Fixed-point binary: signed fractional numbers

Place values after the point

After the binary point the place values continue halving: 1/2 (0.5), 1/4 (0.25), 1/8 (0.125).

For an 8-bit two's complement number with 3 bits after the point, the place values are:

-16, 8, 4, 2, 1 . 1/2, 1/4, 1/8

Example: 01011.011 = 8 + 2 + 1 + 1/4 + 1/8 = 11.375.

The position of the point is agreed in advance; it is not stored.

Negative fixed-point numbers

If the MSB is 1 the number is negative. Either add the place values (with the MSB negative), or negate (invert, add 1) to find the positive value.

Example: 11010.110 = -16 + 8 + 2 + 1/2 + 1/4 = -5.25.

Converting denary to fixed point

1. Write the positive value using the place values, for example 5.375 = 4 + 1 + 1/4 + 1/8 = 00101.011.

2. If the number is negative, invert and add 1: -5.375 = 11010.101.

Characteristics: fixed precision (with 3 fraction bits the smallest step is 1/8); numbers that are not multiples of the smallest step, such as 0.1, cannot be stored exactly. Moving the point right gives a bigger range but less precision; moving it left gives more precision but a smaller range.

Worked example

A 10-bit two's complement fixed-point number has 3 bits after the point: 0100111.011. Convert it to denary.

1. Place values: -64, 32, 16, 8, 4, 2, 1 . 1/2, 1/4, 1/8
2. The MSB is 0, so the number is positive. Integer part 0100111 = 32 + 4 + 2 + 1 = 39
3. Fraction part 011 = 1/4 + 1/8 = 0.375

Answer: 39.375 (39 3/8)

Check yourself

1. An 8-bit two's complement fixed-point number has 3 bits after the point: 00110.110. What is it in denary?
   Show answer

   6.750. 4 + 2 + 1/2 + 1/4 = 6.75.

2. An 8-bit two's complement fixed-point number has 3 bits after the point: 10101.100. What is it in denary?
   Show answer

   -10.500. -16 + 4 + 1 + 1/2 = -10.5.

3. An 8-bit two's complement fixed-point number has 4 bits after the point: 1111.1000. What is it in denary?
   Show answer

   -0.5000. Place values -8, 4, 2, 1 . 1/2, 1/4, 1/8, 1/16: -8 + 4 + 2 + 1 + 1/2 = -0.5.

4. Write 3.625 as an 8-bit two's complement fixed-point number with 3 bits after the point (include the point).
   Show answer

   00011.101. 3.625 = 2 + 1 + 1/2 + 1/8 = 00011.101.

5. Write -2.25 as an 8-bit two's complement fixed-point number with 3 bits after the point (include the point).
   Show answer

   11101.110. +2.25 = 00010.010; invert and add 1: 11101.110.

6. With 3 bits after the binary point, what is the smallest positive value that can be represented?
   Show answer

   0.125. The last place value is 1/8 = 0.125.

7. What is the largest value of an 8-bit two's complement fixed-point number with 3 bits after the point?
   Show answer

   15.875. 01111.111 = 15 + 7/8 = 15.875.

8. What happens if the binary point is moved one place to the right in a fixed-point format with the same number of bits?
   • The range stays the same and the precision increases
   • The range decreases and the precision increases
   • Nothing changes because the point is not stored
   • The range increases and the precision decreases
   Show answer

   The range increases and the precision decreases. More integer bits give a bigger range; fewer fraction bits give coarser steps.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Computer Science specification (2026)
• Pearson Edexcel IAL Computer Science IAS sample assessment materials (Unit 1, WCP01)

Last updated 8 November 2026. Spotted a mistake? Tell us.