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 A new set of numbers is necessary to extend tetration to real exponents. marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 03/19/2015, 12:47 AM (This post was last modified: 03/19/2015, 02:57 AM by marraco.) $a =\, ^{1}{a} \, =\, ^{\frac{n}{n}}{a} \,=\, ^{\frac{1}{n_n}+\frac{1}{n_{n-1}}+...+\frac{1}{n_1}}{a} \,=\, ^{n}({^{\frac{1}{n}}{a}})$ 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. « Next Oldest | Next Newest »

 Messages In This Thread A new set of numbers is necessary to extend tetration to real exponents. - by marraco - 03/11/2015, 07:56 PM RE: A new set of numbers is necessary to extend tetration to real exponents. - by marraco - 03/12/2015, 06:04 PM RE: A new set of numbers is necessary to extend tetration to real exponents. - by marraco - 03/12/2015, 10:58 PM RE: A new set of numbers is necessary to extend tetration to real exponents. - by marraco - 03/12/2015, 11:10 PM RE: A new set of numbers is necessary to extend tetration to real exponents. - by sheldonison - 03/14/2015, 10:37 PM RE: A new set of numbers is necessary to extend tetration to real exponents. - by marraco - 03/19/2015, 12:47 AM RE: A new set of numbers is necessary to extend tetration to real exponents. - by sheldonison - 03/19/2015, 09:59 PM RE: A new set of numbers is necessary to extend tetration to real exponents. - by marraco - 03/19/2015, 10:45 PM

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