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PART I. THE REAL NUMBERS - UH

PART I. THE REAL NUMBERSThis material assumes that you are already familiar with the real number system and the represen-tation of the real NUMBERS as points on the real THE NATURAL NUMBERS AND INDUCTIONLetNdenote the set of natural NUMBERS (positive integers).Axiom:IfSis a nonempty subset ofN, thenShas a least element. That is, there is anelementm Ssuch thatm nfor alln :A set which has the property that each non-empty subset has a least element is said to bewell-ordered. Thus, the axiom tells us that the natural NUMBERS are a subset the following properties:1. 1 S, Simpliesk+1 S,thenS= :SupposeS6=N.

Consequently, q2 =2k2 and so q is even. Thus p and q have the common factor 2, a contradiction. Other examples of irrational numbers are √ m where m is any rational number which is not a perfect square, 3 √ m where m is any rational number which is not a perfect cube, etc. Also, the numbers π and e are irrational. Definition 2. Let S be a ...

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Transcription of PART I. THE REAL NUMBERS - UH

1 PART I. THE REAL NUMBERSThis material assumes that you are already familiar with the real number system and the represen-tation of the real NUMBERS as points on the real THE NATURAL NUMBERS AND INDUCTIONLetNdenote the set of natural NUMBERS (positive integers).Axiom:IfSis a nonempty subset ofN, thenShas a least element. That is, there is anelementm Ssuch thatm nfor alln :A set which has the property that each non-empty subset has a least element is said to bewell-ordered. Thus, the axiom tells us that the natural NUMBERS are a subset the following properties:1. 1 S, Simpliesk+1 S,thenS= :SupposeS6=N.

2 LetT=N S. ThenT6= . Letmbe the least element inT. Thenm 1/ T. Therefore,m 1 Swhich implies that (m 1) + 1 =m S, a :LetSbe a subset ofNsuch Ifk m S, thenk+1 ,S={n N:n m}.ExampleProve that 1 + 2 + 22+23+ +2n 1=2n 1 for alln the set of integers for which the statement is 20=1=21 1, 1 that the positive integerk +21+ +2k 1+2k=(20+21+ +2k 1)+2k=2k 1+2k=2 2k 1=2k+1 ,k+1 have shown that 1 Sand thatk Simpliesk+1 S. It follows thatScontains all thepositive Prove that 1 + 2 + 3 + +n=n(n+1)2for alln Prove that 12+22+32+ +n2=n(n+ 1)(2n+1)6for alln Letrbe a real numberr6= 1. Prove that1+r+r2+r3+ +rn=1 rn+11 alln N4.

3 Prove that 1 + 2n 3nfor alln Prove that1 1+1 2+1 3+ +1 n nfor alln Prove that(1 122)(1 132) (1 1n2)=n+12nfor alln True or False: IfSis a non-empty subset ofN, then there exists an elementm Ssuchthatm kfor allk ORDERED FIELDSLetRdenote the set of real NUMBERS . The setR, together with the operations of addition (+)and multiplication ( ), satisfies the following axioms:Addition:A1. For allx, y R,x+y R(addition is a closed operation).A2. For allx, y R,x+y=y+x(addition is commutative)A3. For allx, y, z R,x+(y+z)=(x+y)+z(addition is associative).A4. There is a unique number 0 such thatx+0=0+xfor allx R.

4 (0 is theadditive identity.)A5. For eachx R, there is a unique number x Rsuch thatx+( x)=0. ( xis theadditive inverse ofx.)Multiplication:M1. For allx, y R,x y R(multiplication is a closed operation).2M2. For allx, y R,x y=y x(multiplication is commutative)M3. For allx, y, z R,x (y z)=(x y) (multiplication is associative).M4. There is a unique number 1 such thatx 1=1 xfor allx R. (1 is themultiplicativeidentity.)M5. For eachx R,x6= 0, there is a unique number 1/x=x 1 Rsuch thatx (1/x)=1.(1/xis themultiplicative inverse ofx.)Distributive Law:D. For allx, y, z R,x (y+z)=x y+x non-empty setStogether with two operations, addition and multiplication which sat-isfies A1-A5, M1-M5, and D is called afield.

