Tugas 1 (analisis Real)

Analisis Real (Tugas 1) Nama : Syaiful Rohman ( 107017001306 ) Teguh Membara ( 107017000966 ) Qosim Nur Hidayat ( 107017000931 ) Pendidikan Matematika 5B

Lemma 1.2.1 Misalkan S ⊂ N dan S ≠ Ø, maka S memiliki unsur terkecil yaitu terdapat no Є S,sehingga no ≤ n, ∀nЄS. 1. Misal S = { 1, 2, 3, 4} no = 1

1≤ 2 1≤ 4 1≤ 3 1≤ 1

no ≤ n

2. Misal S = { 3, 5, 7, 9} no = 3

3≤ 5 3≤ 7 3≤ 9 3≤ 3

no ≤ n

3. Misal S = { 4, 7, 9, 12, 15 } no = 4

4 ≤ 7 4 ≤ 9 4 ≤ 12 4 ≤ 15 4 ≤ 4 no ≤ n

4. Misal S = { 2, 3, 6, 7, 9, 10 } no = 2

2 ≤ 3 2 ≤ 6 2 ≤ 7 2 ≤ 9 2 ≤ 10 2 ≤ 2 no ≤ n

5. Misal S = { 1, 3, 5, 9, 13, 16 } no = 1

1 ≤ 3 1 ≤ 5 1 ≤ 9 1 ≤ 13 1 ≤ 16 1 ≤ 1 no ≤ n

Lemma 1.2.2 Jika x, y Є Q dan x < y maka terdapat z Є Q, sehingga x < z < y. 1. Misal m1 = 1 m2 = 3 n1 = 2 n2 = 4

x= z=

m1 1 = n1 2

y=

m2 3 = n2 4

x + y m1 n2 + m2 n1 1.4 + 3.2 5 = = = 2 2n1 n2 2.2.4 8

x
1 5 3 < < 2 8 4

2. Misal m1 = 5 m2 = 7 n1 = 6 n2 = 8

x= z=

m1 5 = n1 6

y=

m2 7 = n2 8

x + y m1 n 2 + m2 n1 5.8 + 7.6 41 = = = 2 2n1 n 2 2.6.8 48

x
5 41 7 < < 6 48 8

3. Misal m1 = 9 m2 = 11 n1 = 10 n2 = 12

x= z=

m1 9 = n1 10

y=

m 2 11 = n 2 12

x
x + y m1 n 2 + m2 n1 9.12 + 10 .11 109 9 109 11 = = = < < sehingga x < z < y  2 2n1 n2 2.10 .12 120 10 120 12

4. Misal m1 = 13 m2 = 15 n1 = 14 n2 = 16

x= z=

m1 13 = n1 14

y=

m 2 15 = n 2 16

x
x + y m1 n2 + m2 n1 13.16 + 15.14 209 13 209 15 = = = < < sehingga x < z < y  2 2n1 n 2 2.14.16 224 14 224 16

5. Misal m1 = 5 m2 = 9 n1 = 7 n2 = 11

x= z=

m1 5 = n1 7

y=

m2 9 = n2 11

x
x + y m1 n 2 + m2 n1 5.11 + 9.7 59 5 59 9 = = = < < sehingga x < z < y  2 2n1 n 2 2.7.11 77 7 77 11

Lemma 1.2.3 (Sifat Archimedes) Jika x Є Q, maka terdapat n Є Z. sehingga x < n. 1. Misalkan

P=2 q=5  x=

p 2 = q 5

p.q> 0 2.5>0  10>0 1

1

q ≥ 1  5≥ 1  q ≤ 1 ⇒ 5 ≤ 1 p

2 Mengakibatkan q ≤ p  ≤ 2 5

n = p + 1  2+1 =3 x
2 ≤ 3 (terbukti) 5

2. Misalkan

P=1 q=5 

p.q> 0 1.5>0  5>0 1

1 Mengakibatkan q ≤ p  ≤1 5 p

n = p + 1  1+1 =2 x
1 ≤ 2 (terbukti) 5

3. Misalkan

P=3 q=4  x=

p 3 = q 4

p.q> 0 3.4>0  12>0 1

p

n = p + 1  3+1 =4 3 ≤ 3 (terbukti) 4

4. Misalkan

1

q ≥ 1  5≥ 1  q ≤ 1 ⇒ 5 ≤ 1

3 Mengakibatkan q ≤ p  ≤ 3 4

x
1

q ≥ 1  5≥ 1  q ≤ 1 ⇒ 5 ≤ 1

p 1 x= = q 5

P=5 q=2  x=

p.q> 0 5.2>0  10>0

p 5 = q 2

1

1

q ≥ 1  5≥ 1  q ≤ 1 ⇒ 5 ≤ 1 p

5 Mengakibatkan q ≤ p  ≤ 5 2

n = p + 1  5+1 =6 x
5 ≤ 6 (terbukti) 2

5. Misalkan

P=4 q=1  x=

p.q> 0 4.1>0  4>0

p 4 = q 1

1

1

q ≥ 1  5≥ 1  q ≤ 1 ⇒ 5 ≤ 1

4 Mengakibatkan q ≤ p  ≤ 4 1 p

n = p + 1  4+1 =5 x
4 ≤ 3 (terbukti) 1

Teorema 1.4.1 ( Sifat Archimedes) Untuk setiap x, y Є R dan x > 0, terdapat n Є N, sehingga nx > y. 1. misal x=2 y=2 n=2 maka nx > y  2.2 > 2  4 > 2 2. misal x=4 y=-2 n=3 maka nx > y  3.4> -2  12> -2 3. misal x=

1 y=2 n=6 2

maka nx > y  6 .

