The Algebraic and Order Properties of R

THE REAL NUMBERS

Section 2.1 The Algebraic and Order Properties of R

CHAPTER 2

algebra, the system of real numbers is a ‘‘field’’ with respect to addition and multiplication.

The basic properties listed in 2.1.1 are known as the field axioms. A binary operation associates with each pair (a,b) a unique elementB(a,b), but we will use the conventional notations ofaþbandabwhen discussing the properties of addition and multiplication.

2.1.1 Algebraic Properties of R On the setR of real numbers there are two binary operations, denoted byþandand calledadditionandmultiplication, respectively. These operations satisfy the following properties:

(A1) aþb¼bþafor alla,b inR (commutative property of addition);

(A2) ðaþbÞ þc¼aþ ðbþcÞfor alla,b,cinR (associative property of addition);

(A3) there exists an element 0 in R such that 0þa¼a andaþ0¼a for all a inR (existence of a zero element);

(A4) for eachainRthere exists an elementainRsuch thataþ ðaÞ ¼0 andðaÞ þ a¼0 (existence of negative elements);

(M1) ab¼ba for alla,binR (commutative property of multiplication);

(M2) ðabÞ c¼a ðbcÞfor alla,b,cinR (associative property of multiplication);

(M3) there exists an element 1 inRdistinct from0 such that 1a¼aanda1¼afor all ainR (existence of a unit element);

(M4) for eacha6¼0 inR there exists an element 1=ainR such thata ð1=aÞ ¼1 and ð1=aÞ a¼1 (existence of reciprocals);

(D) a ðbþcÞ ¼ ðabÞ þ ðacÞandðbþcÞ a¼ ðbaÞ þ ðcaÞ for alla, b, cin R (distributive property of multiplication over addition).

These properties should be familiar to the reader. The first four are concerned with addition, the next four with multiplication, and the last one connects the two operations.

The point of the list is that all the familiar techniques of algebra can be derived from these nine properties, in much the same spirit that the theorems of Euclidean geometry can be deduced from the five basic axioms stated by Euclid in hisElements. Since this task more properly belongs to a course in abstract algebra, we will not carry it out here. However, to exhibit the spirit of the endeavor, we will sample a few results and their proofs.

We first establish the basic fact that the elements 0 and 1, whose existence were asserted in (A3) and (M3), are in fact unique. We also show that multiplication by 0 always results in 0.

2.1.2 Theorem (a) If z and a are elements inR with zþa¼a,then z¼0.

(b) If u and b6¼0are elements inR with ub¼b,then u¼1.

Proof. (a) Using (A3), (A4), (A2), the hypothesiszþa¼a, and (A4), we get z¼zþ0¼zþ ðaþ ðaÞÞ ¼ ðzþaÞ þ ðaÞ ¼aþ ðaÞ ¼0:

(b) Using (M3), (M4), (M2), the assumed equalityub¼b, and (M4) again, we get

u¼u1¼u ðb ð1=bÞÞ ¼ ðubÞ ð1=bÞ ¼b ð1=bÞ ¼1:

aþa0¼a1þa0¼a ð1þ0Þ ¼a1¼a:

Therefore, we conclude from (a) thata0¼0. Q.E.D.

We next establish two important properties of multiplication: the uniqueness of reciprocals and the fact that a product of two numbers is zero only when one of the factors is zero.

2.1.3 Theorem (a) If a6¼0 and b inR are such that ab¼1, then b¼1=a. (b) If ab¼0,then either a¼0or b¼0.

Proof. (a) Using (M3), (M4), (M2), the hypothesisab ¼1, and (M3), we have

b¼1b¼ ðð1=aÞ aÞ b¼ ð1=aÞ ðabÞ ¼ ð1=aÞ 1¼1=a:

(b) It suffices to assumea6¼0 and prove thatb¼0. (Why?) We multiplyabby 1=aand apply (M2), (M4), and (M3) to get

ð1=aÞ ðabÞ ¼ ðð1=aÞ aÞ b¼1b¼b:

Sinceab¼0, by 2.1.2(c) this also equals

ð1=aÞ ðabÞ ¼ ð1=aÞ 0¼0:

Thus we have b¼0. Q.E.D.

These theorems represent a small sample of the algebraic properties of the real number system. Some additional consequences of the field properties are given in the exercises.

