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OCR H446 1.4.1 Binary floating-point representation
Part 5 of 11 · H446 1.4.1 · Data types
OCR's floating-point format for H446 1.4.1 uses a two's-complement mantissa and a two's-complement exponent, not IEEE 754, so students need the format the paper actually assesses. Working from audio and sensor values, this worksheet fixes where the binary point sits and then has students both interpret and construct values at a 6-bit mantissa and 4-bit exponent.
Students will:
- locate the mantissa and exponent fields and state where the binary point sits
- interpret a stored value as a signed mantissa multiplied by a power of two
- represent a denary value at a stated mantissa and exponent width
- read a two's-complement exponent as signed rather than as an unsigned pattern
- explain how mantissa width and exponent width affect precision and range differently
Inside: 8 explanation cells, 3 multiple-choice questions, 1 fill-in-the-blanks cell, 1 written answer and 2 number answers. 22 marks, about 45 to 55 minutes.
Series: H446 1.4.1 · Data types, part 5 of 11.
Shared by Coding PathwayVerified teacher
- 15 cells
- About 45 minutes
- CC BY-SA 4.0
- Shared 31 Aug 2026
- Updated 3 Sept 2026
Preview
The whole resource, exactly as a class sees it. Answers and marking are held back.
Binary floating-point representation
Audio and sensor values may need fractions and a wide range. OCR uses a two’s-complement mantissa and two’s-complement exponent, not IEEE 754.
By the end, you will be able to
- locate and interpret both fields
- read the mantissa as a signed binary fraction
- apply the signed power-of-two exponent
- construct a representation and distinguish range from precision
Reactivate: two’s-complement signed integers and negative powers such as 1/2, 1/4 and 1/8.
Read mantissa × 2^exponent
The binary point is immediately after the mantissa sign bit. Interpret the mantissa as a two's-complement fraction, interpret the exponent as a two's-complement integer, then multiply by the stated power of two.
Positive example with a 6-bit mantissa and 4-bit exponent: 010100 0010 means 0.10100₂ × 2² = 2.5₁₀.
Negative example: 101000 1111 means -0.11000₂ × 2⁻¹. The mantissa is -0.75, the exponent is -1, and the represented value is -0.375. The exponent bits are signed; reading 1111 as unsigned 15 would be a representation error.
See the two fields before calculating
Worked model
For mantissa 010100 and exponent 0010:
- binary point after the sign bit gives 0.10100₂ = 1/2 + 1/8 = 0.625;
- exponent 0010₂ is +2;
- multiply by 2²: 0.625 × 4 = 2.5.
Sense check: a positive mantissa and positive exponent must give a positive value larger than the mantissa fraction.
In OCR's floating-point representation, where is the binary point placed in the mantissa?
- AImmediately after the sign bit
- BBefore the sign bit
- CImmediately before the last bit
- DIts position is stored in the exponent bits
Interpret each floating-point value in denary. The mantissa and exponent both use two's complement.
a)What is the denary value of the floating point number with mantissa 100000 and exponent 0100?
b)What is the denary value of the floating point number with mantissa 100001 and exponent 1101?
c)What is the denary value of the floating point number with mantissa 100000 and exponent 0011?
d)What is the denary value of the floating point number with mantissa 101001 and exponent 0001?
Represent a value
Choose an exponent that brings the significant binary digits into the mantissa. Store the fractional mantissa and the signed exponent at the stated widths. Different unnormalised encodings can represent the same value; the next worksheet will select a standard form.
Guided representation plan
To represent 5.5₁₀, first write 101.1₂. Move the binary point so the mantissa is a signed fraction: 0.1011₂ × 2³. Then encode the mantissa and exponent at the stated widths.
Keep three lines of working: ordinary binary value → fractional mantissa × power → fixed-width fields. This prevents the common mistake of treating the exponent bits as an unsigned shift count. A fixed number of bits provides only finitely many patterns, so not every real value can be exact; values between available patterns must be approximated.
Represent each denary value using a 6-bit two's-complement mantissa and 4-bit two's-complement exponent.
a)Represent 3.125 as a normalised floating point number.
b)Represent -6.5 as a normalised floating point number.
c)Represent 6 as a normalised floating point number.
d)Represent -0.1640625 as a normalised floating point number.
A learner reads the exponent 1110 as denary 14. What is the error?
- AThe exponent should always be read as sign-and-magnitude.
- BThe exponent should be read as two's complement, so 1110 is -2.
- CThe exponent is appended to the mantissa before conversion.
- DThe exponent cannot contain a 1 sign bit.
Which statement is accurate for this OCR worksheet?
- AThe exponent uses an IEEE 754 bias.
- BThe mantissa uses sign-and-magnitude.
- CBoth mantissa and exponent use two's complement.
- DThe exponent is always positive.
Independent transfer
Apply the mantissa-versus-exponent distinction to a changed field width.
A sensor uses more bits for its mantissa but keeps the exponent width unchanged. Explain one likely benefit and one limit that remains.
Think separately about precision and range.
Students type their answer here.
Closed-book checkpoint
Complete each sentence from memory. There is no answer bank and correctness is held for teacher review.
Retrieval check
Draw and label the mantissa and exponent fields, state where the binary point sits, and explain why extra mantissa bits improve precision while extra exponent bits extend range.