1.1 Data Representation - A-Level / IGCSE Computer Science

Learning Objectives

By the end of this lesson, you will be able to:

Understandhow and why computers use binaryto represent all forms of data
Understand thedenary(base 10),binary(base 2) andhexadecimal(base 16) number systems
Convert betweenpositive denary and positive binary
Convert betweenpositive denary and positive hexadecimal
Convert betweenpositive hexadecimal and positive binary
Understand how and whyhexadecimalis used as a beneficial method of data representation, includingMAC addresses,colour codes,URLsanderror codes
Add twopositive 8-bit binary integers
Understand the concept ofoverflowand why it occurs in binary addition
Performlogical binary shiftson a positive 8-bit binary integer and understand the effect this has on the value
Usetwo's complementto represent positive and negative 8-bit binary integers
Understand themeasurement of computer memory sizes(bit, nibble, byte, KB/KiB, MB/MiB, GB/GiB, TB/TiB, PB/PiB)

Key Terms

Bit

A single binary digit — either 0 or 1. The smallest unit of data.

Nibble

A group of 4 bits (half a byte). One nibble = one hexadecimal digit.

Byte

A group of 8 bits. The standard unit for storing a single character.

Denary (Decimal)

The everyday base-10 number system using digits 0–9.

Binary

The base-2 number system using only 0 and 1.

Hexadecimal

The base-16 number system using 0–9 and A–F. One hex digit represents 4 bits.

Most Significant Bit (MSB)

The left-most bit in a binary number — it has the largest place value.

Least Significant Bit (LSB)

The right-most bit in a binary number — it has the smallest place value (1).

Overflow

An error that occurs when the result of a calculation needs more bits than the register can hold.

Logical Shift

Moving bits left or right. Left shift multiplies by 2; right shift divides by 2.

Two's Complement

A method of representing negative numbers in binary. The MSB carries a negative weight.

Sign Bit

The MSB in a signed binary number: 0 = positive, 1 = negative.

MAC Address

A 48-bit unique hardware address written as 6 pairs of hexadecimal digits.

HTML Colour Code

A 6-digit hexadecimal code (#RRGGBB) representing red, green and blue intensities.

Kibibyte (KiB)

1024 bytes (2¹⁰ bytes). Used for measuring internal memory like RAM.

Kilobyte (KB)

1000 bytes (10³ bytes) in decimal. Commonly used for storage sizes.

Memory Dump

A printout of memory contents, usually shown in hexadecimal to help trace errors.

Register

A small, fast storage area inside the CPU that holds binary values being processed.

URL Encoding

Special characters in a URL are converted into hexadecimal codes prefixed with %.

1. Why Computers Use Binary

Any form of data needs to be converted tobinaryto be processed by a computer. Data is processed usinglogic gatesand stored inregisters.

Key Ideas

Computers are made up of millions of tinyswitches(transistors).
A switch that isONis represented by1.
A switch that isOFFis represented by0.
Switches uselogic gatesthat can only havetwo states(1 or 0).
Because there are only two possible states, all data must be converted tobinarybefore a computer can understand and process it.
Converting data to binary allows computers to process it at incredible speed, perform complex calculations and store vast amounts of data efficiently.

Secondary Storage Examples

Magnetic hard drivesuseNorth and South polarityto represent a 1 or a 0.
Inoptical disks, light hitting a flat area (land) is interpreted as a1and light hitting a bump (pit) is interpreted as a0.

Optical disk encoding

Land = 1 Pit = 0

1 0 0 0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0

Real-Life Analogy: Driving a Car

When driving a car, the accelerator pedal is used to increase the car's speed. If a car was accelerating from 50 mph to 100 mph, the increase would be gradual.

In a computer system, the car is doing either 50 mph (0) or 100 mph (1) — there is no in-between. Trying to change the computer system so that it has more options would beless efficientand requiremore complex parts.

Worked Example

Question:Explain why computers process data in binary format. [2]

Answer:

Computers process data using logic gates... [1]
... that can only have two states (1/0) [1]

Exam tip: Mark schemes expect "logic gates + two states (1/0)". Phrases like "computers understand binary" do not get the mark.

Activity 1: Binary in Everyday Devices

A student says: "Computers use binary because it is easier for humans to read." Explain why this statement is incorrect. In your answer, refer to logic gates, switches and the two states of a computer. [4 marks]

Solution:

[1] Computers do not use binary because it is easier for humans — binary is actually harder for humans to read than denary.
[1] Computers use binary because they are made of millions of switches (transistors) that can only be ON (1) or OFF (0).
[1] Data is processed using logic gates that only have two possible states (1 or 0).
[1] Therefore all data must be converted to binary before a computer can process it.

Check Your Understanding: Why Binary?

1. What two states can a computer switch be in? [2 marks]

[1]ON — represented by the binary digit 1
[1]OFF — represented by the binary digit 0
Additional:These two states are the basis of all data storage and processing in a computer.

2. How is data stored on a magnetic hard drive? [2 marks]

[1]Magnetic hard drives use North and South polarity to represent binary values
[1]One polarity represents a 1 and the other represents a 0

3. How is data stored on an optical disk? [2 marks]

[1]Light hitting a flat area (land) is interpreted as a 1
[1]Light hitting a bump (pit) is interpreted as a 0

4. Explain why a computer cannot process denary numbers directly. [2 marks]

[1]Computers are made from logic gates that only have two states (1 or 0)
[1]Denary has ten digits (0–9), which cannot be represented directly by a two-state switch, so data must be converted to binary first

5. What is the main advantage of binary over having more than two states? [2 marks]

[1]Having only two states is simpler and more reliable — it is easy to tell if a switch is ON or OFF
[1]Using more states would require more complex components and would be less efficient, and small voltage changes could cause errors

6. A student says: "All data is stored in a computer as text." Explain why this is wrong. [3 marks]

[1]All data (text, images, sound, video) is stored as binary (1s and 0s)
[1]Different types of data use different encoding methods (e.g. ASCII for text, pixels for images)
[1]The computer does not know what the data "means" — it just processes the binary patterns

2. The Denary, Binary & Hexadecimal Number Systems

Anumber systemis a way of representing amounts or quantities of something. Different number systems use differentbases. The base tells you how many different digits are used in that system.

