on floating-point numbers

[Hope Rouselle]

"Peter J. Holzer" <hjp-python at hjp.at> writes:

> On 2021-09-05 03:38:55 +1200, Greg Ewing wrote:
>> If 7.23 were exactly representable, you would have got
>> 723/1000.
>> 
>> Contrast this with something that *is* exactly representable:
>> 
>> >>> 7.875.as_integer_ratio()
>> (63, 8)
>> 
>> and observe that 7875/1000 == 63/8:
>> 
>> >>> from fractions import Fraction
>> >>> Fraction(7875,1000)
>> Fraction(63, 8)
>> 
>> In general, to find out whether a decimal number is exactly
>> representable in binary, represent it as a ratio of integers
>> where the denominator is a power of 10, reduce that to lowest
>> terms,
>
> ... and check if the denominator is a power of two. If it isn't (e.g.
> 1000 == 2**3 * 5**3) then the number is not exactly representable as a
> binary floating point number.
>
> More generally, if the prime factorization of the denominator only
> contains prime factors which are also prime factors of your base, then
> the number can be exactle represented (unless either the denominator or
> the enumerator get too big). So, for base 10 (2*5), all numbers which
> have only powers of 2 and 5 in the denominator (e.g 1/10 == 1/(2*5),
> 1/8192 == 1/2**13, 1/1024000 == 1/(2**13 * 5**3)) can represented
> exactly, but those with other prime factors (e.g. 1/3, 1/7,
> 1/24576 == 1/(2**13 * 3), 1/1024001 == 1/(11 * 127 * 733)) cannot.
> Similarly, for base 12 (2*2*3) numbers with 2 and 3 in the denominator
> can be represented and for base 60 (2*2*3*5), numbers with 2, 3 and 5.

Very grateful to these paragraphs.  They destroy all the mystery.