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 extension of the Ackermann function to operators less than addition JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 11/06/2011, 08:06 PM I realize now we have to create a second axiom in order that the series of operators become the true Ackermann function. Consider the possibility that: $a\,\,\bigtriangleup_2\,\,2=a$ we still have the result that $a + (a\,\,\bigtriangleup_2\,\,2) = a\,\, \bigtriangleup_2\,\, (2 + 1) = a\,\,\bigtriangleup_2\,\,3$ the only difference is that $S(2) = 2$ and in return we get $a\,\,\bigtriangleup_2\,\,b = a \cdot (b-1)$ So in order to get the true Ackermann function we must make the second assertion: $n\,\,\in\,\,\mathbb{Z};n \ge 2$ $S(n) = 1$ This is actually the equivalent to the iteration axiom: $a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_\sigma\,\,(a_2\,\,\bigtriangleup_\sigma\,\,(a_3\,\, \bigtriangleup_\sigma \,\,...(a_{n-1}\,\,\bigtriangleup_\sigma\,\,a_n)$ though only true for integer values of sigma greater than or equal to two. Therefore we define the Ackermann function from axioms: $\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b$ such that $a\,\,\bigtriangleup_\sigma\,\,(a\,\,\bigtriangleup_{\sigma + 1}\,\,b) = a\,\,\bigtriangleup_\sigma\,\,(b+1)$ $a\,\,\bigtriangleup_\sigma\,\,S(\sigma) = a$ $n\,\,\in\,\,\mathbb{Z}\,\,;\,\,n \ge 2$ $S(n) = 1$ $a\,\,\bigtriangleup_1\,\,b = a + b$ Does anyone see any modifications necessary? « Next Oldest | Next Newest »

 Messages In This Thread extension of the Ackermann function to operators less than addition - by JmsNxn - 11/06/2011, 01:31 AM RE: extension of the Ackermann function to operators less than addition - by JmsNxn - 11/06/2011, 04:56 PM RE: extension of the Ackermann function to operators less than addition - by JmsNxn - 11/06/2011, 08:06 PM

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