CAIE 9618 · Paper 3 · Topic 3.1.3

Floating Point
Representation

Mantissa · Exponent · Normalisation · Precision vs Range · Errors

CSZone Cambridge International AS & A Level Computer Science 9618

Floating Point Structure

Mantissa × 2^Exponent

Floating point represents very large or very small real numbers by splitting storage into two fields: a mantissa (the significant digits) and an exponent (the scale/power of 2).

Example: 8-bit float — 5-bit mantissa | 3-bit exponent (both in two's complement)

0

1

0

1

      MMMMM
      
      EEE

MANTISSA

01100 in two's complement = 0.1100 (normalised binary fraction). Value = +¾. Number of bits controls precision.

EXPONENT

011 in two's complement = +3. Scales mantissa by 2³. Number of bits controls range of representable values.

Normalisation

Maximum Precision from Available Bits

Normalisation ensures the mantissa uses all its bits as efficiently as possible — no leading redundant bits. In normalised form, the first two bits of the mantissa must differ (one is the sign bit, the next must be the opposite).

NORMALISED — first two bits differ

          Positive: 0.1... (starts with 01)

          Negative: 1.0... (starts with 10)

NOT NORMALISED — first two bits same

          0.01... (starts 00) → shift left

          1.11... (starts 11) → shift left

NORMALISING A NUMBER

1. Shift mantissa left until first two bits differ
2. Decrease exponent by 1 for each left shift
3. Result: same value, maximum precision

E.g. 0.00110 × 2³ is NOT normalised
→ 0.11000 × 2¹ IS normalised (shifted 2 left, exponent -2)

Precision vs Range

The Bit Allocation Trade-off

With a fixed total number of bits, every bit added to the mantissa improves precision but reduces range — and vice versa.

MORE MANTISSA BITS → MORE PRECISION

More significant digits in the fractional part

Smaller rounding errors when approximating fractions

Important for scientific calculations

MORE EXPONENT BITS → MORE RANGE

Larger maximum and minimum values representable

Important for astronomy, physics — very large/small numbers

FLOATING POINT ERRORS

Rounding error — value cannot be stored exactly; nearest representable value used (e.g. 0.1 in binary)

Overflow — result too large for exponent field; exception raised

Underflow — result too small (close to zero) for exponent field; stored as zero

Truncation — bits beyond mantissa length simply dropped

Exam Practice

Cambridge-style questions

Question 1

A 16-bit floating point number uses 10 bits for the mantissa and 6 bits for the exponent, both stored in two's complement. (a) State what is meant by normalisation of a floating-point number. (b) State the effect of increasing the number of mantissa bits from 10 to 12 (reducing exponent bits from 6 to 4). [4]

2

(a) Normalisation means adjusting the mantissa and exponent so that the mantissa is in its most precise form — for a positive number the mantissa begins with 01, for a negative number with 10. This ensures no leading redundant bits waste precision.

1

(b) Increasing mantissa bits increases precision — more significant digits can be stored, reducing rounding errors.

1

However, reducing exponent bits decreases range — smaller maximum and minimum values can be represented, increasing the risk of overflow or underflow.

Common Mistakes

Don't lose easy marks

1

Saying normalisation "saves memory" or "makes numbers smaller" — normalisation does NOT change what value is stored. It changes HOW the value is encoded to maximise precision within the fixed number of bits. The value stored is identical.

2

Confusing overflow and underflow — overflow = result too large for the exponent (e.g. trying to store 10³⁰⁰ with only 4 exponent bits). Underflow = result too close to zero (e.g. trying to store 10⁻³⁰⁰). Not "overflow = too many bits".

3

Giving the wrong normalisation condition — positive mantissa starts with 01 (not 0 alone). Negative mantissa starts with 10 (not 1 alone). The sign bit and the first fractional bit must differ — this is the key rule.

Topic Summary — 3.1.3

What You Need to Know

STRUCTURE

Value = Mantissa × 2^Exponent
Both stored in two's complement
More mantissa bits → more precision
More exponent bits → more range

ERRORS

Rounding — can't store exactly
Overflow — exponent too large
Underflow — result too close to zero

NORMALISATION

Maximises precision by removing leading redundant bits.
Positive: mantissa starts 01...
Negative: mantissa starts 10...
Shift mantissa left → decrease exponent by same amount.

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3.1.4

Data Compression

Lossless · Lossy · RLE · Huffman · JPEG · MP3

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