andydude Wrote:If we choose to hold \( a [N-1] b = a [N] (\text{hy}N\text{log}_a(b) + 1) \) to be true for all N, then the natural consequence is that \( a [0] b = b + 1 \) for all a ... .
If we choose to hold \( a [N-1] a = a [N] 2 \) true for all N, then we get \( a [0] b = b + 2 \) 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.