Denary

Base 10
Digits: 0–9
Used by humans in everyday life
Each column is a power of 10

10³ 10² 10¹ 10⁰

1000 100 10 1

Binary

Base 2
Digits: 0 and 1
Used by computers
Each column is a power of 2

2³ 2² 2¹ 2⁰

8 4 2 1

Hexadecimal

Base 16
Digits: 0–9 and A–F
Used by humans as shorthand for binary
Each column is a power of 16

16¹ 16⁰

16 1

Hexadecimal Reference Table

Denary	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
Binary	0000	0001	0010	0011	0100	0101	0110	0111	1000	1001	1010	1011	1100	1101	1110	1111
Hex	0	1	2	3	4	5	6	7	8	9	A	B	C	D	E	F

Exam Tips & Tricks

All conversions at IGCSE are capped at16 bits.
The largest denary number that can be represented using 16 bits is65,535(binary 1111111111111111).
One hexadecimal digit represents exactly four bits (one nibble).
Write out the hex digits (0–F) during the exam to avoid confusion.

Activity 2: Recognising Number Systems

Look at the following values. State whether each onemustbe binary, hexadecimal, or could be either. Explain each answer.

1010
2F
1001
7C
0

Solution:

1010— Could be either. All digits are 0s and 1s, so it is a valid binary number. It is also a valid hexadecimal number (using only digits 0 and 1).
2F— Must be hexadecimal. The letter F cannot appear in a binary number.
1001— Could be either. Valid binary (only 0s and 1s) and also valid hex.
7C— Must be hexadecimal. The letter C cannot appear in binary.
0— Could be either. It is valid in every number system.

Check Your Understanding: Number Systems

1. What base is the denary number system? [1 mark]

[1]Base 10 — it uses ten digits (0–9)

2. Why is hexadecimal described as base 16? [2 marks]

[1]Hexadecimal uses 16 different digits: 0–9 and A–F
[1]Each digit position represents a power of 16

3. How many bits are in a nibble? [1 mark]

[1]4 bits (half a byte). One nibble corresponds to exactly one hexadecimal digit.

4. What is the largest denary number that can be represented in 8 bits? [1 mark]

[1]255 (binary 11111111). If you try to represent 256, an overflow error occurs.

5. State one example of a denary number and its equivalent hexadecimal value. [2 marks]

[1]Any correct pair, for example: denary 15 = hex F
[1]Another correct pair, for example: denary 28 = hex 1C or denary 255 = hex FF

6. Which number system do humans use, and which do computers use? [2 marks]

[1]Humans use denary (base 10)
[1]Computers use binary (base 2). Hexadecimal is used by humans as a shorthand for binary, but computers translate hex back to binary before processing.

3. Measurement of the Size of Computer Memories

Bit

Each individual 1 or 0 is known as a bit.

Example:0is one bit.

Nibble

A nibble is 4 bits, or half a Byte.

Example:1101is one nibble.

Byte

Each group of 8 bits is known as a Byte.

Example:11001100is one byte.

Decimal (SI) Units

Name of memory size	Equivalent denary value (bytes)
1 kilobyte (1 KB)	1 000
1 megabyte (1 MB)	1 000 000
1 gigabyte (1 GB)	1 000 000 000
1 terabyte (1 TB)	1 000 000 000 000
1 petabyte (1 PB)	1 000 000 000 000 000
1 exabyte (1 EB)	1 000 000 000 000 000 000
1 zettabyte (1 ZB)	1 000 000 000 000 000 000 000
1 yottabyte (1 YB)	1 000 000 000 000 000 000 000 000

Binary (IEC) Units

Name of memory size	Number of bytes	Equivalent denary value (bytes)
1 kibibyte (1 KiB)	2¹⁰	1 024
1 mebibyte (1 MiB)	2²⁰	1 048 576
1 gibibyte (1 GiB)	2³⁰	1 073 741 824
1 tebibyte (1 TiB)	2⁴⁰	1 099 511 627 776
1 pebibyte (1 PiB)	2⁵⁰	1 125 899 906 842 624
1 exbibyte (1 EiB)	2⁶⁰	1 152 921 504 606 846 976
1 zebibyte (1 ZiB)	2⁷⁰	1 180 591 620 717 411 303 424
1 yobibyte (1 YiB)	2⁸⁰	1 208 925 819 614 629 174 706 176

The Difference Between Kilobytes and Kibibytes

Akibibyte (KiB)is the formal name for1024 bytes(2¹⁰).
Akilobyte (KB)is1000 bytes(10³) according to SI naming conventions.
In practice, powers of 1024 bytes tend to be used forRAM(internal memory).
Powers of 1000 tend to be used forselling hard disks and other persistent storage.
Internal memories (such as RAM) should be measured using the IEC system.A 64 GiB RAM could therefore store 64 × 2³⁰ bytes of data (68 719 476 736 bytes).

Quick Conversion Tip

To convert between IEC units, multiply or divide by 1024 each time. For example, to go from KiB to MiB, divide by 1024. To go from GiB to MiB, multiply by 1024.

Activity 3: Memory Size Calculations

Answer the following questions. Show all working.

