c2

- 2.1 -

Chapter 2Problem Solutions

2.1.

Decimal Binary Ternary Octal Hexadecimal

32 100000 1012 40 20

33 100001 1020 41 21

34 100010 1021 42 22

35 100011 1022 43 23

36 100100 1100 44 24

37 100101 1101 45 25

38 100110 1102 46 26

39 100111 1110 47 27

40 101000 1111 50 28

41 101001 1112 51 29

42 101010 1120 52 2A

43 101011 1121 53 2B

44 101100 1122 54 2C

45 101101 1200 55 2D

46 101110 1201 56 2E

47 101111 1202 57 2F

48 110000 1210 60 30

49 110001 1211 61 31

50 110010 1212 62 32

- 2.2 -

2.2.Decimal Quaternary Quinary Duodecimal

0 0 0 0

1 1 1 1

2 2 2 2

3 3 3 3

4 10 4 4

5 11 10 5

6 12 11 6

7 13 12 7

8 20 13 8

9 21 14 9

10 22 20 A

11 23 21 B

12 30 22 10

13 31 23 11

14 32 24 12

15 33 30 13

16 100 31 14

17 101 32 15

18 102 33 16

19 103 34 17

20 110 40 18

21 111 41 19

22 112 42 1A

23 113 43 1B

24 120 44 20

25 121 100 21

26 122 101 22

27 123 102 23

28 130 103 24

29 131 104 25

30 132 110 26

31 133 111 27

- 2.3 -

2.3. (a) 1 1 (b) 1111 (c) 11 11 1 1101 1001 1100.01

+ 1001_____ + 111_____ + 101.11________ 10110 10000 10010.00

(d) 111 11 (e) 111 (f) 1 11 11.011 111.01 0.1110+ 10.111_______ + 11.10_______ +0.1011______ 110.010 1010.11 1.1001

2.4. (a) 00 (b) 001 (c) 0110 1 1/1/01 1/1/0/01.1 1/0/0/1/.0/0- 110____ - 1011.0_______ - 11.01_______ 111 1110.1 101.11

(d) 0 0 (e) 01 0 (f) 0 1 1/01/.01 1/0/0.11/01 1011/.0/0- 10.10______ - 11.1010________ - 10.11_______

10.11 1.0011 1000.01

2.5. (a) 10111 (b) 11011 × 110_____ × 1011_____ 00000 11011 10111 11011 10111 ________ 0000010001010 11011 _________

100101001

(c) 1010 (d) 101.1 ×1.01____ ×11.01_____ 10 10 1 011 000 0 00 001010 _______ 101 11100.10 1011 _________

10001.111

2.6. (a) 10.1________ (b) 110011____________1010)11001.0 10101)10000101111 -1010____ - 10101______ 101 0 11000 -101 0_____ -10101_____ 0 11111

-10101_____ 10101 -10101_____ 0

- 2.4 -

2.6. (continued)

(c) 110.101___________ (d) 101 1.01____________110)100111.110 101.1 )111101.1 11 - 110____ -1011____ 111 10001 -110___ - 1011_____ 11 1 110 1 -11 0____ -101 1_____ 110 1 0 11 -110___ -1 0 11______ 0 0

2.7. (a) 1 1 (b) 11 1 (c) 120 2012.0 220.12 2/0/1/02+ 1102.1_______ + 121.20_______ -12121_____ 10121.1 1112.02 211

(d) 012 1 (e) 120.21 (f) 12202 1/2/0/02/.12 × 122______ × 21.2_____- 2121.20________ 1011 12 10211 1 2110.22 10111 2 12202

12021 _________ 102111 _________100221.02 1122000.1

(g) 210.2_________ (h) 1 2.02___________1012)221200.1 212.1 )11111.2 12

-2101____ - 2121_____ 1110 1220 2 -1012____ -1201 2______ 210 1 12 0 12 -210 1_____ -12 0 12_______ 0 0

2.8. (a) Quaternary

a+b b

0 1 2 3

a

0 0 1 2 3

1 1 2 3 1:0

2 2 3 1:0 1:1

3 3 1:0 1:1 1:2

Notation: x:y denotes carry digit x and sum digit y.

- 2.5 -

2.8. (continued)

a-b b

0 1 2 3

a

0 0 1:3 1:2 1:1

1 1 0 1:3 1:2

2 2 1 0 1:3

3 3 2 1 0

Notation: x:y denotes borrow digit x and difference digit y.

a×b b

0 1 2 3

a

0 0 0 0 0

1 0 1 2 3

2 0 2 1:0 1:2

3 0 3 1:2 2:1

Notation: x:y denotes carry digit x and product digit y.

(b) Octal

a+b b

0 1 2 3 4 5 6 7

a

0 0 1 2 3 4 5 6 7

1 1 2 3 4 5 6 7 1:0

2 2 3 4 5 6 7 1:0 1:1

3 3 4 5 6 7 1:0 1:1 1:2

4 4 5 6 7 1:0 1:1 1:2 1:3

5 5 6 7 1:0 1:1 1:2 1:3 1:4

6 6 7 1:0 1:1 1:2 1:3 1:4 1:5

7 7 1:0 1:1 1:2 1:3 1:4 1:5 1:6

Notation: x:y denotes carry digit x and sum digit y.

