Inverse confusion about floating point precision

[Dan Bishop]

Skip Montanaro wrote:
> I understand why the repr() of float("95.895") is
"95.894999999999996".
> What I don't understand is why if I multiply the best approximation
to
> 95.895 that the machine has by 10000 I magically seem to get the lost
> precision back.  To wit:
>
>     % python
>     Python 2.3.4 (#12, Jul  2 2004, 09:48:10)
>     [GCC 3.3.2] on sunos5
>     Type "help", "copyright", "credits" or "license" for more
information.
>     >>> 95.895
>     95.894999999999996
>     >>> 95.895 * 10000
>     958950.0
>
> Why isn't the last result "958949.99999999996"?  IOW, how'd I get
back the
> lost bits?

You were just lucky.

The floating-point representation of 95.895 is exactly 6748010722917089
* 2**-46.

Multiplying by 10000 gives you 67480107229170890000 * 2**-46.  But
floats can have only 53 significant bits, so this gets normalized to
8237317776998399.658203125 * 2**-33 and rounded to 8237317776998400 *
2**-33, which happens to be exactly equal to 958950.

For analogy, consider a decimal calculator with only 3 significant
digits.  On this calculator, 1/7=0.143, an error of 1/7000.
Multiplying 0.143 by 7 gives 1.001, which is rounded to 1.00, and so
you get an exact answer for 1/7*7 despite roundoff error in the
intermediate step.