IGCSE · Practice questions · Two's complement for negative numbers

Below zero.

Six original Cambridge-style questions on 8-bit two's complement, with worked solutions.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Minus 128 first.

Place values: −128, 64, 32, 16, 8, 4, 2, 1.
Sign: leftmost bit 1 means negative, 0 means positive.
Make a negative: write the positive in 8 bits, flip every bit, add 1.
Check: add up the place values, starting from −128.

Convert

[2 marks]

Convert −27 to an 8-bit two's complement integer. Show your working.

27 = 00011011; flipped = 11100100. ✓
Add 1: 11100101. ✓

Convert

[2 marks]

Convert the 8-bit two's complement integer 11001110 to denary.

−128 + 64 + 8 + 4 + 2. ✓
= −50. ✓

Explain

[2 marks]

Explain how you can tell whether an 8-bit two's complement integer is positive or negative.

Look at the most significant (leftmost) bit, which has a place value of −128. ✓
If it is 1 the number is negative; if it is 0 the number is positive (or zero). ✓

Recall

[2 marks]

(a) State the range of values that can be stored as an 8-bit two's complement integer. (b) Explain why +200 cannot be stored.

(a) −128 to +127. ✓
(b) 200 is greater than the largest value, +127. ✓

Convert

[2 marks]

Give the 8-bit two's complement representation of (a) +84 and (b) −84.

(a) 01010100. ✓
(b) 10101100 (flip 01010100 to 10101011, then add 1). ✓

Analysis

[3 marks]

A student represents −12 in 8-bit two's complement as 10001100. Explain the error, give the correct answer, and state what the student's answer actually represents.

The student only changed the leftmost bit to 1, as if it were just a minus sign. ✓
Correct: 12 = 00001100; flip 11110011; add 1: 11110100. ✓
10001100 = −128 + 12 = −116. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.