Transcript ANALISIS VARIABEL REAL I

ANALISIS VARIABEL REAL I
BAB II
SISTEM 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. 2.1.1
Sistem 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
(associative property 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
(associative property 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 a≠0 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 b≠0 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 b≠0
≠0
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
=u
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:
(i) The equation a + x = b has a unique
solution x = (-a) + b;
(ii) If a≠0, 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) + b
0 + x1 = (-a) + b
→x
1
= (-a) + b,
Thus x = x1 , this is the meaning of that unique.
The proof of part (b) is left
as an exercise.⌂
Theorem 2.1.5
If a is any element of R, Then:
a) a.0 =0
b) (-1).a = -a
c) - (-a) = a
d) (-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 = 0
b) 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 2.1.6
•
Let a, b, c be elements of R.
a) If a≠0, then 1/a≠0 and 1/(1/a) = a.
b) If a.b = a.c and a≠0, then b = c
c) If a.b = 0, then either a=0 or b=0
Proof:
• a) We are given a≠0, 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
1/a and apply the associative
, we
a.b=a.c by
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 a≠0 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 a≠0.
IS DEFINED BY
•
•
•
•
•
Subtruction: a – b = a +(-b)
Division: a/b = a.(1/b)
Multiplication: a.b = ab
a.a =aa=a2,. (a2).a =(a2)a=a3
We agree to adopt the convention that a0=1 and
a1 =a for any a∈R
• Also, if a≠0, a-1=1/a and if n∈N 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), a≠0
Exercise
• 1. Solve the following equation:
A. 2x + 5 = 8
B. x2=2x
C. (x-1)(x+2)=0
Proof that if a,b∈R, then:
(a)
–(a+b)=(-a) + (-b)
Proof: -(a+b)= -1.(a+b)
= (-1).a + (-1).b
= (-a) + (-b)
• b) (-a).(-b) = a.b
Proof: (-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 a≠0, 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) implies
1/(-a) = -(1/a)
If a≠0 and b≠0, show that
1/(ab)=(1/a).(1/b)
• Proof:
ab.(1/ab) = 1, since
ab.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 Numbers
• Elements of R that can be written in form b/a where a,b∈Z
and a≠0 are called rational numbers.The set of all rational
numbars denoted by the standard notation Q
• The 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 apparent
• In 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 2.1.7
•
There 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 R
• 2.2.1 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:
a∈P, a=0, -a∈P
The set {-a| a∈P} strictly negative
• Definisi 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 a≥0.
If –a is in P, we say that a is stricly negative real number
and we write a<0.
If -a is either in P or 0, we say that a is a negative real
number and we write a≤0
2.2.3 Definition: Let a,b be elements of R
(i)
If a - b∈P, then we write a> b or
b<a
(ii) If
a - b∈P∪{0},
Then we write a≥b
or b≤a
For notational convenience, we shall
write a<b<c, to mean that both a<b
and b<c are stisfied.
Properties of the Order
• “Or rules of inequalities”
Theorem 2.2.4: Let a,b,c be elements of R
(a) If a>b and b>c,then a>c
(b) Exactly one of the following holds: a>b,
a=b, a<b
(c) If a ≥ b dan b ≥ a,Then a = b
Proof:
(a) If a - b∈P,andb-c∈Pimplies (a-b) + (b-c)=
a-c belongs to P. Henc a>c
(b) By the trichotomy property 2.2.1 (iii), exactly
one of the following possibilities occurs: a b∈P, a – b =0, -(a – b)=b-a∈P
(c) If a≠b, then a - b≠0, so from part (b)
we have either a - b∈P or b-a∈P:That is
either a>b or b>a. In either case, one of
the hypotheses is contradicted. There
fore we must have a=b
Theorem2.2.5
(a) If a∈R and a≠0, then a2>0
(b) 1>0
(c) If n∈N, then n>0
Theorem 2.2.6
• Theorem 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 c<0, then ca<cb
(e) If a>0, then 1/a>0
(f) If a<0, then 1/a<0
Theorem 2.2.7
• If a and b are in R and if a>b, then
a>½(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>½(a+b)>b.
