ANALISIS VARIABEL REAL IBAB IISISTEM BILANGAN REAL

MERUPAKAN HASIL PERADABAB MANUSIA YANG SANGAT MENAKJUBKAN. TANPA SISTEM INI TAK MUNGKIN IPTEK DAPAT BERKEMBANG SEPERTI SEKARANG INI. DARI SUDUT PERKEMBANGAN MATEMATIKA SAJA SISTEM BILANGAN REAL BANYAK MEMBERI IDE, INSPIRASI BAGI PERKEMBANGAN DAN PENGEMBANGAN MATEMATIKA ITU SENDIRI.

CONTOH: TEORI FUNGSI, TEORI OPERATOR, TEORI UKURAN DAN INTEGRAL, TEORI KODE, DLL BERKEMBANGAN DENGAN MEMANFAATKAN SIFAT-SIFAT DASAR BILANGAN REAL. OLEH KARENA ITU SECARA SINGKAT DAN KOMPREHENSIF BAB INI AKAN MEMUAT PEMBICARAAN TENTANG SISTEM BILANGAN REAL SECARA ALJABAR.

2.1 Sifat-sifat Dasar Bilangan Nyata Sistem bilangan real, notasi = R, Sifat Dasar dari struktur aljabar R

Definisi. 2.1.1Sistem bilangan real R adalah suatu sistem aljabar yang terhadap operasi penjumlahan (+) dan perkalian (.) mempunyai sifat-sifat sebagai berikut:

(A1) for each a, b in R there exactly an element c in R such that a+b = c (property of closed)

(A2) a+b = b + a for all a and b in R (commutative property of addition);

(A3) (a + b) + c = a +( b + c) for all a,b, c in R (associativeproperty of addition);

(A4) There exist an element 0 in R such that 0 + a = a + 0 = a for all a in R (existence of a zero element); (A5) for each a in R there exist en element a in R such that a + (-a) = 0 and (-a) + a = 0 (existence of negative elements);

The system of R is a group commutative in addition operation.

(M1) for each a, b in R there exactly an element c in R such that a.b = c (property of closed)

(M2) a.b = b .a for all a and b in R (commutative property of multiplication);

(M3) (a . b) . c = a .( b . c) for all a,b, c in R (associativeproperty of multiplication );

(M4) There exist an element 1 in R such that 1. a = a .1 = a for all a in R (existence of a unit element); (M5) for each a0 in R there exist en element 1/a in R such that a .(1/a) = 1, and (1/a) .a = 1 (existence of reciprocals elements);

The system of R {0}is a group commutative in multiplication operation.

(D) Distributive Property Of multiplication over addition

a.(b+c) = (a.b) + (a.c) and (b + c) . a = (b . a) + (c .a), for all a, b, and c in R

Proof:(a) The hypothesis is that z + a = a. we add the element a whose existence is given in (A5), to both sides. Then using (A5), the hypothesis, (A3), and (A5) in succession, we obtain:0 = a +(-a) = (z + a) + (-a) = z + (a+(-a)) = Z + 0 = z.

Theorem 2.1.2(a) If z and a are elements of R such that z + a = a, then z = 0

(b) If u and b0 are elements of R such that u.b = b, than u = 1.

The proof of part (b) is left as an exercise. Note that the hypothesis b0 is crucial. The hypothesis is that u.b=b, we multiply the element 1/b whose existence is given in (M5), to both sides. Then using (M5), The hypothesis, (M3), and (M5) in succession, we obtain: 1 = b. (1/b) = (u.b).(1/b) =u.(b.(1/b)) =u.1 =u0

Theorem 2.1.3(a) If a anf b are elements of R such that a + b =0, then b = -a

(b) If a 0 and b element of R such that a.b = 1, then b = 1/a

The proof of (a) and (b) is left as an exercise

Teorema 2.1.4 Let a, b be arbitrary elements of R. Then:

The equation a + x = b has a unique solution x = (-a) + b;If a0, the equation a.x = b has the unique solution x = (1/a).b

Proof:(i) since a + ((-a) + b)=(a + (-a))+b = 0 + b = b by properties above, we see that x = (-a) + b, is solution of the equation a + x =b.