The operation ofsubtractionis defined byab:¼aþ ðbÞfora,binR. Similarly, divisionis defined fora,binRwithb6¼0 bya=b:¼a ð1=bÞ. In the following, we will use this customary notation for subtraction and division, and we will use all the familiar properties of these operations. We will ordinarily drop the use of the dot to indicate multiplication and writeabforab. Similarly, we will use the usual notation for exponents and writea²foraa,a³forða²Þa; and, in general, we definea^nþ¹:¼ ðaⁿÞaforn2N. We agree to adopt the convention thata¹¼a. Further, ifa6¼0, we writea⁰¼1 anda¹ for 1=a, and ifn2N, we will writeaⁿforð1=aÞⁿ, when it is convenient to do so. In general, we will freely apply all the usual techniques of algebra without further elaboration.

Rational and Irrational Numbers

We regard the setNof natural numbers as a subset ofR, by identifying the natural number n2Nwith then-fold sum of the unit element 12R. Similarly, we identify 02Zwith the zero element of 02R, and we identify then-fold sum of1 with the integern. Thus, we considerN andZto be subsets ofR.

Elements ofRthat can be written in the formb=awherea;b2Zanda6¼0 are called rational numbers. The set of all rational numbers inR will be denoted by the standard notationQ. The sum and product of two rational numbers is again a rational number (prove this), and moreover, the field properties listed at the beginning of this section can be shown to hold forQ.

The fact that there are elements inRthat are not inQis not immediately apparent. In the sixth century B.C. the ancient Greek society of Pythagoreans discovered that the diagonal of a square with unit sides could not be expressed as a ratio of integers. In view of the Pythagorean Theorem for right triangles, this implies that the square of no rational number can equal 2. This discovery had a profound impact on the development of Greek mathematics. One consequence is that elements ofR that are not inQ became known as irrational numbers, meaning that they are not ratios of integers. Although the word

‘‘irrational’’ in modern English usage has a quite different meaning, we shall adopt the standard mathematical usage of this term.

2.1 THE ALGEBRAIC AND ORDER PROPERTIES OFR 25

We will now prove that there does not exist a rational number whose square is 2. In the proof we use the notions of even and odd numbers. Recall that a natural number isevenif it has the form 2nfor somen2N, and it isoddif it has the form 2n1 for somen2N. Every natural number is either even or odd, and no natural number is both even and odd.

2.1.4 Theorem There does not exist a rational number r such that r²¼2.

Proof. Suppose, on the contrary, thatpandqare integers such thatðp=qÞ²¼2. We may assume thatpandqare positive and have no common integer factors other than 1. (Why?) Since p²¼2q², we see thatp²is even. This implies thatpis also even (because ifp¼ 2n1 is odd, then its squarep²¼2ð2n²2nþ1Þ 1 is also odd). Therefore, sincepand q do not have 2 as a common factor, thenq must be an odd natural number.

Sincepis even, thenp¼2mfor somem2N, and hence 4m² ¼2q², so that 2m²¼q². Therefore,q² is even, and it follows thatq is an even natural number.

Since the hypothesis thatðp=qÞ²¼2 leads to the contradictory conclusion thatqis

both even and odd, it must be false. Q.E.D.

The Order Properties ofR

The ‘‘order properties’’ ofRrefer to the notions of positivity and inequalities between real numbers. As with the algebraic structure of the system of real numbers, we proceed by isolating three basic properties from which all other order properties and calculations with inequalities can be deduced. The simplest way to do this is to identify a special subset ofR by using the notion of ‘‘positivity.’’

2.1.5 The Order Properties ofR There is a nonempty subsetPofR, called the set of positive real numbers, that satisfies the following properties:

(i) Ifa,bbelong toP, then aþbbelongs to P. (ii) Ifa,bbelong toP, then abbelongs toP.

(iii) Ifa belongs toR, then exactly one of the following holds:

a2P; a¼0; a2P:

The first two conditions ensure the compatibility of order with the operations of addition and multiplication, respectively. Condition 2.1.5(iii) is usually called the Trichotomy Property, since it dividesR into three distinct types of elements. It states that the setfa:a2Pgofnegativereal numbers has no elements in common with the set Pof positive real numbers, and, moreover, the setR is the union of three disjoint sets.

Ifa2P, we writea>0 and say thatais apositive(or astrictly positive) real number.

Ifa2P[ f0g, we writea0 and say thatais anonnegativereal number. Similarly, if a2P, we writea<0 and say thatais anegative(or astrictly negative) real number.

Ifa2P[ f0g, we writea0 and say thatais anonpositivereal number.

The notion of inequality between two real numbers will now be defined in terms of the setPof positive elements.