- 2.6 -

2.8. (continued)

a-b b

0 1 2 3 4 5 6 7

a

0 0 1:7 1:6 1:5 1:4 1:3 1:2 1:1

1 1 0 1:7 1:6 1:5 1:4 1:3 1:2

2 2 1 0 1:7 1:6 1:5 1:4 1:3

3 3 2 1 0 1:7 1:6 1:5 1:4

4 4 3 2 1 0 1:7 1:6 1:5

5 5 4 3 2 1 0 1:7 1:6

6 6 5 4 3 2 1 0 1:7

7 7 6 5 4 3 2 1 0

Notation: x:y denotes borrow digit x and difference digit y.

a×b b

0 1 2 3 4 5 6 7

a

0 0 0 0 0 0 0 0 0

1 0 1 2 3 4 5 6 7

2 0 2 4 6 1:0 1:2 1:4 1:6

3 0 3 6 1:1 1:4 1:7 2:2 2:5

4 0 4 1:0 1:4 2:0 2:4 3:0 3:4

5 0 5 1:2 1:7 2:4 3:1 3:6 4:3

6 0 6 1:4 2:2 3:0 3:6 4:4 5:2

7 0 7 1:6 2:5 3:4 4:3 5:2 6:1

Notation: x:y denotes carry digit x and product digit y.

- 2.7 -

2.8. (continued)

(c) Hexadecimal

a+b b

0 1 2 3 4 5 6 7 8 9 A B C D E F

a

0 0 1 2 3 4 5 6 7 8 9 A B C D E F

1 1 2 3 4 5 6 7 8 9 A B C D E F 1:0

2 2 3 4 5 6 7 8 9 A B C D E F 1:0 1:1

3 3 4 5 6 7 8 9 A B C D E F 1:0 1:1 1:2

4 4 5 6 7 8 9 A B C D E F 1:0 1:1 1:2 1:3

5 5 6 7 8 9 A B C D E F 1:0 1:1 1:2 1:3 1:4

6 6 7 8 9 A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5

7 7 8 9 A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6

8 8 9 A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7

9 9 A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8

A A B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9

B B C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A

C C D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A 1:B

D D E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A 1:B 1:C

E E F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A 1:B 1:C 1:D

F F 1:0 1:1 1:2 1:3 1:4 1:5 1:6 1:7 1:8 1:9 1:A 1:B 1:C 1:D 1:E

Notation: x:y denotes carry digit x and sum digit y.

- 2.8 -

2.8. (continued)

a-b b

0 1 2 3 4 5 6 7 8 9 A B C D E F

a

0 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5 1:4 1:3 1:2 1:1

1 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5 1:4 1:3 1:2

2 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5 1:4 1:3

3 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5 1:4

4 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6 1:5

5 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7 1:6

6 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8 1:7

7 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9 1:8

8 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A 1:9

9 9 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B 1:A

A A 9 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C 1:B

B B A 9 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D 1:C

C C B A 9 8 7 6 5 4 3 2 1 0 1:F 1:E 1:D

D D C B A 9 8 7 6 5 4 3 2 1 0 1:F 1:E

E E D C B A 9 8 7 6 5 4 3 2 1 0 1:F

F F E D C B A 9 8 7 6 5 4 3 2 1 0

Notation: x:y denotes borrow digit x and difference digit y.

- 2.9 -

2.8. (continued)

a×b b

0 1 2 3 4 5 6 7 8 9 A B C D E F

a

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 0 1 2 3 4 5 6 7 8 9 A B C D E F

2 0 2 4 6 8 A C E 1:0 1:2 1:4 1:6 1:8 1:A 1:C 1:E

3 0 3 6 9 C F 1:2 1:5 1:8 1:B 1:E 2:1 2:4 2:7 2:A 2:D

4 0 4 8 C 1:0 1:4 1:8 1:C 2:0 2:4 2:8 2:C 3:0 3:4 3:8 3:C

5 0 5 A F 1:4 1:9 1:E 2:3 2:8 2:D 3:2 3:7 3:C 4:1 4:6 4:B

6 0 6 C 1:2 1:8 1:E 2:4 2:A 3:0 3:6 3:C 4:2 4:8 4:E 5:4 5:A

7 0 7 E 1:5 1:C 2:3 2:A 3:1 3:8 3:F 4:6 4:D 5:4 5:B 6:2 6:9

8 0 8 1:0 1:8 2:0 2:8 3:0 3:8 4:0 4:8 5:0 5:8 6:0 6:8 7:0 7:8

9 0 9 1:2 1:B 2:4 2:D 3:6 3:F 4:8 5:1 5:A 6:3 6:C 7:5 7:E 8:7

A 0 A 1:4 1:E 2:8 3:2 3:C 4:6 5:0 5:A 6:4 6:E 7:8 8:2 8:C 9:6

B 0 B 1:6 2:1 2:C 3:7 4:2 4:D 5:8 6:3 6:E 7:9 8:4 8:F 9:A A:5

C 0 C 1:8 2:4 3:0 3:C 4:8 5:4 6:0 6:C 7:8 8:4 9:0 9:C A:8 B:4

D 0 D 1:A 2:7 3:4 4:1 4:E 5:B 6:8 7:5 8:2 8:F 9:C A:9 B:6 C:3

E 0 E 1:C 2:A 3:8 4:6 5:4 6:2 7:0 7:E 8:C 9:A A:8 B:6 C:4 D:2

F 0 F 1:E 2:D 3:C 4:B 5:A 6:9 7:8 8:7 9:6 A:5 B:4 C:3 D:2 E:1

Notation: x:y denotes carry digit x and product digit y.

