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How are real numbers stored in floating-point form, and what does normalisation achieve?

Floating-point representation: storing a real number as a mantissa and an exponent in two's complement, evaluating a stored floating-point value, normalising a mantissa, and the trade-off between range and precision.

An Eduqas Component 2 answer on floating-point representation: storing a real number as a two's complement mantissa and exponent, evaluating a stored floating-point value, normalising a mantissa, and the trade-off between range and precision.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A floating-point number stores a real number in two parts: the mantissa (the significant digits, a two's complement fraction with the binary point after the sign bit) and the exponent (a two's complement integer saying how far to move the binary point). The value is the mantissa shifted by the exponent: a positive exponent moves the point right (multiply by $2^{\text{exponent}}$ ), a negative exponent moves it left. To evaluate a stored number, read the exponent as the shift and apply it to the mantissa. Normalising adjusts the mantissa so it has no wasted leading bits, a normalised positive mantissa begins $0.1...$ and a normalised negative mantissa begins $1.0...$ , with the exponent adjusted to keep the value unchanged; this maximises precision for the available bits and gives each number a unique representation. For a fixed total of bits there is a trade-off: more exponent bits give greater range, more mantissa bits give greater precision.

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What this dot point is asking

Eduqas wants you to explain how a real number is stored as a mantissa and exponent in two's complement, evaluate a stored floating-point value, normalise a mantissa, and explain the trade-off between range and precision. The classic questions are evaluating a stored value and normalising one, both without a calculator.

The answer

Storing a real number: mantissa and exponent

Evaluating a stored floating-point number

Normalisation and the range-precision trade-off

Normalise a floating-point number

Normalise the positive mantissa $0.0011010$ (binary point after the sign bit) with exponent $0101$ ( $= +5$ ).

step 1: Identify why it is not normalised

A normalised positive mantissa must start $0.1...$ , but this one starts $0.001...$ , so it has redundant leading zeros wasting precision.

step 2: Shift the mantissa left

Shift the mantissa left by 2 places to remove the leading zeros: $0.0011010$ becomes $0.1101000$ , which now starts $0.1...$ as required.

step 3: Adjust the exponent to compensate

Shifting the mantissa left by $2$ multiplies it by $2^2$ , so to keep the value unchanged the exponent must decrease by 2: $+5 - 2 = +3$ , which is $0011$ .

step 4: State the normalised form

The normalised number is mantissa $0.1101000$ with exponent $0011$ . The value is unchanged, but the mantissa now uses all its bits for significant figures, maximising precision.

Examples in context

Floating point is how computers store and calculate with real numbers, scientific measurements, money in some systems, graphics coordinates, so the range-precision trade-off is a real engineering decision (single versus double precision). Normalisation is exactly why the IEEE floating-point standards used in every modern CPU keep numbers in a unique, maximally precise form. The two's complement skills from the previous dot point are used here for both the mantissa and the exponent, and rounding errors in floating point explain why 0.1 + 0.2 is not exactly 0.3 in many languages.

Try this

Q1. State the leading bits of a normalised positive mantissa and a normalised negative mantissa. [2 marks]

Cue. Positive: $0.1...$ ; negative: $1.0...$ .

Q2. A mantissa represents $0.5$ and the exponent is $+2$ . What is the denary value? [1 mark]

Cue. $0.5 \times 2^2 = 0.5 \times 4 = 2$ .

Q3. For a fixed number of bits, what is the effect of giving more bits to the exponent? [1 mark]

Cue. It increases the range of magnitudes representable but reduces precision (fewer mantissa bits).

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20205 marksA floating-point number uses an 8-bit two's complement mantissa and a 4-bit two's complement exponent, with the binary point after the first mantissa bit. The mantissa is

0.110\,0000

(using the convention

0.1100000

) and the exponent is

0011

. Calculate the denary value, showing your working.

Show worked answer →

Read the exponent (up to 1): exponent $0011 = +3$ , so the binary point moves $3$ places to the right.

Read the mantissa (up to 1): the mantissa represents $0.1100000_2 = 0.5 + 0.25 = 0.75$ in denary (with the point after the sign bit).

Apply the exponent (up to 2): moving the point $3$ places right multiplies by $2^3 = 8$ , so the value is $0.75 \times 8 = 6$ (equivalently the bits become $110.0000_2 = 6$ ).

State the result (up to 1): the denary value is $+6$ .

Markers reward reading the exponent as the shift, reading the mantissa as a fraction, applying the shift (multiply by $2^{\text{exponent}}$ ), and the final value.

Eduqas 20215 marksExplain what it means to normalise a floating-point number, state the leading bits of a normalised positive and negative mantissa, and explain the trade-off between range and precision.

Show worked answer →

Normalisation (up to 2 marks): adjusting the mantissa so it has no redundant leading bits, shifting it and changing the exponent to keep the value the same; this maximises the precision held in the available mantissa bits and gives each value a unique representation.

Leading bits (up to 1 mark): a normalised positive mantissa begins $0.1...$ and a normalised negative mantissa begins $1.0...$ (the first bit after the point differs from the sign bit).

Range versus precision (up to 2 marks): for a fixed total number of bits, giving more bits to the exponent increases the range of magnitudes that can be represented but leaves fewer mantissa bits, reducing precision; giving more bits to the mantissa increases precision but reduces range.

Markers reward the remove-redundant-bits definition, the correct leading bits, and the more-exponent-bits-more-range versus more-mantissa-bits-more-precision trade-off.

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Sources & how we know this

WJEC Eduqas GCE AS/A Level Computer Science specification (from 2015) — Eduqas (2015)