On Fri, Jun 24, 2016 at 01:33:46AM +0200, Timon Gehr via Digitalmars-d wrote: > On 24.06.2016 00:53, H. S. Teoh via Digitalmars-d wrote: > > > >Because 0^^0 = 1, and 1 is representable. > > > > > > > >E.g. n^^m counts the number of functions from an m-set to an n-set, > > > >and there is exactly one function from {} to {}. > > This argument only works for discrete sets. > > No, it works for any cardinals n and m.
It doesn't. What is the meaning of m-set and n-set when m and n are real numbers? How many elements are in a pi-set? How many functions are there from a pi-set to an e-set? Your "number of functions" argument only works if m and n are discrete numbers. > > If n and m are reals, you'd need a different argument. > > > > I don't want to argue this at all. x^^0 is an empty product no matter > what set I choose x and 0 from. And 0^^x is a non-empty product when x != 0. So why should we choose the limit of x^^0 as opposed to the limit of 0^^x? > 0^^0 = 1 is the only reasonable convention, and this is absolutely > painfully obvious from where I stand. What context are you using 'pow' > in that would suggest otherwise? When computing the limit of x^y as x->0? > Also, Andrei's implementation explicitly works on integers anyway. Even for integers, the limit of x^y as x->0 is 0. My point is that the choice is an arbitrary one. It doesn't arise directly from the mathematics itself. I understand that 0^0 is chosen to equal 1 in order for certain theorems to be more "aesthetically beautiful", just like 0! is chosen to equal 1 because it makes the definition of factorial "nicer". But it's still an arbitrary choice. T -- Those who don't understand Unix are condemned to reinvent it, poorly.