Operations on inf



On Fri, 9 Mar 2007, Jay Belanger wrote:
> "Andrey G. Grozin" <A.G.Grozin at inp.nsk.su> writes:
>> 1^inf = exp(inf*log(1)) = exp(inf*0) = exp(und) = und
> I don't know why this is convincing.
> Given a choice between preserving the identity
>  1^x = 1 for all x
> or the identity
>  a^b = exp(b*log(a)) for some a, all b
> I would think the former is more basic and worthy of being preserved.
Following your reasonong,
0*x = 0 for all x
also should be preserved.

I think that the status of the following two rules is *exactly* the same:

0*inf = und
1^inf = und

The second one follows from the first one by taking exp().
The first one follows from the second one by taking log().

If, following your line of reasoning that 1 is *exactly* 1, not something 
that tends to 1 in the limit being considered, we declare that 0 is 
*exactly* 0, not something that tends to 0, we have to conclude that

0*inf = 0

I don't think this is good.

Andrey