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14 questions

Unit 2 · Number System

2. Answer these questions

What is a number system?

Answer

A number system is a way of writing numbers: a set of symbols (digits) and rules for using them to express numbers and to do calculations with them. Each number system has its own base, the number of digits it uses. For example, the decimal number system uses the ten digits 0 to 9, and the binary number system, which computers use, uses 0 and 1.

Define the base or radix of a number system.

Answer

The base or radix of a number system is the total number of different digits (symbols) it uses. It also decides the place values: each position is worth the base times the position to its right. Decimal has base 10 (0–9), binary base 2 (0, 1), octal base 8 (0–7) and hexadecimal base 16 (0–9, A–F). The base is written as a small number after a number, as in (101)₂.

List the different types of number systems.

Answer

Number system	Base	Digits used
Binary	2	0, 1
Octal	8	0 to 7
Decimal	10	0 to 9
Hexadecimal	16	0 to 9 and A to F

What is the hexadecimal number system? Write the symbols used in it.

Answer

The hexadecimal number system is the base-16 number system. It uses sixteen symbols: the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and the six letters A, B, C, D, E, F, which stand for 10, 11, 12, 13, 14 and 15. Its place values are powers of 16 (1, 16, 256, …). One hex digit stands for exactly four bits, so programmers use hex as a short way of writing binary, for example in web colors such as #FF8800, memory addresses and MAC addresses.

3. Calculate the following as indicated

Perform the following binary additions:

i.(11110)₂ + (1001)₂
ii.(1011)₂ + (1001)₂
iii.(101011)₂ + (11011)₂
iv.(1010)₂ + (110)₂
v.(101001)₂ + (1110)₂
vi.(100001)₂ + (100011)₂
vii.(100111)₂ + (11010)₂
viii.(110001)₂ + (100101)₂

Answer

Each sum is set out in columns, with the carries above and the decimal values on the right as a check.

i. (11110)₂ + (1001)₂ = (100111)₂

carry  1 1         1 1 1 1 0    30     +     1 0 0 1     9       -----------       1 0 0 1 1 1    39

ii. (1011)₂ + (1001)₂ = (10100)₂

carry  1   1 1         1 0 1 1    11     +   1 0 0 1     9       ---------       1 0 1 0 0    20

iii. (101011)₂ + (11011)₂ = (1000110)₂

carry  1 1 1   1 1         1 0 1 0 1 1    43     +     1 1 0 1 1    27       -------------       1 0 0 0 1 1 0    70

iv. (1010)₂ + (110)₂ = (10000)₂

carry  1 1 1         1 0 1 0    10     +     1 1 0     6       ---------       1 0 0 0 0    16

v. (101001)₂ + (1110)₂ = (110111)₂

carry    1       1 0 1 0 0 1    41     +     1 1 1 0    14       -----------       1 1 0 1 1 1    55

vi. (100001)₂ + (100011)₂ = (1000100)₂

carry  1       1 1         1 0 0 0 0 1    33     +   1 0 0 0 1 1    35       -------------       1 0 0 0 1 0 0    68

vii. (100111)₂ + (11010)₂ = (1000001)₂

carry  1 1 1 1 1         1 0 0 1 1 1    39     +     1 1 0 1 0    26       -------------       1 0 0 0 0 0 1    65

viii. (110001)₂ + (100101)₂ = (1010110)₂

carry  1         1         1 1 0 0 0 1    49     +   1 0 0 1 0 1    37       -------------       1 0 1 0 1 1 0    86

Perform the following binary subtractions:

i.(1100)₂ − (1001)₂
ii.(1001)₂ − (110)₂
iii.(11101)₂ − (1010)₂
iv.(101100)₂ − (10011)₂
v.(11111)₂ − (10110)₂
vi.(110011)₂ − (10100)₂
vii.(100100)₂ − (1110)₂
viii.(1000001)₂ − (10101)₂

