📄 Paper 1 · 1.1 Data Representation
1.1.3 Floating-Point Representation
Cambridge 9618 · International A Level Computer Science · ~15 min read
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Why Floating-Point?

Fixed-point binary can only represent a limited range of values with a fixed precision. Floating-point representation allows computers to represent very large numbers, very small numbers, and fractional numbers using a fixed number of bits by separating the value into a mantissa and an exponent.

The general form is: Value = Mantissa × BaseExponent

In binary floating-point: Value = Mantissa × 2Exponent

Structure of a Floating-Point Number

ComponentRoleEffect of increasing bits
MantissaStores the significant digits of the numberGreater precision — more decimal places
ExponentStores the power of 2 (scales the mantissa)Greater range — can represent larger/smaller numbers

The exponent is typically stored in two's complement, allowing negative exponents (for fractions) and positive exponents (for large numbers).

Normalised Floating-Point

A floating-point number is normalised when the mantissa is in a standard form that maximises precision. For binary floating-point:

  • Positive normalised: Mantissa starts with 0.1... (the bit after the binary point is 1)
  • Negative normalised: Mantissa starts with 1.0... (the bit after the binary point is 0)

Normalisation ensures there is no wasted precision from leading zeros (or ones for negatives).

Example — Representing a value in floating-point

Suppose we have an 8-bit mantissa and 4-bit exponent (all two's complement):

To represent the denary value 5.5:

  1. 5.5 in binary = 101.1
  2. Normalise: shift to get 0.1011 × 2³ (shift left by 3 positions)
  3. Mantissa: 01011000 (8 bits, with binary point after bit 7)
  4. Exponent: 3 = 0011 (4-bit two's complement)

Precision and Range

PropertyDetermined byTo improve it
PrecisionNumber of mantissa bitsIncrease mantissa bits
RangeNumber of exponent bitsIncrease exponent bits

Trade-off

With a fixed total number of bits, allocating more bits to the mantissa increases precision but decreases range, and vice versa. This is a fundamental design trade-off.

Rounding Errors and Underflow/Overflow

  • Rounding error: Some values cannot be represented exactly in floating-point — the stored value is an approximation. E.g. 0.1 in binary is a repeating fraction.
  • Overflow: The result is too large for the exponent to represent.
  • Underflow: The result is too small (too close to zero) for the exponent to represent — it rounds to zero.
Exam tip: Cambridge 9618 requires understanding of precision vs. range trade-offs. For normalisation questions: positive numbers must start with 0.1, negatives with 1.0. The exponent tells you how many places to shift the binary point. More mantissa bits = more precision (more significant figures). More exponent bits = greater range.
⚠️ Common Mistakes
  • Confusing precision (mantissa bits) with range (exponent bits)
  • Thinking a longer mantissa increases the range — it does not
  • Forgetting normalised positive numbers start with 0.1 (not 0.0 or 1.0)
  • Not applying two's complement when the exponent is negative
  • Confusing overflow (too large) with underflow (too small/close to zero)
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What's covered in this video

  • • Floating-point format: mantissa and exponent explained
  • • Normalisation of positive and negative binary floating-point numbers
  • • Precision vs. range trade-off and how bit allocation affects each
  • • Rounding errors, overflow and underflow in floating-point
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Worksheet — 1.1.3 Floating-Point Representation

8 questions · instantly marked · Cambridge 9618 standard

Q1Explain the role of (a) the mantissa and (b) the exponent in a floating-point number representation.[4]
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(a) The mantissa stores the significant digits/precision of the number [1]; more mantissa bits gives greater precision [1]. (b) The exponent stores the power of 2 by which the mantissa is multiplied [1]; more exponent bits gives a greater range of representable values [1].
Q2State what is meant by a normalised floating-point number for (a) a positive value and (b) a negative value.[2]
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(a) Positive: mantissa starts with 0.1 (binary point immediately followed by a 1) [1]; (b) Negative: mantissa starts with 1.0 (binary point immediately followed by a 0) [1].
Q3A floating-point number has a mantissa of 01101000 and an exponent of 0100 (both in two's complement). What is the denary value represented? Show your working.[3]
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Mantissa = 0.1101000 (binary point after first bit) [1]; Exponent = 4 [1]; Shift binary point 4 places right: 01101.000 = 13 in denary [1].
Q4Explain the trade-off between precision and range when choosing how many bits to allocate to the mantissa versus the exponent, given a fixed total number of bits.[3]
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More bits for the mantissa → greater precision (more significant figures, finer detail) but fewer bits for the exponent → smaller range [1]; More bits for the exponent → greater range (larger/smaller values representable) but fewer bits for the mantissa → less precision [1]; It is not possible to maximise both simultaneously with a fixed number of bits [1].
Q5What is meant by a rounding error in the context of floating-point representation?[2]
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A rounding error occurs when a value cannot be represented exactly in binary floating-point [1]; the stored value is an approximation — the difference between the true value and the stored approximation [1].
Q6Distinguish between overflow and underflow in floating-point arithmetic.[2]
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Overflow: the result is too large to be represented — the exponent value is too large [1]; Underflow: the result is too close to zero to be represented — rounds to zero [1].
Q7A system uses 16 bits total for floating-point numbers. Suggest one way to increase the maximum representable value and explain any disadvantage.[2]
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Allocate more bits to the exponent (e.g. increase from 4 to 6 bits) [1]; disadvantage: fewer bits for the mantissa, so less precision — numbers stored with fewer significant figures / more rounding errors [1].
Q8The mantissa of a normalised positive binary floating-point number must start with 0.1. Explain why this rule maximises precision.[2]
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It ensures that no bit positions in the mantissa are wasted on leading zeros [1]; every bit contributes to the precision of the stored value — the first fractional bit is always 1, so the most significant bit position is always used [1].
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Mini Test — 1.1.3 Floating-Point Representation

10 questions · 10 marks · 10 minutes

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