How many bits to represent these numbers precisely?

Question

Consider the following numbers:

$$19=10011_b, 12.75=1100.11_b, 7.125=111.001_b$$

What is the minimum number of bits necessary to represent the above three numbers precisely?

A system like the IEEE floating point standard should look like

$$\pm1.b_1\dots b_5 \times 2^p$$

where $2\le p\le 4$ so that there is one sign bit, one "free" bit (the leading $1$), and five bits for $b_1\dots b_5$. Then we need eight values for $p$, so that gives additional three bits. So does this mean that to represent the above numbers one would need at least a 9-bit floating point system? I think it's somewhat dubious, seems like too many bits would be needed for these simple numbers. Where's my mistake?

Answer 1

The answer depends on what you want to represent and the data format.

• To represent them precisely requires zero bits, under the data format that "this format represents 19, 12.75, and 7.125".

• If you need a that contains a value that is one of these three numbers, then two bits is enough, using the data format "00 means 19, 01 means 12.75, and 10 means 7.125".

• If you need an array of five such values, you can make do with 8 bits, by reinterpreting the entries as a base-3 numeral for a number $0 \leq n < 243$, which can be stored in binary using 8 bits.

• If you need a IEEE-like floating point format capable of exactly representing these three values under the usual interpretation of such formats, then you actually need 10 bits: you forgot the sign bit for the exponent.

• You could modify the IEEE-format in various ways as well. For example, you might choose a format where "the sign is always positive and the exponents are always in the range $1 \leq e \leq 4$". Then seven bits would suffice.

Answer 2

$\quad\quad 19=10011_b, \;\;12.75=1100.11_b, \;\;7.125=111.001_b$

There are $\,2^8\,$ distinct values between $\,00000.000_b\,$ and $\,11111.111_b\,$, which take $\,8\,$ bits to represent. Add a sign bit, and that's the $\,9\,$ bits you counted.

However, if you literally just need to represent positive binary values between $\,111.001_b\,$ and $\,10011.000_b\,$, then that's just $\,12 \cdot 2^3=96\,$ distinct values, which would only require $\,\lceil \log_2 96 \rceil=7\,$ bits with the appropriate encoding.