To establish it is only solution, suppose that x1 is any solution of this equation; then:a + x1=b

If we add a to both sides , we get: (-a) + (a + x1)= (-a) + b, or((-a)+a) + x1=(-a) + b0 + x1 = (-a) + b x1 = (-a) + b,Thus x = x1 , this is the meaning of that unique.

The proof of part (b) is leftas an exercise.

Theorem 2.1.5If a is any element of R, Then:a.0 =0(-1).a = -a- (-a) = a(-1).(-1) = 1

Proof:a). We know that a.1= a, so a + a. 0=a.1 + a.0 = a. ( 1+0)=a.1=a Since 0 is unique, by theorem 2.1.2 (a), we conclude that: a.0 = 0b) a + (-1).a= a.1+ (-1). a =a.(1+(-1))= a. 0 = 0. Thus from Theorem 2.1.3 (a) we conclude that : (-1).a = -a

c) By (A5) we have: (-a) +a =0. According to the uniqueness assertion of theorem 2.1.3(a), it follows that: a = -(-a)

d) In part (b), substitute a = -1.Then (-1).(-1) = -(-1). Hence, the assertion follows from part (c) with a = 1

Theorem 2.1.6Let a, b, c be elements of R.

If a0, then 1/a0 and 1/(1/a) = a.If a.b = a.c and a0, then b = cIf a.b = 0, then either a=0 or b=0

Proof:a) We are given a0, so that 1/a exists. If 1/a = 0, then 1=a.(1/a)=a.0=0, contrary to (M4). Thus (1/a) 0 and since (1/a).a=1, the uniqueness asserted in theorem 2.1.3(b) implies 1/(1/a) = a

b) If we multiply both sides of the equation a.b=a.c by 1/a and apply the associative , we get:((1/a) . a).b=((1/a) . a).c. Thus , 1.b = 1.c which is the same as b = c.

c). It suffices to assume a0 and deduce that b=0. (Why?). Since a.b = 0=a.0, we apply part (b) to equation a.b = a.0 to conclude that b = 0 if a0.

IS DEFINED BYSubtruction: a b = a +(-b)Division: a/b = a.(1/b)Multiplication: a.b = aba.a =aa=a2,. (a2).a =(a2)a=a3 We agree to adopt the convention that a0=1 and a1 =a for any aRAlso, if a0, a-1=1/a and if nN we write a-n for (1/a)n

Prove that if a,b in R, then:. (a) (a+b)=(-a) + (-b)(b). (-a).(-b)=a.b 1/(-a) = -(1/a), a0

Exercise1. Solve the following equation:A. 2x + 5 = 8B. x2=2xC. (x-1)(x+2)=0

Proof that if a,bR, then: (a+b)=(-a) + (-b) Proof: -(a+b)= -1.(a+b) = (-1).a + (-1).b = (-a) + (-b)

b) (-a).(-b) = a.bProof: (-a).(-b) =(-1).a. (-1).b = (-1). (a.(-1)).b=(-1).((-1).a).b=(-1).(-1).(a).b=1.a.b=a.b

c) If a0, then 1/(-a) = -(1/a)Proof: (-a).1/(-a) = 1, since (-a). -(1/a)=(-a).(-1).(1/a) =(-1).a.(-1).(1/a)=(-1).((-1).a).(1/a)=((-1).(-1)).a.(1/a)=(1.a).(1/a)=a.(1/a)=1 since (-a).1/(-a) = 1, the uniqueness asserted in theorem 2.1.3(b) implies1/(-a) = -(1/a)

If a0 and b0, show that 1/(ab)=(1/a).(1/b)Proof:ab.(1/ab) = 1, sinceab.1/(a).1/(b) =a.b. 1/(a).1/(b) =a.( b.1/(a)).1/(b) =(a.1/(a)).(b.1/(b)) =1.1 = 1 we conclude that : (1/ab) = 1/(a).1/(b)

Prove that: -(a/b)= -a/(b)Proof:-(a/b) = (-1).(a/b) = (-1).a.(1/b) = ((-1).a).(1/b) =(-a).(1/b), since a/b = a.(1/b), thus (-a).(1/b)=-a/(b).