Answer

Each is set out in columns with the decimal values on the right, then the borrows, then a check by adding.

i. (1100)₂ − (1001)₂ = (11)₂

       1 1 0 0    12     − 1 0 0 1     9       -------           1 1     3

Borrowing: the 1s column borrows from the 4s (the 0 in the 2s becomes 1). Check: 11 + 1001 = 1100 ✓

ii. (1001)₂ − (110)₂ = (11)₂

       1 0 0 1    9     −   1 1 0    6       -------           1 1    3

Borrowing: the 2s column borrows from the 8s (the 0 in the 4s becomes 1). Check: 11 + 110 = 1001 ✓

iii. (11101)₂ − (1010)₂ = (10011)₂

       1 1 1 0 1    29     −   1 0 1 0    10       ---------       1 0 0 1 1    19

Borrowing: the 2s column borrows from the 4s. Check: 10011 + 1010 = 11101 ✓

iv. (101100)₂ − (10011)₂ = (11001)₂

       1 0 1 1 0 0    44     −   1 0 0 1 1    19       -----------         1 1 0 0 1    25

Borrowing: the 1s column borrows from the 4s (the 0 in the 2s becomes 1); the 16s column borrows from the 32s. Check: 11001 + 10011 = 101100 ✓

v. (11111)₂ − (10110)₂ = (1001)₂

       1 1 1 1 1    31     − 1 0 1 1 0    22       ---------         1 0 0 1     9

No column needs to borrow. Check: 1001 + 10110 = 11111 ✓

vi. (110011)₂ − (10100)₂ = (11111)₂

       1 1 0 0 1 1    51     −   1 0 1 0 0    20       -----------         1 1 1 1 1    31

Borrowing: the 4s column borrows from the 16s (the 0 in the 8s becomes 1); the 16s column borrows from the 32s. Check: 11111 + 10100 = 110011 ✓

vii. (100100)₂ − (1110)₂ = (10110)₂

       1 0 0 1 0 0    36     −     1 1 1 0    14       -----------         1 0 1 1 0    22

Borrowing: the 2s column borrows from the 4s; the 4s column borrows from the 32s (the 0s in the 8s and 16s become 1s). Check: 10110 + 1110 = 100100 ✓

viii. (1000001)₂ − (10101)₂ = (101100)₂

       1 0 0 0 0 0 1    65     −     1 0 1 0 1    21       -------------         1 0 1 1 0 0    44

Borrowing: the 4s column borrows from the 64s (the 0s in the 8s, 16s and 32s become 1s). Check: 101100 + 10101 = 1000001 ✓

4. Convert the given numbers as indicated

Decimal to binary:
i. (56)₁₀ ii. (78)₁₀ iii. (123)₁₀ iv. (345)₁₀ v. (540)₁₀ vi. (572)₁₀ vii. (546)₁₀ viii. (1098)₁₀ ix. (2103)₁₀ x. (445)₁₀

Answer

Divide by 2 again and again (“r” is the remainder), then read the remainders from the bottom up.