2.9. (a) 11 1 (b) 0233 (c) 1111 1 31213(4) 1/3/0/0/12(4) 466735(8)+ 23102(4)_________ - 33321(4)_________ + 375627(8)__________ 120321(4) 30031(4) 1064564(8)

(d) 6775 (e) 1 1 (f) C5 AF4 7/0/0/6/05(8) 8C9F65(16) D/6/2B/0/5/3(16)-356742(8)_________ +374B27(16)__________ -47E3C89(16)___________ 321643(8) C3EA8C(16) 8E473CA(16)

- 2.10 -

2.10. (a) 101101.1(2) = 1×105 + 0×104 + 1×103 + 1×102 + 0×101

+ 1×100 + 1×10-1

/ 1×25 + 0×24 + 1×23 + 1×22 + 0×21

+ 1×20 + 1×2-1

= 32 + 0 + 8 + 4 + 0 + 1 + 0.5

= 45.5(10)

(b) 110111.101(2) = 1×105 + 1×104 + 0×103 + 1×102 + 1×101

+ 1×100 + 1×10-1 + 0×10-2 + 1×10-3

/ 1×25 + 1×24 + 0×23 + 1×22 + 1×21

+ 1×20 + 1×2-1 + 0×2-2 + 1×2-3

= 32 + 16 + 0 + 4 + 2 + 1 + 0.5 + 0

+ 0.125

= 55.625(10)

(c) 2110(3) = 2×103 + 1×102 + 1×101 + 0×100

/ 2×33 + 1×32 + 1×31 +1×30

= 54 + 9 + 3 + 0

= 66(10)

(d) 12021.1(3) = 1×104 + 2×103 + 0×102 + 2×101 + 1×100

+ 1×10-1

/ 1×34 + 2×33 + 0×32 + 2×31 + 1×30 + 1×3-1

= 81 + 54 + 0 + 6 + 1 + 0.333...

= 142.333...(10)

(e) 362(8) = 3×102 + 6×101 + 2×100

/ 3×82 + 6×81 + 2×80

= 192 + 48 + 2

= 242(10)

(f) 1475.2(8) = 1×103 + 4×102 + 7×101 + 5×100 + 2×10-1

/ 1×83 + 4×82 + 7×81 + 5×80 + 2×8-1

= 512 + 256 + 56 + 5 + 0.25

= 829.25(10)

- 2.11 -

2.10. (continued)

(g) 2C3(16) = 2×102 + C×101 + 3×100

/ 2×162 + 12×161 + 3×160

= 512 + 192 + 3

= 707(10)

(h) AD.E(16) = A×101 + D×100 + E×10-1

/ 10×161 + 13×160 + 14×16-1

= 160 + 13 + 0.875

= 173.875(10)

2.11. (a) 42(10) = 4×101 + 2×100

/ 100×10101 + 10×10100

= 101000 + 10

= 101010(2)

(b) 78.5(10) = 7×101 + 8×100 + 5×10-1

/ 111×10101 + 1000×10100 + 101×1010-1

= 1000110 + 1000 + 0.1

= 1001110.1(2)

(c) 201(3) = 2×102 + 0×101 + 1×100

/ 10×112 + 0×111 + 1×110

= 10010 + 0 + 1

= 10011(2)

(d) 21.2(3) = 2×101 + 1×100 + 2×10-1

/ 10×111 + 1×110 + 10×11-1

= 110 + 1 + 0.101010...

= 111.101010...(2)

(e) 204(8) = 2×102 + 0×101 + 4×100

/ 10×10002 + 0×10001 + 100×10000

= 10000000 + 0 + 100

= 10000100(2)

- 2.12 -

2.11. (continued)

(f) 56.3(8) = 5×101 + 6×100 + 3×10-1

/ 101×10001 + 110×10000 + 11×1000-1

= 101000 + 110 + 0.011

= 101110.011(2)

2.12. (a) 11010(2) = 1×104 + 1×103 + 0×102 + 1×101 + 0×100

/ 1×24 + 1×23 + 0×22 + 1×21 + 0×20

= 1×121 + 1×22 + 0×11 + 1×2 + 0×1

= 121 + 22 + 0 + 2 + 0

= 222(3)

(b) 73.2(8) = 7×101 + 3×100 + 2×10-1

/ 21×221 + 10×220 + 2×22-1

= 2002 + 10 + 0.020202...

= 2012.020202...(3)

(c) 75(10) = 7×101 + 5×100

/ 21×1011 + 12×1010

= 2121 + 12

= 2210(3)

(d) 3D(16) = 3×101 + D×100

/ 10×1211 + 111×1210

= 1210 + 111

= 2021(3)

- 2.13 -

2.13. Using the polynomial method of converting 225(b) to

decimal:

225(b) = 2×102 + 2×101 + 5×100

/ (2×b2 + 2×b1 + 5×b0)(10) = 89(10)

Solving the quadradic 2b2+2b-84=0 for b:

b = +6 or -7

Since the base of a number system is defined as the number

of symbols in the system, the base must be positive.

