not really. It's rather, any factor with a denominator that is not in the part of the set 2^n, (or with decimal floating point, 2^n * 5^m)
They should be paying attention to the lumberjack project
Posted May 31, 2012 0:19 UTC (Thu) by nybble41 (subscriber, #55106) [Link]
> not really. It's rather, any factor with a denominator that is not in the part of the set 2^n, (or with decimal floating point, 2^n * 5^m)
I was referring to the prime factors; any number of the form 2^n * 5^m with naturals n and m (!= 0) has prime factors 2 and 5. A number with a prime factor of 3 or 7, for example, would not terminate in either binary or decimal. However, your description is equivalent.
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