"Ph. Marek" <[EMAIL PROTECTED]> writes:
>> This is obviously some new definition of Inf of which I was not
>> previously aware.
> Well, after reading my sentence one more, I see what may have caused
> some troubles. Inf is not in N; but *in my understanding* it fits
> naturally as an extension to N, that is, Inf is (or can be) integer
> as is "after" N...
>
> This won't be written in math books, I know.
Actually, it's discussed to death in math books, but math books deal
with Inf in ways Perl isn't prepared to do. If you want to treat
infinities as numbers, then you have to be prepared to have different
infinities.
>> Also, if that were the case, 0..Inf would be a finite list. (It is
>> trivial to prove that 0..N is a finite list with finite cardinality
>> for all natural numbers N. So if you set N equal to Inf, 0..Inf would
>> have finite cardinality, if Inf is a natural number.)
>
> If I extend the natural numbers N with Inf to a new set NI (N with
> Inf)
The problem is, NI is not a group with respect to addition for any
definition of addition of which I am aware. Translated from mathese
to English, that means NI isn't terribly useful or meaningful or
interesting. J and R and C (i.e., the integers and the reals and the
complex numbers) are much more worthy of consideration, because they
form groups with respect to addition.
It is possible to construct a group that includes infinities, but NI
isn't it, and for Perl purposes it doesn't seem necessary.
--
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