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Understanding number systems is essential for OCR J277 Section 2.6 (Data Representation). Computers store and process all data using binary (base-2), but humans normally use denary (base-10, also called decimal). You must be able to convert fluently between these two systems.
The denary number system uses ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Each position in a denary number represents a power of 10.
| Position | Thousands (10^3) | Hundreds (10^2) | Tens (10^1) | Units (10^0) |
|---|---|---|---|---|
| Example: 4527 | 4 | 5 | 2 | 7 |
| Value | 4000 | 500 | 20 | 7 |
4527 = (4 x 1000) + (5 x 100) + (2 x 10) + (7 x 1) = 4527
This is the system you use every day, so it feels natural. Computers, however, use binary because their electronic circuits can only distinguish between two states: on (1) and off (0).
The binary number system uses only two digits: 0 and 1. Each position represents a power of 2. A single binary digit is called a bit.
| Position | 128 (2^7) | 64 (2^6) | 32 (2^5) | 16 (2^4) | 8 (2^3) | 4 (2^2) | 2 (2^1) | 1 (2^0) |
|---|---|---|---|---|---|---|---|---|
| Example: 10110011 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
To convert to denary: 128 + 32 + 16 + 2 + 1 = 179
flowchart LR
Denary((Denary - base 10)) -->|successive division by 2| R[Read remainders bottom to top]
Denary -->|place value subtraction| P[128 64 32 16 8 4 2 1]
R --> Binary((Binary - base 2))
P --> Binary
Binary -->|sum place values where bit = 1| Denary
Binary --- Hardware[Maps to transistor on/off states]
Use the successive division method (divide by 2 and record remainders):
Example: Convert 156 to binary
| Division | Quotient | Remainder |
|---|---|---|
| 156 / 2 | 78 | 0 |
| 78 / 2 | 39 | 0 |
| 39 / 2 | 19 | 1 |
| 19 / 2 | 9 | 1 |
| 9 / 2 | 4 | 1 |
| 4 / 2 | 2 | 0 |
| 2 / 2 | 1 | 0 |
| 1 / 2 | 0 | 1 |
Read remainders from bottom to top: 156 = 10011100
Alternatively, use the place value method: find the largest power of 2 that fits and subtract:
OCR Exam Tip: The place value method is usually quicker in the exam and less prone to error. Always double-check by converting your binary answer back to denary.
Write out the place values and add up the columns where there is a 1.
Example: Convert 11010110 to denary
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|
| 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
128 + 64 + 16 + 4 + 2 = 214
| Reason | Explanation |
|---|---|
| Electronic circuits | Transistors have two states: on (1) and off (0) |
| Reliability | It is easy to distinguish between two voltage levels, reducing errors |
| Simplicity | Binary arithmetic is simpler to implement in hardware |
| Boolean logic | Binary maps directly to true/false logic operations |
| Term | Definition |
|---|---|
| Bit | A single binary digit (0 or 1) |
| Nibble | 4 bits |
| Byte | 8 bits |
| Denary | Base-10 number system (0-9) |
| Binary | Base-2 number system (0, 1) |
| Place value | The value of a digit based on its position |
OCR Exam Tip: The maximum value for an n-bit binary number is 2^n - 1. For 8 bits, this is 2^8 - 1 = 255. You should memorise this.
This worked example walks through the full denary-to-binary-to-denary cycle so you can verify your answer in the exam.
Step 1: Denary to binary using the place value method.
Write the place values: 128, 64, 32, 16, 8, 4, 2, 1.
Result: 10101101
Step 2: Binary to denary check.
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 |
128 + 32 + 8 + 4 + 1 = 173. The conversion is verified.
Mini-challenge: Convert 218 to 8-bit binary using the same method, then check by converting back. (Answer: 11011010 = 128 + 64 + 16 + 8 + 2 = 218.)
Exam-style question (5 marks): Convert the denary number 165 into 8-bit binary. Show your working, and explain why a computer uses binary rather than denary to store this value.
165 in binary is 10100101. Computers use binary because they only understand 0 and 1.
Examiner comment: The conversion is correct but no working is shown, so method marks are at risk if the answer were wrong. The explanation is true but very thin — it does not link to hardware.
165 - 128 = 37, so 1 in the 128 column. 37 - 32 = 5, so 1 in the 32 column. 5 - 4 = 1, so 1 in the 4 column. 1 - 1 = 0, so 1 in the 1 column. Answer: 10100101. Computers use binary because electronic circuits have two states (on and off), which match the digits 1 and 0.
Examiner comment: Working is shown clearly and the link to two states (on/off) is made. Could be improved by mentioning that two distinct voltage levels are easier to distinguish reliably.
Using the place value method: 165 = 128 + 32 + 4 + 1, giving the columns 1, 0, 1, 0, 0, 1, 0, 1. Final answer: 10100101. Verification: 128 + 32 + 4 + 1 = 165. Computers use binary because transistors operate as switches with two stable states (on and off, represented as high/low voltage). Two states can be distinguished reliably with minimal error, which makes binary arithmetic both simple to implement in hardware and resistant to noise. Binary also maps directly onto Boolean logic (true/false), which underpins all processor instructions.
Examiner comment: Full marks. The candidate shows correct working, verifies the answer, and gives a precise hardware-level explanation that links transistors, voltage, reliability, and Boolean logic.
This content is aligned with OCR GCSE Computer Science (J277) specification section 1.2 Memory and storage (data representation). For the most accurate and up-to-date information, please refer to the official OCR specification document.