2.1.6 Definition Leta,bbe elements ofR. (a) Ifab2P, then we writea>b orb<a. (b) Ifab2P[ f0g, then we writeabor ba.

The Trichotomy Property 2.1.5(iii) implies that for a;b2R exactly one of the following will hold:

a>b; a¼b; a<b:

Therefore, if bothab andba, thena¼b. For notational convenience, we will write

a<b<c

to mean that both a<b and b<c are satisfied. The other ‘‘double’’ inequalities ab<c,abc, and a<bcare defined in a similar manner.

To illustrate how the basic Order Properties are used to derive the ‘‘rules of inequalities,’’ we will now establish several results that the reader has used in earlier mathematics courses.

2.1.7 Theorem Let a, b, c be any elements ofR. (a) If a>b and b>c , then a>c.

(b) If a>b,then aþc>bþc. (c) If a>b and c>0,then ca>cb.

If a>b and c <0,then ca<cb.

Proof. (a) Ifab2Pand bc2P, then 2.1.5(i) implies that ðabÞ þ ðbcÞ ¼ acbelongs to P. Hencea >c.

(b) Ifab2P, thenðaþcÞ ðbþcÞ ¼ab is inP. Thusaþc>bþc.

On the other hand, ifc<0, thenc2P, so thatcbca¼ ðcÞðabÞis inP. Thus

cb>cawhenc<0. Q.E.D.

It is natural to expect that the natural numbers are positive real numbers. This property is derived from the basic properties of order. The key observation is that the square of any nonzero real number is positive.

2.1.8 Theorem

(a) If a2R and a6¼0,then a²>0.

(b) 1>0.

Proof. (a) By the Trichotomy Property, ifa6¼0, then eithera2Pora2P. Ifa2P, then by 2.1.5(ii), we havea²¼aa2P. Also, ifa2P, thena²¼ ðaÞðaÞ 2P. We conclude that ifa 6¼0, thena²>0.

(b) Since 1¼1², it follows from (a) that 1>0.

(c) We use Mathematical Induction. The assertion forn¼1 is true by (b). If we suppose the assertion is true for the natural numberk, thenk2P, and since 12P, we havekþ12Pby 2.1.5(i). Therefore, the assertion is true for all natural numbers. Q.E.D.

It is worth noting thatno smallest positive real number can exist. This follows by observing that ifa>0, then since¹₂>0 (why?), we have that

0<¹₂a<a:

2.1 THE ALGEBRAIC AND ORDER PROPERTIES OFR 27

Thus if it is claimed thatais the smallest positive real number, we can exhibit a smaller positive number¹₂a.

This observation leads to the next result, which will be used frequently as a method of proof. For instance, to prove that a numbera0 is actually equal to zero, we see that it suffices to show thata is smaller than an arbitrary positive number.

2.1.9 Theorem If a2R is such that0a<efor everye>0, then a¼0.

Proof. Suppose to the contrary thata>0. Then if we takee0:¼¹₂a, we have 0<e0<a. Therefore, it is false thata<efor everye>0 and we conclude that a¼0. Q.E.D.

Remark It is an exercise to show that ifa2Ris such that 0aefor everye>0, thena¼0.

The product of two positive numbers is positive. However, the positivity of a product of two numbers does not imply that each factor is positive. The correct conclusion is given in the next theorem. It is an important tool in working with inequalities.

2.1.10 Theorem If ab>0, then either (i) a>0and b>0,or

(ii) a<0and b<0.

Proof. First we note thatab>0 implies thata6¼0 andb6¼0. (Why?) From the Trichotomy Property, eithera>0 ora<0. Ifa>0, then 1=a>0, and thereforeb¼ ð1=aÞ ðabÞ>0.

Similarly, ifa<0, then 1=a<0, so thatb¼ ð1=aÞðabÞ<0. Q.E.D.

2.1.11 Corollary If ab<0,then either (i) a<0and b>0,or

(ii) a>0and b<0.

Inequalities

We now show how the Order Properties presented in this section can be used to ‘‘solve’’

certain inequalities. The reader should justify each of the steps.

2.1.12 Examples (a) Determine the setAof all real numbersxsuch that 2xþ36.

We note that we have^y

x2A () 2xþ36 () 2x3 () x³₂: ThereforeA¼x2R:x³₂

(b) Determine the set B:¼ fx2R:x²þx>2g.

