Floating Point Binary Numbers (OCR A Level Computer Science): Revision Note
Exam code: H446
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

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

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