On Mon, Jun 28, 2010 at 9:54 PM, Alex Rufon <[email protected]> wrote:
> I actually smiled when he wrote:
> In general, when an Iverson-bracketed statement is false, we want it to 
> evaluate into
> a "very strong 0," namely a zero so strong that it annihilates anything it is 
> multiplied by-even if
> that other factor is undefined.

I wonder if there are many significant cases where this matters which
do not involve 0 divided by 0?  [I imagine there are, but I am drawing
a blank when I try to think of them.]

(I was also wondering about 1.17 where he apparently ignores
negative numbers.)

-- 
Raul
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