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OCR H446 1.4.1 Floating-point normalisation
Part 6 of 11 · H446 1.4.1 · Data types
Several mantissa and exponent pairs can encode the same number, so H446 1.4.1 asks for the normalised form that makes full use of a finite mantissa. Students first learn to recognise a normalised value from its first two mantissa bits, for positive and negative values, then shift and compensate so the represented value is preserved.
Students will:
- decide whether a two's-complement mantissa is normalised, whether positive or negative
- explain why the first two mantissa bits must differ
- shift a mantissa and move the exponent in the opposite direction to preserve the value
- normalise values using a 6-bit mantissa and a 4-bit exponent
- diagnose what happens when a mantissa is shifted and the exponent is left unchanged
Inside: 7 explanation cells, 2 multiple-choice questions, 1 fill-in-the-blanks cell, 1 written answer and 2 number answers. 24 marks, about 45 to 55 minutes.
Series: H446 1.4.1 · Data types, part 6 of 11.
Shared by Coding PathwayVerified teacher
- 13 cells
- About 45 minutes
- CC BY-SA 4.0
- Shared 31 Aug 2026
- Updated 3 Sept 2026
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The whole resource, exactly as a class sees it. Answers and marking are held back.
Floating-point normalisation
Different mantissa/exponent pairs can encode the same value. Normalisation selects a form that uses the finite mantissa efficiently.
By the end, you will be able to
- recognise positive and negative normalised mantissas
- shift a mantissa while compensating in the exponent
- prove that the represented value is preserved
- explain how normalisation supports precision
Reactivate: OCR’s two’s-complement mantissa and exponent fields.
Recognition model: the normalised form
With OCR's two's-complement fractional mantissa, the first two mantissa bits must differ:
- a positive normalised mantissa begins 01;
- a negative normalised mantissa begins 10.
Shift the mantissa and adjust the exponent in the opposite direction so the represented value does not change.
Why the first two bits must differ
A positive two’s-complement fraction has sign bit 0; normalised positive values begin 01 so the first fractional place is used. A negative normalised fraction begins 10. Prefixes 00 or 11 contain a redundant sign-extension bit that can be shifted out.
Worked proof that the value is preserved
001100 0010 represents 0.01100₂ × 2² = 0.375 × 4 = 1.5. Shift the mantissa left once and decrease the exponent once: 011000 0001 represents 0.11000₂ × 2¹ = 0.75 × 2 = 1.5. The pattern is now normalised and the value is unchanged.
Normalisation uses the available significant places and gives a consistent form for comparing or processing equal values.
Which 6-bit mantissa is normalised under OCR's two's-complement convention?
- A001101
- B110100
- C000111
- D010110
Decide whether each two's-complement mantissa is normalised.
a)Is the floating point number with mantissa 001110 and exponent 0100 normalised?
b)Is the floating point number with mantissa 011001 and exponent 0001 normalised?
c)Is the floating point number with mantissa 100111 and exponent 0001 normalised?
d)Is the floating point number with mantissa 001100 and exponent 1110 normalised?
e)Is the floating point number with mantissa 011010 and exponent 0010 normalised?
Preserve the value
If the mantissa moves left by one place, its numerical value doubles, so decrease the exponent by one to compensate. If it moves right by one place, increase the exponent by one. Keep both fields at their fixed widths.
Worked positive value: 001101 0010 becomes 011010 0001: the mantissa shifts left once and the exponent falls from 2 to 1. Worked negative value: 110100 0010 becomes 101000 0001. In both cases the first two mantissa bits now differ and the represented denary value is unchanged.
Preserve-value equation
If a mantissa is shifted left two places, it becomes four times as large. Decreasing the exponent by 2 multiplies by 2⁻², cancelling that factor: 4 × 1/4 = 1.
For each guided attempt, write above the fields: mantissa factor ×2 or ÷2; exponent correction ∓1; new prefix 01 or 10; value unchanged.
Normalise each value using a 6-bit mantissa and 4-bit exponent.
a)Normalise the floating point number with mantissa 110110 and exponent 1111.
b)Normalise the floating point number with mantissa 001000 and exponent 1111.
c)Normalise the floating point number with mantissa 110111 and exponent 0011.
d)Normalise the floating point number with mantissa 111000 and exponent 0110.
e)Normalise the floating point number with mantissa 001100 and exponent 0000.
A learner shifts a mantissa left by two places but leaves the exponent unchanged. Explain the effect on the represented value and state the exponent correction needed.
State the factor caused by the shift, then compensate for it.
Students type their answer here.
Why is a floating-point value normally normalised?
- ATo use the available mantissa bits for significant digits
- BTo remove its sign
- CTo make every exponent positive
- DTo convert it to fixed point
Closed-book checkpoint
Complete each sentence from memory. There is no answer bank and correctness is held for teacher review.
Retrieval check
Explain why 001101 is unnormalised, how to repair it, and why the exponent moves in the opposite direction.