Binary and two’s complement worksheet

01 / Understand the representation

The bits stay the same. Their interpretation matters.

A bit is 0 or 1. In an unsigned eight-bit word, the leftmost position is worth 128. In eight-bit two’s complement, that same position has weight −128. The remaining seven weights are unchanged. Always state the word length and whether the number is signed.

To negate a representable value in a fixed word, invert every bit and add 1, retaining the stated width. Check the result using signed weights. The smallest negative value is the exception: +128 does not fit in eight-bit two’s complement, as question 12 explores.

  00010010   +18
+ 11100001   −31
──────────
  11110011   −13

Check overflow against the signed range.

Signed overflow means the exact answer lies outside −128 to +127. When adding two positive operands, a negative stored result signals overflow; when adding two negative operands, a non-negative stored result signals overflow. Adding opposite signs cannot overflow the signed range. A carry out is a separate unsigned-arithmetic observation.

02 / Try it independently

Twelve questions. Keep your working.

Use eight-bit two’s complement whenever the question specifies a signed interpretation. Hexadecimal is labelled explicitly. Write your answer on paper or in your own notes before opening a solution. Each question is worth 2 marks.

A printed copy includes writing space and a separate answer key after the questions. Keep the answer pages aside until you have finished.

1. Q01 Read an unsigned word

   2 marks

   Convert unsigned binary 10110110 to denary. Show the place values you add.

   Check answer and working for question 1

   182

   The 1 bits select 128, 32, 16, 4 and 2. Their sum is 128 + 32 + 16 + 4 + 2 = 182. This question specifies unsigned interpretation.

   Original FocusED self-check scheme · 2 marks

   • 1 mark: a correct place-value expansion using the set bits.
   • 1 mark: the denary result 182.

2. Q02 Keep all eight positions

   2 marks

   Represent denary 93 as eight-bit unsigned binary. Show a place-value decomposition or an equivalent conversion method.

   Check answer and working for question 2

   01011101

   93 = 64 + 16 + 8 + 4 + 1. Under the weights 128, 64, 32, 16, 8, 4, 2, 1 this gives 0, 1, 0, 1, 1, 1, 0, 1. The leading zero keeps the stated eight-bit width.

   Original FocusED self-check scheme · 2 marks

   • 1 mark: a correct conversion method, such as 64 + 16 + 8 + 4 + 1.
   • 1 mark: the complete eight-bit result 01011101.

3. Q03 Use hexadecimal groups

   2 marks

   Convert hexadecimal A7 to eight-bit binary. Make the two four-bit groups clear.

   Check answer and working for question 3

   1010 0111

   Hexadecimal A represents ten, so its four-bit group is 1010. Hexadecimal 7 becomes 0111. Joining the groups gives 10100111; do not drop the zero at the start of the second group.

   Original FocusED self-check scheme · 2 marks

   • 1 mark: A becomes the four-bit group 1010.
   • 1 mark: 7 becomes 0111, giving 1010 0111 overall.

4. Q04 Interpret a negative value

   2 marks

   Interpret 11101100 as an eight-bit two’s-complement integer. Show how you obtain the denary value.

   Check answer and working for question 4

   −20

   Use −128 + 64 + 32 + 8 + 4 = −20. Alternatively, invert 11101100 to obtain 00010011 and add 1 to obtain 00010100, whose magnitude is 20; the original word is negative.

   Original FocusED self-check scheme · 2 marks

   • 1 mark: a valid signed-weight calculation or invert-and-add-one method.
   • 1 mark: the signed denary result −20.

5. Q05 Encode a negative value

   2 marks

   Represent −45 in eight-bit two’s complement. Show the negation steps from +45.

   Check answer and working for question 5

   11010011

   +45 is 00101101. Invert to obtain 11010010, then add 1 to obtain 11010011. Checking the signed weights gives −128 + 64 + 16 + 2 + 1 = −45.

   Original FocusED self-check scheme · 2 marks

   • 1 mark: +45 = 00101101 and a valid inversion/add-one method.
   • 1 mark: the final eight-bit word 11010011.

6. Q06 State the signed range

   2 marks

   State the smallest and largest denary integers representable in eight-bit two’s complement.

   Check answer and working for question 6

   Smallest −128; largest +127

   The smallest word is 10000000, worth −128. The largest is 01111111, worth 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127. The range is asymmetric because zero is included and each of the 256 patterns represents one value.

   Original FocusED self-check scheme · 2 marks

   • 1 mark: the minimum is −128.
   • 1 mark: the maximum is +127.

7. Q07 Subtract by adding a negative

   2 marks

   Calculate 37 − 52 using eight-bit two’s complement. Give the stored result bits and their signed denary interpretation.

