is upper bounded, lower bounded, nd if they re find their infimum nd/or supremum: ➀ {( 2) n : n N}.

34 28 Chpter 1 ➁ {1/n : n Z, n 0}. ➂ {1 + ( 1) n : n N}. ➃ {1/n + ( 1) n : n N}. ➄ {n/(2n + 1) : n N}. ➅ {x : x 2 1 < 3}. Exercise 1.13 Let A, B be non-empty sets of rel numbers. We define A + B = { + b : A, b B} A B = { b : A, b B} A B = { b : A, b B}. Find A + B, A B nd A A in ech of the following cses: ➀ A = {1, 2, 3} nd B = { 1, 2, 1/2, 1/2}. ➁ A = B = [0, 1]. Exercise 1.14 Let A be non-empty set of rel numbers. Is ech of the following sttements necessrity true? Prove it or give counter exmple: ➀ A + A = {2} A. ➁ A A = {0}. ➂ x A A, x 0. Exercise 1.15 In ech of the following items, A, B re non-empty subsets of complete ordered field. For the moment, you cn only use the xiom of completeness for the lest upper bound. ➀ Suppose tht A is lower bounded nd set B = {x : x A}. Prove tht B is upper bounded, tht A hs n infimum, nd sup B = inf A. ➁ Show tht if A is upper bounded nd B is lower bounded, then sup(a B) = sup A inf B. ➂ Show tht if A, B re lower bounded then inf(a + B) = inf A + inf B. ➃ Suppose tht A, B re upper bounded. It is necessrily true tht sup(a B) = sup A sup B? ➄ Show tht if A, B only contin non-negtive terms nd re upper bounded, then sup(a B) = sup A sup B.

35 Rel numbers 29 Definition 1.8 Let A F be subset of n ordered field. It is sid to hve mximum if there exists n element M A which is n upper bound for A. We denote M = mx A. It is sid to hve minimum if there exists n element m A which is lower bound for A. We denote m = min A. (7 hrs, 2010) (7 hrs, 2013) Proposition 1.24 If set A hs mximum, then the mximum is the lest upper bound, i.e., sup A = mx A. Similrly, if A hs minimum, then the minimum is the gretest lower bound, inf A = min A. Proof : Let M = mx A. 14 Then M is n upper bound for A, nd for every ɛ > 0, A M > M ɛ, tht is ( ɛ > 0)( A)( > M ɛ), which proves tht M is the lest upper bound. Exmples: Wht re the mxim (if they exist) in the following exmples: ➀ [ 5, 15] (nswer: 15). ➁ [ 5, 15) (nswer: none). ➂ [ 5, 15] {20} (nswer: 20). ➃ [ 5, 15] (17, 18) (nswer: none). ➄ [ 5, ) (nswer: none). ➅ {1 1/n : n N} (nswer: none). 14 We emphsize tht not every (non-empty) set hs mximum. The sttement is tht if mximum exists, then it is lso the supremum.

36 30 Chpter 1 (5 hrs, 2009) Proposition 1.25 Every finite set in n ordered field hs minimum nd mximum. Proof : The proof is by induction on the size of the set. Corollry 1.8 Every finite set in n ordered field hs lest upper bound nd gretest lower bound. Consider now the field Q of rtionl numbers, nd its subset A = {x Q : 0 < x, x 2 < 2}. This set is bounded from bove, s 2, for exmple, is n upper bound. Indeed, if x A, then x 2 < 2 < 2 2, which implies tht x < 2. A n upper bound But does A hve lest upper bound? Proposition 1.26 Suppose tht A hs lest upper bound, then (sup A) 2 = 2.

