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# Class Note for MATH 300 at UMass(6)

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Date Created: 02/06/15

The real numbers1 November 26 2007 0926pm What are the real numbers As explained in the introduction to Chapter 3 there are two approaches to answering this question One approach would be to construct the set R of real numbers starting with the set N of natural numbers Along the way we would need to construct the set Z of integers and the set Q of rational numbers The other approach7the one followed here7is not to say what the real numbers are but only to describe how they behave In other words our approach is to state in the form of axioms the fundamental properties of the set of real numbers Our axiomatic approach does not of course establish that any set having these properties actually exists To establish that one needs to take a constructive approach But the axiomatic approach does uniquely characterize the set of real numbers in the sense that any two sets having these properties are essentially the same in technical terms they are order isomor phicquot as ordered elds This essential uniqueness is established in Section 34 However in this course we shall not look into the matter of uniqueness of R The axioms for R can be summarized as follows R is an ordered eld that includes N has the Archimededn Ordering Property and has the Nested Interval Property To say that R is a eld means that there are operations of addition and multiplication in R with the usualquot algebraic properties These algebraic properties are listed in Axioms 314 In particular each x E R has a negative 726 for which 6 7x 0 and each nonzero y E R has a multiplicative inverse y 1 for which 3134 1 Then subtraction and division of real numbers may be de ned by x 7 y x 73 and for y at 0 2631 xy l Beyond those basic properties we shall assume without proof all the familiar algebraic prop erties listed in Proposition 3112 To say that the eld R is ordered means that there is a relation lt in R having the usualquot algebraic properties such as those listed in Proposition 324 Among these properties are the following For eachx E R exactly one of the relationsO ltxx0andx lt0holds ForallxyERxlty a 0lty7x Forallxy Rif0ltxand0ltythen0ltxyand0ltxy From just these three properties all the other familiar properties of order for real numbers as listed in Proposition 324 can be deduced Here s an example For all x y z E R xlty gtxzltyz In fact assume 6 lt y This means 0 lt y 7 x Buty 7 x y z 7 x 2 Thus 0 lt yz7xzThismeansxzltyz Exercise 1 Using just the properties of ordering listed above together with the usual algebraic properties of addition and multiplication deduce a IfO lt x then 7x lt O b Ifx lt y then 73 lt 7x c 0 lt1 When we say here that R includes N we do not mean merely that N C R Rather we mean also that 1Copyright 2007 Murray Eisenberg All Rights Reserved if mn E N then their sum m n and product m n as elements of R are the same as their sum and product respectively as natural numbers as de ned in Example 1210 2 and Exercise 1211 3 the number 0 E N is the identity element for addition in the eld R and the number 1 E N is the identity element for multiplication in the eld R and if m n E N then the relation m lt 11 holds for m and n as elements of R if and only if it holds for them as elements of N as de ned in Exercise 1211 2 With N as a subset of R the set Z of all integers and the set Q of all rational numbers may be de ned as ZNu7nneN QmeNn Nn O The Archimedean Ordering Property of the ordered eld R is the following For every E E R with E gt 0 and for each c E R there exists a positive integer n such that 115 gt c This property may be expressed by saying that if you take a real number 6 no matter how large and a positive real number S no matter how small then if you add E to itself enough times the sum will be greater than c For more about the Archimedean Ordering Property see Section 32 Proposition 2 Let c E R Then there exists a unique 11 E Z for which n s c lt n 1 Proof Existence Case i c 2 0 By the Archimedean Ordering Property there exists some k E N such that k 1 gt c By the WellOrdering Principle for N there is a least such k call it k1 Letn k1 71 Thenn s c lt 11 1 why Case ii 6 lt 0 Then 6 gt 0 and so by what was just proved there exists an integer m with m 5 6 lt m 1 Then nish the existence proof in this case Uniqueness Exercise 1 Theorem 3 Order Density of Q in R Ifa h E R with a lt b there exists some q E Q for which a lt q lt 1 Proof See the proof of Theorem 3227 In that proof just change F to R QF to Q and NF to N D For real numbers a and h as usual the closed interval a h is the set x E R a s x s h The Nested Interval Property of R is the following If an hnnoylyzm is a sequence of closed intervals in R that is decreasing in the sense that mun 7111 C am 71 for each n 0 1 2 then 0an 71 at Q For more about the Nested Interval Property see Section 33 We already know there is no rational