ˆ b = 6

2.14. (a) Remainders Integers

163÷2=81 1(10)/1(2) 0.75×2=1.50 1(10)/1(2)

81÷2=40 1(10)/1(2) 0.50×2=1.00 1(10)/1(2) 40÷2=20 0(10)/0(2) 20÷2=10 0(10)/0(2) 10÷2= 5 0(10)/0(2) 5÷2= 2 1(10)/1(2) 2÷2= 1 0(10)/0(2) 1÷2= 0 1(10)/1(2)

ˆ 163.75(10) / 10100011.11(2)

Remainders Integers

202÷2=101 0(10)/0(2) 0.9×2=1.8 1(10)/1(2)101÷2= 50 1(10)/1(2) 0.8×2=1.6 1(10)/1(2) 50÷2= 25 0(10)/0(2) 0.6×2=1.2 1(10)/1(2) 25÷2= 12 1(10)/1(2) 0.2×2=0.4 0(10)/0(2) 12÷2= 6 0(10)/0(2) 0.4×2=0.8 0(10)/0(2) 6÷2= 3 0(10)/0(2) 3÷2= 1 1(10)/1(2) 1÷2= 0 1(10)/1(2)

ˆ 202.9(10) / 11001010.111001100...

(2)

- 2.14 -

2.14. (continued)

(b) Remainders Integers

163÷3=54 1(10)/1(3) 0.75×3=2.25 2(10)/2(3)

54÷3=18 0(10)/0(3) 0.25×3=0.75 0(10)/0(3) 18÷3= 6 0(10)/0(3) 6÷3= 2 0(10)/0(3) 2÷3= 0 2(10)/2(3)

ˆ 163.75(10) / 20001.2020...

(3)

Remainders Integers

202÷3=67 1(10)/1(3) 0.9×3=2.7 2(10)/2(3) 67÷3=22 1(10)/1(3) 0.7×3=2.1 2(10)/2(3) 22÷3= 7 1(10)/1(3) 0.1×3=0.3 0(10)/0(3) 7÷3= 2 1(10)/1(3) 0.3×3=0.9 0(10)/0(3) 2÷3= 0 2(10)/2(3)

ˆ 202.9(10) / 21111.22002200...

(3)

(c) Remainders Integers

163÷8=20 3(10)/3(8) 0.75×8=6.00 6(10)/6(8)

20÷8= 2 4(10)/4(8) 2÷8= 0 2(10)/2(8)

ˆ 163.75(10) / 243.6(8)

Remainders Integers

202÷8=25 2(10)/2(8) 0.9×8=7.2 7(10)/7(8) 25÷8= 3 1(10)/1(8) 0.2×8=1.6 1(10)/1(8) 3÷8= 0 3(10)/3(8) 0.6×8=4.8 4(10)/4(8)

0.8×8=6.4 6(10)/6(8) 0.4×8=3.2 3(10)/3(8)

ˆ 202.9(10) / 312.714631463...

(3)

(d) Remainders Integers

163÷16=10 3(10)/3(16) 0.75×16=12.00 12(10)/C(16)

10÷16= 0 10(10)/A(16)ˆ 163.75(10) / A3.C(16)

- 2.15 -

2.14. (continued)

Remainders Integers

202÷16=12 10(10)/A(16) 0.9×16=14.4 14(10)/E(16)

12÷16= 0 12(10)/C(16) 0.4×16= 6.4 6(10)/6(16)ˆ 202.9(10) / CA.E66

...(16)

2.15. (a) Remainders

11100010÷11=1001011 1(2)/1(3) 1001011÷11= 11001 0(2)/0(3) 11001÷11= 1000 1(2)/1(3) 1000÷11= 10 10(2)/2(3) 10÷11= 0 10(2)/2(3)

Integers

0.1101×11=10.0111 10(2)/2(3)0.0111×11= 1.0101 1(2)/1(3)0.0101×11= 0.1111 0(2)/0(3)0.1111×11=10.1101 10(2)/2(3)

ˆ 11100010.1101(2) / 22101.21022102...

(3)

(b) Remainders

11100010÷1000=11100 10(2)/2(8) 11100÷1000= 11 100(2)/4(8) 11÷1000= 0 11(2)/3(8)

Integers

0.1101×1000=110.1000 110(2)/6(8)0.1000×1000=100.0000 100(2)/4(8)

ˆ 11100010.1101(2) / 342.64(8)

(c) Remainders

11100010÷1010=10110 110(2)/6(10) 10110÷1010= 10 10(2)/2(10) 10÷1010= 0 10(2)/2(10)

- 2.16 -

2.15. (continued)

Integers

0.1101×1010=1000.0010 1000(2)/8(10)0.0010×1010= 1.0100 1(2)/1(10)0.0100×1010= 10.1000 10(2)/2(10)0.1000×1010= 101.0000 101(2)/5(10)

ˆ 11100010.1101(2) / 226.8125(10)

(d) Remainders

11100010÷10000=1110 10(2)/2(16) 1110÷10000= 0 1110(2)/E(16)

Integers

0.1101×10000=1101.0000 1101(2)/D(16)ˆ 11100010.1101(2) / E2.D(16)

2.16. (a) Remainders Integers

10112÷2=1202 1(3)/1(2) 0.1×2=0.2 0(3)/0(2)

1202÷2= 212 1(3)/1(2) 0.2×2=1.1 1(3)/1(2) 212÷2= 102 1(3)/1(2) 102÷2= 12 1(3)/1(2) 12÷2= 2 1(3)/1(2) 2÷2= 1 0(3)/0(2) 1÷2= 0 1(3)/1(2)

ˆ 10112.1(3) / 1011111.0101...

(2)

(b) Remainders Integers

10112÷22=102 21(3)/7(8) 0.1×22= 2.2 2(3)/2(8)

102÷22= 1 10(3)/3(8) 0.2×22=12.1 12(3)/5(8) 1÷22= 0 1(3)/1(8)

ˆ 10112.1(3) / 137.2525...

