Unsigned binary — AQA A-Level Computer Science
Unsigned binary is a base-2 numerical representation system used in digital computers to represent non-negative integers using only positional bits of 0 and 1. For an n-bit unsigned binary number, values range strictly from 0 to 2^n - 1, meaning an 8-bit byte represents denary values from 0 up to 255 without allocating a sign bit.
Unsigned binary explained
Unsigned binary represents only non-negative whole numbers.
Read the full explanation
Every bit contributes to the magnitude, so an 8-bit unsigned pattern can represent 0 to 255. Signed binary reserves part of the pattern to indicate sign, usually by using the most significant bit as a sign indicator in two's complement form. This reduces the magnitude range but allows negative values to be represented. For example, the pattern 11111111 is 255 as unsigned but −1 as an 8-bit two's complement signed value. Being able to explain this contrast, including what happens to the range and how the same bit pattern can have different meanings, is the core of the statement.
Your focus
- Know the difference between unsigned binary and signed binary.
Which command word is this?
Give a concise, factual answer without explanation or working, such as stating a final binary bit pattern or a numerical maximum.
Unsigned binary exam tips
Topic Overview
Unsigned binary is the fundamental system used by computers to store non-negative whole numbers using base-2 place values. Because every bit contributes purely to the magnitude of the number rather than indicating a positive or negative sign, all available bits directly increase the maximum representable value.
Understanding unsigned binary is essential for mastering digital arithmetic logic, computer architecture, memory addressing, and data communication. It highlights the hardware boundaries of computing registers and illustrates how physical bit limits produce critical errors such as overflow during program execution.
Key Concepts
- →Positional weighting: In base-2, each place value doubles moving from right to left (1, 2, 4, 8, 16, 32, 64, 128, etc.).
- →Range and capacity: An n-bit unsigned binary sequence encodes 2^n unique combinations, representing integers strictly from 0 to 2^n - 1.
- →Fundamental addition rules: 0+0=0, 1+0=1, 1+1=0 (carry 1), and 1+1+1=1 (carry 1).
- →Arithmetic overflow: Occurs when an addition operation creates a carry out of the most significant bit that cannot be stored within the allocated register width.
Marking Points
- States that unsigned binary represents only zero and positive whole numbers.
- States that signed binary can represent negative as well as positive numbers.
- Explains that signed binary uses one bit (usually the most significant bit) to indicate sign.
- Recognises that the same bit pattern can have different values depending on whether it is interpreted as signed or unsigned.
Examiner Tips
- 💡Name the range difference explicitly when comparing the two systems.
- 💡Use a concrete 8-bit pattern to illustrate the contrast.
- 💡State clearly which bit indicates sign in signed binary.
- 💡Always count the bits in your final written answer to ensure it matches the register width requested in the question (e.g., pad with leading zeros to make a full 8-bit byte).
- 💡Clearly write down intermediate carry bits above each column during binary addition; examiners frequently award partial method marks for correct carry propagation.
- 💡When explaining overflow, always mention both the numerical result and the register capacity in bits to secure full explanation marks.
Common Mistakes
- Describes unsigned binary as able to store negative numbers; correct the idea by stating that unsigned binary is restricted to zero and positive values.
- Says signed binary has a larger range than unsigned for the same number of bits; correct the idea by noting that signed binary trades magnitude for the ability to represent negatives.
- Treats the sign bit as just another magnitude bit; correct the idea by explaining that in two's complement it indicates sign and affects the value of the remaining bits.
- Assuming an 8-bit unsigned binary register can store numbers up to 256; it stores 256 unique values, but the maximum value is 255 because 0 is the first value.
- Mistaking unsigned binary for signed binary (two's complement) and treating a leading 1 as a negative sign rather than its full positive place value.
- Believing that an overflow error immediately causes a system halt or crash; in reality, hardware registers wrap around and retain only the lower bits while setting an internal CPU status flag.
Revision Plan
- 1Step 1: Memorise the powers of 2 from 2^0 (1) to 2^8 (256) and practice writing place value grid headers automatically.
- 2Step 2: Practice converting denary integers up to 255 into 8-bit unsigned binary patterns and back using repeated subtraction and place value addition.
- 3Step 3: Complete timed drills on 8-bit binary addition, paying close attention to carry propagation and multi-bit carries.
- 4Step 4: Solve past paper questions requiring you to calculate ranges for arbitrary bit lengths (e.g., 10-bit, 12-bit) and explain overflow conditions.
Exam Question Types
- 📋Denary-binary conversions: Translating integers between denary and fixed-width unsigned binary patterns.
- 📋Binary addition: Performing columnar binary addition with carry bits clearly shown.
- 📋Range calculations: Calculating the minimum, maximum, and total number of values that can be stored in an n-bit register.
- 📋Short-answer explanations: Explaining the nature, cause, and consequences of an overflow error in a specific processor register.
Command Word Expectations (AQA)
Give a concise, factual answer without explanation or working, such as stating a final binary bit pattern or a numerical maximum.
Provide all intermediate working and calculations step-by-step to demonstrate how the final binary or denary value was reached.
Set out the cause and effect clearly; for instance, explaining why an overflow occurs by referencing register limits, carry bits, and data loss.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: Add the two 8-bit unsigned binary integers 11001100 and 01100111. State whether an overflow error occurs and justify your answer.
- 1.Step 1: Align the two 8-bit numbers vertically and add from right to left (least significant bit to most significant bit). Apply binary addition rules: 0+1=1, 1+1=0 (carry 1), 1+1+1=1 (carry 1).
- 2.Step 2: Add column 1 (1s): 0 + 1 = 1. Add column 2 (2s): 0 + 1 = 1. Add column 3 (4s): 1 + 1 = 0 (carry 1). Add column 4 (8s): 1 + 0 + 1 = 0 (carry 1).
- 3.Step 3: Add column 5 (16s): 0 + 0 + 1 = 1. Add column 6 (32s): 0 + 1 = 1. Add column 7 (64s): 1 + 1 = 0 (carry 1). Add column 8 (128s): 1 + 0 + 1 = 0 (carry 1).
- 4.Step 4: Evaluate the 9th carry bit. The raw sum is 100110011 in base 2 (denary 307).
- 5.Step 5: Check register limits: An 8-bit register has a maximum unsigned value of 2^8 - 1 = 255. Because the result (307) exceeds 255, the ninth bit cannot be accommodated in the 8-bit register.
Question: Convert the denary integer 213 into an 8-bit unsigned binary number, showing your method.
- 1.Step 1: Write down the 8-bit positional place values: 128, 64, 32, 16, 8, 4, 2, 1.
- 2.Step 2: Compare 213 to 128: 213 >= 128, so place a 1 in the 128 column. Remainder: 213 - 128 = 85.
- 3.Step 3: Compare 85 to 64: 85 >= 64, so place a 1 in the 64 column. Remainder: 85 - 64 = 21.
- 4.Step 4: Compare 21 to 32: 21 < 32, so place a 0 in the 32 column. Compare 21 to 16: 21 >= 16, so place a 1 in the 16 column. Remainder: 21 - 16 = 5.
- 5.Step 5: Compare 5 to 8: 5 < 8, so place a 0 in the 8 column. Compare 5 to 4: 5 >= 4, so place a 1 in the 4 column. Remainder: 5 - 4 = 1.
- 6.Step 6: Compare 1 to 2: 1 < 2, so place a 0 in the 2 column. Compare 1 to 1: 1 >= 1, so place a 1 in the 1 column. Remainder: 0.