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 (2**N)

as

J ~= 2**N / 10

and recalling that J has exactly 53 bits (is >= 2**52 but < 2**53), the best value for N is 56:

>>> 2**52
4503599627370496
>>> 2**53
9007199254740992
>>> 2**56/10
7205759403792793

That is, 56 is the only value for N that leaves J with exactly 53 bits. The best possible value for J is then that quotient rounded:

>>> q, r = divmod(2**56, 10)
>>> r
6

Since the remainder is more than half of 10, the best approximation is obtained by rounding up:

>>> q+1
7205759403792794

Therefore the best possible approximation to 1/10 in 754 double precision is that over 2**56, or

7205759403792794 / 72057594037927936

Note that since we rounded up, this is actually a little bit larger tha