How many bytes are in 3 KiB?
A hard drive is advertised as 2 TB. How many bytes is this using decimal (SI) units?
A computer has 16 GiB of RAM. How many bytes is this using IEC units?
Explain why a 1 TB hard drive shows less than 1 TB in Windows File Explorer (which uses IEC units).

Solution:

3 × 1024 =3 072 bytes
2 × 1 000 000 000 000 =2 000 000 000 000 bytes
16 × 2³⁰ = 16 × 1 073 741 824 =17 179 869 184 bytes
The manufacturer uses decimal (1 TB = 1 000 000 000 000 bytes), but the operating system uses binary (1 TiB = 1 099 511 627 776 bytes). 1 000 000 000 000 ÷ 1 099 511 627 776 ≈ 0.909 TiB, so it appears as about 909 GiB.

Check Your Understanding: Memory Measurement

1. How many bits are there in a byte? [1 mark]

[1]8 bits = 1 byte

2. How many bytes are in 1 KiB? [1 mark]

[1]1 024 bytes (2¹⁰)

3. Why is IEC (binary) measurement preferred for internal memory such as RAM? [2 marks]

[1]Internal memory is built using powers of 2 (e.g. 8, 16, 32, 64 GiB), so IEC units match the physical structure
[1]The IEC system is more accurate for measuring internal memory because 1 KiB = 1024 bytes exactly

4. How many bits are in 3 bytes? [1 mark]

[1]24 bits (3 × 8)

5. What is the difference between 1 KB and 1 KiB? [2 marks]

[1]1 KB = 1 000 bytes (decimal/SI units)
[1]1 KiB = 1 024 bytes (binary/IEC units)

6. A computer has 64 GiB of RAM. How many bytes is this? [2 marks]

[1]64 × 2³⁰
[1]= 68 719 476 736 bytes

4. Converting Between Number Systems

Interactive Number System Converter

Type a value into any box and the other boxes will update automatically. Try converting denary 45, hex B5 or binary 10110101.

Denary (Base 10)

Binary (Base 2)

Hexadecimal (Base 16)

Tip:If a value contains only 0s and 1s it could be binary or hex. If it contains A–F it must be hex.

A) Converting from Denary to Binary

Method: Place Value Table

Write out the binary column headings. Find the first heading larger than your denary number. Use the headings to the right of it.

Example: Convert 45 to binary

The first heading bigger than 45 is 64, so use: 32 16 8 4 2 1

32 fits into 45 once (remainder 13) → 1

16 fits into 13 zero times → 0

8 fits into 13 once (remainder 5) → 1

4 fits into 5 once (remainder 1) → 1

2 fits into 1 zero times → 0

1 fits into 1 once → 1

Denary 45 = 101101 in binary

Method: Repeated Division by 2

Repeatedly divide the denary number by 2, writing down the remainder each time. Read the remainders from bottom to top.

Example: Convert 203 to binary

203 ÷ 2 = 101 remainder 1

101 ÷ 2 = 50 remainder 1

50 ÷ 2 = 25 remainder 0

25 ÷ 2 = 12 remainder 1

12 ÷ 2 = 6 remainder 0

6 ÷ 2 = 3 remainder 0

3 ÷ 2 = 1 remainder 1

Reading bottom to top: 11001011

B) Converting from Binary to Denary

Method

Write out the column headings for the number of bits given (from right to left). Add together any column heading that has a 1 under it.

Example 1: Convert 1011 to denary

8 4 2 1

1 0 1 1

1×8 + 0×4 + 1×2 + 1×1 = 11

Binary 1011 = 11 in denary

Example 2: Convert 01100011 to denary

128 64 32 16 8 4 2 1

0 1 1 0 0 0 1 1

1×64 + 1×32 + 1×2 + 1×1 = 99

Binary 01100011 = 99 in denary

Example 3: Convert 01110001110100 to denary

8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1

0 1 1 1 0 0 0 1 1 1 0 1 0 0

4096 + 2048 + 1024 + 64 + 32 + 16 + 4 = 7284

Binary 01110001110100 = 7284 in denary

C) Converting from Denary to Hexadecimal

Method 1: Denary → Binary → Hex

Example: Convert 28 to hex

28 in binary = 00011100

Split into nibbles: 0001 | 1100

0001 = 1, 1100 = 12 = C

Denary 28 = 1C in hex

Method 2: Divide by 16

Example: Convert 2004 to hex

2004 ÷ 16 = 125 remainder 4

125 ÷ 16 = 7 remainder 13 (D)

7 ÷ 16 = 0 remainder 7

Read remainders bottom to top: 7 D 4

Denary 2004 = 7D4 in hex

D) Converting from Hexadecimal to Denary

Method 1: Hex → Binary → Denary

Example: Convert B9 to denary

B = 11 = 1011, 9 = 1001

Join: 10111001

1×128 + 1×32 + 1×16 + 1×8 + 1×1 = 185

Hex B9 = 185 in denary

Method 2: Multiply by 16

Example: Convert 79 to denary

7 × 16 = 112

112 + 9 = 121

Hex 79 = 121 in denary

Example: Convert 45A to denary

4 × 256 = 1024

5 × 16 = 80

A (10) × 1 = 10

Hex 45A = 1114 in denary

E) Converting from Binary to Hexadecimal

Split the binary number into groups of 4 bits (nibbles) starting from the right. If the last group has fewer than 4 bits, fill in with 0s from the left. Convert each nibble to its hexadecimal digit.

Example 1: Convert 10110111 to hex

Split: 1011 | 0111

1011 = 11 = B, 0111 = 7

Binary 10110111 = B7 in hex

Example 2: Convert 00111001 to hex

Split: 0011 | 1001

0011 = 3, 1001 = 9

Binary 00111001 = 39 in hex

F) Converting from Hexadecimal to Binary

Take each hexadecimal digit and write down the 4-bit binary code that corresponds to it. Put the groups together.