Rational NumbersElements of R that can be written in form b/a where a,bZ and a0 are called rational numbers.The set of all rational numbars denoted by the standard notation QThe sum and product of two rational numbers is again a rational numbers. (prove this)The fact that there are elements of R that are not in Q is not immediately apparentIn the sixth century B.C. the ancient Greek Pythagoreans discovered that the diagonal of square with unite sides could not be expressed as a ratio of interger. In view of the Pythagorean Theorem for right triangles, this implies that the square of no rational number can equal 2.

This discovery had considerable imfact on the development of Greek mathematics. One consequence is that elements of R that are not in Q become known as irrational numbers.

Theorem 2.1.7There does not exist a rational number r such that r2 = 2

Proof: Seppose, on this contrary, that p and q are integer that (p/q)2 = 2. It may be assumed that p and q have no common integer factors other than 1. (why?).

Since p2 =2q2, we see that p2 is even. This implies that p is even (for if p = 2k +1 is odd, then p2 =4k2 + 4k +1= 2(k2 + 2k) +1 is also odd.

Therefore, q must be an odd integer. However, p = 2m for some integer m because p is even, and hence 4m2 =2q2 so that 2m2 =q2. As above, it follows that q is even integer, and we have arrived at a contradiction to the fact that no integer is both even and odd.

2.2 The Order Properties Of R2.2.1 The order properties of R. There is non-empty subset P of R, called the set of stricly positive real numbers, that satisfies the following properties:(i) If a,b belong to P, then a + b belongs to P (ii) If a,b belong to P, then ab belongs to P (iii) If a belong to R, then exactly one of the following holds: aP, a=0, -aP

The set {-a| aP} strictly negativeDefinisi 2.2.2: If a is in P, we say that a is strictly positive real number and we write a>0.

If a is either in P or 0, we say that a is a positive real number and we write a0.

If a is in P, we say that a is stricly negative real number and we write a

2.2.3 Definition: Let a,b be elements of R(i) If a - bP, then we write a> b or b

Properties of the OrderOr rules of inequalitiesTheorem 2.2.4: Let a,b,c be elements of R If a>b and b>c,then a>cExactly one of the following holds: a>b, a=b, a

Proof:If a - bP,andb-cPimplies (a-b) + (b-c)= a-c belongs to P. Henc a>cBy the trichotomy property 2.2.1 (iii), exactly one of the following possibilities occurs: a - bP, a b =0, -(a b)=b-aPIf ab, then a - b0, so from part (b) we have either a - bP or b-aP:That is either a>b or b>a. In either case, one of the hypotheses is contradicted. There fore we must have a=b

Theorem2.2.5If aR and a0, then a2>01>0If nN, then n>0

Theorem 2.2.6Theorem 2.2.6a,b,c,d be elements of R(a) If a>b, Then a+c>b+c (b) If a>b and c>d, Then a+c>b+d (c) If a>b and c>0, Then ca>cb (d) If a>b and c0 (f) If a

Theorem 2.2.7If a and b are in R and if a>b, then a>(a+b)>b

Bukti: Karena a>b maka a + a > a+b, atau 2a> a+b. Karena a>b maka a+b>b+b atau a+b>2b. Akibatnya 2a>a+b>2b. Jadi a>(a+b)>b.

