Nope.
Put another way, consider a proper, irreducible fraction of the form N/D. The value is exactly representable in binary (meaning with a finite number of bits to the right of the binary point) if an only if D = 2^p, where p = 1, 2, 3, 4, 5, ….
Nope.
Put another way, consider a proper, irreducible fraction of the form N/D. The value is exactly representable in binary (meaning with a finite number of bits to the right of the binary point) if an only if D = 2^p, where p = 1, 2, 3, 4, 5, ….