plasmacel - 2 years ago 89
C++ Question

# Invertability of IEEE 754 floating-point division

What is the invertability of the IEEE 754 floating-point division? I mean is it guaranteed by the standard that if

`double y = 1.0 / x`
then
`x == 1.0 / y`
, i.e.
`x`
can be restored precisely bit by bit?

The cases when
`y`
is
`infinity`
or
`NaN`
are obvious exceptions.

Yes, there are IEEE 754 double-precision(*) values `x` that are such `x != 1.0 / (1.0 / x)`.
It is easy to build an example of a normal value with this property by hand: the one that's written `0x1.fffffffffffffp0` in C99's hexadecimal notation for floating-point values is such that `1.0 / (1.0 / 0x1.fffffffffffffp0) == 0x1.ffffffffffffep0`. It was natural to expect `0x1.fffffffffffffp0` to be a counter-example because `1.0 / 0x1.fffffffffffffp0` falls at the beginning of a binade, where floating-point numbers are less dense, so a larger relative error had to happen on the innermost division. More precisely, `1.0 / 0x1.fffffffffffffp0` falls just above the midpoint between `0.5` and its double-precision successor, so that `1.0 / 0x1.fffffffffffffp0` is rounded up to the successor of 0.5, with a large relative error.
In decimal `%.16e` format, `0x1.fffffffffffffp0` is `1.9999999999999998e+00` and `0x1.ffffffffffffep0` is `1.9999999999999996e+00`.