   Check answer and working for question 7

   11110001, representing −15

   +37 is 00100101. +52 is 00110100, so −52 is 11001100. Adding 00100101 + 11001100 gives 11110001. Decode −128 + 64 + 32 + 16 + 1 = −15. The exact result fits, so no signed overflow occurs.

   Original FocusED self-check scheme · 2 marks

   • 1 mark: the eight-bit result 11110001 from signed subtraction.
   • 1 mark: its signed denary interpretation −15.

8. Q08 Recognise positive signed overflow

   2 marks

   An eight-bit two’s-complement processor adds +100 and +40. Give its stored eight-bit result and explain whether signed overflow occurs.

   Check answer and working for question 8

   10001100; signed overflow occurs

   01100100 + 00101000 = 10001100. The exact denary answer is 140, outside the maximum +127. The stored pattern decodes to −116: two positive operands have produced a negative signed result.

   Original FocusED self-check scheme · 2 marks

   • 1 mark: the stored result 10001100.
   • 1 mark: signed overflow, justified by 140 exceeding +127 or by two positive operands producing a negative stored result.

9. Q09 Recognise negative signed overflow

   2 marks

   Add −60 and −75 in eight-bit two’s complement. Retain the low eight result bits, and explain whether signed overflow occurs.

   Check answer and working for question 9

   01111001; signed overflow occurs

   −60 is 11000100 and −75 is 10110101. Their full binary sum is 1 01111001; the stored eight bits are 01111001, which decode to +121. The exact result −135 is below −128, so signed overflow occurred. The carry out does not repair the stored signed value.

   Original FocusED self-check scheme · 2 marks

   • 1 mark: the stored eight bits 01111001.
   • 1 mark: signed overflow, justified by −135 being below −128 or by two negative operands producing a positive stored result.

10. Q10 Preserve the value in a wider word

    2 marks

    Extend eight-bit two’s-complement 11101100 to sixteen bits without changing its value. Explain which bits you add on the left.

    Check answer and working for question 10

    11111111 11101100

    Repeat the original sign bit, which is 1, in the eight new leftmost positions. The sixteen-bit word is 1111111111101100 and still represents −20. Adding zeros on the left would instead make the signed sixteen-bit value +236.

    Original FocusED self-check scheme · 2 marks

    • 1 mark: the sixteen-bit result 1111111111101100.
    • 1 mark: repeat the sign bit 1 on the left to preserve the negative value.

11. Q11 Separate carry from signed overflow

    2 marks

    Adding 11111111 and 00000001 produces a carry out and stored bits 00000000. Interpret the operands as eight-bit two’s complement: state the signed calculation and whether signed overflow occurred.

    Check answer and working for question 11

    −1 + 1 = 0; no signed overflow

    11111111 represents −1, not unsigned 255, under the stated signed interpretation. The calculation is −1 + 1 = 0, exactly represented by 00000000. The carry out would signal an out-of-range result for unsigned 255 + 1, but it does not by itself establish signed overflow.

    Original FocusED self-check scheme · 2 marks

    • 1 mark: the signed calculation −1 + 1 = 0.
    • 1 mark: no signed overflow because the exact signed answer 0 fits; a carry out alone is insufficient.

12. Q12 Explain the smallest-value exception

    2 marks

    A student negates the eight-bit two’s-complement word 10000000 by inverting and adding 1, then claims the result represents +128. Give the resulting eight bits and explain the error.

    Check answer and working for question 12

    10000000 again; +128 cannot be represented in eight-bit two’s complement

    Inversion gives 01111111; adding 1 gives 10000000. The original value is −128, while its mathematical negative +128 lies outside the eight-bit signed range. The retained word still decodes as −128, so this fixed-width negation cannot represent the required positive result.

    Original FocusED self-check scheme · 2 marks

    • 1 mark: invert-and-add-one retains 10000000.
    • 1 mark: +128 exceeds the signed maximum +127, so the word cannot mean +128 in this format.

03 / Review the reasoning

Use the marks to find your next practice step.

Award each listed mark once, using the stated criterion. A different correct method can earn the corresponding method mark. Add the twelve question scores for a total out of 24. This is a learning check, not a predicted examination grade.

• Conversions: if questions 1–3 were difficult, write every place value and preserve four-bit hexadecimal groups.
• Negative values: for questions 4–7, 10 and 12, check the word length, sign weight and signed range.
• Overflow: for questions 8, 9 and 11, compare the exact signed answer with the permitted range before interpreting a carry.

Choose one error, explain why it happened, then retry that question without looking. Correct reasoning is more useful than remembering one bit pattern.