Converting a positive denary number to two's complement
Name:
Date:
Class:
A
Quick Questions
Shade in the A or B circle for each question.
1. Which is 5 written in 8-bit two's complement?
2. What is the leftmost (sign) bit of every positive number in 8-bit two's complement?
3. Which is 65 written in 8-bit two's complement?
4. Can the denary number 200 be stored as a positive number in 8-bit two's complement?
5. Which is 100 written in 8-bit two's complement?
B
Match the Pairs
Draw a line connecting each item on the left to its match on the right.
| Left column | Right column | |
|---|---|---|
| 100101101 | AThe denary number 45 in 8-bit two's complement, with a leading 0 sign bit | |
| 201111111 | BThe binary is not padded to 8 bits. It should be 00001101, so that the leftmost sign bit is 0. | |
| 3Denary 5 is 101 in binary, so it is padded with leading zeros to fill 8 bits | C130 is above 127, so it cannot be stored in 8-bit two's complement. A leading 1 would mean a negative number (-126). | |
| 4Denary 100 is written as ordinary binary (64 + 32 + 4) with the leftmost sign bit set to 0 | D127, the largest positive number that 8 bits can store in two's complement | |
| 513 = 1101 | E01100100 | |
| 6130 = 10000010 | F00000101 |
C
Definitions
positive two's complement
D
Fill in the Blanks
Complete each line of code by filling in the missing parts.
1.
65 in 8-bit two's complement: _1000001
2.
127 in 8-bit two's complement = ________
3.
13 = 8 + 4 + 1, so 8 bits are ________
4.
Pad 1101 to 8 bits: ____1101
5.
IF number > ___ THEN
OUTPUT "Too big for 8 bits"
ENDIF
E
Spot the Error
Shade in the letter of the line that contains an error.
Question 1
A
Number: 45
B
45 = 32 + 8 + 4 + 1
C
8-bit: 10101101
D
Sign bit is 0, so positive
Question 2
A
Number: 100
B
100 = 64 + 32 + 4
C
Binary: 1100100 (7 bits)
D
Pad to 8 bits: 11100100
E
Leftmost bit is the sign bit
Question 3
A
Number: 127
B
Largest positive 8-bit value
C
127 = 11111111
D
Sign bit is 0
Question 4
A
Number: 5
B
5 = 4 + 1
C
Binary: 101
D
Pad to 8 bits: 10100000
E
Leftmost bit is 0
Question 5
A
IF number <= 255 THEN
B
Write ordinary binary
C
Pad with 0s to 8 bits
D
ELSE
E
OUTPUT "Cannot store"
Answer Key
Converting a positive denary number to two's complement
Section A: Quick Questions
Q1
A — 00000101
Q2
B — 0
Q3
B — 01000001
Q4
A — No, the largest positive value is 127
Q5
A — 01100100
Section B: Match the Pairs
Q1
00101101 → The denary number 45 in 8-bit two's complement, with a leading 0 sign bit
Q2
01111111 → 127, the largest positive number that 8 bits can store in two's complement
Q3
Denary 5 is 101 in binary, so it is padded with leading zeros to fill 8 bits → 00000101
Q4
Denary 100 is written as ordinary binary (64 + 32 + 4) with the leftmost sign bit set to 0 → 01100100
Q5
13 = 1101 → The binary is not padded to 8 bits. It should be 00001101, so that the leftmost sign bit is 0.
Q6
130 = 10000010 → 130 is above 127, so it cannot be stored in 8-bit two's complement. A leading 1 would mean a negative number (-126).
Section C: Definitions
Q1
positive two's complement — A denary number from 0 to 127 is written as ordinary binary, padded with zeros to 8 bits so the leftmost sign bit is 0. Numbers above 127 cannot be stored.Section D: Fill in the Blanks
Q1
65 in 8-bit two's complement: 01000001Q2
127 in 8-bit two's complement = 01111111Q3
13 = 8 + 4 + 1, so 8 bits are 00001101Q4
Pad 1101 to 8 bits: 00001101Q5
IF number > 127 THEN
OUTPUT "Too big for 8 bits"
ENDIFSection E: Spot the Error
Q1
C — A positive number must have a leading sign bit of 0. 45 in 8 bits is 00101101. Starting with 1 would make it a negative number.
Q2
D — Padding adds 0s on the left, not 1s. The padded 8-bit value must be 01100100, so the sign bit stays 0 and the value is still 100.
Q3
C — 127 is 01111111 in 8-bit two's complement. 11111111 has a sign bit of 1 and means -1.
Q4
D — The padding zeros go on the left of the binary number, giving 00000101. Adding zeros on the right changes the value.
Q5
A — In 8-bit two's complement the largest positive number is 127, not 255. The condition should be number <= 127.