On Thu, 24 Aug 2023, David Hough CFP wrote: > ... in the case of overflow, both results are the same, infinity. > That means, in this case, as with an exact zero result, |x.h| = |x.t|. Yes and Yes. > But I still suppose that the algorithm I proposed never yields |x.h| < |x.t|. Yes. Looks like it. Thanks - Damian