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OCR H446 1.4.1 Signed integers

Part 3 of 11 · H446 1.4.1 · Data types

Negative values have to fit the same fixed-width field as positive ones, and H446 1.4.1 covers both sign-and-magnitude and two's complement. Using a game that records temperature and score changes, the tasks keep all eight columns visible so the invert-and-add-one method and the signed place-value check are both shown.

Students will:

  • represent and interpret 8-bit sign-and-magnitude values
  • convert a signed denary integer to 8-bit two's complement using invert and add one
  • check a two's-complement pattern with signed place values starting at minus 128
  • compare the two representations on range and on how zero is stored
  • explain why a bit pattern cannot be interpreted until its encoding is named

Inside: 7 explanation cells, 1 multiple-choice question, 1 fill-in-the-blanks cell, 1 written answer and 4 number answers. 28 marks, about 40 to 50 minutes.

Series: H446 1.4.1 · Data types, part 3 of 11.

Shared by Coding PathwayVerified teacher

  • 14 cells
  • About 45 minutes
  • CC BY-SA 4.0
  • Shared 31 Aug 2026
  • Updated 3 Sept 2026

Preview

The whole resource, exactly as a class sees it. Answers and marking are held back.

Signed integers

A game records temperature change and score adjustments, so the same fixed-width field must represent positive and negative whole numbers.

By the end, you will be able to

  • distinguish sign-and-magnitude from two’s complement
  • represent and interpret signed values in both systems
  • state the range and number of zero representations
  • check an answer using fixed-width place values

Reactivate: an 8-bit unsigned pattern has place values 128 to 1.

Worked model: two interpretations

In 8-bit sign-and-magnitude, the leftmost bit stores the sign and the remaining seven bits store magnitude. This creates both positive zero and negative zero.

In 8-bit two's complement, place values are -128, 64, 32, 16, 8, 4, 2, 1. There is one zero and the range is -128 to +127.

The leading bit has a different role in two signed representations Do not apply the wrong interpretation Sign-and-magnitude 10100101 sign: negativeseven-bit magnitude: 37 Two's complement signed place values:−128, 64, 32, 16, 8, 4, 2, 1

The left bit changes meaning

Comparing signed integer fieldsSign and magnitudesign7-bit magnitudeTwo’s complement−12864, 32, 16, 8, 4, 2, 1The left bit has a different meaning in each representation.

Worked example: interpret 10000101

  • In sign-and-magnitude, sign 1 means negative and magnitude 0000101 is 5, so the value is −5.
  • In two’s complement, add signed place values: −128 + 4 + 1 = −123.

The bits are identical; the interpretation creates different values. Always name the representation before calculating.

Multiple choice1 mark

What is the range of an 8-bit two's-complement integer?

  • A-127 to +127
  • B-128 to +128
  • C0 to 255
  • D-128 to +127
Number systems4 marks

Represent each signed denary integer as 8-bit sign-and-magnitude.

a)Write 12 as an 8-bit sign and magnitude number.

Binary

b)Write 111 as an 8-bit sign and magnitude number.

Binary

c)Write 60 as an 8-bit sign and magnitude number.

Binary

d)Write −108 as an 8-bit sign and magnitude number.

Binary
Number systems3 marks

Interpret each 8-bit sign-and-magnitude pattern as a signed denary integer.

a)What is the denary value of the sign and magnitude number 11010000?

Binary
11010000
DenaryNot answered

b)What is the denary value of the sign and magnitude number 11110001?

Binary
11110001
DenaryNot answered

c)What is the denary value of the sign and magnitude number 01010110?

Binary
01010110
DenaryNot answered

Two's-complement method

To form a negative value from its positive 8-bit pattern: invert every bit, then add 1. To interpret a pattern, you can add its signed place values. Always keep the stated width.

Worked example: represent -37. Start with +37 = 00100101. Invert to 11011010, then add 1 to obtain 11011011. Check using signed place values: -128 + 64 + 16 + 8 + 2 + 1 = -37. The leading 1 is a -128 place value here; it is not a separate sign bit.

Worked model: represent −18 in eight-bit two’s complement

  1. Write +18: 00010010.
  2. Invert every bit: 11101101.
  3. Add 1: 11101110.
  4. Check using signed place values: −128 + 64 + 32 + 16 + 8 + 4 + 2 = −18.

For guided practice, keep all eight columns visible and perform the signed-place-value check after the invert-and-add method.

Number systems8 marks

Represent each signed denary integer in 8-bit two's complement. Record the method.

a)Write −7 as an 8-bit two's complement number.

Flip the bits
Add one
Binary

b)Write −44 as an 8-bit two's complement number.

Flip the bits
Add one
Binary

c)Write −103 as an 8-bit two's complement number.

Flip the bits
Add one
Binary

d)Write −38 as an 8-bit two's complement number.

Flip the bits
Add one
Binary
Number systems4 marks

Interpret each negative 8-bit two's-complement value in denary.

a)What is the denary value of the two's complement number 11100110?

Binary
11100110
DenaryNot answered

b)What is the denary value of the two's complement number 11100101?

Binary
11100101
DenaryNot answered

c)What is the denary value of the two's complement number 10110101?

Binary
10110101
DenaryNot answered

d)What is the denary value of the two's complement number 10110011?

Binary
10110011
DenaryNot answered
Written answer4 marks

Compare sign-and-magnitude with two's complement for representing signed 8-bit integers. Include zero and range in your answer.

Write linked differences using the same criterion for both representations.

Students type their answer here.

Closed-book checkpoint

Complete each sentence from memory. There is no answer bank and correctness is held for teacher review.

Fill in the blanks4 marks
In sign-and-magnitude, the leading bit stores the completion 1 and the remaining bits store the completion 2. In 8-bit two's complement, the leading place value is completion 3. Two's complement has completion 4 representation of zero.

Closed-book retrieval

State the 8-bit range and zero count for each representation, then explain why 10000101 cannot be interpreted until the encoding is named.