(2)

(c) Remainders Integers

10112÷101=100 12(3)/5(10) 0.1×101=10.1 10(3)/3(10) 100÷101= 0 100(3)/9(10)

ˆ 10112.1(3) / 95.33...

(10)

- 2.17 -

2.16. (continued)

(d) Remainders Integers

10112÷121=12 120(3)/F(16) 0.1×121=12.1 12(3)/5(16) 12÷121= 0 12(3)/5(16)

ˆ 10112.1(3) / 5F.55...

(16)

2.17. (a) 7 7 1 . 1 7 2(8) 8 8 8 8 8 8 000111111001.001111010(2)

9 9 9 9 9 1 F 9 . 3 D(16)

(b) 1 2 1 3 . 4(8) 8 8 8 8 8

001010001011.1000(2)

9 9 9 9 2 8 B . 8(16)

(c) 2 7 4 2 . 2 2 6(8) 8 8 8 8 8 8 8

010111100010.010010110(2)

9 9 9 9 9 5 E 2 . 4 B(16)

(d) 3 4 4 6 . 5(8) 8 8 8 8 8

011100100110.1010(2)

9 9 9 9 7 2 6 . A(16)

2.18. (a) 3 7 . 5(8) 9 9 9

011111.101(2)

- 2.18 -

2.18. (continued)

(b) 4 5 . 1(8) 9 9 9

100101.001(2)

(c) 6 1 . 3(8) 9 9 9

110001.011(2)

(d) 7 2 4 . 0 6(8) 9 9 9 9 9

111010100.000110(2)

2.19. (a) 1 C . 3(16) 9 9 9

00011100.0011(2)

(b) F 2 . C(16) 9 9 9

11110010.1100(2)

(c) 4 5 0 . B(16) 9 9 9 9

010001010000.1011(2)

(d) 8 E A . 5 9(16) 9 9 9 9 9

100011101010.01011001(2)

- 2.19 -

2.20. Algorithm to convert between base-3 and base-9:

1. Form blocks of 2 digits of the base-3 number starting

at the radix point and working both right and left,

adding leading and trailing 0's if necessary.

2. Replace each block of 2 digits by its equivalent digit

in base-9.

The conversion for 21021.112(3) is

021021.1120(3) 9 9 9 9 9 2 3 7 . 4 6(9)

2.21. (a) Let N= be the r's-complement of N. Therefore,

N= = rn - N

The r's-complement of N= is

rn - N= = rn - (rn - N) = (rn - rn) + N = N

(b) Let N- be the (r-1)'s-complement of N. Therefore,

N- = rn - r-m - N

The (r-1)'s-complement of N- is

rn - r-m - N- = rn - r-m - (rn - r-m - N)

= (rn - rn) + (r-m - r-m) + N

= N

2.22. 1's-complements:

(a) 108 - 10-0 - 10111011 = 01000100

(b) 109 - 10-0 - 101110100 = 010001011

(c) 106 - 10-0 - 101100 = 010011

(d) 107 - 10-0 - 0110101 = 1001010

(e) 103 - 10-2 - 010.11 = 101.00

(f) 105 - 10-3 - 11011.100 = 00100.011

(g) 106 - 10-3 - 100101.101 = 011010.010

- 2.20 -

2.22. (continued)

(h) 107 - 10-3 - 1010110.110 = 0101001.001

2's-complements:

(a) 108 - 10111011 = 01000101

(b) 109 - 101110100 = 010001100

(c) 106 - 101100 = 010100

(d) 107 - 0110101 = 1001011

(e) 103 - 010.11 = 101.01

(f) 105 - 11011.100 = 00100.100

(g) 106 - 100101.101 = 011010.011

(h) 107 - 1010110.110 = 0101001.010

2.23. 9's-complements:

(a) 106 - 10-0 - 285302 = 714697

(b) 105 - 10-0 - 39040 = 60959

(c) 106 - 10-0 - 059637 = 940362

(d) 106 - 10-0 - 610500 = 389499

(e) 104 - 10-2 - 7142.89 = 2857.10

(f) 104 - 10-4 - 5263.4580 = 4736.5419

(g) 104 - 10-3 - 0283.609 = 9716.390

(h) 103 - 10-4 - 134.5620 = 865.4379

10's-complements:

(a) 106 - 285302 = 714698

(b) 105 - 39040 = 60960

(c) 106 - 059637 = 940363

(d) 106 - 610500 = 389500

- 2.21 -

2.23. (continued)

(e) 104 - 7142.89 = 2857.11

(f) 104 - 5263.4580 = 4736.5420

(g) 104 - 0283.609 = 9716.391

(h) 103 - 134.5620 = 865.4380

2.24. r's-complements:

(a) 104 - 0120.21(3) = 2102.02(3)

(b) 103 - 101.120(3) = 121.110(3)

(c) 103 - 241.03(5) = 203.42(5)

(d) 103 - 031.240(5) = 413.210(5)

(e) 103 - 407.270(8) = 370.510(8)

(f) 104 - 0156.0037(8) = 7621.7741(8)

(g) 103 - 83D.9F(16) = 7C2.61(16)

(h) 105 - 0070C.B6E(16) = FF8F3.492(16)

(r-1)'s-complements:

(a) 104 - 10-2 - 0120.21(3) = 2102.01(3)

(b) 103 - 10-3 - 101.120(3) = 121.102(3)

