plasmacel - 10 months ago 49

C++ Question

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`

`x == 1.0 / y`

`x`

The cases when

`y`

`infinity`

`NaN`

Answer

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`

.

(*) there is no reason for the inverse function to have the property in the question for any of the IEEE 754 format

Source (Stackoverflow)