01/09/2016, 06:20 AM
(01/08/2016, 06:26 PM)marraco Wrote: .... No proof, but numerically:
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{(^{r+1}a)^{^{-(r+1)}a}=(^{-r}a)^{^{r}a}} \)
ok, first lets define y and z as follows:
\( y=\; ^{-(r+1)}a \;\;\; \)
\( z=\; ^{r}a \;\;\; \)
Then substitute these values of y and z into the Op's equation above, noting that
\( ^{(r+1)}a = a^z\;\;\;\; ^{-r}a = a^y \)
\( (a^z)^{y}=(a^y)^z\;\;\; \) This is the Op's equations with the substitutions
\( (a^z)^{y}=(a^y)^z=a^{(y\cdot z)}\;\;\; \) this equation holds for all values of a,y,z
- Sheldon