Example: Convert 5F to binary

5 = 0101, F (15) = 1111

Join: 01011111

Hex 5F = 01011111 in binary

Example: Convert 26 to binary

2 = 0010, 6 = 0110

Join: 00100110

Hex 26 = 00100110 in binary

Activity 4: Conversions Practice

Convert the following. Show your working.

Denary 165 → binary
Binary 01111001 → denary
Denary 227 → hexadecimal
Hex 4A → denary
Binary 11000011 → hex
Hex 9C → binary
IP address values: Convert denary 167 and 214 to 8-bit binary.

Solution:

165 = 128 + 32 + 4 + 1 =10100101
01111001 = 64 + 32 + 16 + 8 + 1 =121
227 ÷ 16 = 14 (E) remainder 3 →E3
4 × 16 = 64; 64 + 10 =74
Split: 1100 | 0011 → C3 =C3
9 = 1001, C (12) = 1100 →10011100
167 = 10100111, 214 = 11010110

Check Your Understanding: Conversions

1. Convert denary 115 to 8-bit binary. [1 mark]

[1]01110011 (64 + 32 + 16 + 2 + 1 = 115)

2. Convert binary 11001010 to denary. [1 mark]

[1]128 + 64 + 8 + 2 = 202

3. Convert denary 490 to hexadecimal. [2 marks]

[1]490 ÷ 16 = 30 remainder 10 (A)
[1]30 ÷ 16 = 1 remainder 14 (E); 1 ÷ 16 = 0 remainder 1 →1EA

4. Convert hex FF to denary. [1 mark]

[1](16 × 15) + (1 × 15) = 240 + 15 = 255

5. Convert binary 11110111 to hexadecimal. [2 marks]

[1]Split: 1111 | 0111
[1]1111 = F, 0111 = 7 →F7

6. Convert hex A9 to binary and denary. [3 marks]

[1]A = 1010, 9 = 1001 → binary 10101001
[1]10 × 16 = 160
[1]160 + 9 = 169 in denary

5. Uses of Hexadecimal

In Computer Science, hexadecimal is often preferred when working with large values. It takesfewer digitsto represent a given value in hexadecimal than in binary. One hexadecimal digit corresponds to4 bits(one nibble) and can represent16 unique values(0–F).

Why is hexadecimal used?

The more bits there are in a binary number, the harder it makes it for a human to read.
Numbers with more bits are more prone to errors when being copied.
Hexadecimal is shorter and easier to remember than binary.

Exam tip: Use phrases like "fewer digits = easier for humans to read" or "less chance of copying errors".

MAC Addresses

AMAC address(Media Access Control address) is a number thatuniquely identifies a deviceon a network.
It refers to thenetwork interface card (NIC)which is part of the device.
It is rarely changed so that a particular device can always be identified no matter where it is.
A MAC address is usually made up of48 bitswhich are shown as6 groups of two hexadecimal digits.
The first half (NN-NN-NN) is themanufacturer identity number.
The second half (DD-DD-DD) is theserial numberof the device.

Example MAC address:

00-1C-B3-4F-25-FE

Manufacturer code: 001CB3 (Apple Corporation)

Serial number: 4F25FE

Equivalent binary: 48 bits — writing this out would be very error-prone!

Colour Codes

Hex can be used to represent colours on web pages and image-editing programs using the format#RRGGBB(RR = reds, GG = greens, BB = blues).
The# symbolindicates that the number has been written in hex format.
This system usestwo hex digits for each colour, e.g. #FF6600.
The values for each colour run between00 and FF. In binary, 00 is 0000 0000 and FF is 1111 1111.
That provides256 possible valuesfor each of the three colours.
That gives a total spectrum of 256 reds × 256 greens × 256 blues — which is over16 million coloursin total.
#FF0000is the purest red — red only, no green or blue.
Blackis #000000 — no red, no green and no blue.
Whiteis #FFFFFF.

#FF0000

Red

#00FF00

Green

#0000FF

Blue

#FF6600

Orange

RGB model:An orange colour that is #FF6600 in hex would be 255, 102, 0 in RGB.

Error Codes & Memory Dumps

Error codesare often shown as hexadecimal values.
These numbers refer to thememory location of the errorand are usually automatically generated by the computer.
When memory contents are output to a printer or monitor, this is known as amemory dump.
A program developer can look at each hexadecimal code and determine where the error lies.
The value on the far left shows thememory location, so it is possible to find out exactly where in memory the fault occurs.
This is much more manageable using hexadecimal rather than binary — it is a very powerful fault-tracing tool.

Example memory dump (hex):

00990F60 54 68 69 73 20 69 73 20 61 6E 20 65 78 61 6D 70 6C 65 20 6F 66

00990F77 61 20 6D 65 6D 6F 72 79 20 64 75 6D 70 20 66 72 6F 6D 20 61

00990E8E 74 79 70 69 63 61 6C 20 63 6F 6D 70 75 74 65 72 20 20 6D 85

This is far easier to read than the equivalent binary!

URLs

A URL can only contain standard characters (a–z and A–Z), numbers (0–9) and some special symbols.
If a URL needs to include a character outside of this set, it is converted into ahexadecimal code.
Hexadecimal codes included in a URL are prefixed with a% sign.

IPv6 Addresses

AnIPv4 addressis a 32-bit number written in denary or hexadecimal form, e.g. 109.108.158.1 (or 77.76.9e.01 in hex).
AnIPv6 addressis a 128-bit number broken down into 16-bit chunks, represented by hexadecimal numbers.
IPv6 uses acolon (:)rather than a decimal point (.) as used in IPv4.