1 > 2  3> 2 2

4. misal x=

1 y=-5 n=4 4

maka nx > y  4 . 5. misal x=

1 > -5  1> -5 4

3 y=1 n=8 2

3 > 1  12> 1 2

maka nx > y  8.

Teorema 1.4.2 Untuk setiap x, y Є R dan x < y, terdapat p Є Q, sehingga x < p < y. 1.

misal x=

1 5 y= n=5 2 4

n (y-x) > 1 5(

5 1 15 − )>1  >1 4 2 4

…………………….…………..(1)

m1 = 4 m2 = 4 - m2 < nx < m1  -4 < 5.

1 5 < 4  -4 < < 4 ………………… (2) 2 2

M=3 m-1 ≤ nx ≤ m 3-1 ≤ 5. maka di peroleh

5.

1 5 ≤ 32≤ ≤ 3 ………………………(3) 2 2

nx < m ≤ nx + 1 < ny

1 1 5 < 3 ≤ 5. + 1 < 5. 2 2 4

P=

m m 1 3 5 x<
5 <3 ≤ 2

7 25 < 2 4

(terbukti)

2.

misal x=

3 9 y= n=5 2 4

n (y-x) > 1 5(

9 3 15 − )>1  >1 4 2 4

…………………….…………..(1)

m1 = 3 m2 = 3 - m2 < nx < m1  -3 < 5.

3 5 < 3  -3 < < 3 ………………… (2) 2 2

M=8 m-1 ≤ nx ≤ m 8-1 ≤ 5.

3 15 ≤ 87≤ ≤ 8 ………………………(3) 2 2

maka di peroleh nx < m ≤ nx + 1 < ny

5.

3 3 9 < 8≤ 5. + 1 < 5. 2 2 4 15 <8≤ 2

P=

17 45 < 2 4

m m 3 8 9 x<
(terbukti)

3.

.missal x=

3 9 y= n=3 4 4

n (y-x) > 1 2(

9 3 − )>1  3 >1 4 4

…………………….…………..(1)

m1 = 3 m2 = 3 - m2 < nx < m1  -3 < 3.

3 9 < 3  -3 < < 3 ………………… (2) 4 4

M=3 m-1 ≤ nx ≤ m 7-1 ≤ 3.

3 9 ≤ 82≤ ≤ 3………………………(3) 4 4

maka di peroleh nx < m ≤ nx + 1 < ny

2.

3 3 9 < 3≤ 2. + 1 < 2. 4 4 4 6 <3≤ 4

P=

4.

10 18 < 4 4

m m 3 3 9 x<
misal x=

1 5 y= n=3 2 4

(terbukti)

n (y-x) > 1 4(

5 1 12 − )>1  >1 4 2 4

…………………….…………..(1)

m1 = 4 m2 = 4 - m2 < nx < m1  -4 < 3.

1 1 < 4  -4 < 3. < 4 ………………… (2) 2 2

M=2 m-1 ≤ nx ≤ m 2-1 ≤ 3.

1 3 ≤ 21≤ ≤ 2 ………………………(3) 2 2

maka di peroleh nx < m ≤ nx + 1 < ny

3.

1 1 5 < 2 ≤ 3. + 1 < 3. 2 2 4 3 <2 ≤ 2

P=

5.

5 15 < 2 4

m m 1 3 5 x<
missal x=

(terbukti)

1 1 y= n=3 8 2

n (y-x) > 1 3(

1 1 9 − )>1  >1 2 8 8

…………………….…………..(1)

m1 = 5 m2 = 5 - m2 < nx < m1  -5 < 3. M=1

1 3 < 5  -5 < < 5………………… (2) 8 8

m-1 ≤ nx ≤ m 1-1 ≤ 3.

1 3 ≤ 20≤ ≤ 1………………………(3) 8 8

maka di peroleh nx < m ≤ nx + 1 < ny

3.

1 1 1 < 1 ≤ 3. + 1 < 3. 8 8 2 3 <1 ≤ 8

P=

11 3 < 8 2

m m 1 1 1 x<
(terbukti)

Teorema 1.4.3 Untuk Setiap a Є R, a > 0 dan n Є N, terdapat x Є R, sehingga xn = a. 1. Misal a = 4 dan n = 5 sehingga ∃ xЄR

Sehingga x5 = 4

 x = 5 4 = 1,319507911

xЄI

2. Misal a = 6 dan n = 7 sehingga ∃ xЄR

Sehingga x7 = 6

 x = 7 6 = 1,291708342

xЄI

3. Misal a = 8 dan n = 9 sehingga ∃ xЄR

Sehingga x9 = 8

 x = 9 8 = 1,259921050

xЄI

4. Misal a = 10 dan n = 11 sehingga ∃ xЄR

Sehingga x11 = 10  x = 11 10 = 1,23284673

xЄI

5. Misal a = 12 dan n = 13 sehingga ∃ xЄR

Sehingga x13 = 12  x = 13 12 = 1,23007551

xЄI