37 Rel numbers 31 Proof : We ssume tht A hs lest upper bound, nd denote Clerly, 1 < α < 2. α = sup A. Suppose, by contrdiction, tht α 2 < 2. Relying on the Archimeden property, we choose n to be n integer lrge enough, such tht ( ) 1 n > mx α, 6. 2 α 2 We re going to show tht there exists rtionl number r > α, such tht r 2 < 2, which mens tht α is not even n upper bound for A contrdiction: (α + 1/n) 2 = (α) n α + 1 n 2 < α α (use 1/n < α) n < α n (use α < 2) which mens tht α + 1/n A. < α α 2 = 2, (use 6/n < 2 α 2 ) Similrly, suppose tht α 2 > 2. This time we set n > 2α α 2 2. We re going to show tht there exists rtionl number r < α, such tht r 2 > 2, which mens tht r is n upper bound for A smller thn α gin, contrdiction: (α 1/n) 2 = α 2 2 n α + 1 n 2 > α 2 2 n α (omit 1/n2 ) > α α 2 = 2, (use 2α/n > 2 α 2 ) which mens tht, indeed, α 1/n is n upper bound for A You my sk yourself: how could I hve ever guessed such proof? Tke for exmple the ssumption tht α 2 > 2. We wnt to show tht there exists n upper bound for A tht is smller

38 32 Chpter 1 (8 hrs, 2013) Since we sw tht there is no r Q such tht r 2 = 2, it follows tht there is no supremum to this set in Q. We will now see tht completeness gurntees the existence of lowest upper bound: Proposition 1.27 In complete ordered field every non-empty set tht is upperbounded hs lest upper bound. Proof : Let A F be non-empty set tht is upper-bounded. Define B = {b F : b is n upper bound for A}, which by ssumption is non-empty. Clerly, By the xiom of completeness, ( A b B)( b). ( c F) : ( A b B)( c b). Clerly c is n upper bound for A smller or equl thn every other upper bound, hence c = sup A. (8 hrs, 2010) The following proposition is n immedite consequence: thn α, contrdicting the fct tht it is the lest upper bound. Let s try to find n n sufficiently lrge such tht α 1/n is n upper bound for A, nmely, such tht (α 1/n) 2 > 2. We expnd: ( α 1 ) 2 = α 2 2α n n + 1 n > 2 + ( α 2 2 ) 2α. 2 } {{ n} needs to be positive Similrly, when we tke the ssumption α 2 < 2 we wnt to show tht there exists n element of A tht is greter thn α. Let s try to find n n sufficiently lrge such tht α + 1/n is in A, nmely, such tht (α + 1/n) 2 < 2. We expnd: ( α + 1 n ) 2 = α 2 + 2α n + 1 n 2 < 2 + ( α 2 2 ) + 2α n + 1 n 2 } {{ } needs to be negtive.

39 Rel numbers 33 Proposition 1.28 In complete ordered field every non-empty set tht is bounded from below hs gretest lower bound (n infimum). Proof : Let m be lower bound for A, i.e., Consider the set Since ( A)(m ). B = {( ) : A}. ( A)(( m) ( )), it follows tht ( m) is upper bound for B. By the xiom of completeness, B hs lest upper bound, which we denote by M. Thus, ( A)(M ( )) nd ( ɛ > 0)( A) : (( ) > M ɛ). This mens tht ( A)(( M) ) nd ( ɛ > 0)( A) : ( < ( M) + ɛ). This mens tht ( M) = inf A. With tht, we define the set of rel numbers s complete ordered field, which we denote by R. It turns out tht this defines the set uniquely, up to relbeling of its elements (i.e., up to n isomorphism). 16 The xiom of completeness cn be used to prove the Archimeden property. In other words, the xiom of completeness solves t the sme time the problems pointed out in the previous section. Proposition 1.29 (Archimeden property) In complete ordered field N is not bounded from bove. 16 Strictly speking, we should prove tht such n niml exists. Constructing the set of rel numbers from the set of rtionl numbers is beyond the scope of this course. You re strongly recommended, however, to red bout it. See for exmple the construction of Dedekind cuts.

40 34 Chpter 1 Proof : Suppose N ws bounded. Then it would hve lest upper bound M, which is, in prticulr, nd upper bound: Since n N implies n + 1 N, i.e., ( n N) (n M). ( n N) ((n + 1) M), ( n N) (n (M 1)). This mens tht M 1 is lso n upper bound for N, contrdicting the fct tht M is the lest upper bound. (9 hrs, 2011) We next prove n importnt fcts bout the density (צפיפות ) of rtionl numbers within the rels. Proposition 1.30 (The rtionl numbers re dense in the rels) Let x, y R, such tht x < y. There exists rtionl number q Q such tht x < q < y. In forml nottion, ( x, y R : x < y)( q Q) : (x < q < y). Proof : Since the nturl numbers re not bounded, there exists n m Z Q such tht m < x. Also there exists nturl number n N such tht 1/n < y x. Consider now the sequence of rtionl numbers, r k = m + k, k = 0, 1, 2,.... n From the Archimeden property follows tht there exists k such tht r k = m + k n > x. Let k be the smllest such number (the exists smllest k becuse every finite set hs minimum), i.e., r k > x nd r k 1 x.