Example IPv6 address:

a8ff:7a88:ffc3:0ff:3d21:2c95:6c7b:d0a

Activity 5: Hexadecimal in the Real World

Answer the following questions.

Describe two applications of the hexadecimal system. [4]
A MAC address is 00-1C-B3-4F-25-FE. What part of this address identifies the manufacturer? [1]
Explain why hexadecimal is used for MAC addresses instead of binary. [2]
What colour is #000000? What about #FFFFFF? [2]

Solution:

Any two of: MAC addresses, colour codes, URLs, error codes, memory dumps, IPv6 addresses. (2 marks each, 1 for naming, 1 for explaining.)
00-1C-B3— this is the manufacturer ID.
MAC addresses are 48 bits long. Binary would be 48 digits (very hard to read and prone to errors). Hexadecimal uses only 12 digits, so it is shorter, easier to read and less error-prone.
#000000 is black (no red, no green, no blue). #FFFFFF is white (all colours at full intensity).

Check Your Understanding: Uses of Hex

1. How many hexadecimal digits are there in a typical MAC address? [1 mark]

[1]12 hexadecimal digits (48 bits)

2. Why is hex used for HTML colour codes? [2 marks]

[1]Hex is shorter than binary (6 digits instead of 24 bits for a colour)
[1]It is easier for web designers to read and copy without making errors

3. What does the # symbol mean in a hex colour code? [1 mark]

[1]It indicates that the number has been written in hexadecimal format

4. Give two examples of where hexadecimal is used in computing. [2 marks]

[1]Any one of: MAC addresses, colour codes, URLs, error codes, memory dumps, IPv6 addresses
[1]Another from the list above

5. How many colours can be represented using HTML hex codes? [2 marks]

[1]256 possible values for each colour channel (red, green, blue) — from 00 to FF
[1]256 × 256 × 256 = over 16 million colours in total

6. Why is hexadecimal preferred over binary for memory dumps? [2 marks]

[1]Hex is much shorter and easier for programmers to read than long strings of binary
[1]It is easier to spot errors and find memory locations when the data is written in hex

6. Binary Addition and Overflow

The 5 Golden Rules of Binary Addition

Binary Addition	Binary Answer	Working
0 + 0	0	no carry
0 + 1	1	no carry
1 + 0	1	no carry
1 + 1	0	carry 1 (1 + 1 = 2 = 10)
1 + 1 + 1	1	carry 1 (1 + 1 + 1 = 3 = 11)

Start from therightmost digitand move left. Carry over when the sum of a column is greater than 1.

Overflow Explained

Anoverflow erroroccurs when the result of a binary additionexceeds the available bits.
For an 8-bit register, the maximum value that can be stored is255(binary 11111111).
An overflow error will occur if the value is greater than 255 in an 8-bit register.
A computer or device has a predefined limit that it can represent or store, for example 16-bit.
An overflow error occurs when a value outside this limit should be returned.

Worked Examples

Example 1: 00100111 + 01001010

00100111

+ 01001010

---------

01110001

Column 1: 1 + 0 = 1 (no carry)
Column 2: 1 + 1 = 0 (carry 1)
Column 3: 1 + 0 + 1 = 0 (carry 1)
Column 4: 0 + 1 + 1 = 0 (carry 1)
Column 5: 0 + 0 + 1 = 1 (no carry)
Column 6: 1 + 0 = 1 (no carry)
Column 7: 0 + 1 = 1 (no carry)
Column 8: 0 + 0 = 0 (no carry)

Example 2: 126 + 62

126 = 01111110

62 = 00111110

---------

10111100

10111100 = 128 + 32 + 16 + 8 + 4 = 188, which is the same as 126 + 62.

Overflow Example: 11111111 + 00000001

11111111 (255)

+ 00000001 (1)

---------

100000000 (would need a 9th bit)

In an 8-bit register, the result is00000000— but the correct answer (256) cannot be represented. This is anoverflow error.

Interactive Binary Addition

Enter two 8-bit binary numbers and see the sum with carries.

First binary number (8 bits)

Second binary number (8 bits)

Activity 6: Binary Addition Practice

Carry out the following binary additions. State whether any cause an overflow.

00011101 + 01100110
00100111 + 00111111
00101110 + 01001101
01110111 + 00111111
00111100 + 00110011
11111111 + 00000001

Solution:

00011101 + 01100110 =10000011(29 + 102 = 131)
00100111 + 00111111 =01100110(39 + 63 = 102)
00101110 + 01001101 =01111011(46 + 77 = 123)
01110111 + 00111111 =10110110(119 + 63 = 182)
00111100 + 00110011 =01101111(60 + 51 = 111)
11111111 + 00000001 =00000000withoverflow error(255 + 1 = 256, needs 9 bits)

Check Your Understanding: Binary Addition

1. What is 1 + 1 in binary? [2 marks]

[1]1 + 1 = 10 in binary
[1]The result is 0 with a carry of 1 to the next column

2. What is the largest denary number that can be stored in an 8-bit register? [1 mark]

[1]255 (binary 11111111)

3. Why does an overflow error occur? [2 marks]

[1]The result of a calculation is too large to fit in the available number of bits
[1]The result needs more bits than the register can hold (e.g. 256 needs 9 bits but an 8-bit register only has 8)

4. Add 00101110 + 01001101. [2 marks]

[1]Correct addition process with carries
[1]= 01111011 (46 + 77 = 123)

5. A student adds 11111111 + 00000001 and writes the answer as 100000000. Explain why this answer is not valid in an 8-bit register. [2 marks]

[1]100000000 has 9 bits, but the register can only store 8 bits
[1]This is an overflow error — the correct value (256) cannot be represented in 8 bits

6. Add 126 and 62 in binary and check that the result matches denary addition. [3 marks]

[1]126 = 01111110, 62 = 00111110
[1]Binary sum = 10111100
[1]10111100 = 188, and 126 + 62 = 188 — the results match

7. Logical Binary Shifts

Alogical binary shiftis how a computer system performs basicmultiplication and divisionon non-negative values (0 and positive numbers). Binary digits are moved left or right a set number of times.