We rewrite the inequality so that Theorem 2.1.10 can be applied. Note that x2B () x²þx2>0 () ðx1Þðxþ2Þ>0:

Therefore, we either have (i)x1>0 andxþ2>0, or we have (ii)x1<0 and xþ2<0. In case (i) we must have bothx>1 andx>2, which is satisfied if and only

yThe symbol()should be read ‘‘if and only if.’’

ifx>1. In case (ii) we must have bothx<1 andx<2, which is satisfied if and only ifx<2.

We conclude thatB¼ fx2R:x>1g [ fx2R:x<2g. (c) Determine the set

C:¼ x2R :2xþ1 xþ2 <1

: We note that

x2C () 2xþ1

xþ2 1<0 () x1 xþ2<0:

Therefore we have either (i)x1<0 and xþ2>0, or (ii)x1>0 andxþ2<0.

(Why?) In case (i) we must have bothx<1 andx>2, which is satisfied if and only if 2<x<1. In case (ii), we must have bothx>1 andx<2, which is never satisfied.

We conclude thatC¼ fx2R:2<x<1g. &

The following examples illustrate the use of the Order Properties ofRin establishing certain inequalities. The reader should verify the steps in the arguments by identifying the properties that are employed.

It should be noted that the existence of square roots of positive numbers has not yet been established; however, we assume the existence of these roots for the purpose of these examples. (The existence of square roots will be discussed in Section 2.4.)

2.1.13 Examples (a) Leta0 andb0. Then ð1Þ a<b () a²<b² () ffiffiffi

< ffiffiffi pb

We consider the case wherea>0 andb>0, leaving the casea¼0 to the reader. It follows from 2.1.5(i) thataþb>0. Sinceb²a²¼ ðbaÞðbþaÞ, it follows from 2.1.7(c) that ba>0 implies thatb²a²>0. Also, it follows from 2.1.10 thatb²a²>0 implies thatba>0.

Ifa>0 andb>0, thenpffiffiffia>0 and ffiffiffi pb

>0. Sincea¼ðpffiffiffiaÞ²andb¼ ffiffiffi pb 2

, the second implication is a consequence of the first one whenaandbare replaced by ffiffiffi

pa ffiffiffi and

, respectively.

We also leave it to the reader to show that ifa0 andb0, then ð1⁰Þ ab () a²b² () ffiffiffi

pa ffiffiffi

(b) Ifaandbare positive real numbers, then theirarithmetic meanis¹₂ðaþbÞand their geometric meanis ffiffiffiffiffi

pab

. TheArithmetic-Geometric Mean Inequalityfora,bis

ð2Þ ffiffiffiffiffi

pab

¹₂ðaþbÞ with equality occurring if and only ifa¼b.

To prove this, note that if a>0;b>0, and a6¼b, then ffiffiffi pa

>0; ffiffiffi pb

>0, and ffiffiffia

p 6¼ ffiffiffi pb

. (Why?) Therefore it follows from 2.1.8(a) that ffiffiffi pa

ffiffiffi pb

>0. Expanding this square, we obtain

a2 ffiffiffiffiffi pab

þb>0; whence it follows that

ffiffiffiffiffi pab

<¹₂ðaþbÞ:

2.1 THE ALGEBRAIC AND ORDER PROPERTIES OFR 29

Therefore (2) holds (with strict inequality) whena6¼b. Moreover, ifa¼bð>0Þ, then both sides of (2) equala, so (2) becomes an equality. This proves that (2) holds fora>0;b>0.

On the other hand, suppose that a>0;b>0 and that ffiffiffiffiffi pab

¼¹₂ðaþbÞ. Then, squaring both sides and multiplying by 4, we obtain

4ab¼ ðaþbÞ²¼a²þ2abþb²; whence it follows that

0¼a²2abþb² ¼ ðabÞ²:

But this equality implies thata¼b. (Why?) Thus, equality in (2) implies thata¼b. Remark The general Arithmetic-Geometric Mean Inequality for the positive real numbers a1;a2;. . .;a_n is

ð3Þ ða1a2 anÞ¹⁼ⁿa1þa2þ þan

with equality occurring if and only ifa1 ¼a2¼ ¼an. It is possible to prove this more general statement using Mathematical Induction, but the proof is somewhat intricate. A more elegant proof that uses properties of the exponential function is indicated in Exercise 8.3.9 in Chapter 8.

