Manufacturers often use 1000 for marketing purposes. A “500 GB” hard drive advertised by the Manufacturer is actually about 465 GiB when measured in binary. The IEC prefixes KiB, MiB, GiB refer Specifically to powers of 1024.
Binary addition follows the same rules as denary addition, but with only two digits:
0 + 0 0 + 0 0 + 0 0 + 1 0 + 1 0 + 1 1 + 0 1 + 0 1 + 0 1 + 1 1 + 1 1 + 1 0 1 1 10 (0 carry 1)
Worked Example. 0110 + 0101 0110 + 0101 0110 + 0101
Worked Example. 1101 + 1011 1101 + 1011 1101 + 1011
Worked Example with overflow. 1100 + 1100 1100 + 1100 1100 + 1100
The result (11000) requires 5 bits, but we are working with 4-bit numbers. The leftmost 1 is lost, Giving an incorrect result of 1000. This is called overflow and occurs when the result of an Operation exceeds the number of bits available.
To subtract, add the two’s complement of the number being subtracted.
Two’s complement: Invert all bits (1’s complement), then add 1.
Worked Example. Calculate 6 − 3 6 - 3 6 − 3
6 = 0110$$3 = 0011
Two’s complement of 3: invert 0011 → 1100 0011 \to 1100 0011 → 1100 1 → 1101 1 \to 1101 1 → 1101
0110 + 1101 = 10011 0110 + 1101 = 10011 0110 + 1101 = 10011
Discard the overflow bit: 0011 = 3 0011 = 3 0011 = 3
Worked Example. Calculate 3 − 6 3 - 6 3 − 6
3 = 0011$$6 = 0110 .
Two’s complement of 6: invert 0110 → 1001 0110 \to 1001 0110 → 1001 1 → 1010 1 \to 1010 1 → 1010
0011 + 1010 = 1101 0011 + 1010 = 1101 0011 + 1010 = 1101
In two’s complement, 1101 = − 8 + 4 + 0 + 1 = − 3 1101 = -8 + 4 + 0 + 1 = -3 1101 = − 8 + 4 + 0 + 1 = − 3
Worked Example. Calculate 15 − 8 15 - 8 15 − 8
15 = 01111$$8 = 01000 .
Two’s complement of 8: invert 01000 → 10111 01000 \to 10111 01000 → 10111 1 → 11000 1 \to 11000 1 → 11000
01111 + 11000 = 100111 01111 + 11000 = 100111 01111 + 11000 = 100111
Discard the overflow bit: 00111 = 7 00111 = 7 00111 = 7
Proof that two’s complement negation works. For an n n n x x x 0 < x < 2 n − 1 0 \lt x \lt 2^{n-1} 0 < x < 2 n − 1 y y y x x x y = 2 n − x y = 2^n - x y = 2 n − x x + y = x + ( 2 n − x ) = 2 n x + y = x + (2^n - x) = 2^n x + y = x + ( 2 n − x ) = 2 n N N N y = − x y = -x y = − x ■ \blacksquare ■
A logical left shift moves all bits to the left by a specified number of positions. Zeros fill The vacated positions on the right. Each left shift multiplies by 2.
Example. 00001100 00001100 00001100 00110000 00110000 00110000
Formal justification. Shifting left by k k k 2 k 2^k 2 k 2 k 2^k 2 k
A logical right shift moves all bits to the right. Each right shift divides by 2 (integer Division).
Example. 00001100 00001100 00001100 00000110 00000110 00000110
Worked Example. What is 00010110 00010110 00010110
00010110 = 22 00010110 = 22 00010110 = 22 10110000 = 176 10110000 = 176 10110000 = 176 22 × 2 3 = 22 × 8 = 176 22 \times 2^3 = 22 \times 8 = 176 22 × 2 3 = 22 × 8 = 176
Hexadecimal (base 16) uses digits 0—9 and letters A—F (where A=10, B=11, …, F=15).
Hex is used as a shorthand for binary because:
4 binary digits = 1 hex digit Large binary numbers are easier to read and write in hex Hex is easier for humans to work with than long strings of 0s and 1s Binary Hex Denary 0000 0 0 0001 1 1 0010 2 2 0011 3 3 0100 4 4 0101 5 5 0110 6 6 0111 7 7 1000 8 8 1001 9 9 1010 A 10 1011 B 11 1100 C 12 1101 D 13 1110 E 14 1111 F 15
Binary to hex: Group the binary digits into groups of 4 (from the right), then convert each Group.