Corrolary 2.2.8If aR and a>0, Then a>a>0Proof: Take b=0, in 2.2.7if a>b, then a>(a+b)>b, so if b=0, thena>(a+0)>0, atau a>a>0

Theorem 2.2.9If aR is such that 0a0, maka menurut teorema 2.2.8 a>1/2a>0. Jika kita ambil 0=1/2a, maka kita akan peroleh a>o>0 sehingga adalah salah bahwa a< untuk setiap >0. Karena kita telah misalkan a>0 maka terjadi kontradiksi. Jadi a=0

Theorem 2.2.10If ab>0, then either (i) a>0 and b>0, or (ii) a0 maka berlaku 1/a.(ab)>0 atau b>0. Jadi (i) a>0 and b>0

Untuk a

Corrolary 2.2.11If ab0 and b

Examples 2.2.12(a) determine the set A of real numbers x such that 2x + 3 6(b) Determine the set B = {xR|x2+x>2}(c) Determine the set : C={xR|(2x+1)/(x+2)

(2x+1)/(x+2) - 1

Example 2.2.13(a) If 0

(b). Jika a,bP maka ab (a+b)(The Arithmetic-Geometric Mean Inequality)Bukti: Karena a,bP dan ab maka a>0, dan b>0, serta a b, maka (a -b)2 >0, atau(a - 2ab + b)>0a + b > 2ab (a+b)>ab. Sedangkan bila a = b, makaBerlaku: (a+a)=aa, atau (2a)=a a=a. Jadi, Jika a,bP maka ab (a+b).Atau bila (a+b)=ab, maka (a+b)2=4ab atau a2 -2ab +b2 =0 (a-b)2=0. Jadi a=b.

Remark: The general Arithmetic-Geometric Mean Inequality a1, a2, , anP is:(a1 a2 an)1/n (a1 + a2 + +an)/nIngat sama dengannya berlaku untuk a1 = a2 = =anPembuktiannya menggunakan induksi matematika dan cukup rumit. The more elegent proof that uses properties of the exponential function will be indicate in Chapter 7.

(c).Bernoullis inequalityIf x>-1, then (1+x)n1+nx, for all nN.Bukti: (Induksi matematika)Untuk n = 1 maka (1+x)1 (1+1x)Asumsikan berlaku untuk bilangan asli n, yaitu: (1+x)n1+nx (ingat 1+x>0) dan sekarang kita buktikan berlaku untuk bilangan asli n+1. (1+x)n+1 =(1+x)n (1+x)(1) Karena (1+x)n1+nx maka (1+x)n (1+x)(1+nx)(1+x).Akibatnya(1+x)n+1(1+nx)(1+x), atau(1+x)n+1 1+x +nx +nx2(1+x)n+1 1+(n+1)x +nx2 (1+x)n+1 1+(n+1)x . Jadi betul berlaku untuk semua n anggota bilangan asli n

(d). Cauch`s InequalityIf a1, a2, , an and b1, b2, , bn are real number, then: (a1b1 + +anbn)2 (a12 ++an2) (b12 ++bn2) Bukti: Untuk sembarang t R berlakulah: (a1 - tb1 )2 + (a2 tb2 )2 + + (an - tbn )2 0. If we expand the squares, we obtain:

A 2ABt + Ct2 0, where A= a12 ++an2 , B= a1b1 + +anbn , and C= b12 ++bn2. Karena A 2ABt + Ct2 positif atau nol untuk setiap t R, maka mestilah D 0, atau D= (2B)2 - 4AC 0, atau B2 AC . Jadi (a1b1 + +anbn)2 (a12 ++an2) (b12 ++bn2)

(e) The Triangle InequalityIf a1, a2, , an and b1, b2, , bn are real number, then: [(a1 + b1 )2 + + (an + bn )2]1/2 (a12 ++an2)1/2 + (b12 ++bn2)1/2 Bukti: (a1 + b1 )2 + + (an + bn )2 = A +2B +C A +2B +C A + 2AC + C = (A + C)2, Oleh Karena B = a1b1 + +anbn AC = (a12 ++an2) (b12 ++bn2), atau (a1b1 + +anbn)2 (a12 ++an2) (b12 ++bn2). Jadi: [(a1 + b1 )2 + + (an + bn )2]1/2 (a12 ++an2)1/2 + (b12 ++bn2)1/2