(c) 103 - 10-2 - 241.03(5) = 203.41(5)

(d) 103 - 10-3 - 031.240(5) = 413.204(5)

(e) 103 - 10-3 - 407.270(8) = 370.507(8)

(f) 104 - 10-4 - 0156.0037(8) = 7621.7740(8)

(g) 103 - 10-2 - 83D.9F(16) = 7C2.60(16)

(h) 105 - 10-3 - 0070C.B6E(16) = FF8F3.491(16)

- 2.22 -

2.25. Using 1's-complements:

(a) N1 = 1101101 6 N1 = 1101101

N2 = 0110110 6 + N-2 = + 1001001_________

1 0110110 + 1 _______ 0110111

(b) N1 = 010100 6 N1 = 010100

N2 = 110000 6 + N-2 = + 001111______

100011

(c) N1 = 10010.00111 6 N1 = 10010.00111

N2 = 00101.11000 6 + N-2 = + 11010.00111_____________

1 01100.01110 + 1___________ 01100.01111

(d) N1 = 10110.0100 6 N1 = 10110.0100

N2 = 01011.1101 6 + N-2 = + 10100.0010____________

1 01010.0110 + 1__________ 01010.0111

(e) N1 = 01101.1011 6 N1 = 01101.1011

N2 = 10110.1100 6 + N-2 = + 01001.0011__________

10110.1110

(f) N1 = 00101.100100 6 N1 = 00101.100100

N2 = 11010.010011 6 + N-2 = + 00101.101100____________

01011.010000

Using 2's-complements:

(a) N1 = 1101101 6 N1 = 1101101

N2 = 0110110 6 + N=2 = + 1001010_________

1/ 0110111

(b) N1 = 010100 6 N1 = 010100

N2 = 110000 6 + N=2 = + 010000______

100100

- 2.23 -

2.25. (continued)

(c) N1 = 10010.00111 6 N1 = 10010.00111

N2 = 00101.11000 6 + N=2 = + 11010.01000_____________

1/ 01100.01111

(d) N1 = 10110.0100 6 N1 = 10110.0100

N2 = 01011.1101 6 + N=2 = + 10100.0011____________

1/ 01010.0111

(e) N1 = 01101.1011 6 N1 = 01101.1011

N2 = 10110.1100 6 + N=2 = + 01001.0100__________

10110.1111

(f) N1 = 00101.100100 6 N1 = 00101.100100

N2 = 11010.010011 6 + N=2 = + 00101.101101____________

01011.010001

2.26. Using 1's-complements:

(a) N1 = 0s10110 6 N1 = 0s10110

N2 = 0s01101 6 + N-2 = + 1s10010_________

1 0s01000 + 1 _______ 0s01001

(b) N1 = 0s010111 6 N1 = 0s010111

N2 = 0s110100 6 + N-2 = + 1s001011________

1s100010

(c) N1 = 0s110.1001 6 N1 = 0s110.1001

N2 = 0s011.0100 6 + N-2 = + 1s100.1011____________

1 0s011.0100 + 1__________ 0s011.0101

(d) N1 = 0s10101.1 6 N1 = 0s10101.1

N2 = 0s10101.1 6 + N-2 = + 1s01010.0_________

1s11111.1

- 2.24 -

2.26. (continued)

(e) N1 = 0s101.11000 6 N1 = 0s101.11000

N2 = 0s010.01011 6 + N-2 = + 1s101.10100_____________

1 0s011.01100 + 1___________ 0s011.01101

(f) N1 = 0s010111.10101 6 N1 = 0s010111.10101

N2 = 0s111010.11000 6 + N-2 = + 1s000101.00111______________

1s011100.11100

Using 2's-complements:

(a) N1 = 0s10110 6 N1 = 0s10110

N2 = 0s01101 6 + N=2 = + 1s10011_________

1/ 0s01001

(b) N1 = 0s010111 6 N1 = 0s010111

N2 = 0s110100 6 + N=2 = + 1s001100________

1s100011

(c) N1 = 0s110.1001 6 N1 = 0s110.1001

N2 = 0s011.0100 6 + N=2 = + 1s100.1100____________

1/ 0s011.0101

(d) N1 = 0s10101.1 6 N1 = 0s10101.1

N2 = 0s10101.1 6 + N=2 = + 1s01010.1___________

1/ 0s00000.0

(e) N1 = 0s101.11000 6 N1 = 0s101.11000

N2 = 0s010.01011 6 + N=2 = + 1s101.10101_____________

1/ 0s011.01101

(f) N1 = 0s010111.10101 6 N1 = 0s010111.10101

N2 = 0s111010.11000 6 + N=2 = + 1s000101.01000______________

1s011100.11101

- 2.25 -

2.27. Using 9's-complements:

(a) N1 = 7842 6 N1 = 7842

N2 = 3791 6 + N-2 = + 6208______

1 4050 + 1 ____ 4051

(b) N1 = 265 6 N1 = 265

N2 = 894 6 + N-2 = + 105___

370

(c) N1 = 508.3 6 N1 = 508.3

N2 = 094.7 6 + N-2 = + 905.2_______

1 413.5 + 1_____ 413.6

(d) N1 = 073.68 6 N1 = 0.73.68

N2 = 538.90 6 + N-2 = + 461.09______

534.77

(e) N1 = 427.08 6 N1 = 427.08

N2 = 089.30 6 + N-2 = + 910.69________

1 337.77 + 1______ 337.78

(f) N1 = 0804.20 6 N1 = 0804.20

N2 = 3621.47 6 + N-2 = + 6378.52_______

7182.72

Using 10's-complements:

(a) N1 = 7842 6 N1 = 7842

N2 = 3791 6 + N=2 = + 6209______

1/ 4051

(b) N1 = 265 6 N1 = 265

N2 = 894 6 + N=2 = + 106___

371

- 2.26 -

2.27. (continued)

(c) N1 = 508.3 6 N1 = 508.3

N2 = 094.7 6 + N=2 = + 905.3_______

1/ 413.6

(d) N1 = 073.68 6 N1 = 073.68

N2 = 538.90 6 + N=2 = + 461.10______

534.78

(e) N1 = 427.08 6 N1 = 427.08

N2 = 089.30 6 + N=2 = + 910.70________

1/ 337.78

(f) N1 = 0804.20 6 N1 = 0804.20

N2 = 3621.47 6 + N=2 = + 6378.53_______

7182.73

2.28. Using 9's-complements:

(a) N1 = 0s546 6 N1 = 0s546

N2 = 0s232 6 + N-2 = + 1s767_______

1 0s313 + 1 _____ 0s314

(b) N1 = 0s384 6 N1 = 0s384

N2 = 0s726 6 + N-2 = + 1s273_____

1s657

(c) N1 = 0s326.4 6 N1 = 0s326.4

N2 = 0s087.2 6 + N-2 = + 1s912.7_________

1 0s239.1 + 1_______ 0s239.2

(d) N1 = 0s076.23 6 N1 = 0s076.23

N2 = 0s209.40 6 + N-2 = + 1s790.59________

1s866.82

- 2.27 -

2.28. (continued)

(e) N1 = 0s406.9 6 N1 = 0s406.9

N2 = 0s406.9 6 + N-2 = + 1s593.0_______

1s999.9

(f) N1 = 0s063.40 6 N1 = 0s063.40

N2 = 0s240.36 6 + N-2 = + 1s759.63________

1s823.03

Using 10's-complements:

(a) N1 = 0s546 6 N1 = 0s546

N2 = 0s232 6 + N=2 = + 1s768_______

1/ 0s314

(b) N1 = 0s384 6 N1 = 0s384

N2 = 0s726 6 + N=2 = + 1s274_____

1s658

(c) N1 = 0s326.4 6 N1 = 0s326.4

N2 = 0s087.2 6 + N=2 = + 1s912.8_________

1/ 0s239.2

(d) N1 = 0s076.23 6 N1 = 0s076.23

N2 = 0s209.40 6 + N=2 = + 1s790.60________

1s866.83

(e) N1 = 0s406.9 6 N1 = 0s406.9

N2 = 0s406.9 6 + N=2 = + 1s593.1_________

1/ 0s000.0

(f) N1 = 0s063.40 6 N1 = 0s063.40

N2 = 0s240.36 6 + N=2 = + 1s759.64________

1s823.04

- 2.28 -

2.29. A=0s1000110 A==1s0111010

B=1s1010011 B==0s0101101

(a) A+B (b) A-B = A+B=

A = 0s1000110 A = 0s1000110

+ B = + 1s1010011___________ + B= = + 0s0101101_________

1/ 0s0011001 0s1110011

(c) B-A = B+A=

(d) -A-B = A=+B=

B = 1s1010011 A= = 1s0111010

+ A= = + 1s0111010___________ + B

= = + 0s0101101_________

1/ 1s0001101 1s1100111

2.30. A=0s601.7 A==1s398.3

B=1s754.2 B==0s245.8

(a) A+B (b) A-B = A+B=

A = 0s601.7 A = 0s601.7

+ B = + 1s754.2_________ + B= = + 0s245.8_______

1/ 0s355.9 0s847.5

(c) B-A = B+A=

(d) -A-B = A=+B=

B = 1s754.2 A= = 1s398.3

+ A= = + 1s398.3_________ + B

= = + 0s245.8_______

1/ 1s152.5 1s644.1

2.31. A=0s1010110 A-=1s0101001

B=1s1101100 B-=0s0010011

(a) A+B (b) A-B = A+B-

A = 0s1010110 A = 0s1010110

+ B = + 1s1101100___________ + B- = + 0s0010011_________

1 0s1000010 0s1101001 + 1_________

0s1000011

- 2.29 -

2.31. (continued)

(c) B-A = B+A- (d) -A-B = A-+B-

B = 1s1101100 A- = 1s0101001

+ A- = + 1s0101001___________ + B- = + 0s0010011_________ 1 1s0010101 1s0111100 + 1_________ 1s0010110

2.32. A=0s418.5 A-=1s581.4

B=1s693.0 B-=0s306.9

(a) A+B (b) A-B = A+B-

A = 0s418.5 A = 0s418.5

+ B = + 1s693.0_________ + B= = + 0s306.9_______

1 0s111.5 0s725.4 + 1_______

0s111.6

(c) B-A = B+A- (d) -A-B = A-+B-

B = 1s693.0 A- = 1s581.4

+ A- = + 1s581.4_________ + B- = + 0s306.9_______ 1 1s274.4 1s888.3 + 1_______ 0s274.5

2.33. (a) 100001010011 (b) 011001000010 (c) 101110000110

8 5 3 8 5 3 8 5 3

(d) 100100010000010101000 (e) 100100101000110

8 5 3 8 5 3

2.34. (a) 8 9 5 8 3

(b) 5 6 2 5 0

10001001010110000011

(c) 7 2 6 1

- 2.30 -

2.35.

