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 the logical hierarchy tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/08/2009, 04:16 PM here i present what is according to me the " logical hierarchy " i found it important to say , because it appears often on math forums and is usually stated or ordered in a way i disagree with. no tetration or non-commutativity. no ' popularized ' ackermann or buck. but plain good old logic in my humble opinion. most imporant is the existance of a single neutral element f_n ( a , neutral ) = f_n ( neutral , a ) = a for all a ! 1) a + b 2) a * b 3) a ^ log(b) to see how i arrived at 3 : a ^ log(b) = b ^ log(a) = exp( log(a) * log(b) ) 4) exp ( log(a) ^ log(log(b)) ) to see how i arrived at 4 : note that 3) is used upon log(a) and log(b). etc etc note that the neutral elements are 1) addition -> 0 2) multiplication -> 1 3) a ^ log(b) -> e 4) -> e^e 5) -> e^e^e 6) -> e^e^e^e etc bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/08/2009, 10:25 PM (This post was last modified: 02/08/2009, 10:25 PM by bo198214.) Ya this hierarchy was already considered. The main observation is that the real numbers with operations $a+b$ and $a*b$ are isomorphic to the positive real numbers with $a*b$ and $a^{\log(b)}$ (The isomorphism is $\exp$). I.e. we dont add really something new. Each two consecutive operations are isomorphic (i.e. behave completely the same as) to + and *. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/08/2009, 10:45 PM bo198214 Wrote:Ya this hierarchy was already considered. The main observation is that the real numbers with operations $a+b$ and $a*b$ are isomorphic to the positive real numbers with $a*b$ and $a^{\log(b)}$ (The isomorphism is $\exp$). I.e. we dont add really something new. Each two consecutive operations are isomorphic (i.e. behave completely the same as) to + and *. right. but some people always insist on a + b a * b a ^ b which is wrong. glad we agree. thread closed ? regards tommy1729 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/08/2009, 10:53 PM tommy1729 Wrote:but some people always insist on a + b a * b a ^ b which is wrong. Its not wrong its a different hierarchy. It follows the pattern: a[n+1](b+1)=a[n](a[n+1]b) where [n] is the the nth operation. For example: a[2](b+1)=a[1](a[2]b) which corresponds to a*(b+1)=a+a*b a[3](b+1)=a[2](a[3]b) which corresponds to a^(b+1)=a*(a^b) and so tetration [4] satisfies: a[4](b+1)=a[3](a[4]b) « Next Oldest | Next Newest »

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