Converting between number bases — AQA GCSE Computer Science
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Converting between number bases involves translating numerical values between denary (base 10), binary (base 2), and hexadecimal (base 16). In AQA GCSE Computer Science, students must master place-value tables and nibble-splitting methods to convert accurately between these representations.
Converting between number bases explained
Eight binary columns give place values 128, 64, 32, 16, 8, 4, 2 and 1, so any decimal from 0 to 255 can be written with exactly eight bits.
Read the full explanation
To convert, work left to right: if the place value fits into the remaining amount, write 1 and subtract it; otherwise write 0. For 200, 128 fits (remainder 72), 64 fits (remainder 8), 32 and 16 do not, 8 fits (remainder 0), and the last three columns are 0, giving 11001000. Check by adding the place values of the 1s back to the original number. The limits matter: 0 is 00000000 and 255 is 11111111, so nine bits would be needed for 256.
Your focus
- Convert a decimal value between 0 and 255 into an 8-bit binary pattern.
- Explain the place values used in an 8-bit binary number.
- Verify a binary conversion by summing the place values of the set bits.
Which command word is this?
Translate a given value from one number representation to another. Marks are awarded for the correct final answer, but working out should always be shown.
Converting between number bases exam tips
Topic Overview
Number base conversion is the process of translating values across binary (base 2), denary (base 10), and hexadecimal (base 16) systems. Computers natively operate using binary switches (transistors representing 0 or 1), whereas humans naturally calculate using denary. Hexadecimal bridges this gap by offering a concise, human-readable shorthand for binary data.
Mastering these conversions is fundamental to GCSE Computer Science as it underpins how computers store data types, execute instructions, and represent colours or MAC addresses. Being fluent in conversions allows students to understand low-level architecture, memory management, and debugging without error.
Key Concepts
- →Base systems define the number of unique digits available: base 2 uses 0 and 1; base 10 uses 0 to 9; base 16 uses 0 to 9 and A to F.
- →Place values increase as powers of the base moving right-to-left: base 2 powers (1, 2, 4, 8, 16, 32, 64, 128) and base 16 powers (1, 16, 256).
- →A nibble (4 binary bits) maps directly to exactly one hexadecimal digit (e.g. 1111 in binary is F in hexadecimal), enabling straightforward conversions.
- →To convert between denary and hexadecimal easily, convert through binary as an intermediate stepping stone.
Marking Points
- Recognises the eight place values 128, 64, 32, 16, 8, 4, 2 and 1 and that eight bits span 0 to 255.
- Converts a given decimal correctly by subtracting place values from left to right and recording 1 or 0 in each column.
- Verifies an answer by summing the place values of the 1 bits to recover the original decimal.
- Identifies boundary cases: 0 as 00000000 and 255 as 11111111, and explains why 256 needs a ninth bit.
- Keeps all eight positions, including leading zeros, so the result is a full 8-bit pattern.
Examiner Tips
- 💡Write the place values 128 to 1 above your working before converting, so each bit is placed deliberately.
- 💡Always check by adding the place values of the 1s; this catches most slips quickly.
- 💡For multiple-choice items, eliminate options with more than eight bits or with a value-sum that does not match the target decimal.
- 💡Always pad binary answers with leading zeros if the question specifies an 8-bit byte (e.g. write 00101101 rather than 101101).
- 💡Double-check your addition when converting binary to denary by adding the active place values twice to eliminate simple arithmetic errors.
- 💡When converting binary to hex, group bits into sets of 4 starting from the least significant bit (rightmost) towards the left, adding leading zeros if needed.
Common Mistakes
- Writing bits right to left in the wrong order: the correction is to fix the place values above the columns first, largest on the left, then fill each column.
- Dropping leading zeros so 5 becomes 101 rather than 00000101: the correction is to always show all eight positions for a 0 to 255 value.
- Adding place values instead of subtracting as each bit is set: the correction is to track the running remainder and only set a bit when its place value fits.
- Thinking hexadecimal ends at 16 or uses 16 digits up to '16': Hexadecimal uses digits 0-9 and letters A-F, representing values 0 to 15. The value 16 is written as '10' in hex.
- Dividing by 2 incorrectly or reading the remainders top-to-bottom instead of bottom-to-top when using repeated division to convert denary to binary.
- Splitting an 8-bit byte into two nibbles unevenly (e.g. 5 bits and 3 bits) instead of two equal groups of 4 bits from right to left.
Revision Plan
- 1Day 1-2: Master binary-to-denary and denary-to-binary conversions using 8-bit place-value grids.
- 2Day 3-4: Learn hexadecimal digits (0-9, A-F) and practice 4-bit nibble to single hex character mapping.
- 3Day 5-6: Practise full two-digit hex to denary and denary to hex conversions, using binary as an intermediate step.
- 4Day 7: Complete timed AQA past-paper questions focusing on precision, showing all workings, and padding to 8 bits.
Exam Question Types
- 📋Direct short-answer conversions (e.g. 'Convert the binary byte 11001010 to hexadecimal'): 1-2 marks requiring accuracy or nibble splitting.
- 📋Working-out structured questions (e.g. 'Convert 77 to an 8-bit binary number, showing all your working'): 2 marks assessing both method and final byte.
- 📋Contextual application questions (e.g. explaining why HTML colour codes or MAC addresses use hexadecimal instead of binary): 2-3 marks focusing on readability and error reduction.
Command Word Expectations (AQA)
Translate a given value from one number representation to another. Marks are awarded for the correct final answer, but working out should always be shown.
Give a concise, single-word or short-phrase answer without explanation (e.g. stating the hex equivalent of a given 4-bit pattern).
Set out reasons or mechanisms, such as explaining why hexadecimal is preferred over binary by computer programmers (e.g. takes up less screen space, easier for humans to read, reduces transcription errors).
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: Convert the denary number 183 into an 8-bit binary number. Show your working.
- 1.Step 1: Write down the 8-bit place values: 128, 64, 32, 16, 8, 4, 2, 1.
- 2.Step 2: Compare 183 with each place value from left to right. 183 >= 128, so place a 1 in 128 (remainder 183 - 128 = 55).
- 3.Step 3: 55 < 64, place a 0 in 64. 55 >= 32, place a 1 in 32 (remainder 55 - 32 = 23).
- 4.Step 4: 23 >= 16, place a 1 in 16 (remainder 23 - 16 = 7). 7 < 8, place a 0 in 8.
- 5.Step 5: 7 >= 4, place a 1 in 4 (remainder 7 - 4 = 3). 3 >= 2, place a 1 in 2 (remainder 3 - 2 = 1). 1 >= 1, place a 1 in 1 (remainder 0).
- 6.Step 6: Combine the bits: 10110111.
Question: Convert the hexadecimal number 3E into denary. Show your working.
- 1.Step 1: Identify the place values for a two-digit hexadecimal number: 16 (sixteens) and 1 (units).
- 2.Step 2: Determine the denary value of each hexadecimal digit. The digit '3' represents 3. The digit 'E' represents 14 (since A=10, B=11, C=12, D=13, E=14).
- 3.Step 3: Multiply each digit by its place value: (3 * 16) + (14 * 1).
- 4.Step 4: Calculate the sum: 48 + 14 = 62.