Left Shift

A left shiftmultipliesa binary number by 2 (×2) per shift
Bits shift left; the leftmost bit(s) may be lost (overflow)
Empty positions on the right are filled with0
A left shift of 2 places multiplies the number by 4 (×4)

Right Shift

A right shiftdividesa binary number by 2 (÷2) per shift
Bits shift right; the rightmost bit(s) are lost
Empty positions on the left are filled with0
A right shift of 2 places divides the number by 4 (÷4)

Left Shift Example: 21 → 42 → 84 → 168

Original: 21 = 00010101

128 64 32 16 8 4 2 1

0 0 0 1 0 1 0 1

Left shift 1: 00101010 = 42 (21 × 2¹)

0 0 1 0 1 0 1 0

Left shift 2: 01010100 = 84 (21 × 2²)

0 1 0 1 0 1 0 0

Left shift 3: 10101000 = 168 (21 × 2³)

1 0 1 0 1 0 0 0

Note:If we shift 21 four places left, the leftmost 1-bit is lost, causing an error — this exceeds the maximum shifts for an 8-bit register.

Right Shift Example: 200 → 100 → 50 → 25

Original: 200 = 11001000

1 1 0 0 1 0 0 0

Right shift 1: 01100100 = 100 (200 ÷ 2¹)

0 1 1 0 0 1 0 0

Right shift 2: 00110010 = 50 (200 ÷ 2²)

0 0 1 1 0 0 1 0

Right shift 3: 00011001 = 25 (200 ÷ 2³)

0 0 0 1 1 0 0 1

Note:Shifting four places right loses the rightmost 1-bit and gives an incorrect result (12 instead of 12.5).

Overflow in Binary Shifts

Overflow happens when a 1 is shifted out of themost significant bit (MSB)on the left in a logical left shift.
This means important data is lost, which can seriously change the number's value.
At IGCSE, you usually only need to spot an overflow when a 1 is shifted out on the left.

Interactive Logical Shift

Enter an 8-bit binary number, then shift it left or right.

Binary number (8 bits)

Activity 7: Logical Shift Practice

Write 24 as an 8-bit register.
Show the result of a logical shift 2 places to the left.
Show the result of a logical shift 3 places to the right.
Write down the denary value of 01101000.
Shift 01101000 three places to the right and comment on your result.
Write down the denary value of 00001111.
Shift 00001111 four places to the left and comment on your result.

Solution:

24 =00011000
Left shift 2:01100000= 96 (24 × 4)
Right shift 3:00000011= 3 (24 ÷ 8)
01101000 =104
Right shift 3:00001101= 13 (104 ÷ 8). The rightmost 1-bit is lost, but the result is correct because 104 is divisible by 8.
00001111 =15
Left shift 4:11110000= 240 (15 × 16). This is still valid because no 1-bits were shifted out of the MSB.

Check Your Understanding: Logical Shifts

1. What happens to a binary number when it is shifted one place to the left? [2 marks]

[1]Each bit is moved one place to the left
[1]The value is multiplied by 2; a 0 is inserted at the rightmost position

2. What happens to a binary number when it is shifted two places to the right? [2 marks]

[1]Each bit is moved two places to the right
[1]The value is divided by 4; the two rightmost bits are lost and 0s are inserted on the left

3. Shift 00010101 three places to the right. What is the new denary value? [2 marks]

[1]00000010
[1]21 ÷ 8 = 2 (the value 21 is not divisible by 8, so data is lost)

4. When does overflow occur during a logical left shift? [2 marks]

[1]When a 1 is shifted out of the most significant bit (MSB) on the left
[1]The lost 1-bit means important data is lost, changing the value incorrectly

5. What is the effect of a left shift of 3 places on a binary number? [1 mark]

[1]The number is multiplied by 8 (2³)

6. A student says "shifting left then right by the same number of places always returns the original number." Is this correct? Explain. [2 marks]

[1]Not always — if a 1-bit is shifted out of the register during the shift, that data is lost permanently
[1]This means the reverse shift cannot restore the original value, so the statement is incorrect for numbers where overflow/bit loss occurs

8. Two's Complement (Representing Negative Numbers)

Up until now we have assumed all binary numbers have positive values. To represent both positive and negative numbers, we usetwo's complement. In two's complement, the leftmost bit is designated themost significant bit (MSB)and has anegative value.

Signed Magnitude

Leftmost bit = sign bit
0 = positive, 1 = negative
Remaining bits = magnitude
Has two representations of zero (+0 and −0) — a problem

One's Complement

Negative numbers formed byinvertingevery bit of the positive version
0 becomes 1 and 1 becomes 0
Still has two representations of zero (+0 = 0000, −0 = 1111)

Two's Complement

MSB has anegative value(−128 in 8-bit)
Formed by inverting bits of the positive version andadding 1
Only one representation of zero — most common method used by computers

Two's Complement in 8 Bits

Column Values

−128

If the MSB is 1, the number is negative. If the MSB is 0, the number is positive.

