andydude Wrote:If we choose to hold to be true for all N, then the natural consequence is that for all a ... .

If we choose to hold true for all N, then we get for a == b. Both cannot be true, so we must make a choice.

Hm, the first one (demanding that a[N-1](a[N]b)=a[N](b+1) for all real b and integer N) seems to be the pure approach.

Interestingly the derivation of a previous operation seems not to depend on an intial condition like a[N]1=a (which is not true for N=1, but for all N>1).

But we just could conclude from a[N-1](a[N]b)=a[N](b+1) that a[0]b=b+1. And we can further conclude from

a[-1](a[0]b)=a[0](b+1)

that

a[-1](b+1)=a[0](b+1)

hence

a[-1]b=a[0]b=b+1 moreover by induction that a[N]b=b+1 for all N<0.