i.(56)₁₀: 56 ÷ 2 = 28 r 0; 28 ÷ 2 = 14 r 0; 14 ÷ 2 = 7 r 0; 7 ÷ 2 = 3 r 1; 3 ÷ 2 = 1 r 1; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (111000)₂
ii.(78)₁₀: 78 ÷ 2 = 39 r 0; 39 ÷ 2 = 19 r 1; 19 ÷ 2 = 9 r 1; 9 ÷ 2 = 4 r 1; 4 ÷ 2 = 2 r 0; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (1001110)₂
iii.(123)₁₀: 123 ÷ 2 = 61 r 1; 61 ÷ 2 = 30 r 1; 30 ÷ 2 = 15 r 0; 15 ÷ 2 = 7 r 1; 7 ÷ 2 = 3 r 1; 3 ÷ 2 = 1 r 1; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (1111011)₂
iv.(345)₁₀: 345 ÷ 2 = 172 r 1; 172 ÷ 2 = 86 r 0; 86 ÷ 2 = 43 r 0; 43 ÷ 2 = 21 r 1; 21 ÷ 2 = 10 r 1; 10 ÷ 2 = 5 r 0; 5 ÷ 2 = 2 r 1; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (101011001)₂
v.(540)₁₀: 540 ÷ 2 = 270 r 0; 270 ÷ 2 = 135 r 0; 135 ÷ 2 = 67 r 1; 67 ÷ 2 = 33 r 1; 33 ÷ 2 = 16 r 1; 16 ÷ 2 = 8 r 0; 8 ÷ 2 = 4 r 0; 4 ÷ 2 = 2 r 0; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (1000011100)₂
vi.(572)₁₀: 572 ÷ 2 = 286 r 0; 286 ÷ 2 = 143 r 0; 143 ÷ 2 = 71 r 1; 71 ÷ 2 = 35 r 1; 35 ÷ 2 = 17 r 1; 17 ÷ 2 = 8 r 1; 8 ÷ 2 = 4 r 0; 4 ÷ 2 = 2 r 0; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (1000111100)₂
vii.(546)₁₀: 546 ÷ 2 = 273 r 0; 273 ÷ 2 = 136 r 1; 136 ÷ 2 = 68 r 0; 68 ÷ 2 = 34 r 0; 34 ÷ 2 = 17 r 0; 17 ÷ 2 = 8 r 1; 8 ÷ 2 = 4 r 0; 4 ÷ 2 = 2 r 0; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (1000100010)₂
viii.(1098)₁₀: 1098 ÷ 2 = 549 r 0; 549 ÷ 2 = 274 r 1; 274 ÷ 2 = 137 r 0; 137 ÷ 2 = 68 r 1; 68 ÷ 2 = 34 r 0; 34 ÷ 2 = 17 r 0; 17 ÷ 2 = 8 r 1; 8 ÷ 2 = 4 r 0; 4 ÷ 2 = 2 r 0; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (10001001010)₂
ix.(2103)₁₀: 2103 ÷ 2 = 1051 r 1; 1051 ÷ 2 = 525 r 1; 525 ÷ 2 = 262 r 1; 262 ÷ 2 = 131 r 0; 131 ÷ 2 = 65 r 1; 65 ÷ 2 = 32 r 1; 32 ÷ 2 = 16 r 0; 16 ÷ 2 = 8 r 0; 8 ÷ 2 = 4 r 0; 4 ÷ 2 = 2 r 0; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (100000110111)₂
x.(445)₁₀: 445 ÷ 2 = 222 r 1; 222 ÷ 2 = 111 r 0; 111 ÷ 2 = 55 r 1; 55 ÷ 2 = 27 r 1; 27 ÷ 2 = 13 r 1; 13 ÷ 2 = 6 r 1; 6 ÷ 2 = 3 r 0; 3 ÷ 2 = 1 r 1; 1 ÷ 2 = 0 r 1. Remainders from the bottom up: (110111101)₂

Binary to decimal:
i. (1101)₂ ii. (1010)₂ iii. (10010)₂ iv. (10110)₂ v. (101001)₂ vi. (11100111)₂ vii. (111100)₂ viii. (10010011)₂ ix. (1011100)₂ x. (100110)₂

Answer

Multiply each digit by its place value (a power of 2) and add.