Lowest integer: −128
Highest integer: +127

Converting to Denary

Example: 10010011

1 0 0 1 0 0 1 1

−128 + 16 + 2 + 1 =−109

Example: 11111111

−128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 =−1

Converting Negative Denary to Two's Complement

Method 1: Put 1s in Correct Places

Convert −79 to 8-bit two's complement

−79 is the same as −128 + 49

49 = 32 + 16 + 1

1 0 1 1 0 0 0 1

Check: −128 + 32 + 16 + 1 = −79 ✓

Method 2: Flip and Add 1

Convert −67 to 8-bit two's complement

Write +67 in binary: 01000011

Invert the bits: 10111100

Add 1: 10111101

−67 = 10111101

Quick Method: Copy from the Right, then Invert

To represent −76:

Write +76 in binary: 01001100

Starting from the LSB (right), copy bits up to and including the first 1: ...100

Invert the remaining bits to the left: 10110100

−76 = 10110100

Interactive Two's Complement Converter

Enter a denary number between −128 and 127 to see its 8-bit two's complement representation.

Denary value (−128 to 127)

Activity 8: Two's Complement Practice

Convert the following negative denary numbers into 8-bit two's complement. Use both methods where possible.

−18
−31
−47
−63
−88
−100
−127

Then convert the following two's complement numbers to denary:

11001110
10111111
11101111
10000111
11111111

Solution:

Denary to two's complement:

−18 =11101110(18 = 00010010, invert = 11101101, +1 = 11101110)
−31 =11100001(31 = 00011111, invert = 11100000, +1 = 11100001)
−47 =11010001(47 = 00101111, invert = 11010000, +1 = 11010001)
−63 =11000001(63 = 00111111, invert = 11000000, +1 = 11000001)
−88 =10101000(88 = 01011000, invert = 10100111, +1 = 10101000)
−100 =10011100(100 = 01100100, invert = 10011011, +1 = 10011100)
−127 =10000001(127 = 01111111, invert = 10000000, +1 = 10000001)

Two's complement to denary:

11001110 = −128 + 64 + 8 + 4 + 2 =−50
10111111 = −128 + 32 + 16 + 8 + 4 + 2 + 1 =−65
11101111 = −128 + 64 + 32 + 8 + 4 + 2 + 1 =−17
10000111 = −128 + 4 + 2 + 1 =−121
11111111 =−1

Check Your Understanding: Two's Complement

1. What is the range of values that can be stored in an 8-bit two's complement register? [2 marks]

[1]Lowest value: −128
[1]Highest value: +127

2. Convert −1 to 8-bit two's complement. [2 marks]

[1]+1 = 00000001
[1]Invert = 11111110, add 1 =11111111

3. Convert 10110100 (two's complement) to denary. [2 marks]

[1]MSB is 1, so number is negative: −128
[1]−128 + 32 + 16 + 4 =−76

4. What is the difference between one's complement and two's complement? [2 marks]

[1]In one's complement, you simply invert all the bits of the positive number
[1]In two's complement, you invert the bits and then add 1; this removes the duplicate zero representation

5. What does the MSB represent in two's complement? [1 mark]

[1]In 8-bit two's complement, the MSB represents −128 (a negative place value)

6. Convert −104 to 8-bit two's complement. [2 marks]

[1]+104 = 01101000
[1]Invert = 10010111, add 1 =10011000

9. Hexadecimal Arithmetic

Just like denary and binary, we can perform arithmetic operations on hexadecimal numbers. You may be asked to add, subtract or multiply hex values in the exam.

Hex Addition

Add the denary values of each digit. If the total is 16 or more, carry 1 to the next column.

C + D = ?

C = 12, D = 13

12 + 13 = 25

25 in hex = 19 (1 sixteen + 9 units)

Hex Subtraction

Subtract the denary values. If the top digit is smaller, borrow 16 from the next column.

8B − 7A = ?

8B = 139, 7A = 122

139 − 122 = 17

17 in hex = 11

Hex Multiplication

Multiply the denary values. Convert the result back to hex.

AA × A = ?

AA = 170, A = 10

170 × 10 = 1700

1700 in hex = 6A4

Hexadecimal Addition: 8B + AA = 135

8B = 1000 1011 = 139 in denary

AA = 1010 1010 = 170 in denary

139 + 170 = 309

309 in hex: 309 ÷ 16 = 19 remainder 5; 19 ÷ 16 = 1 remainder 3; 1 ÷ 16 = 0 remainder 1 →135

Activity 9: Hexadecimal Calculations

Addition:

A + E = ?

D + A = ?

E7 + AB = ?

9F + CB = ?

A5 + EF = ?

99 + EE = ?

Subtraction:

F − B = ?

D − A = ?

E6 − 3B = ?

C2 − A9 = ?

5E − 1C = ?

D1 − 3F = ?

Multiplication:

2 × A = ?

F × F = ?

11 × C = ?

7D × 0A = ?

D8 × 6 = ?

32 × 12 = ?

Solution (selected):

Addition:

A + E = 10 + 14 = 24 =18
D + A = 13 + 10 = 23 =17
E7 + AB = 231 + 171 = 402 =192
9F + CB = 159 + 203 = 362 =16A
A5 + EF = 165 + 239 = 404 =194
99 + EE = 153 + 238 = 391 =187

Subtraction:

F − B = 15 − 11 = 4 =4
D − A = 13 − 10 = 3 =3
E6 − 3B = 230 − 59 = 171 =AB
C2 − A9 = 194 − 169 = 25 =19
5E − 1C = 94 − 28 = 66 =42
D1 − 3F = 209 − 63 = 146 =92

Multiplication:

2 × A = 2 × 10 = 20 =14
F × F = 15 × 15 = 225 =E1
11 × C = 17 × 12 = 204 =CC
7D × 0A = 125 × 10 = 1250 =4E2
D8 × 6 = 216 × 6 = 1296 =510
32 × 12 = 50 × 18 = 900 =384

Check Your Understanding: Hexadecimal Arithmetic

1. What is C + D in hexadecimal? [2 marks]

[1]C = 12, D = 13; 12 + 13 = 25 in denary
[1]25 in hex = 19

2. What is 8B − 7A in hexadecimal? [2 marks]

[1]8B = 139, 7A = 122; 139 − 122 = 17
[1]17 in hex = 11

3. What is AA × A in hexadecimal? [2 marks]

[1]AA = 170, A = 10; 170 × 10 = 1700
[1]1700 in hex = 6A4

4. What is 15 in hexadecimal? [1 mark]

[1]F (since 15 in denary is F in hex)

5. A student calculates 29 + C8 in hex and gets E1. Is this correct? Show your working. [3 marks]

[1]29 in hex = 41 in denary; C8 in hex = 200 in denary
[1]41 + 200 = 241 in denary
[1]241 in hex = F1. So E1 is not correct — the correct answer is F1.