Worked Example. Convert 110101101011 to hex.
Group: 1101 0110 1011
Convert: D 6 B
Result: D6B
Worked Example. Convert 10110 to hex.
Pad with leading zeros: 0001 0110
Convert: 1 6
Result: 16
Hex to binary: Convert each hex digit to 4 binary digits.
Worked Example. Convert 3F7 to binary.
3 = 0011, F = 1111, 7 = 0111
Result: 001111110111
Hex to denary: Multiply each hex digit by its positional value (16 0 , 16 1 , 16 2 16^0, 16^1, 16^2 1 6 0 , 1 6 1 , 1 6 2
Worked Example. Convert 2AF to denary.
2 × 16 2 + 10 × 16 1 + 15 × 16 0 = 2 × 256 + 10 × 16 + 15 = 512 + 160 + 15 = 687 2 \times 16^2 + 10 \times 16^1 + 15 \times 16^0 = 2 \times 256 + 10 \times 16 + 15 = 512 + 160 + 15 = 687 2 × 1 6 2 + 10 × 1 6 1 + 15 × 1 6 0 = 2 × 256 + 10 × 16 + 15 = 512 + 160 + 15 = 687
Denary to hex: Repeatedly divide by 16 and record remainders.
Worked Example. Convert 300 to hex.
300 / 16 = 18 300 / 16 = 18 300/16 = 18 18 / 16 = 1 18 / 16 = 1 18/16 = 1 1 / 16 = 0 1 / 16 = 0 1/16 = 0
Result: 12C
Colour codes: In HTML/CSS, colours are represented as #RRGGBB (e.g. #FF0000 = red, #00FF00 = green, #0000FF = blue). Each pair is a hex value from 00 to FF (0 to 255 in denary).MAC addresses: 48-bit addresses represented as 12 hex digits (e.g. 00:1A:2B:3C:4D:5E)Error codes: Memory addresses and error messages often use hexAssembly language: Machine code instructions are shown in hexWhy hex for colours? Each colour channel (red, green, blue) uses 8 bits (0-255). Two hex digits Represent 8 bits exactly, so #RRGGBB gives precise control over each channel in a compact format.
Worked Example. What colour is #40FF00?
Red = 40 hex = 64 denary (moderate red), Green = FF hex = 255 denary (maximum green), Blue = 00 hex = 0 denary (no blue). This is a bright yellow-green colour.
Worked Example. What is the denary value of the MAC address byte 3C?
3 × 16 + 12 = 48 + 12 = 60 3 \times 16 + 12 = 48 + 12 = 60 3 × 16 + 12 = 48 + 12 = 60
A character set is a collection of characters that a computer can recognise. Each character is Assigned a unique binary code.
Character Set Description Characters ASCII American Standard Code for Information Interchange 128 characters (7 bits) Extended ASCII Extended version 256 characters (8 bits) Unicode Universal character set Over 140,000 characters (up to 32 bits)
ASCII uses 7 bits, giving 2 7 = 128 2^7 = 128 2 7 = 128
0—31: Control characters (non-printable, e.g. Carriage return, tab) 32—127: Printable characters (letters, digits, punctuation, symbols) Key ASCII values:
Character ASCII Code Character ASCII Code A 65 a 97 B 66 b 98 Z 90 z 122 0 48 9 57 Space 32 ! 33
Useful pattern: The ASCII code for ‘a’ is 97, for ‘b’ is 98, …, for ‘z’ is 122. The code for ‘A’ is 65, for ‘B’ is 66, …, for ‘Z’ is 90. The difference between uppercase and lowercase is 32. So to convert ‘A’ to ‘a’, add 32 to its ASCII code.
Worked Example. What is stored in binary for the text “Hi”?
H = 72 = 1001000 i = 105 = 1101001
“Hi” in 8-bit ASCII: 01001000 01101001
Worked Example. How many bytes are needed to store the word “COMPUTER” in ASCII?
Each character is 1 byte (8 bits). “COMPUTER” has 8 characters, so 8 bytes.