41 Rel numbers 35 Hence, which completes the proof. x < r k = r k n x + 1 n < y, 1/" q If we strt #om n integer poin$ on the le% of x nd proceed by rtionl steps of size less thn &y'x(, then w) rech rtionl point between x nd y.! x y The following proposition provides useful fct bout complete ordered fields. We will use it lter in the course. Proposition 1.31 החתכים )) ((למת Let A, B F be two non-empty sets in complete ordered field, such tht ( A b B)( b). Then the following three sttements re equivlent: ➀ (!M F) : ( A b B)( M b). ➁ sup A = inf B. ➂ ( ɛ > 0)( A, b B) : (b < ɛ). Comment: The clim is not tht the three items follow from the given dt. The clim is tht ech sttement implies the two other. Proof : Suppose tht the first sttement holds. This mens tht M is both n upper bound for A nd lower bound for B. Thus, sup A M inf B.

42 36 Chpter 1 This implies t once tht sup A inf B. If sup A < inf B, then ny number m in between would stisfy < m < b for ll A, b B, contrdicting the ssumption, hence 1 implies 2. By the properties of the infimum nd the supremum, Tht is, i.e., ( ɛ > 0)( A, b B) such tht Thus, 2 implies 3. ( ɛ > 0)( A) : ( > sup A ɛ/2) ( ɛ > 0)( b B) : (b < inf B + ɛ/2). ( ɛ > 0)( A) : ( + ɛ/2 > sup A) ( ɛ > 0)( b B) : (b ɛ/2 < inf B), b < (inf B + ɛ/2) (sup A ɛ/2) = ɛ. It remins to show tht 3 implies 1. Suppose M ws not unique, i.e., there were M 1 < M 2 such tht ( A b B)( M 1 < M 2 b). Set m = (M 1 + M 2 )/2 nd ɛ = (M 2 M 1 ). Then, ( A, b B) ( m ɛ/2, b m + ɛ/2), } {{ } b (m+ɛ/2) (m ɛ/2)=ɛ i.e., ( ɛ > 0) : ( A b B)(b > ɛ), i.e., if 1 does not hold then 3 does not hold. This concludes the proof. (7 hrs, 2009) (10 hrs, 2013) Exercise 1.16 In this exercise you my ssume tht 2 is irrtionl. ➀ Show tht if + b 2 = 0, where, b Q, then = b = 0. ➁ Conclude tht + b 2 = c + d 2, where, b, c, d Q, then = c nd b = d. ➂ Show tht if + b 2 Q, where, b Q, then b = 0.

43 Rel numbers 37 ➃ Show tht the irrtionl numbers re dense, nmely tht there exists n irrtionl number between every two. ➄ Show tht the set { + b 2 :, b Q} with the stndrd ddition nd multipliction opertions is field. Exercise 1.17 Show tht the set is dense. {m/2 n : m Z, n N} Exercise 1.18 Let A, B R be non-empty sets. Prove or disprove ech of the following sttements: ➀ If A B nd B is bounded from bove then A is bounded from bove. ➁ If A B nd B is bounded from below then A is bounded from below. ➂ If every b B is n upper bound for A then every A is lower bound form B. ➃ A is bounded from bove if nd only if A Z is bounded from bove. Exercise 1.19 Let A R be non-empty set. Determine for ech of the following sttements whether it is equivlent, contrdictory, implied by, or implying the sttement tht s = sup A, or none of the bove: ➀ s for ll A. ➁ For every A there exists r R, such tht < t < s. ➂ For every x R such tht x > s there exists n A such tht s < < x. ➃ For every finite subset B A, s mx B. ➄ s is the supremum of A {s}. 1.8 Rtionl powers We defined the integer powers recursively, x 1 = x x k+1 = x x k. We lso define for x 0, x 0 = 1 nd x n = 1/x n.

