02/14/2008, 10:58 PM
But see Gianfranco, the postulate
aob = max{a,b}+1 for a!=b
aob = a+2=b+2 for a=b
does not follow from your initial rule
aoa=a+2
nor does it follow from the extended set of rules
ao(a+x)=a+x+1, x>0
I am not completely sure whether it follows from the last rules, but even if it would, the restriction x>0 is arbitrary, imho.
Finding a unique continuous extension is already as we have seen from tetration a nearly hopeless task. But *without* continuity as you propose for zeration, I think you can construct arbitrary nasty solutions to the functional equation ao(a+x)=a+x+1.
aob = max{a,b}+1 for a!=b
aob = a+2=b+2 for a=b
does not follow from your initial rule
aoa=a+2
nor does it follow from the extended set of rules
ao(a+x)=a+x+1, x>0
I am not completely sure whether it follows from the last rules, but even if it would, the restriction x>0 is arbitrary, imho.
Finding a unique continuous extension is already as we have seen from tetration a nearly hopeless task. But *without* continuity as you propose for zeration, I think you can construct arbitrary nasty solutions to the functional equation ao(a+x)=a+x+1.