Corrolary 2.2.8
• If a∈R and a>0, Then a>½a>0
Proof: Take b=0, in 2.2.7
if a>b, then a>½(a+b)>b, so if b=0, then
a>½(a+0)>0, atau
a>½a>0
Theorem 2.2.9
• If a∈R is such that 0≤a<ε for every
strictly positive ε in R, then a=0
Bukti:Misal a>0, maka menurut teorema
2.2.8 a>1/2a>0. Jika kita ambil ε0=1/2a,
maka kita akan peroleh a>εo>0
sehingga adalah salah bahwa a< ε untuk setiap
ε>0. Karena kita telah misalkan a>0
maka terjadi kontradiksi. Jadi a=0
Theorem 2.2.10
• If ab>0, then either (i) a>0 and b>0, or (ii) a<0 and b<0
• Bukti: jika ab>0 maka a≠0 dan b≠0.
Bila a≠0 maka a<0 atau a>0. Untuk a>0 maka berlaku
1/a.(ab)>0 atau b>0. Jadi (i) a>0 and b>0
Untuk a<0 maka berlaku 1/a (ab) <0 atau b<0. Jadi
berlaku (ii) a<0 and b<0
Corrolary 2.2.11
• If ab<0, then either (i) a<0 and b>0, or (ii)
a>0 and b<0
• Buktikanlah !
Examples 2.2.12
• (a) determine the set A of real numbers x
such that 2x + 3 ≤6
• (b) Determine the set B = {x∈R|x2+x>2}
• (c) Determine the set :
C={x∈R|(2x+1)/(x+2)<1}
• (2x+1)/(x+2) - 1<0
(2x+1)/(x+2) – (x+2)/(x+2)<0
(x-1)/(x+2)<0……
(jika ab<0 maka a<0 dan b>0, atau a>0
dan b<0)
(x-1).1/(x+2)<0
Example 2.2.13
• (a) If 0<a<b then a2<b2 and √a < √b
Bukti: (b-a), dan (b+a) ∈P
Berarti (b+a)(b-a) ∈P, atau
b2-a2 ∈P . Jadi a2< b.
Karena (√b -√a) (√b+ √a)=b-a, maka
(√b -√a) ∈P, dan (√b+ √a) ∈P . Jadi
√b >√a.
(b). Jika a,b∈P maka √ab ≤ ½(a+b)
(The Arithmetic-Geometric Mean Inequality)
• Bukti: Karena a,b∈P dan a≠b maka √a>0, dan √b>0,
serta √a ≠√b, maka (√a -√b)2 >0, atau
(a - 2√ab + b)>0
⇔a + b > 2√ab
⇔ ½(a+b)>√ab. Sedangkan bila a = b, maka
Berlaku: ½(a+a)=√aa, atau
⇔ ½(2a)=a
⇔ a=a. Jadi, Jika a,b∈P maka √ab ≤ ½(a+b).
Atau bila ½(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, …, an∈P is:
• (a a … a ) ≤ (a + a +… +a )/n
• Ingat sama dengannya berlaku untuk
1
2
n
1/n
1
2
n
a1 = a2 =… =an
Pembuktiannya menggunakan induksi matematika
dan cukup rumit.
The more elegent proof that uses properties of the
exponential function will be indicate in Chapter 7.
(c).Bernoulli’s inequality
• If x>-1, then (1+x)n≥1+nx, for all n∈N.
Bukti: (Induksi matematika)
Untuk n = 1 maka (1+x)1 ≥ (1+1x)
Asumsikan berlaku untuk bilangan asli n, yaitu:
(1+x)n≥1+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)n≥1+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 Inequality
• If 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 Inequality
• If 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 + 2√AC + 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