44 38 Chpter 1 Proposition 1.32 (Properties of integer powers) Let x, y 0 nd m, n Z, then ➀ x m x n = x m+n. ➁ (x m ) n = x mn. ➂ (xy) n = x n y n. ➃ 0 < α < β nd n > 0 implies α n < β n. ➄ 0 < α < β nd n < 0 implies α n > β n. ➅ α > 1 nd n > m implies α n > α m. ➆ 0 < α < 1 nd n > m implies α n < α m. Proof : Do it. (10 hrs, 2011) Definition 1.9 Let x > 0. An n-th root of x is positive number y, such tht y n = x. It is esy to see tht if x hs n n-th root then it is unique (since y n = z n, would imply tht y = z). Thus, we denote it by either n x, or by x 1/n. Theorem 1.1 (Existence nd uniqueness of roots) ( x > 0 n N)(!y > 0) : (y n = x). (10 hrs, 2010) Proof : Consider the set S = {z 0 : z n < x}. This is non-empty set (it contins zero) nd bounded from bove, since if x 1 nd z S then z n < x 1, which implies tht z 1 if x > 1 nd z S then z n < x < x n, which implies tht z < x,

45 Rel numbers 39 It follows tht regrdless of the vlue of x, z S implies tht z mx(1, x), nd the ltter is hence n upper bound for S. As consequence of the xiom of completeness, there exists unique y = sup S, which is the nturl suspect for being n n-th root of x. Indeed, we will show tht y n = x. Clim: y is positive Indeed, 0 < x 1 + x < 1, hence ( x ) n x < 1 + x 1 + x < x. Thus, x/(1 + x) S, which implies tht 0 < x 1 + x y. Clim: y n x Suppose, by contrdiction tht y n < x. We will show by contrdiction tht y is not n upper bound for S by showing tht there exists n element of S tht is greter thn y. We will do it by showing the existence of n ɛ > 0 such tht y ( 1 ɛ S i.e. y ) n < x. 1 ɛ This mens tht we look for n ɛ > 0 stisfying y n x < (1 ɛ)n. Since we cn choose ɛ > 0 t will, we cn tke it smller thn 1, in which cse (1 ɛ) n (1 ɛn) If we choose ɛ sufficiently smll such tht 1 ɛn > yn x, then forteriori (1 ɛ) n > y n /x, nd this will be stisfied if we choose ɛ < 1 yn /x, n which is possible becuse the right hnd side is positive. Thus, we found number greter thn y, whose n-th power is less thn x, i.e., in S. This contrdicts the ssumption tht y is n upper bound for S.

46 40 Chpter 1 Clim: y n x Suppose, by contrdiction tht y n > x. This time we will show tht there exists positive number less thn y whose n-th power is greter then x, i.e., it is n upper bound for S, contrdicting the ssumption tht y is the lest upper bound for S. To find such number we will show tht there exists n ɛ > 0, such tht [(1 ɛ)y] n > x, i.e., (1 ɛ) n > x y n. Once gin, we cn ssume tht ɛ < 1, in which cse by the Bernoulli inequlity (1 ɛ) n (1 ɛn), so tht if (1 ɛn) > x y n, then forteriori (1 ɛ) n > x/y n. Thus, we need 0 < ɛ < 1 to stisfiy ɛ < 1 x/yn, n which is possible becuse the right hnd side is positive. From the two inequlities follows (by trichotomy) tht y n = x. Hving shown tht every positive number hs unique n-th root, we my proceed to define rtionl powers. Definition 1.10 Let r = m/n Q, then for ll x > 0 x r = (x m ) 1/n. There is one little deliccy with this definition. Recll tht rtionl numbers do not hve unique representtion s the rtio of two integers. We thus need to show tht the bove definition is independent of the representtion 17. In other words, if d = bc, with, b, c, d Z, b, d 0, then for ll x > 0 x /b = x c/d. 17 This is not obvious. Suppose we wnted to define the notion of even/odd for rtionl numbers, by sying tht rtionl number is even if its numertor is even. Such s definition would imply tht 2/6 is even, but 1/3 (which is the sme number!) is odd.

