01/08/2016, 11:05 PM
(01/08/2016, 06:26 PM)marraco Wrote: This gives a natural way to define °a for the Z branch. °a should be the value that makes:I retract that. The last equation is satisfied no matter what the value of °a is.
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{^{-1}a\,^{^1a}\,=\,^1a\,\,^{^{-1}a}\,=\,^{0}a\,^{^0a}\, \; or \; (log_a{^0a})\,^{(a^{(^0a))}=(a^{(^0a)})\,^{(log_a{^0a})} \)
then it also satisfies:
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{(^{r+(n+1)}a)^{^{-(r+(n+1))}a}=(^{-(r+n)}a)^{^{r+n}a}} \)
I have the result, but I do not yet know how to get it.

