Floating Point Binary Numbers (OCR A Level Computer Science): Revision Note

Exam code: H446

Robert Hampton

Written by: Robert Hampton

Reviewed by: James Woodhouse

Updated on

Floating Point Binary

What is floating point binary?

  • Floating point binary addresses the limitations of fixed-point binary in representing a wide array of real numbers

  • It allows for both fractional and whole-number components

  • It accommodates extremely large and small numbers by adjusting the floating point

  • It optimises storage and computational resources for most applications

Mantissa and exponent

  • In A Level Computer Science, the mantissa and exponent are the two main components of a floating point number:

    • Mantissa

      • The part of the number that holds the actual digits of the value

      • It represents the precision of the number but does not determine its scale

    • Exponent

      • Controls how far the binary point moves, effectively scaling the number up or down

      • A larger exponent moves the point to the right (making a larger number), while a smaller exponent moves it to the left (making a smaller number)

  • For example, in standard scientific notation, the decimal number 3.14 × 10³ has:

    • Mantissa: 3.14 (the significant digits)

    • Exponent: 3 (which shifts the decimal point three places to the right)

  • Floating point binary works similarly but in base-2

  • Instead of multiplying by powers of 10, it multiplies by powers of 2 using a binary exponent

Representation of floating point

  • The appearance of floating-point binary is mostly the same except for the presence the decimal point

Binary representation of 7.875 as a floating point number, showing bit values for 16, 8, 4, 2, 1, 0.5, 0.25, and 0.125.
  • In the example above, an 8-bit number can represent a whole number and fractional elements

  • The point is always placed between the whole and fractional values

  • The consequence of floating point binary is a significantly reduced maximum value

  • The benefit of floating point binary is increased precision

Representation of negative floating point

  • Negative numbers can also be represented in floating point form using two's Complement

  • The MSB is used to represent the negative offset of the number, and the bits that follow it are used to count upwards

  • The fractional values are then added to the whole number

Binary to decimal conversion, showing powers of two and fractional components resulting in the decimal value -9.875.

Converting Denary to Floating Point

Denary to floating point binary

Example: Convert 6.75 to floating point binary

Step 1: Represent the number in fixed point binary. 

-8

4

2

1

.

0.5

0.25

0

1

1

0

.

1

1

Step 2: Move the decimal point.

0

.

1

1

0

1

1

Step 3: Calculate the exponent 

The decimal point has moved three places to the left and therefore has an exponent value of three

-4

2

1

0

1

1

Step 4: Calculate the final answer:

Mantissa: 011011

Exponent: 011

Converting Floating Point to Denary

Binary floating point to denary

Example: Convert this floating point number to denary:

  • Mantissa - 01100

  • Exponent - 011

Step 1: Write out the binary number. 

0

.

1

1

0

0

Step 2: Work out the exponent value.

The exponent value is 3. 

-4

2

1

0

1

1

Step 3: Move the decimal point three places to the right. 

-8

4

2

1

.

0.5

0

1

1

0

.

0

Step 4: Calculate the final answer: 6

Normalising Floating Point Binary

  • A floating point number is said to be normalised when it starts with 01 or 10

Why normalise?

  • Ensures a consistent format for floating point representation

  • Makes arithmetic and comparisons more straightforward

Steps to normalise a floating point number

  1. Shift the decimal point left or right until it starts with a 01 or 10

  2. Adjust the exponent value accordingly as you move the decimal point

    • Moving the point to the left increases the exponent and vice versa

  • Example

    • Before normalisation:

      • Mantissa = 0.0011

      • Exponent = 0010 (2)

  • After normalisation:

    • Mantissa = 0.1100

      • Decimal point has moved 2 places to right so it starts with 01

    • Exponent = 0000 (0)

      • Exponent has been reduced by 2

Decode a normalised floating point binary number

  • In A Level Computer Science, you may be given a floating point number made up of two parts:

    • A 4-bit mantissa (in two’s complement)

    • A 4-bit exponent (also in two’s complement)

  • For example:

    • 1101 1111

    • Mantissa = 1101

    • Exponent = 1111

Step 1: Convert the exponent to denary

  • The exponent is stored in 4-bit two’s complement, so:

    • If the first bit is 0 → it's a positive number

    • If the first bit is 1 → it's negative

  • Example:

Exponent: 1111 → two’s complement = -1

Step 2: Convert the mantissa to binary

  • The mantissa is also in 4-bit two’s complement

  • Convert it to a denary number.

  • Then convert it to a binary value, assuming the binary point goes just after the first bit (because it's normalised)

  • Example:

Mantissa: 1101 → two’s complement = -3

Binary of 3 = 011

So -3 in normalised binary = -0.110 (we assume a leading 1 is implied)

Step 3: Shift the binary point

  • Now shift the binary point by the exponent value.

    • If exponent is positive, move the point to the right

    • If exponent is negative, move it to the left

  • Example:

Start with: -0.110

Exponent = -1

Shift the point 1 place left → -0.0110

Step 4: Convert to denary

  • Now convert the final binary number into denary

-0.0110 =

0 × ½ = 0

1 × ¼ = 0.25

1 × ⅛ = 0.125

0 × 1/16 = 0

Final value = -0.25 - 0.125 = -0.375

  • Answer: –0.375

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Robert Hampton

Author: Robert Hampton

Expertise: Computer Science Content Creator

Rob has over 16 years' experience teaching Computer Science and ICT at KS3 & GCSE levels. Rob has demonstrated strong leadership as Head of Department since 2012 and previously supported teacher development as a Specialist Leader of Education, empowering departments to excel in Computer Science. Beyond his tech expertise, Robert embraces the virtual world as an avid gamer, conquering digital battlefields when he's not coding.

James Woodhouse

Reviewer: James Woodhouse

Expertise: Computer Science & English Subject Lead

James graduated from the University of Sunderland with a degree in ICT and Computing education. He has over 14 years of experience both teaching and leading in Computer Science, specialising in teaching GCSE and A-level. James has held various leadership roles, including Head of Computer Science and coordinator positions for Key Stage 3 and Key Stage 4. James has a keen interest in networking security and technologies aimed at preventing security breaches.