Re: Is infinity equal to infinity?

Michael Rubenstein writes:


>>However, if the numerator converges
>>on some non-zero value c, the ratio is +/-infinity. Therefore,
>>in IEEE arithmetic x/0 is +/-infinity for any finite, non-zero x.
>
>let x(i) = 1; y(i) = (-1)^i/i
>
>then lim(i->inf) x(i) = 1
> lim(i->inf) y(i) = 0
> lim(i->inf) x(i) / y(i) = ?
>
>[I think the word "continuous" got lost somewhere in there. -John]


Actually it was not the idea of continuity that got lost. A similar
example could be given taking x(t) = 1 and y(t) = sin(t)/t. Note that
y(t) is continuous for all nonzero real t, and lim(t->inf) y(t) = 0 from
a standard mathematical viewpoint, in which 0 is *unsigned*.


What actually got lost was the fact that in IEEE 754 arithmetic, all
zeros are *signed*. Therefore in both the quoted example and in my
example, if zeros must be signed, then lim y does not exist! Of
course lim x/y does not exist either, but that fact then ceases to be
problematic.


This all goes to show that the projective mode, having one unsigned 0
and one unsigned infinity, does have its uses. In that mode (which
unfortunately does not appear in IEEE 754 itself, although it did appear
in some drafts of the standard), all of the limits discussed above exist.


David Cantrell
--

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