ð4Þ ð1þxÞⁿ1þnx for all n2N

The proof uses Mathematical Induction. The casen¼1 yields equality, so the assertion is valid in this case. Next, we assume the validity of the inequality (4) fork2Nand will deduce it forkþ1. Indeed, the assumptions thatð1þxÞ^k1þkx and that 1þx>0 imply (why?) that

ð1þxÞ^kþ¹ ¼ ð1þxÞ^k ð1þxÞ

ð1þkxÞ ð1þxÞ ¼1þ ðkþ1Þxþkx² 1þ ðkþ1Þx:

Thus, inequality (4) holds forn¼kþ1. Therefore, (4) holds for alln2N. _&

Exercises for Section 2.1

1. Ifa;b2R, prove the following.

(a) Ifaþb¼0, thenb¼ a, (b) ðaÞ ¼a,

2. Prove that ifa;b2R, then

(a) ðaþbÞ ¼ ðaÞ þ ðbÞ, (b) ðaÞ ðbÞ ¼ab,

3. Solve the following equations, justifying each step by referring to an appropriate property or theorem.

(a) 2xþ5¼8; (b) x²¼2x;

4. Ifa2Rsatisfiesaa¼a, prove that eithera¼0 ora¼1.

5. Ifa6¼0 andb6¼0, show that 1=ðabÞ ¼ ð1=aÞð1=bÞ.

6. Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational numberssuch thats²¼6.

7. Modify the proof of Theorem 2.1.4 to show that there does not exist a rational numbertsuch that t²¼3.

8. (a) Show that ifx,yare rational numbers, thenxþyandxyare rational numbers.

(b) Prove that ifxis a rational number andyis an irrational number, thenxþyis an irrational number. If, in addition,x6¼0, then show thatxyis an irrational number.

9. LetK:¼ sþt ffiffiffi p2

:s;t2Q

. Show thatKsatisfies the following:

(a) Ifx1;x22K, thenx1þx22Kandx1x22K.

(b) Ifx6¼0 andx2K, then 1=x2K.

(Thus the setKis asubfieldofR. With the order inherited fromR, the setKis an ordered field that lies betweenQ andR.)

10. (a) Ifa<bandcd, prove thataþc<bþd. (b) If 0<a<band 0cd, prove that 0acbd. 11. (a) Show that ifa>0, then 1=a>0 and 1=ð1=aÞ ¼a.

(b) Show that ifa<b, thena<¹₂ðaþbÞ<b.

12. Leta,b,c,dbe numbers satisfying 0<a<bandc<d<0. Give an example whereac<bd, and one wherebd<ac.

13. Ifa;b2R, show thata²þb²¼0 if and only ifa¼0 andb¼0.

14. If 0a<b, show that a²ab<b². Show by example that it does not follow that a²<ab<b².

15. If 0<a<b, show that (a)a< ffiffiffiffiffi pab

<b, and (b) 1=b<1=a. 16. Find all real numbersxthat satisfy the following inequalities.

(a) x²>3xþ4; (b) 1<x²<4;

17. Prove the following form of Theorem 2.1.9: Ifa2Ris such that 0aefor everye>0, then a¼0.

18. Leta;b2R, and suppose that for everye>0 we haveabþe. Show thatab.

19. Prove that ¹₂ðaþbÞ2

¹₂ða²þb²Þfor alla;b2R. Show that equality holds if and only if a¼b.

20. (a) If 0<c<1, show that 0<c²<c<1.

(b) If 1<c, show that 1<c<c².

21. (a) Prove there is non2Nsuch that 0<n<1. (Use the Well-Ordering Property ofN.) (b) Prove that no natural number can be both even and odd.

22. (a) Ifc>1, show thatcⁿcfor alln2N, and thatcⁿ>cforn>1.

(b) If 0<c<1, show thatcⁿcfor alln2N, and thatcⁿ<cforn>1.

23. Ifa>0;b>0;andn2N, show thata<bif and only ifaⁿ<bⁿ. [Hint:Use Mathematical Induction.]

24. (a) Ifc>1 andm;n2N, show thatc^m>cⁿif and only ifm>n.

(b) If 0<c<1 andm;n2N, show thatc^m<cⁿif and only ifm>n.

25. Assuming the existence of roots, show that ifc>1, thenc¹^=m<c¹⁼ⁿif and only ifm>n. 26. Use Mathematical Induction to show that if a2R and m;n2N, then a^mþn¼a^maⁿ and

ða^mÞ ¼a^mn.

2.1 THE ALGEBRAIC AND ORDER PROPERTIES OFR 31

Dalam dokumen Robert G. Bartle, Donald R. Sherbert - Introduction to Real Analysis-John Wiley & Sons (2011) (Halaman 39-48)