i.(1101)₂ = 1 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰ = 8 + 4 + 0 + 1 = 13
ii.(1010)₂ = 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ = 8 + 0 + 2 + 0 = 10
iii.(10010)₂ = 1 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ = 16 + 0 + 0 + 2 + 0 = 18
iv.(10110)₂ = 1 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 0 × 2⁰ = 16 + 0 + 4 + 2 + 0 = 22
v.(101001)₂ = 1 × 2⁵ + 0 × 2⁴ + 1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰ = 32 + 0 + 8 + 0 + 0 + 1 = 41
vi.(11100111)₂ = 1 × 2⁷ + 1 × 2⁶ + 1 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰ = 128 + 64 + 32 + 0 + 0 + 4 + 2 + 1 = 231
vii.(111100)₂ = 1 × 2⁵ + 1 × 2⁴ + 1 × 2³ + 1 × 2² + 0 × 2¹ + 0 × 2⁰ = 32 + 16 + 8 + 4 + 0 + 0 = 60
viii.(10010011)₂ = 1 × 2⁷ + 0 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰ = 128 + 0 + 0 + 16 + 0 + 0 + 2 + 1 = 147
ix.(1011100)₂ = 1 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 1 × 2³ + 1 × 2² + 0 × 2¹ + 0 × 2⁰ = 64 + 0 + 16 + 8 + 4 + 0 + 0 = 92
x.(100110)₂ = 1 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 0 × 2⁰ = 32 + 0 + 0 + 4 + 2 + 0 = 38

Decimal to octal:
i. (69)₁₀ ii. (216)₁₀ iii. (767)₁₀ iv. (79)₁₀ v. (443)₁₀ vi. (413)₁₀ vii. (765)₁₀ viii. (1334)₁₀ ix. (1825)₁₀ x. (2783)₁₀

Answer

Divide by 8 again and again, then read the remainders from the bottom up.

i.(69)₁₀: 69 ÷ 8 = 8 r 5; 8 ÷ 8 = 1 r 0; 1 ÷ 8 = 0 r 1. Remainders from the bottom up: (105)₈
ii.(216)₁₀: 216 ÷ 8 = 27 r 0; 27 ÷ 8 = 3 r 3; 3 ÷ 8 = 0 r 3. Remainders from the bottom up: (330)₈
iii.(767)₁₀: 767 ÷ 8 = 95 r 7; 95 ÷ 8 = 11 r 7; 11 ÷ 8 = 1 r 3; 1 ÷ 8 = 0 r 1. Remainders from the bottom up: (1377)₈
iv.(79)₁₀: 79 ÷ 8 = 9 r 7; 9 ÷ 8 = 1 r 1; 1 ÷ 8 = 0 r 1. Remainders from the bottom up: (117)₈
v.(443)₁₀: 443 ÷ 8 = 55 r 3; 55 ÷ 8 = 6 r 7; 6 ÷ 8 = 0 r 6. Remainders from the bottom up: (673)₈
vi.(413)₁₀: 413 ÷ 8 = 51 r 5; 51 ÷ 8 = 6 r 3; 6 ÷ 8 = 0 r 6. Remainders from the bottom up: (635)₈
vii.(765)₁₀: 765 ÷ 8 = 95 r 5; 95 ÷ 8 = 11 r 7; 11 ÷ 8 = 1 r 3; 1 ÷ 8 = 0 r 1. Remainders from the bottom up: (1375)₈
viii.(1334)₁₀: 1334 ÷ 8 = 166 r 6; 166 ÷ 8 = 20 r 6; 20 ÷ 8 = 2 r 4; 2 ÷ 8 = 0 r 2. Remainders from the bottom up: (2466)₈
ix.(1825)₁₀: 1825 ÷ 8 = 228 r 1; 228 ÷ 8 = 28 r 4; 28 ÷ 8 = 3 r 4; 3 ÷ 8 = 0 r 3. Remainders from the bottom up: (3441)₈
x.(2783)₁₀: 2783 ÷ 8 = 347 r 7; 347 ÷ 8 = 43 r 3; 43 ÷ 8 = 5 r 3; 5 ÷ 8 = 0 r 5. Remainders from the bottom up: (5337)₈

Octal to decimal:
i. (124)₈ ii. (242)₈ iii. (333)₈ iv. (763)₈ v. (103)₈ vi. (451)₈ vii. (3401)₈ viii. (1045)₈ ix. (437)₈ x. (611)₈
The textbook prints ix. as (438)₈, but 8 is not an octal digit, so the book’s number doesn’t work; (437)₈ is used here.