(a) (b) (c) (d)

Decimal

Digit7635- 8324- 834-2- 864-1-

0 0000 0000 0000 0000

1 0101 0111 0101 0111

2 1001 0010 1011 0110

3 0010 0100 0100 1011

4 0111 1001 1010 1010

5 1011 0110 1111 0101

6 0100 1011 1001 0100

7 1000 1101 1110 1001

8 1101 1000 1000 1000

9 0110 1111 1101 1111

The 8324- and 864-1- codes are self-complementing.

2.36. In order to represent the decimal integer 1 in a code

having all positive weights, one of the weights must be a

1; while to represent the decimal integer 2, the weight of

2 is needed or another weight of 1 so that 2 can be

represented by 1+1. Since there are no negative weights,

the decimal integer 9 can only be represented if the sum of

the weights is at least equal to 9.

- 2.31 -

2.37. Assume two weights exceed 4. Then, only two weights, say,

wi1 and wi2, are less than or equal to 4. With these two

weights it is necessary to encode the decimal digits 0, 1,

2, 3, and 4 by assigning binary digits 0's and 1's to the

ai's in the formula a1wi1+a2wi2. Since there are only four

ways of making this assignment, the five decimal digits can

not be represented. Thus, no more than one weight can

exceed 4.

2.38. Divide the weights of the code into two groups: those which

are used in the representation of the decimal digit 0 and

those which are not used in this representation. Let the

sum of the weights in the first group be Ewi and the sum

of the weights in the second group be Ewj. Clearly,

Ewi=0. By the definition of a self-complementing code,

those weights which are not used to represent the decimal

digit 0 must be used to represent the decimal digit 9.

Hence, Ewj=9. Combining these results, Ewi+Ewj=9.

2.39. Assume three weights are negative and only one weight is

positive. There can exist at most 9 non-negative

combinations of the four weights. Therefore it is

impossible to represent the ten non-negative digits which

must be coded. Hence, at most two weights can be negative.

2.40. (a) 011100101101100110000

9 6 0

(b) 101100001010111011001

X + Y

(c) 1000011110111111001001100101

C o d e

- 2.32 -

2.41. (a) 101100101011100000110011

2 8 3

which is represented by B2B833 in hexadecimal.

(b) 010110101011110110110001

Z = 1

which is represented by 5ABDB1 in hexadecimal.

(c) 01000010011010010111010011110011

B i t s

which is represented by 426974F3 in hexadecimal.

2.42. (a)

Decimaldigit

7 6 5 4 3 2 1 Position

2 4 2 p3 1 p2 p1 Format

0 0 0 0 0 0 0 0

1 0 0 0 0 1 1 1

2 0 0 1 1 0 0 1

3 0 0 1 1 1 1 0

4 0 1 0 1 0 1 0

5 1 0 1 0 1 0 1

6 1 1 0 0 0 0 1

7 1 1 0 0 1 1 0

8 1 1 1 1 0 0 0

9 1 1 1 1 1 1 1

- 2.33 -

2.42. (continued)

(b)

Decimaldigit

9 8 7 6 5 4 3 2 1 Position

b5 p4 b4 b3 b2 p3 b1 p2 p1 Format

0 1 1 1 0 0 1 0 1 0

1 0 0 0 0 1 1 1 1 0

2 0 0 0 1 0 1 1 0 1

3 0 0 0 1 1 0 0 1 1

4 0 0 1 0 0 1 1 0 0

5 0 0 1 0 1 0 0 1 0

6 0 0 1 1 0 0 0 0 1

7 1 1 0 0 0 0 1 1 0

8 1 1 0 0 1 1 0 0 0

9 1 1 0 1 0 1 0 1 1

2.43.

12 11 10 9 8 7 6 5 4 3 2 1 Position

b8 b7 b6 b5 p4 b4 b3 b2 p3 b1 p2 p1 Format

(a) 1 1 1 0 1 0 0 1 0 1 1 1

(b) 0 1 0 1 0 1 0 0 1 0 0 1

(c) 1 0 0 1 0 0 1 0 0 1 0 0

2.44. (a) 7 6 5 4 3 2 1 Position0 0 1 1 0 0 0 Received code group

c*3c*2c*1=001 which indicates position 1 in error.

Transmitted code group: 0011001

(b) 7 6 5 4 3 2 1 Position1 1 1 1 0 0 0 Received code group

c*3c*2c*1=000 which indicates no error.

Transmitted code group: 1111000

- 2.34 -

2.44. (continued)

(c) 7 6 5 4 3 2 1 Position1 1 0 1 1 0 0 Received code group

c*3c*2c*1=110 which indicates position 6 in error.

Transmitted code group: 1001100

2.45. (a) Position: 11 10 9 8 7 6 5 4 3 2 1

Format: poverall

b6b5p4b4b3b2p3b1p2p1Code group: 1 1 0 1 0 1 0 1 1 1 1

(b) If errors occur in bit positions 2 and 9, the received

code group is 11110101101. Using the received code

group, the overall parity bit is correct. The binary

check number is c*4c*3c*2c*1=1011. Since c*4c*3c*2c*1…0000 and

the overall parity bit is correct, a double error has

occurred.

2.46. Upon comparing each pair of code groups, the minimum

distance of this code is found to be three. Therefore, it

can be used for double-error detection or single-error

correction.