Floating-Point Conversions (Cambridge International A-Level Computer Science)

Denary to floating-point binary

How do you convert denary to floating-point binary?

• These examples assume:

  • 12-bit mantissa (two's complement, binary point after the leftmost bit)
  • 4-bit exponent (two's complement)

1. Convert the magnitude to binary (integer part + fractional part)

2. e.g. 5 = 101₂, 0.25 = 0.01₂, so 5.25 = 101.01₂

3. Normalise into the form 0.xxxxx × 2ⁿ

4. move the binary point until the first two bits differ; the number of places moved is the exponent

5. Build the positive mantissa: write 0, then the fraction bits, then pad with zeros on the right

6. If the number is negative, take the two's complement of the whole positive mantissa (flip the bits and add 1)

7. Convert the exponent to two's complement in the required number of bits

8. Write the final answer as Mantissa and Exponent

Example: +5.25

• Convert to binary: 5.25 = 101.01₂
• Normalise: 101.01 = 0.10101 × 2³
• Mantissa: 0 + fraction 10101, padded = 010101000000
• Exponent: +3 = 0011
• Result: Mantissa 010101000000, Exponent 0011

Example: −5.25

• Work out the positive version first (as above): positive mantissa = 010101000000
• Take the two's complement of the whole mantissa:

• The exponent is unchanged: +3 = 0011
• Result: Mantissa 101011000000, Exponent 0011
• The mantissa begins 10, confirming it is a normalised negative number

Floating-point binary to denary

How do you convert floating-point binary to denary?

• These examples assume a 12-bit two's complement mantissa and a 4-bit two's complement exponent

1. Read the exponent and convert it from two's complement to denary
2. Check the first bit of the mantissa: 0 = positive, 1 = negative
3. If the mantissa is negative, take its two's complement (flip the bits and add 1) to find the magnitude of the fraction
4. Write the fraction as 0.<fraction bits> and apply the exponent by moving the binary point (right if positive, left if negative)
5. Apply the sign: if the first mantissa bit was 1, make the final result negative

Example (positive)

• Mantissa: 010101000000

• Exponent: 0011

• Exponent: 0011 = +3

• First bit: 0 → positive

• Fraction: 0.10101

• Apply exponent: 0.10101 × 2³ = 101.01₂ = 5.25

• Result: +5.25

Example (negative)

• Mantissa: 101011000000

• Exponent: 0011

• Exponent: 0011 = +3

• First bit: 1 → negative

• Two's complement of the mantissa to find the magnitude:

• Fraction (magnitude): 0.10101
• Apply exponent: 0.10101 × 2³ = 101.01₂ = 5.25
• Apply sign: negative → −5.25
• Result: −5.25

Summary

Exam Hint

• The whole mantissa is two's complement. For a negative number (mantissa starts with 1), take the two's complement of the mantissa to find the magnitude, then make the final answer negative.
• Both the mantissa and the exponent are two's complement.
• Quicker check: the leftmost mantissa bit carries a negative place value, so 1.01011 = −1 + 0.34375 = −0.65625, then × 2³ = −5.25.
• Show the working the mark scheme expects: the binary conversion, the two's complement step for negatives, and the exponent (the number of places the point moves).

Worked Example Numbers are stored in two different computer systems using floating-point representation.

System 1 uses:

• 10 bits for the mantissa
• 6 bits for the exponent
• two's complement form for both the mantissa and the exponent

System 2 uses:

• 8 bits for the mantissa
• 8 bits for the exponent
• two's complement form for both the mantissa and the exponent

Calculate the normalised floating-point representation of 113.75 in each system. Show your working.

[4]

Answer

• Convert 113.75 to binary: 1110001.11 (1 mark)
• Normalise and find the exponent: 0.111000111 × 2⁷, so the point moves 7 places, exponent = 7 (1 mark for evidence of exponent 7)
• System 1: exponent 7 = 000111

Mantissa	Exponent
0111000111	000111

• System 2: exponent 7 = 00000111 (the mantissa is truncated to 8 bits, losing precision)

Mantissa	Exponent
01110001	00000111