pertemuan 4(sistem bilangan)
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Sistem BilanganSistem BilanganSistem BilanganSistem Bilangan
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Terdapat 4 sistem bilangan yaitu :
1. Bilangan Desimal (10)
2. Bilangan Biner(2)
3. Bilangan Oktal(8)
4. Bilangan Hexadesimal(16)
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Sistem Bilangan Desimal
• Bentuk nilai suatu bil.desimal dapat berupa integer desimal (decimal integer) atau pecahan decimal (decimal fraction)
• Integer Desimal adalah nilai desimal yang bulat.
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contoh 357 artinya :
3 x 102 = 3005 x 101 = 507 x 100 = 7 -------- + 357
absolute value
Position value
Absolute value : nilai mutlak dari masing-masing digit.Position value : bobot dari masing-masing digit tergantung
dari letak/ posisinya.
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Pecahan Desimal adalah nilai desimal yang mengandung nilai pecahan di belakang koma.contoh 173,25 artinya :
1 x 102 = 1007 x 101 = 703 x 100 = 32 x 10-1 = 0,25 x 10-2 = 0,05
----------- + 173,25
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Penjumlahan Sistem Bilangan Desimal
Contoh : 458 + 67 = …….(10)
11458 67----- +525 8+7=15, 15/10=5 carry of (di bawa) 1
1+5+6=12, 12/10=2 carry of 11+4=5
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Pengurangan Sistem Bilangan DesimalContoh : 524 - 78 = …….(10)
524 78----- -446 4-8=x, borrow of (pinjam) 1->10, 10+4-8=14-8=6
2 diambil 1 tinggal 1-7=x, 10+1-7=11-7=45-1=4
Perkalian Sistem Bilangan DesimalContoh : 57 x 24 = …….(10)
57 34 ----- x 228 4x7=28, 28/10=2 sisa 8 171 4x5=20+2=22, 22/10=2 sisa 2------- + 3x7=21, 21/10=2 sisa 1 1938 3x5=15+2=17, 17/10=1 sisa 7
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Pembagian Sistem Bilangan Desimal
Contoh : 125 : 5 = …….(10)
5/ 125 \ 25 10 ---- - 25 25 ---- - 0
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Sistem Bilangan Biner
• Bentuk nilai suatu bil.biner dapat berupa integer biner (binary integer) atau pecahan biner (binary fraction)
• Integer Biner adalah nilai biner yang bulat.
contoh 1001 artinya :
1302 01 10= (1x23)+(0x22)+(0x21)+(1x20) = (1x8)+(0x4)+(0x2)+(1x1) = 8+0+0+1 = 910
1 0 0 11 x 20 = 10 x 21 = 00 x 22 = 01 x 23 = 8
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Position value biner
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Pecahan biner
contoh 0.111 artinya :
1 x 2-1 = 1/2 atau 0.51 x 2-2 = 1/4 atau 0.251 x 2-3 = 1/8 atau 0.125
----------- + 0.875(10)
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Penjumlahan Sistem Bilangan Biner
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Pengurangan Sistem Bilangan Biner
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Perkalian Sistem Bilangan Biner
Prinsip : 0 x 0 = 0 0 x 1 = 0 1 x 0 = 0 1 x 1 = 1
Contoh : 101 x 11 = …….(2)
101 11----- x101
101 ------- + 1111
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Pembagian Sistem Bilangan Biner
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Sistem Bilangan Oktal• Bentuk nilai suatu bil.oktal dapat berupa integer octal
(octal integer) atau pecahan oktal (octal fraction)• Integer Oktal adalah nilai oktal yang bulat.
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Position value oktal
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Penjumlahan Sistem Bilangan Oktal
Contoh : 376(8) + 45(8) = …….(8)
11376 45------ +443 5+6=11, 11/8=1 sisa 3
1+7+4=12, 12/8=1 sisa 4
1+3=4
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Pengurangan Sistem Bilangan Oktal
Contoh : 4352(8) - 764(8) = …….(8)
4352 764------ -3366 2-4=x, bo 1, 1->8, 8+2-4=10-4=6
5-1=4, 4-6=x, bo 1, 1->8, 8+4-6=12-6=6
3-1=2, 2-7=x, bo 1, 1->8, 8+2-7=10-7=3
4-1=3
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Perkalian Sistem Bilangan Oktal
Contoh : 56(8) x 43(8) = …….(8)
5643
------ x 212 3x6=18, 18/8=2 sisa 2 270 3x5=15+2=17, 17/8=2 sisa 1 ------- + 4x6=24, 24/8=3 sisa 0 3112 4x5=20+3=23, 23/8=2 sisa 7
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Pembagian Sistem Bilangan Oktal
Contoh : 250(8) : 14(8) = …….(8)
14 / 250 \ -> 14 14 6 ------ - ----- x 110 110 <-- 6x4=24,24/8=3
sisa 0 110 6x1=6+3=9, 9/8=1
sisa 1 ------ -
0
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Sistem Bilangan Hexadesimal• Bentuk nilai suatu bil.hexa dapat berupa integer hexa (hexa
integer) atau pecahan hexa (hexa fraction)• Integer Hexa adalah nilai hexa yang bulat.
contoh 152B(16) artinya : 152B(16) = (1x163) + (5x162) +(2x161) +(Bx160)
= (1x4096) + (5x256) + (2x16) + (11x1)
= 4096 + 1280 + 32 + 11= 5419(10)
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Penjumlahan Sistem Bilangan Hexa
Contoh : 176(16) +8C(16) = …….(16)
176 8C
------ +202 6+C=6+12=18, 18/16=1 sisa 2
7+8=15+1=16, 16/16=1 sisa 01+1=2
Pengurangan Sistem Bilangan Hexa
Contoh : B435(16) – A7D(16) = …….(16)
B435 A7D ------ - A9B8 5-D=5-13=x, bo 1, 1->16, 16+5-13=21-
13=8 3-1=2, 2-7=x, bo 1, 1->16, 16+2-7=18-
7=11 (B) 4-1=3, 3-A=3-10=x, bo 1, 1->16, 16+3-
10=19-10=9 B-1=11-1=10 (A)
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Perkalian Sistem Bilangan Hexa
Contoh : 5C(16) x 76(16) = …….(16)
5C 76 ------ x 228 6xC=6x12=72, 72/16=4 sisa
8 284 6x5=30+4=34, 34/16=2 sisa
2 ------- + 7xC=7x12=84, 84/16=5 sisa
4 2A68 7x5=35+5=40, 40/16=2 sisa
8
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Pembagian Sistem Bilangan Hexa
Contoh : 1224(16) : 1B(16) = …….(16)
1B / 1224 \AC -> 1B 10E A ------ - ----- x 144 10E <-- AxB=10x11=110,110/16=6 sisa
14(E) 144 Ax1=10+6=16, 16/16=1 sisa 0 ------ - -> 1B
0 C ---- x
144 <-- CxB=12x11=132,132/16=8 sisa 4
Cx1=12+8=20, 20/16=1 sisa 4
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Tugas Kerjakan di Rumah :1. 11102 + 10112 =………..2
2. 100112 - 10112 =………..2
3. 1102 x 112 =………..2
4. 1000112 : 1012 =………..2
5. 4678 + 568 =………..8
6. 24318 - 758 =………..8
7. 768 x 438 =………..8
8. 3608 : 148 =………..8
9. 4AB16+7816 =………..16
10. 236416-FA16 =………..16
11. 9716xA416 =………..16
12. D1A16:2B16 =………..16