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 Zeration = inconsistant ? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 10/01/2014, 08:40 AM In the discussions about zerations and hyperoperators I showed , together with many others , that many proposed equations lead to severe paradoxes such as a = a + 1 for all real a. Also I advocated exp^[a](ln^[a](x) + ln^[a](y)) over the usual +,*,^ because it is more consistant with the concepts of commutativity and superfunctions. This , which I related to the tommy's extended distributive property ( https://sites.google.com/site/tommy1729/...e-property ) , gives for a = 0 ( zeration ) the suggestion that zeration is the identity function. Another line of thinking is that the superfunction of x+1 = x+1. f(x) = x+1 f( f^[-1](x) + 1 ) = x + 1 Q.E.D SO expecting zeration different from the id(x) and having a distinction between the successor function (x+1) or having a distributive property seems unrealistic. Also the idea of an analytic zeration seems even more difficult. ALthough I like the Max+ algebra , I do not see a (good) zeration in it. Another thing : To understand tetration you need to understand +,*,^ at least. To understand ^ you need to understand *. To understand * you need to understand +. defining and understanding a hyperoperator seems to require the lower hyperoperators. But with zeration that is A PROBLEM. zeration tries to define itself in terms of HIGHER operators. There are NO lower operators defined than zeration so its seems necc. Any attempt at trying to define zeration with lower hyperoperations , requires those lowers to be defined as well by even lower hyperoperations ? That would PROBABLY give an infinite descent problem. However properties like distributive require use of a lower hyperoperator : a * (b+c) = a*b + a*c. UNLESS , and now we are getting at the heart of the post , a \$0\$ (b + c) = a \$0\$ b + a \$0\$ c. However after many attempts that also seems to lead to paradoxes. What do most proponents of zeration want ? Its USUALLY this for some constant C : a \$0\$ a = a + C The problem then becomes : (2a) \$0\$ (2a) = (a+a) \$0\$ (a+a) = a \$0\$ a + a \$0\$ a + a \$0\$a + a \$0\$ a = 4a + 4C. However we should have gotten 2a \$0\$ 2a = 2a + C. SO it seems \$0\$ , + do not have the distributive property. THEREFORE it seems pretty hard to define a \$0\$ b = b \$0\$ a = ... in a continuous way. Afterall it is typical for math to do computations and definitions based on properties/structures. Zeration seems to lack properties and therefore consistancy. regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread Zeration = inconsistant ? - by tommy1729 - 10/01/2014, 08:40 AM RE: Zeration = inconsistant ? - by MphLee - 10/01/2014, 11:27 AM RE: Zeration = inconsistant ? - by GFR - 10/02/2014, 02:44 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/02/2014, 09:27 PM RE: Zeration = inconsistant ? - by MphLee - 10/02/2014, 10:02 PM RE: Zeration = inconsistant ? - by GFR - 10/04/2014, 09:58 AM RE: Zeration = inconsistant ? - by tommy1729 - 10/02/2014, 10:58 PM RE: Zeration = inconsistant ? - by GFR - 10/03/2014, 11:29 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 12:11 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/02/2014, 11:11 PM RE: Zeration = inconsistant ? - by GFR - 10/03/2014, 11:39 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 12:12 PM RE: Zeration = inconsistant ? - by MphLee - 10/03/2014, 09:20 AM RE: Zeration = inconsistant ? - by tommy1729 - 10/03/2014, 09:32 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/03/2014, 09:41 PM RE: Zeration = inconsistant ? - by GFR - 10/04/2014, 10:19 AM RE: Zeration = inconsistant ? - by MphLee - 10/04/2014, 11:24 AM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 12:16 PM RE: Zeration = inconsistant ? - by MphLee - 10/04/2014, 12:58 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 10:20 PM RE: Zeration = inconsistant ? - by MphLee - 10/05/2014, 03:36 PM

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