Worked Example (Higher Tier). Write the binary for “Cat” in 8-bit ASCII.
C = 67 = 01000011, a = 97 = 01100001, t = 116 = 01110100.
“Cat” = 01000011 01100001 01110100 (3 bytes).
Unicode supports multiple languages and special characters. The first 128 characters of Unicode are The same as ASCII (backwards compatible).
Unicode uses up to 4 bytes per character. Common encodings include UTF-8 (variable length, 1—4 Bytes) and UTF-16.
Why Unicode replaced ASCII: ASCII can only represent 128 characters, which is insufficient for Languages like Chinese, Japanese, Arabic, and Hindi, and cannot represent emojis or mathematical Symbols. Unicode assigns a unique code point to every character in every language.
UTF-8 encoding efficiency. UTF-8 uses 1 byte for ASCII characters, 2 bytes for most Latin-based Scripts, 3 bytes for Asian scripts, and 4 bytes for emojis and rare characters. This means plain English text is stored efficiently in UTF-8, while still supporting the full range of Unicode.
A bitmap image is made up of tiny squares called pixels (picture elements). Each pixel is Assigned a binary value that represents its colour.
Image resolution: The number of pixels in the image (width × \times ×
Colour depth: The number of bits used to represent each pixel.
Colour Depth Number of Colours 1 bit 2 (black and white) 8 bits (1 byte) 256 16 bits 65536 24 bits (3 bytes) 16777216 (True Colour) 32 bits Over 4 billion
Why 2 n 2^n 2 n With n n n 2 n 2^n 2 n 2 24 = 16 , 777 , 216 2^{24} = 16,777,216 2 24 = 16 , 777 , 216
\mathrm{File size (bits) = \mathrm{width \times \mathrm{height \times \mathrm{colour depth
Worked Example. An image is 1920 × \times ×
\mathrm{File size = 1920 \times 1080 \times 24 = 49766400 \mathrm{ bits = \frac{49766400}{8} = 6220800 \mathrm{ bytes = \frac{6220800}{1024 \times 1024} \approx 5.93 \mathrm{ MB
Worked Example. An image is 800 × \times ×
\mathrm{File size = 800 \times 600 \times 32 = 15360000 \mathrm{ bits = 1920000 \mathrm{ bytes \approx 1.83 \mathrm{ MB Worked Example (Higher Tier). A 4-megapixel image has a file size of 3 MB. What is the colour Depth?
4 × 10 6 4 \times 10^6 4 × 1 0 6 × \times × 3 × 1024 × 1024 × 8 3 \times 1024 \times 1024 \times 8 3 × 1024 × 1024 × 8
\mathrm{Colour depth = \frac{3 \times 1048576 \times 8}{4000000} = \frac{25165824}{4000000} \approx 6.29 Bits.
This is not an exact number, which suggests the file may include a header or metadata. Without Metadata, a 4-megapixel image at 24-bit colour depth would be: 4000000 × 24 / 8 = 12000000 4000000 \times 24 / 8 = 12000000 4000000 × 24/8 = 12000000 ≈ 11.44 \approx 11.44 ≈ 11.44
Increasing resolution: Sharper image, larger file sizeIncreasing colour depth: More colours available, larger file sizeDecreasing either: Smaller file size, lower qualityTrade-off. Higher resolution and colour depth produce better images but require more storage and Take longer to transmit. Web designers often compress images to balance quality against file size.
Worked Example. An image is 1024 × \times ×
Original: 1024 × 768 × 16 = 12582912 1024 \times 768 \times 16 = 12582912 1024 × 768 × 16 = 12582912
New: 1024 × 768 × 24 = 18874368 1024 \times 768 \times 24 = 18874368 1024 × 768 × 24 = 18874368
Increase: ( 18874368 − 12582912 ) / 12582912 × 100 = 50 % (18874368 - 12582912) / 12582912 \times 100 = 50\% ( 18874368 − 12582912 ) /12582912 × 100 = 50%
Vector images store images as mathematical descriptions of shapes (lines, curves, colours) Rather than as pixels.
