Normalisation of floating point numbers

Overview

Recall questions on why numbers are normalised, the test for both signs, and normalising positive and negative mantissas without losing the sign.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   What is normalisation of a floating point number?

   Model answer

   Normalisation adjusts a floating point number so that its mantissa is shifted as far left as it will go without changing the value, with the exponent decreased by one for each place shifted. The number represented stays the same, but redundant leading bits are removed.

2. 2

   Why are floating point numbers normalised?

   Model answer

   Two reasons. First, precision: leading bits that duplicate the sign bit carry no information, so shifting them out lets real significant bits occupy the mantissa, giving the most accurate representation the available bits allow. Second, uniqueness: without normalisation the same value could be stored as many different mantissa-and-exponent pairs, which makes comparison and arithmetic awkward; normalisation gives each value exactly one representation.

3. 3

   State the test for whether a mantissa is normalised, for both signs.

   Model answer

   The first two bits of the mantissa must differ. A positive normalised mantissa begins 0.1, and a negative normalised mantissa begins 1.0. If the first two bits are the same — 0.0 or 1.1 — the number is not normalised and the mantissa can be shifted further left.

4. 4

   What range of values does a normalised mantissa take, for a positive and for a negative number?

   Model answer

   A positive normalised mantissa lies in the range 0.5 ≤ m < 1, because it starts 0.1. A negative normalised mantissa lies in the range −1 ≤ m < −0.5, because it starts 1.0. Note that −1 exactly, the pattern 1.0000000, is normalised.

5. 5

   Normalise the floating point number with mantissa 00011010 and exponent 0101 (8-bit mantissa with the point after the sign bit, 4-bit exponent, both two's complement).

   Model answer

   Mantissa 01101000, exponent 0011. The mantissa 0.0011010 begins 0.0, so it is not normalised. Shift it two places left to 0.1101000, which begins 0.1, and reduce the exponent by 2 from +5 to +3 = 0011. Both forms are worth 6.5: 0.203125 × 32 = 6.5 and 0.8125 × 8 = 6.5.

6. 6

   Normalise the floating point number with mantissa 11101000 and exponent 0100 (8-bit mantissa with the point after the sign bit, 4-bit exponent, both two's complement), showing that the value is unchanged.

   Model answer

   Mantissa 10100000, exponent 0010. The original mantissa 1.1101000 begins 1.1, so it is not normalised. Shift left once to 1.1010000 with exponent 3, still 1.1, so shift left again to 1.0100000 with exponent 2 = 0010, which begins 1.0 and is normalised. Checking the value: originally (−1 + 0.8125) × 16 = −0.1875 × 16 = −3, and finally (−1 + 0.25) × 4 = −0.75 × 4 = −3.

Other ways to work this

The same content, four other ways.

What is inside Levo

Six ways to work the same material.