Rational numbers
Reminder: Two's complement
| Decimal | -128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|---|
| 47 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 23 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 |
| -23 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
| Decimal | -128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|---|
| Carry | 1 | 1 | 0 | 1 | 1 | 1 | 1 | |
| 47 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| -23 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
| 24 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 |
Fixed-point
Encode 19.25 and -19.25 in 11-bit (8.3) two's-complement fixed-point notation
| Decimal | -128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | ½ | ¼ | ⅛ |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 19.25 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 |
| -19.25 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
Fixed-point numbers have simple arithmetic but limited range and precision; Floating-point numbers require more complex calculations but have much much greater range and precision
Floating-point
- our specification has non-IEEE-754-compliant floating-point numbers -
Activities
Using 8.3 two's-complement fixed-point notation, encode 3.625
| Decimal | -128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | ½ | ¼ | ⅛ |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 3.5625 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 |
Using 5e4 two's-complement floating-point notation, encode 1.625 and -3.25
| Decimal | -1 | ½ | ¼ | ⅛ | ¹̷₁₆ | Exponent | -8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|---|---|---|---|
| 1.625 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | |
| -3.250 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
Using 4e4 two's-complement floating-point numbers, decode 01010110 and 01001110
| Mantissa | -1 | ½ | ¼ | ⅛ | Exponent | -8 | 4 | 2 | 1 | Decimal |
|---|---|---|---|---|---|---|---|---|---|---|
| 0.625 | 0 | 1 | 0 | 1 | 6 | 0 | 1 | 1 | 0 | 40.000 |
| 0.500 | 0 | 1 | 0 | 0 | -2 | 1 | 1 | 1 | 0 | 0.125 |
Normalisation
What are the advantages of normalisation?
The advantages of normalising floating-point numbers are that it ensures there is a single canonical representation of any representable number, and any normalised floating-point number is guaranteed to have the highest precision possible
P, Q, and R are 7e5 two's-complement floating-point binary numbers.
P = 101100110001, Q = 110100110011
State which of P and Q is normalised. Give a reason for your answer.
- Q is not normalised.
- Its mantissa (1101001) is equal to -0.359375. This is outside the range [-1, -0.5) (and its mantissa is greater than -16) and therefore the number is not normalised
The binary number R is not normalised. Write R in normalised form. You must show your working.
R = 000110100101
- Mantissa = 0001101 = 0.203125
- 0.203125 not in [0.5, 1)
- 0.203125 × 2 = 0.40625
- 0.40625 not in [0.5, 1)
- 0.203125 × 4 = 0.8125
- 0.8125 in [0.5, 1)
- Mantissa = 0.8125 = 0110100
- Exponent = 00101 = 5
- Mantissa was multiplied by 4, so subtract 4
- Exponent = 1 = 00001
- R = 011010000001
Using 5e4 two's-complement floating-point representation, normalise 000110010 and 111000110. Show your working.
- Mantissa = 00011 = 0.1875
- 0.1875 not in [0.5, 1)
- 0.1875 × 2 = 0.375
- 0.375 not in [0.5, 1)
- 0.1875 × 4 = 0.75
- 0.75 in [0.5, 1)
- Mantissa = 0.75 = 01100
- Exponent = 0010 = 2
- Mantissa was shifted by 2, so subtract 2
- Exponent = 0 = 0000
- Normalised number: 011000000
- Mantissa = 11100 = -0.25
- -0.25 not in [-1, -0.5)
- -0.25 × 2 = -0.5
- -0.5 not in [-1, -0.5)
- -0.25 × 4 = -1
- -1 in [-1, -0.5)
- Mantissa = -1 = 10000
- Exponent = 0110 = 6
- Mantissa was shifted by 2, so subtract 2
- Exponent = 4 = 0100
- Normalised number: 100000100
Show the denary number -5.25 in two's-complement floating-point binary form, representing the mantissa and exponent in as few bits as possible
- -101.01 = 1010.11 = 1.01011 × 211
- 101011011 (6.3)