\( a =\, ^{1}{a} \, =\, ^{\frac{n}{n}}{a} \,=\, ^{\frac{1}{n_n}+\frac{1}{n_{n-1}}+...+\frac{1}{n_1}}{a} \,=\, ^{n}({^{\frac{1}{n}}{a}}) \)
So, we can define the n\( ^{^{^{th}}} \) root of "a" this way
\( c \,=\, ({^{\frac{1}{n}}{a}}) \,\Leftrightarrow\, ^nc\,=\, a \)
Tetration exponent product is non commutative. In general:
\( ^n(^{\frac{1}{m}}a) \,\neq\, ^{\frac{1}{m}}(^na) \)
The inequality stands as long as m ≠ n, so if a product is defined to operate on tetration exponents, it is not the common product, or it do not operates on the complex field, or results are multivalued to preserve the equality.
So, we can define the n\( ^{^{^{th}}} \) root of "a" this way
\( c \,=\, ({^{\frac{1}{n}}{a}}) \,\Leftrightarrow\, ^nc\,=\, a \)
Tetration exponent product is non commutative. In general:
\( ^n(^{\frac{1}{m}}a) \,\neq\, ^{\frac{1}{m}}(^na) \)
The inequality stands as long as m ≠ n, so if a product is defined to operate on tetration exponents, it is not the common product, or it do not operates on the complex field, or results are multivalued to preserve the equality.