Advantages of vector images:
Scale without losing quality (no pixelation) Smaller file sizes for simple graphics (logos, icons) Easy to edit individual components Disadvantages:
Not suitable for photographs (too complex) Rendering can be slower for very complex images Bitmap vs Vector comparison:
Feature Bitmap Vector Representation Grid of pixels Mathematical descriptions Scaling Loses quality when enlarged No quality loss at any size File size Depends on resolution Depends on complexity Best for Photographs Logos, icons, diagrams Editing Pixel-level manipulation Edit individual shapes
Sound is an analogue wave. To store it digitally:
The sound wave is sampled at regular intervals (sampling rate ) The amplitude at each sample is measured and stored as a binary value (sample resolution / bit depth ) This process is called analogue to digital conversion (ADC) .
Intuition. Imagine drawing a smooth curve on graph paper by only plotting points at regular Intervals along the x-axis. The more points you plot (higher sample rate) and the more precisely you Measure each y-value (higher bit depth), the more accurately your plotted points will match the Original curve.
Term Definition Sample rate Number of samples taken per second (measured in Hz) Sample resolution (bit depth) Number of bits per sample Bit rate Sample rate × \times ×
\mathrm{File size (bits) = \mathrm{sample rate \times \mathrm{sample resolution \times \mathrm{duration (seconds) \times \mathrm{channels
Worked Example. A 3-minute audio clip is recorded at 44100 Hz with 16-bit resolution (mono). Calculate the file size.
\mathrm{File size = 44100 \times 16 \times 180 = 127008000 \mathrm{ bits = \frac{127008000}{8} = 15876000 \mathrm{ bytes \approx 15.14 \mathrm{ MB
Worked Example. A 2-minute stereo audio clip at 48000 Hz with 24-bit resolution.
\mathrm{File size = 48000 \times 24 \times 120 \times 2 = 276480000 \mathrm{ bits \approx 32.94 \mathrm{ MB
Worked Example (Higher Tier). A 5-minute mono recording at 22050 Hz has a file size of 12.5 MB. What is the bit depth?
\mathrm{File size in bits = 12.5 \times 1024 \times 1024 \times 8 = 104857600 bits.
\mathrm{Bit depth = \frac{104857600}{22050 \times 300} = \frac{104857600}{6615000} \approx 15.85 .
This is closest to 16-bit, which gives: 22050 × 16 × 300 = 105840000 22050 \times 16 \times 300 = 105840000 22050 × 16 × 300 = 105840000 ≈ 12.62 \approx 12.62 ≈ 12.62
Higher sample rate: More faithful reproduction of the original sound, larger fileHigher bit depth: More precise amplitude values, larger fileCD quality: 44100 Hz, 16-bit Studio quality: 96000 Hz, 24-bit Telephone quality: 8000 Hz, 8-bit Nyquist theorem. To accurately reproduce a sound, the sample rate must be at least twice the Highest frequency in the sound. Human hearing ranges up to about 20,000 Hz, so a sample rate of 44,100 Hz (just above 2 × 20000 2 \times 20000 2 × 20000
Proof sketch of the Nyquist theorem. If a signal has maximum frequency f max f_{\max} f m a x f max f_{\max} f m a x f s ≥ 2 f max f_s \ge 2 f_{\max} f s ≥ 2 f m a x f s < 2 f max f_s \lt 2 f_{\max} f s < 2 f m a x ■ \blacksquare ■
Aliasing in practice. If a 30,000 Hz tone is sampled at 44,100 Hz, the reconstructed frequency Is ∣ 44100 − 30000 ∣ = 14100 |44100 - 30000| = 14100 ∣44100 − 30000∣ = 14100
Compression reduces file size, saving storage space and reducing transmission time over networks. Without compression, a 2-hour HD film could require hundreds of gigabytes.
Lossy compression permanently removes some data to reduce file size. The original file cannot be Perfectly reconstructed.
Examples: JPEG (images), MP3 (audio), MP4 (video)
Advantages:
Much smaller file sizes Suitable for images, audio, and video where perfect accuracy is not needed Disadvantages:
Quality loss Not suitable for text or programs (every bit matters) How JPEG works: JPEG exploits the fact that human vision is less sensitive to fine colour detail Than to fine brightness detail. It converts the image to a frequency representation, then discards High-frequency components that are less perceptible. This is why heavily compressed JPEGs show “blocky” artefacts.