Answer

Multiply each digit by its place value (a power of 8) and add.

i.(124)₈ = 1 × 8² + 2 × 8¹ + 4 × 8⁰ = 64 + 16 + 4 = 84
ii.(242)₈ = 2 × 8² + 4 × 8¹ + 2 × 8⁰ = 128 + 32 + 2 = 162
iii.(333)₈ = 3 × 8² + 3 × 8¹ + 3 × 8⁰ = 192 + 24 + 3 = 219
iv.(763)₈ = 7 × 8² + 6 × 8¹ + 3 × 8⁰ = 448 + 48 + 3 = 499
v.(103)₈ = 1 × 8² + 0 × 8¹ + 3 × 8⁰ = 64 + 0 + 3 = 67
vi.(451)₈ = 4 × 8² + 5 × 8¹ + 1 × 8⁰ = 256 + 40 + 1 = 297
vii.(3401)₈ = 3 × 8³ + 4 × 8² + 0 × 8¹ + 1 × 8⁰ = 1,536 + 256 + 0 + 1 = 1,793
viii.(1045)₈ = 1 × 8³ + 0 × 8² + 4 × 8¹ + 5 × 8⁰ = 512 + 0 + 32 + 5 = 549
ix.(437)₈ = 4 × 8² + 3 × 8¹ + 7 × 8⁰ = 256 + 24 + 7 = 287
x.(611)₈ = 6 × 8² + 1 × 8¹ + 1 × 8⁰ = 384 + 8 + 1 = 393

Decimal to hexadecimal:
i. (55)₁₀ ii. (540)₁₀ iii. (225)₁₀ iv. (880)₁₀ v. (2046)₁₀ vi. (2024)₁₀ vii. (6678)₁₀

Answer

Divide by 16 again and again, write a remainder of 10 to 15 as A to F, and read the remainders from the bottom up.

i.(55)₁₀: 55 ÷ 16 = 3 r 7; 3 ÷ 16 = 0 r 3. Remainders from the bottom up: (37)₁₆
ii.(540)₁₀: 540 ÷ 16 = 33 r 12 = C; 33 ÷ 16 = 2 r 1; 2 ÷ 16 = 0 r 2. Remainders from the bottom up: (21C)₁₆
iii.(225)₁₀: 225 ÷ 16 = 14 r 1; 14 ÷ 16 = 0 r 14 = E. Remainders from the bottom up: (E1)₁₆
iv.(880)₁₀: 880 ÷ 16 = 55 r 0; 55 ÷ 16 = 3 r 7; 3 ÷ 16 = 0 r 3. Remainders from the bottom up: (370)₁₆
v.(2046)₁₀: 2046 ÷ 16 = 127 r 14 = E; 127 ÷ 16 = 7 r 15 = F; 7 ÷ 16 = 0 r 7. Remainders from the bottom up: (7FE)₁₆
vi.(2024)₁₀: 2024 ÷ 16 = 126 r 8; 126 ÷ 16 = 7 r 14 = E; 7 ÷ 16 = 0 r 7. Remainders from the bottom up: (7E8)₁₆
vii.(6678)₁₀: 6678 ÷ 16 = 417 r 6; 417 ÷ 16 = 26 r 1; 26 ÷ 16 = 1 r 10 = A; 1 ÷ 16 = 0 r 1. Remainders from the bottom up: (1A16)₁₆

Hexadecimal to decimal:
i. (56)₁₆ ii. (67)₁₆ iii. (558)₁₆ iv. (B74)₁₆ v. (20D3)₁₆ vi. (DEF)₁₆ vii. (6E3)₁₆ viii. (63F)₁₆

Answer

Multiply each digit by its place value (a power of 16; A = 10 … F = 15) and add.

