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1CP2-P-2.1 Two's complement conversion and range
Part 1 of 7 · 1CP2-P-2 · Signed data and encoding
The opening worksheet of the signed data series, converting two's complement in both directions and deriving the range it can hold.
Students will:
- interpret an 8-bit two's complement pattern using signed place values
- explain why 10000000 represents minus 128
- represent a negative denary value at the full 8-bit width
- derive the signed range from the width
- explain why a value outside that range cannot be stored
Inside: 6 explanation cells, 2 multiple-choice questions, 4 number cells, 1 fill-in-the-blanks cell and 2 written answers. 20 marks, about 45 minutes.
Series: 1CP2-P-2 · Signed data and encoding, part 1 of 7.
Shared by Coding PathwayVerified teacher
- 15 cells
- About 45 minutes
- CC BY-SA 4.0
- Shared 17 Aug 2026
Preview
The whole resource, exactly as a class sees it. Answers and marking are held back.
Two's complement conversion and range
This worksheet deepens the signed representation introduced in 1CP2-P-1. In 8-bit two's complement, the most-significant place is −128, so the range is −128 to +127. The same pattern must not be interpreted as unsigned unless the question says so.
1. Interpret with signed place values
| Place | −128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|---|
| Pattern | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 |
11101100 = −128 + 64 + 32 + 8 + 4 = −20. A leading 1 indicates a negative value in two's complement, but the remaining bits contribute their ordinary positive place values.
Interpret these 8-bit two's complement patterns.
Use the displayed signed place values and show each contribution in separate working.
a)What is the denary value of the two's complement number 11110110?
b)What is the denary value of the two's complement number 10101101?
Explain why 10000000 represents −128 rather than zero or +128 in 8-bit two's complement.
Refer to the signed place value of the most-significant bit.
Students type their answer here.
2. Represent a negative value
For −18:
- write positive 18:
00010010; - invert every bit:
11101101; - add one:
11101110; - cross-check: −128 + 64 + 32 + 16 + 8 + 4 + 2 = −18.
The final answer is the add-one result. Do not place signed −128 headings above the positive-magnitude row: that row is ordinary unsigned 18.
Write each denary value as an 8-bit two's complement pattern.
Use the complete positive-magnitude, invert and add-one method shown in the cell.
a)Write −37 as an 8-bit two's complement number.
b)Write −92 as an 8-bit two's complement number.
3. Derive the signed range
For n-bit two's complement:
- minimum =
−2^(n−1); - maximum =
2^(n−1) − 1.
Four bits therefore represent −8 to +7. There is one zero, and the negative side contains one additional value.
- -128
- -16
- +15
- +127
- +255
How should 11111111 be interpreted as 8-bit two's complement?
- A255
- B−127
- C−1
- D0
4. Independent mixed conversion
Remove the staged rows now. Choose signed place values or invert-and-add-one as appropriate, then cross-check against the valid range.
Represent these signed denary values in 8-bit two's complement.
Enter only the final eight-bit patterns.
a)Write −1 as an 8-bit two's complement number.
b)Write −64 as an 8-bit two's complement number.
Interpret these 8-bit two's complement patterns in denary.
Use a reliable method but enter only each final value.
a)What is the denary value of the two's complement number 10000001?
b)What is the denary value of the two's complement number 11010110?
A programmer wants to store −130 in 8-bit two's complement. Explain why this cannot be represented and state the nearest valid boundary.
Use the range rather than attempting to force a pattern.
Students type their answer here.
Which check is valid after representing −64 as 11000000?
- ATreat the pattern as unsigned 192 and stop.
- BAdd signed place values: −128 + 64 = −64.
- CRemove the leading 1 and read 64.
- DChange the most-significant bit to a minus sign.
Route forward
You can now convert signed values and determine the two's-complement range. Logical shifts operate on unsigned patterns next; arithmetic shifts then preserve signed interpretation.