number 6 for which C2 2 One of the consequences of the Nested Interval Property is that there is some real number 6 for which C2 2 or as we shall say that J2 exists With the theory of calculus at our disposal this would be easy to prove see Exercise 4 Exercise 4 Prove that J2 exists by applying the Intermediate Value Theorem to the function f R a Rgivenbyfx xz For the method in Exercise 4 you need to know that the function f is continuous and so you need to know a precise de nition for continuousquot you also need to know have the Intermediate Value Theorem at your disposal So we shall not use such a proof That J exists is proved by an indirect method in Section 33 namely by rst deducing from the Archimedean Ordering Property and the Nested Interval Property that each nonempty subset of R that is bounded above has a least upper bound In this note we give a different proof that directly invokes the Archimedean Ordering Prop erty and the Nested Interval Property In following the proof it may help if you draw a diagram with a line representing R and mark on it the points and intervals constructed Be sure to supply any missing justi cations for steps Theorem 5 There exists a real number 6 for which C2 2 Proof De ne a sequence an bnnoylyzm of closed intervals in R recursively as follows First let 0170 12 and let co the midpoint of 0 70 Next let aoco if 2 lt 65 C0 b0 C5 lt 2 1171 g We cannot have 65 2 because no rational number has square 2 In general once am bu has been constructed let on the midpoint of am km and then let armcn if 2 lt C an1bn1l gun bn if 6 lt 2 Since no rational number has square 2 in fact the case of 2 never arises By the construction amL bml C am km for alln 012 Then an lt 11 s 2 2 E Viki 1 bn an 27 6 for alln 0 1 2 Why By the Nested Interval Property of R there exists some 6 with 00 c 6 ambn n0 We are going to show that C2 2 Just suppose C2 at 2 De ne E 62 7 2 so that E gt 0 By the Archimedean Ordering Property of R there exits a positive integer n for which 1 mlt Why Fix such an 11 Since both C2 and 2 are in ugh 7 then C2 72 s b mi 7 bnmmbwan 1 1 lt 22 Zniz lt E This is impossible because C2 7 2 E 1 Exercise 6 Use the method of the preceding proof to establish that exists Exercise 7 Prove uniqueness of the number C whose existence is guaranteed by Theorem 5 The method of proof of Theorem 5 does not merely establish existence of a real number C whose square is 2 It also provides a method for approximating that number as closely as we wish Indeed look again at the construction of the intervals am bu The number C we want belongs to each of these intervals and is different from the endpoints of each Initially all we know about C is that C 6 0170 1 2 that is 1ltClt2 If we use the midpoint C0 15 as an approximation to C the the error in that approximation7 the size C 7 C0 of the difference between C and the approximation C07satis es C 7 C0 lt 05 2 Next since C5 152 225 gt 2 then 1171 aoCo 115 that is 1ltClt15 half the length of 0 170 1 115 So nowwe knOWC E If we use the midpoint C1 125 as a new approximation to C then the error C 7 C1 in this approximation satis es C 7 C1 lt 025 half the length of 1171 115 Next since C 1252 15625 lt 2 then agbg C1b1 12515 So now we know that C E 125 15 that is 125 lt C lt 15 If we use the midpoint Cg 1125 as a new approximation to C then the error C 7 Cg in this approximation satis es C 7 Cg lt 0125 With each successive bisection of the interval the number C is trapped inside an interval of half the length of the interval used in the previous step and the error in the approximating midpoint is halved This method is known as the Bisection Methodquot Exercise 8 Continue the Bisection Method begun above so as to approximate J2 with an error less than 001 The method of proof used for Theorem 57constructing a nested sequence of closed intervals7 is applicable for the next theorem too Recall the relevant de nitions which we originally gave in N Let A C R An upper bound of A in R is a number 7 E R such that a s b for every a E A The set A is bounded above in R when there exists some upper bound of A in R About N you proved that each nonempty subset A of N that is bounded above in N has a greatest element Such a greatest element g of A is an upper bound of A in N that is no greater than any other upper bound of A in N in other words g is the least element of the set of all upper bounds of A in N So this greatest element g of A is least upper bound of A in N A subset of R that is bounded above in R need not have a greatest element For example the open interval 0 1 in R has no greatest element why But it does have a least upper bound in R namely the real number 1 The following theorem generalizes this example Theorem 9 Order Completeness of R Each nonempty subset of R that is bounded above has a least upper bound Proof See the proof of Theorem 3313 In that proof just change F to R and change the term supremum to least upper boundquot 1

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