Compression ratio comparison:
Format Type Typical Compression Ratio Quality PNG Lossless 2:1 to 3:1 Perfect JPEG Lossy 10:1 to 20:1 Good to poor FLAC Lossless 2:1 Perfect MP3 Lossy 10:1 to 12:1 Good
Lossless compression reduces file size without losing any data. The original file can be perfectly Reconstructed.
Examples: PNG (images), FLAC (audio), ZIP (general files)
Techniques:
Run-length encoding (RLE): Replaces repeated sequences with a count and value. Example: AAAABBBCCD → \to → Huffman coding: Uses shorter codes for more frequent characters and longer codes for less frequent ones.Advantages:
No quality loss Suitable for text, code, and documents Disadvantages:
Smaller reduction in file size compared to lossy Worked Example. Compress the following bitmap row using RLE: WWWWWBBBWWWWWW.
W5 B3 W6
Original: 12 bytes. Compressed: 6 bytes (assuming 1 byte per character and 1 byte per count).
When RLE is ineffective: RLE works well when there are long runs of the same value. For data Like ABABABABAB, RLE produces A1B1A1B1A1B1A1B1A1B1, which is larger than the original.
Worked Example. Compress the following data using RLE: AAAAAAABBBBCCCCCCDDDE.
A7 B4 C6 D3 E1.
Original: 21 bytes. Compressed: 10 bytes. Compression ratio: 21/10 = 2.1:1.
Worked Example (Higher Tier). A bitmap image has 64 pixels per row. Each row consists of 32 Black pixels followed by 32 white pixels. What is the compressed size using RLE (1 byte per count, 1 Byte per value)?
Two runs per row: B32 W32 = 4 bytes per row. If the image is 64 rows tall, compressed size = 64 × \times × 64 × 64 = 4096 64 \times 64 = 4096 64 × 64 = 4096
Huffman coding assigns variable-length codes to characters based on their frequency. More frequent Characters get shorter codes, and no code is a prefix of another (prefix-free property).
Example. For the string “BCCABBDDAECCBBAEDDCC”:
Character Frequency Huffman Code B 5 00 C 6 01 D 4 10 A 3 110 E 2 111
The most frequent character (C, 6 occurrences) gets the shortest code (01). The least frequent (E, 2 Occurrences) gets the longest code (111).
Why prefix-free? If the code for A were “1” and the code for B were “10”, then “10” could be Decoded as either B or A followed by the start of another character. Prefix-free codes eliminate This ambiguity, making decoding unambiguous.
Worked Example. Calculate the total number of bits to encode “BCCABBDDAECCBBAEDDCC” using the Huffman codes above.
5 × 2 + 6 × 2 + 4 × 2 + 3 × 3 + 2 × 3 = 10 + 12 + 8 + 9 + 6 = 45 5 \times 2 + 6 \times 2 + 4 \times 2 + 3 \times 3 + 2 \times 3 = 10 + 12 + 8 + 9 + 6 = 45 5 × 2 + 6 × 2 + 4 × 2 + 3 × 3 + 2 × 3 = 10 + 12 + 8 + 9 + 6 = 45
If fixed-width 3-bit codes were used (5 distinct characters need ⌈ log 2 5 ⌉ = 3 \lceil \log_2 5 \rceil = 3 ⌈ log 2 5 ⌉ = 3 20 × 3 = 60 20 \times 3 = 60 20 × 3 = 60
Savings: ( 60 − 45 ) / 60 × 100 = 25 % (60 - 45) / 60 \times 100 = 25\% ( 60 − 45 ) /60 × 100 = 25%
Huffman tree construction. The Huffman tree is built by repeatedly combining the two Lowest-frequency nodes:
Start with: A(3), B(5), C(6), D(4), E(2) Combine E(2) and A(3) to form EA(5) Combine D(4) and EA(5) to form DEA(9) Combine B(5) and C(6) to form BC(11) Combine DEA(9) and BC(11) to form the root The left branch gets 0 and the right branch gets 1 at each merge. This gives the codes shown in the Table above.
For an n n n
-2^{n-1} \mathrm{ to 2^{n-1} - 1
Bits Range 4 -8 to 7 8 -128 to 127 16 -32768 to 32767 32 -2147483648 to 2147483647
Why this asymmetry? There is one more negative number than positive number because zero is Represented as all zeros (positive). The pattern 100 … 0 100\ldots0 100 … 0 − 2 n − 1 -2^{n-1} − 2 n − 1 2 n − 1 2^{n-1} 2 n − 1
Binary multiplication is performed using long multiplication, similar to denary.