47 Rel numbers 41 This is non-trivil fct. We need to prove tht d = bc implies tht sup{z > 0 : z b < x } = sup{z > 0 : z d < x c }. The rguments goes s follows: (x /b ) bd = (((x ) 1/b ) b ) d = (x ) d = x d = x bc = (x c ) b = (((x c ) 1/d ) d ) b = (x c/d ) bd, hence x /b = x c/d. (8 hrs, 2009) Proposition 1.33 (Properties of rtionl powers) Let x, y > 0 nd r, s Q, then ➀ x r x s = x r+s. ➁ (x r ) s = x rs. ➂ (xy) r = x r y r. ➃ 0 < α < β nd r > 0 implies α r < β r. ➄ 0 < α < β nd r < 0 implies α r > β r. ➅ α > 1 nd r > s implies α r > α s. ➆ 0 < α < 1 nd r > s implies α r < α s. Proof : We re going to prove only two items. Strt with the first. Let r = /b nd s = c/d. Using the (proved!) lws for integer powers, (x /b x c/d ) bd = (x /b ) bd (x c/d ) bd = = x d x bc = x d+bc = (x /b+c/d ) bd, from which follows tht x /b x c/d = x /b+c/d. Tke then the fourth item. Let r = /b. We lredy know tht α < β. Thus it remins to show α 1/b < β 1/b. This hs to be becuse α 1/b β 1/b would hve implied (from the rules for integer powers!) tht α β. (12 hrs, 2013)

48 42 Chpter Rel-vlued powers Delyed to much lter in the course Addendum Existence of non-archimeden ordered fields Does there exist ordered fields tht do not stisfy the Archimeden property? The nswer is positive. We will give one exmple, bering in mind tht some of the rguments rely on lter mteril. Consider the set of rtionl functions. It is esy to see tht this set forms field with respect to function ddition nd multipliction, with f (x) 0 for dditive neutrl element nd f (x) 1 for multiplictive neutrl element. Ignore the fct tht rtionl function my be undefined t finite collection of points. The nturl elements re the constnt functions, f (x) 1, f (x) 2, etc. We endow this set with n order reltion by defining the set P of positive elements s the rtionl functions f tht re eventully positive for x sufficiently lrge. We hce f > g if f (x) > g(x) for x sufficiently lrge. It is esy to see tht the set of nturl elements re bounded, sy, by the element f (x) = x (which obviously belongs to the set). Connected set A set A R will be sid to be connected (קשיר ) if x, y A nd x < z < y implies z A (tht is, every point between two elements in the set is lso n the set). We stte tht (non-empty) connected set cn only be of one of the following forms: {}, (, b), [, b), (, b], [, b] (, ), [, ), (, b), (, b], (, ). Tht is, it cn only be single point, n open, closed or semi-open segments, n open or closed ry, or the whole line. Exercise 1.20 Prove it.

49 Chpter 2 Functions 2.1 Bsic definitions Wht is function? There is stndrd wy of defining functions, but we will delibertely be little less forml thn perhps we should, nd define function s mchine, which when provided with number, returns number (only one, nd lwys the sme for the sme input). In other words, function is rule tht ssigns rel numbers to rel numbers. A function is defined by three elements: ➀ A domin :(תחוּם ) subset of R. The numbers which my be fed into the mchine. ➁ A rnge :(טווח ) nother subset of R. Numbers tht my be emitted by the mchine. We do not exclude the possibility tht some of these numbers my never be returned. We only require tht every number returned by the function belongs to its rnge. ➂ A trnsformtion rule.(העתקה ) The crucil point is tht to every number.(חד ערכיוּת ) in its domin corresponds one nd only one number in its rnge We normlly denote functions by letters, like we do for rel numbers (nd for ny other mthemticl entity). To void mbiguities, the function hs to be defined properly. For exmple, we my denote function by the letter f. If A R is its domin, nd B R is its rnge, we write f : A B ( f mps the set A into the set B). The trnsformtion rule hs to specify wht number in B is ssigned by the