i.(56)₁₆ = 5 × 16¹ + 6 × 16⁰ = 80 + 6 = 86
ii.(67)₁₆ = 6 × 16¹ + 7 × 16⁰ = 96 + 7 = 103
iii.(558)₁₆ = 5 × 16² + 5 × 16¹ + 8 × 16⁰ = 1,280 + 80 + 8 = 1,368
iv.(B74)₁₆ = 11 × 16² + 7 × 16¹ + 4 × 16⁰ = 2,816 + 112 + 4 = 2,932
v.(20D3)₁₆ = 2 × 16³ + 0 × 16² + 13 × 16¹ + 3 × 16⁰ = 8,192 + 0 + 208 + 3 = 8,403
vi.(DEF)₁₆ = 13 × 16² + 14 × 16¹ + 15 × 16⁰ = 3,328 + 224 + 15 = 3,567
vii.(6E3)₁₆ = 6 × 16² + 14 × 16¹ + 3 × 16⁰ = 1,536 + 224 + 3 = 1,763
viii.(63F)₁₆ = 6 × 16² + 3 × 16¹ + 15 × 16⁰ = 1,536 + 48 + 15 = 1,599

Binary to hexadecimal:
i. (1000110)₂ ii. (11001)₂ iii. (1111000)₂ iv. (11110000111)₂ v. (101010110)₂ vi. (1110010110)₂ vii. (11011001)₂ viii. (1001100)₂

Answer

Group the bits in fours from the right, filling the left group with 0s, and write each group as one hex digit. The decimal value is given as a check.

i.(1000110)₂: 0100 0110 = 0100 → 4, 0110 → 6. (46)₁₆ (both are 70)
ii.(11001)₂: 0001 1001 = 0001 → 1, 1001 → 9. (19)₁₆ (both are 25)
iii.(1111000)₂: 0111 1000 = 0111 → 7, 1000 → 8. (78)₁₆ (both are 120)
iv.(11110000111)₂: 0111 1000 0111 = 0111 → 7, 1000 → 8, 0111 → 7. (787)₁₆ (both are 1,927)
v.(101010110)₂: 0001 0101 0110 = 0001 → 1, 0101 → 5, 0110 → 6. (156)₁₆ (both are 342)
vi.(1110010110)₂: 0011 1001 0110 = 0011 → 3, 1001 → 9, 0110 → 6. (396)₁₆ (both are 918)
vii.(11011001)₂: 1101 1001 = 1101 → D, 1001 → 9. (D9)₁₆ (both are 217)
viii.(1001100)₂: 0100 1100 = 0100 → 4, 1100 → C. (4C)₁₆ (both are 76)

Hexadecimal to binary:
i. (D4)₁₆ ii. (643)₁₆ iii. (189)₁₆ iv. (2BF)₁₆ v. (A9F)₁₆ vi. (FACE)₁₆ vii. (FB4)₁₆ viii. (1B2)₁₆

Answer

Write each hex digit as four bits; the 0s at the very front may be dropped.

i.(D4)₁₆: D = 1101, 4 = 0100. (11010100)₂
ii.(643)₁₆: 6 = 0110, 4 = 0100, 3 = 0011. (11001000011)₂
iii.(189)₁₆: 1 = 0001, 8 = 1000, 9 = 1001. (110001001)₂
iv.(2BF)₁₆: 2 = 0010, B = 1011, F = 1111. (1010111111)₂
v.(A9F)₁₆: A = 1010, 9 = 1001, F = 1111. (101010011111)₂
vi.(FACE)₁₆: F = 1111, A = 1010, C = 1100, E = 1110. (1111101011001110)₂
vii.(FB4)₁₆: F = 1111, B = 1011, 4 = 0100. (111110110100)₂
viii.(1B2)₁₆: 1 = 0001, B = 1011, 2 = 0010. (110110010)₂