← IGCSE Computer Science Paper 1

Converting 8-bit two's complement to denary

Add the place values with a 1 under them to get the denary value. A leftmost 1 means -128 is added: 11110110 = -128+64+32+16+4+2 = -10. A leftmost 0 means positive.

Part of: IGCSE Computer Science 0478 — 1.1 Number Systems — Module 31

Understanding Converting 8-bit two's complement to denary in IGCSE Computer Science Paper 1

What it does: The leading 1 gives -128. The 1s under 8 and 1 are added, so -128 + 9 = -119. The leading 0 means -128 is not used, so the result is the sum of the other place values with a 1. The leftmost bit is 0, so the number is positive. Add 8 + 4 + 1 = 13. The leftmost bit is 1, so start with -128. Then add 64 + 32 + 16 + 4 + 2 = 118. -128 + 118 = -10. 246 is the unsigned binary value. Start with -128 and add 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127. -128 + 127 = -1.

Common mistakes: In 8-bit two's complement the leftmost place value is -128, not 128. The line should be Values: -128 64 32 16 8 4 2 1. An 8-bit number needs exactly 8 bits. 0100110 only has 7 and would not add up to 77. It should be Binary: 01001101.

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This module's key terms are also in a printable crossword (Modules 28–41) and word search (Modules 28–58).

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Learning
Duo
In 8-bit two's complement, what is the place value of the leftmost bit?
128-128
Matches
11110110 = 128 + 64 + 32 + 16 + 4 + 2 = 246
↓
The leftmost place value was treated as +128. In 8-bit two's complement it is -128, so the answer is -128 + 64 + 32 + 16 + 4 + 2 = -10.
Applying
Meanings
Term
-128, 64, 32, 16, 8, 4, 2, 1
Meaning
Add the place values with a 1 under them to get the denary value. A leftmost 1 means -128 is added: 11110110 = -128+64+32+16+4+2 = -10. A leftmost 0 means positive.
Practice
11110110 = -128+64+32+16+4+2
Testing
Gaps
, 64, 32, 16, 8, 4, 2, 1
Check Answer
Spots
1Values: 128 64 32 16 8 4 2 1
2Bits: 11110110

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Quick Questions

1. In 8-bit two's complement, what is the place value of the leftmost bit?
A. 128
B. -128 ✓
The leftmost bit is the sign bit and is worth -128. The other place values are 64, 32, 16, 8, 4, 2 and 1.
2. The leftmost bit of an 8-bit two's complement number is 0. What does this tell you?
A. The number is positive ✓
B. The number is negative
A leftmost 0 means -128 is not added, so you only add the positive place values. The number is positive.
3. What is 00001101 in 8-bit two's complement?
A. -13
B. 13 ✓
The leftmost bit is 0, so the number is positive. Add 8 + 4 + 1 = 13.
4. What is 11110110 in 8-bit two's complement?
A. -10 ✓
B. 246
The leftmost bit is 1, so start with -128. Then add 64 + 32 + 16 + 4 + 2 = 118. -128 + 118 = -10. 246 is the unsigned binary value.
5. What is 10000101 in 8-bit two's complement?
A. -123 ✓
B. -5
Start with -128 for the leftmost 1. Then add 4 + 1 = 5. -128 + 5 = -123.

Match the Pairs

00101101 → Leftmost bit is 0, so it is positive: 32 + 8 + 4 + 1 = 45
11111111 → -128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = -1
A leftmost 1 means the number is negative, so start with -128 and add the other place values that have a 1 → 10001001 = -128 + 8 + 1 = -119
A leftmost 0 means the number is positive, so just add the place values that have a 1 → 00110101 = 32 + 16 + 4 + 1 = 53
11110110 = 128 + 64 + 32 + 16 + 4 + 2 = 246 → The leftmost place value was treated as +128. In 8-bit two's complement it is -128, so the answer is -128 + 64 + 32 + 16 + 4 + 2 = -10.
10000110 = -128 + 4 + 2 + 1 = -121 → The 1 place value was added, but the rightmost bit is 0. Only -128, 4 and 2 have a 1 under them, so the answer is -128 + 4 + 2 = -122.

Key Term

-128, 64, 32, 16, 8, 4, 2, 1
Add the place values with a 1 under them to get the denary value. A leftmost 1 means -128 is added: 11110110 = -128+64+32+16+4+2 = -10. A leftmost 0 means positive.

Fill in the Blanks

1. -128, 64, 32, 16, 8, 4, 2, 1
Hint: Type the place value of the leftmost bit in 8-bit two's complement.
2. If the leftmost bit is 0, the number is positive
Hint: Type the word describing a number whose leftmost bit is 0.
3. 00101101 = 32 + 8 + 4 + 1 = 45
Hint: Add the place values that have a 1 under them and type the denary total.
4. 11110110 = -128 + 64 + 32 + 16 + 4 + 2 = -10
Hint: Add all the place values, including the negative one, and type the denary result.
5. 10000011 = -128 + 2 + 1 = -125
Hint: Start with the negative place value, add the other values, and type the denary result.

Spot the Error

Question 1
A Values: 128 64 32 16 8 4 2 1
B Bits: 11110110
C -128+64+32+16+4+2 = -10
Line A — In 8-bit two's complement the leftmost place value is -128, not 128. The line should be Values: -128 64 32 16 8 4 2 1.
Question 2
A Binary: 0100110
B Leftmost 0: positive number
C 64+8+4+1 = 77
Line A — An 8-bit number needs exactly 8 bits. 0100110 only has 7 and would not add up to 77. It should be Binary: 01001101.
Question 3
A Binary: 10000011
B Leftmost 1: start at 128
C -128+2+1 = -125
Line B — When the leftmost bit is 1 the number is negative, so you start at -128, not +128. The line should read Leftmost 1: start at -128.
Question 4
A Binary: 11111111
B Leftmost 1: start at -128
C -128+127 = 1
Line C — -128 plus 127 is -1, not 1. All eight 1s in two's complement give -1. The line should be -128+127 = -1.
Question 5
A Binary: 10101010
B Values: -128 64 32 16 8 4 2 1
C -128+32+8+2
D Answer: 86
Line D — -128+32+8+2 = -128+42 = -86. The answer is negative because the leftmost bit is 1. It should be Answer: -86.

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