Worked Example. Calculate 6 × 3 6 \times 3 6 × 3
6 = 0110 6 = 0110 6 = 0110 3 = 0011 3 = 0011 3 = 0011
0110 (0110 x 1, shifted 0)
0110 (0110 x 1, shifted 1)
Confusing bits and bytes. 1 byte = 8 bits. A 24-bit colour depth is 3 bytes per pixel.Forgetting that 1 KB = 1024 bytes, not 1000. (Though in some contexts, kB = 1000 bytes.)Not counting leading zeros in binary conversions. An 8-bit number must have exactly 8 digits. The number 5 in 8-bit binary is 00000101, not 101.Confusing lossy and lossless compression. Lossy = data permanently lost (JPEG); lossless = no data lost (PNG).Calculating image file size incorrectly. Remember to multiply by the colour depth in bits, then convert to bytes (divide by 8), then to KB (divide by 1024).Forgetting to convert units when calculating file sizes. Divide by 8 for bits to bytes, then by 1024 for KB, then by 1024 for MB.Forgetting to multiply by the number of channels when calculating audio file size. Stereo has 2 channels, so the file size is doubled compared to mono.Assuming RLE always reduces file size. RLE increases file size when there are few or no repeated values.Forgetting to pad binary numbers when converting to hexadecimal. Always group from the right; add leading zeros if necessary.Confusing the MSB in two’s complement with a simple sign bit. In two’s complement, the MSB carries a negative weight of − 2 n − 1 -2^{n-1} − 2 n − 1 Convert the binary number 10110101 to denary and to hexadecimal.
Convert the hexadecimal number 2AF to binary and to denary.
A bitmap image is 1024 × \times ×
A sound file is recorded at 22050 Hz with 8-bit resolution for 2 minutes. Calculate the file size in megabytes.
Use run-length encoding to compress: AAAAABBCCCCCCCCCCDD.
Explain the difference between lossy and lossless compression, giving an appropriate use for each.
Why is the character ‘A’ stored as 1000001 in ASCII? Explain what this binary value represents.
An image file is 2.5 MB. If the colour depth is increased from 8 bits to 24 bits, what will the new file size be?
Explain why hexadecimal is often used to represent binary values in computing.
Calculate the file size of a 30-second stereo (2 channels) audio recording at 48000 Hz with 16-bit resolution.
(Higher Tier) Calculate 1101 + 0111 in binary. Show all working.
(Higher Tier) Calculate 15 − 8 15 - 8 15 − 8
(Higher Tier) Explain the difference between a bitmap image and a vector image. Give an appropriate use case for each.
(Higher Tier) A 4-minute stereo audio file is 25 MB. If the sample rate is 44100 Hz, what is the bit depth?
(Higher Tier) Calculate 9 × 5 9 \times 5 9 × 5
(Higher Tier) The string “MISSISSIPPI” is to be encoded using Huffman coding. Calculate the frequency of each character and determine which character would receive the shortest code.
(Higher Tier) Explain why the two’s complement representation of − 128 -128 − 128 + 128 +128 + 128
(Higher Tier) An image has resolution 2048 × \times ×
Example 1:
A typical exam question on Data Representation requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.
Example 2:
Multi-step problems in Data Representation often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.
A[1_Fundamentals] --> B[Key Concepts]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
D --> G[Real-world usage]
This topic covers the core concepts of data representation, including underlying theory, practical implementation, and key applications.
Key concepts include:
CPU architecture and the fetch-decode-execute cycle memory hierarchy (cache, RAM, virtual) input/output systems operating systems and scheduling interrupts and polling Understanding these concepts thoroughly is essential for both examinations and practical programming, and requires both theoretical knowledge and hands-on practice.
Algebra — Binary and hexadecimal number systems require understanding of base conversion and place value.Atomic Structure — Understanding atoms and electron configuration provides context for how computers store data.Cell Biology — Data storage in biological systems like DNA parallels digital data representation.Ratio and Proportion — Memory addressing and data scaling involve proportional relationships.