2 Apr
2010
2 Apr
'10
4:30 p.m.
On 2 April 2010 02:48, Gottlob Frege <gottlobfrege@gmail.com> wrote:
2. 1/inf = 0 *IS* exact. It is inexact while approaching inf, but finally exactly 0 'at' infinity.
Arguably, since we're dealing in round-toward-0 division, even 1/2 is exactly 0. The question I have then is whether at infinity the remainder somehow manages to disappear. Care to shed some light on that? \forall x > 1, divrem(1,x) = (0, 1), so as x -> inf, it'd still be (0, 1). I think the idea of "inexact" zeros came from the idea that 1/0 would give infinity, where you'd then want a "-0" so that 1/-0 can give negative infinity (like in floating point). I think I was advocating that at one point, but have since some to my senses.