5 The set of real NUMBERS with ordinary addition andmultiplication is an example of a field. The set of rational numbersQ, together with ordinaryaddition and multiplication, is also a field, a sub-field ofR. The set of complex numbersCisanother example of a :There is a subsetPofRthat has the following properties:aIfx, y P, thenx+y , y P, thenx y For eachx Rexactly one of the following holds:x P, x=0, x setPis the set of positive , y R. Thenx<y(read xis less thany ) ify x <yis equivalent toy>x(read yis greater thanx ).P={x R:x>0}.x ymeans eitherx<yorx=y;y xmeans eithery>xory= relation < has the following properties:O1.

6 For allx, y R, exactly one of the following holds:x<y, x=y, x > y. (TrichotomyLaw)O2. For allx, y, z R,ifx<yandy<z, thenx< For allx, y, z R,ifx<y, thenx+z<y+ For allx, y, z R,ifx<yandz>0, thenx z<y z.{R,+, ,<}is anordered system{S,+, ,<}satisfying these 15 axiomsis an ordered field. In particular, the set of rational numbersQ, together with ordinary addition,multiplication and less than , is an ordered field, asubfield , y y+ for every positive number , thenx :Suppose thatx>yand choose =x y2. Thenx<y+ =y+x y2=x+y2<x+x2=x,a contradiction. Therefore,x R. The absolute value ofx, denoted|x|, is given by|x|={x,ifx 0, x,ifx< properties of absolute value are: for anyx, y R.}

7 (1)|x| 0,(2)|xy|=|x||y|,(3)|x+y| |x|+|y|.Exercises True False. Justify your answer by citing a theorem, giving a proof, or giving a counter-example.(a) Ifx, y, z Randx<y, thenxz < yz.(b) Ifx, y Randx<y+ for every positive number , thenx<y.(c) Ifx, y R, then|x y| |x|+|y|.(d) Ifx, y R, then||x| |y|| |x| |y|.2. Prove: If|x y|< for every >0, thenx= Suppose thatx1,x2,..,xnare real NUMBERS . Prove that|x1+x2+ +xn| |x1|+|x2|+ +|xn|. THE COMPLETENESS AXIOMRandQare each ordered fields. What distinguishesRfromQis the completeness axiom. Asyou know,Qis a proper subset ofR; , there are real NUMBERS which are not rational NUMBERS are calledirrational 2.

8 2is not a rational number . In general, ifpis a prime number , then pisnot a rational :Suppose 2 =p/qwherep, q N. Without loss of generality, assume thatp, qhave nointegral factors>1. Nowp2=2q2,sop2is impliespmust be even , sop=2kfor somek N. Consequently,q2=2k2and soqis even . Thuspandqhave the common factor2 , a examples of irrational NUMBERS are mwheremis any rational number which is not aperfect square,3 mwheremis any rational number which is not a perfect cube, etc. Also, thenumbers andeare a subset ofR. A numberu Ris anupper boundofSifs ufor alls S. An elementw Ris alower boundofSifw sfor alls S.

9 If an upperbounduforSis an element ofS, thenuis called themaximum(orlargest element)ofS. Similarly, if a lower boundwforSis an element ofS, thenwis called theminimum(orsmallest element) :Give some examples to illustrate upper bounds, lower bounds, maximum and setS Ris said to bebounded aboveifShas an upper bound;Sisboundedbelowif it has a lower bound. A subsetSofRisboundedif it has both an upper bound and alower Rbe a set that is bounded above. A numberu Ris called the supremum(least upper bound) ofS,denoted bysupS,if it satisfies the ufor alls Ifvis an upper bound forS, thenu Rbe bounded above, and letu=supS.

10 Then, given any positive number , there is an elements Ssuch thatu <s :Suppose there exists an >0 such that the interval (u , u] contains no points u for alls S, which implies thatu is an upper bound forSwhich is lessthanu, a 5.:LetS Rbe a set that is bounded below. A numberu Ris called the infimum(greatest lower bound ) ofSand is denoted byinfSif it satisfies the sfor alls Ifvis a lower bound forS, thenv Completeness AxiomAxiomEvery nonempty subsetSofRthat is bounded above has a least upper bound. That is,ifSis bounded above, then supSexists and is a real set of real numbersRis a complete, ordered, field.)


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