6. What does hexadecimal addition have in common with binary addition? [2 marks]

[1]Both work from right to left
[1]Both involve carrying when the sum exceeds the base (16 for hex, 2 for binary)

Key Takeaways

Computers use binarybecause they are made of switches and logic gates that only have two states: ON (1) and OFF (0). All data must be converted to binary to be processed.
Denaryis base 10,binaryis base 2, andhexadecimalis base 16 (using 0–9 and A–F).
One hex digit = 4 bits (one nibble).Two hex digits = 8 bits (one byte).
When convertingdenary to binary, use place value columns or repeated division by 2. When convertingbinary to denary, add the column values where there is a 1.
Convertbinary to hexby splitting into nibbles and converting each nibble to its hex digit.
Hexadecimal is used forMAC addresses(6 pairs of hex digits),colour codes(#RRGGBB),URLs(% prefix),error codes,memory dumps, andIPv6 addresses.
Inbinary addition: 0+0=0, 0+1=1, 1+0=1, 1+1=0 carry 1, 1+1+1=1 carry 1.
Overflowoccurs when the result of a binary addition exceeds the available bits (e.g. >255 in an 8-bit register).
Alogical left shiftmultiplies by 2 per shift (×2, ×4, ×8). Alogical right shiftdivides by 2 per shift (÷2, ÷4, ÷8).
Overflow in a left shift occurs when a 1 is shifted out of the MSB.
Two's complementrepresents negative numbers. In 8-bit, the MSB has a value of −128. Range: −128 to +127.
To find two's complement of a negative number: write the positive binary, invert the bits, and add 1.
1 KiB = 1024 bytes(binary/IEC) and1 KB = 1000 bytes(decimal/SI). Internal memory (RAM) should be measured using IEC units.

Question Bank

1. Explain why computers process data in binary format. [2 marks]

[1]Computers process data using logic gates...
[1]... that can only have two states (1/0)

2. Convert the denary number 59 into 8-bit binary. [1 mark]

[1]00111011 (32 + 16 + 8 + 2 + 1 = 59)

3. Convert the 12-bit binary number 011010010001 into denary. Show your working. [2 marks]

[1]1024 + 512 + 128 + 16 + 1
[1]= 1681

4. Give the hexadecimal values for the two 8-bit binary values: 01010000 and 00111101. [4 marks]

[1]Split 01010000: 0101 | 0000
[1]0101 = 5, 0000 = 0 →50
[1]Split 00111101: 0011 | 1101
[1]0011 = 3, 1101 = D →3D

5. Describe what is meant by a MAC address. [4 marks]

[1]Used to identify a device on a network
[1]It is a unique address
[1]It is a static address / it does not change
[1]It is set by the manufacturer; the first part is the manufacturer ID and the second part is the serial number

6. Add together the binary values 00011001 and 10001001. Show your carries. [2 marks]

[1]Correct addition with carries shown
[1]= 10100010 (25 + 137 = 162)

7. Explain what is meant by an overflow error and when it occurs. [3 marks]

[1]An overflow error occurs when the result of a calculation is too large to be stored
[1]It happens when the value exceeds the maximum that can be represented with the available bits (e.g. >255 in 8-bit)
[1]For example, 11111111 + 00000001 causes overflow because 256 needs 9 bits

8. Perform a logical shift of 2 places to the left on the binary number 00011000. What is the denary result? [2 marks]

[1]01100000
[1]96 (24 × 4 = 96)

9. Convert −49 into an 8-bit two's complement binary number. [2 marks]

[1]+49 = 00110001; invert = 11001110; add 1 = 11001111
[1]11001111

10. Convert the two's complement number 10000111 into denary. [2 marks]

[1]MSB is 1, so number is negative: −128
[1]−128 + 4 + 2 + 1 =−121

11. Explain why hexadecimal is used for error codes and memory dumps. [2 marks]

[1]Hex is much shorter and easier to read than long strings of binary
[1]It reduces the chance of copying errors and makes it easier to spot where an error has occurred

12. What is the difference between 1 KB and 1 KiB? [2 marks]

[1]1 KB = 1 000 bytes (decimal/SI)
[1]1 KiB = 1 024 bytes (binary/IEC)

13. A stopwatch uses six digits (hours, minutes, seconds). An 8-bit register stores each pair of digits. The registers show: Hours 00000101, Minutes 00011010, Seconds 00110111. Write the denary values shown on the stopwatch. [3 marks]

[1]Hours: 00000101 = 5
[1]Minutes: 00011010 = 26
[1]Seconds: 00110111 = 55

14. Convert the hexadecimal number 5F to denary and binary. [3 marks]

[1]5 × 16 = 80; F (15) × 1 = 15; 80 + 15 = 95 in denary
[1]5 = 0101, F = 1111
[1]Binary = 01011111

15. What would be the effect of shifting the binary number 11001000 four places to the right? Explain your answer. [3 marks]

[1]The result would be 00001100
[1]200 ÷ 16 = 12.5, but the result is only 12
[1]This is because the rightmost 1-bit is lost during the shift, causing an error (loss of data)