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OCR H446 · Component 1 · 1.4.1

Floating Point
Representation

OCR A Level Computer Science · cszone.co.uk
H446 SpecA Level
Learning Objectives

By the end of this topic you will be able to:

Explain what floating point representation is and why it is needed
Define mantissa and exponent and describe how they combine to represent a value
Understand normalisation and why it is used
Explain precision vs range trade-offs in floating point
Floating Point Basics

Why Floating Point?

Fixed point representation stores a fixed number of binary digits before and after a binary point. This limits the range of representable numbers. Floating point separates the number into a mantissa (significant digits) and an exponent (scale factor), allowing a much wider range.
The value = mantissa × 2^exponent. This is analogous to scientific notation in denary: 6.022 × 10²³ — the mantissa is 6.022 and the exponent is 23.
Example format: 8-bit mantissa + 4-bit exponent. The mantissa is a two's complement fraction with an implied binary point after the sign bit.
Mantissa & Exponent

Structure and Calculation

Mantissa: 0.1011 0100    Exponent: 0011 (= +3)
Value = 0.10110100 × 2³
= Shift binary point 3 places right
= 101.10100
= 4 + 1 + 0.5 + 0.125 = 5.625
Positive Exponent
Shift binary point right by the exponent value — makes the number larger.
Negative Exponent
Shift binary point left by the magnitude of the exponent — makes the number smaller (fraction).
Normalisation

Normalised Floating Point

Normalisation ensures the mantissa is in a standard form to maximise precision. A positive normalised mantissa starts with 0.1... (first bit after binary point is 1). A negative normalised mantissa starts with 1.0... (first bit after binary point is 0).
Why normalise? Without normalisation, the same number can be represented multiple ways (wasted bits, lost precision). Normalisation ensures a unique representation with maximum significant bits used.
Normalised (positive)
0.1010 1100 — starts with 0.1 ✓
NOT normalised
0.0010 1100 — leading 0 wastes precision ✗
Precision vs Range

Precision and Range Trade-offs

More Mantissa Bits
Increases precision — more significant digits can be stored, reducing rounding errors. You can represent numbers more accurately, but the total range stays limited by the exponent.
More Exponent Bits
Increases range — the scale factor can be larger/smaller, allowing very large or very tiny numbers. But fewer bits left for mantissa means less precision.
With a fixed total bit count, there is always a trade-off between precision and range. IEEE 754 single precision (32-bit) uses: 1 sign bit + 8 exponent bits + 23 mantissa bits (with implied leading 1). Double precision uses 64 bits for greater precision and range.
Exam Practice
OCR H446 Style · 4 marks
A floating point number uses 8 bits for the mantissa (two's complement, binary point after bit 1) and 4 bits for the exponent (two's complement). The stored value is: Mantissa = 0110 1000, Exponent = 0010. Calculate the denary value. [3 marks] Explain what "normalised" means for this format. [1 mark]
[4 marks]
1
Exponent = 0010 = +2 in two's complement.
1
Mantissa = 0.110 1000 (binary point after first bit). Shift right 2 places → 011.01000
1
011.01000 = 2 + 1 + 0.25 = 3.25
1
Normalised means the bit immediately after the binary point is 1 for a positive number (mantissa starts 0.1…), ensuring the maximum number of significant bits are used and each value has a unique representation.
Common Mistakes

Don't Lose Marks

!
Saying the mantissa is a whole number — in OCR H446 floating point, the mantissa is a binary fraction with the binary point after the first (sign) bit. 0110 1000 means 0.110 1000. Not treating it as a fraction leads to completely wrong answers.
!
Shifting the binary point in the wrong direction — a positive exponent shifts the binary point right (makes number bigger); negative exponent shifts left (makes number smaller). This is the most common calculation error in floating point questions.
!
Saying normalisation increases range — normalisation maximises precision by ensuring all mantissa bits are significant. It does not change the range of representable values. Range is determined by the number of exponent bits.
1.4.1c Complete
Well done! ✓
Floating Point Representation
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