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 Zeration = inconsistant ? tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 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 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 10/01/2014, 11:27 AM (This post was last modified: 10/01/2014, 03:39 PM by MphLee.) Everything depend on the choice of the recursive definition used to define the Hyperoperations sequence. Using a classic definition over the naturals (starting from the addition) leaves an opening for a non-trivial solution for zeration below the addition because if we try to define it starting from the higher Hos. (the addition) we have to use the cutoff subtraction ${-}^{*}$ (not defined if $a\lt n$ ). $a[0]n=(a+((a{-}^{*}n)+1))$ This means that for the recursive definition of the hyperoperation sequence on the naturals we have an infinite ammount of functions that "models" Zeration (some are commutative!) and each of those contains the successor function: in other worlds, those models should behave as the successor function over a restricted subset of $\mathbb{N} \times \mathbb{N}$). Rubtsov and Romerio tried to reach the uniqueness of the solution by introducing a different recursive picewise definition of the Hyperoperations sequence. Inside their operation family (Rubtsov-Romerio Hyperoperation Family) Zeration happens to unique and coincides with the classic Zeration (the one defined by Rubtsov). Anyways taking the Goodstein definition of the Hyperoperations, Zeration is defined as the successor. $G(0,a,n):=n+1$ The problems arise when we want to extend the Hyperoperations to the reals, and we want to extend the validity of the recursion step formula to them: as you noted the sub-function of the successor is the successor itself. $S \circ S \circ S^{\circ -1}=S \circ (S \circ S^{\circ -1})=S \circ Id=S$ And, for every function that commutes with the successor (like the addition), its subfucntion coincides with the successor: $f \circ S \circ f^{\circ -1}=(f \circ S )\circ f^{\circ -1}=(S\circ f )\circ f^{\circ -1}=S\circ( f \circ f^{\circ -1})=S \circ Id=S$ Some choices, thus, "forces", in some sense, Zeration to be trivial (the successor function). But we could aswell change our concept of Hyperoperation and consider the Commutative hyperoperations (like the Bennet's family). In that case there is a binary operations on the real below the addition. About this I have already said something in your thread about the distributive property: Quote:As you know these are a generalization of the Bennet Hyperoperations (Commutative Hyperoperations). I usually use this notation for them because i find it very confortable $a \odot_r^k b:={\exp}_k^{\circ r}(log_k^{\circ r}(a)+log_k^{\circ r}(b))$ Bennet Hyperoperations are a special case of these (with the natural base $a \odot_r^e b$) $\odot_0^{K} =+$ $\odot_1^{K} =\cdot$ and $a \odot_{-1}^{+\infty} b$ is the max operation while $a \odot_{-1}^{0} b$ is the min operation(this limit process is related with the litinov-maslov dequantization of the semifield of non-negative real numbers in to the Tropical semifield $\mathbb{T}_{max}$) from: http://math.eretrandre.org/tetrationforu...hp?tid=520 MathStackExchange account:MphLee GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 10/02/2014, 02:44 PM (This post was last modified: 10/02/2014, 08:30 PM by GFR.) Hi, Tommy 1729! (10/01/2014, 08:40 AM)tommy1729 Wrote: 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. tommy1729 Dear Tommy1729, Hi again! Concerning the supposed inconsistency of zeration, due to the non-validity of the distributive property at the ZER-PLUS levels, because it is definable via higher rank hyperoperations, I should like to start by clarifying that there seems to be a peculiar double misunderstanding in your above-mentioned example. In fact, I suppose by mis-attention, you have mentioned a correct relationship, together with a wrong one, such as: a * (b + c) = a*b + a*c correct! where the stronger operation ‘distributes’ over the weaker, as well as a wrong ‘equality’ such as: a [0] (b + c) = a [0] b + a [0]c, impossible, (first misunderstanding) because [0] is weaker than [+]. Let us see how this works in the framework of the ‘Field’ of ‘Real Numbers’ and, therefore, at the 0 - 3 ranks of the HOH (Hyper-Operations Hierarchical family). Actually, let us start at the 1 / 2 ranks (the PLUS-TIME level), where we have: a * (b + c) = a*b + a*c, correct, because TIMES (stronger) distributes over PLUS, but: a + b*c >< (a + b) * (a +c) the equality would be wrong, because PLUS is weaker. [It would also do it, but in the idempotent Boolean Algebra, not in the ‘Real Field’]. Considering, now, the 0 / 1 ranks (the ZER-PLUS level), we have (sorry for the brackets), with [0]=o: a + (b o c) = (a + b) o (a + c) correct, PLUS ‘dominates’ ZER, but: a o (b + c) >< (a o b) + (a o c) the equality would be wrong, because ZER is ‘dominated’. Example - let’s put: a = 2, b = 4, c = 7, we get: 2 + (4 o 7) = 2 + 8 = 10; and: (2 + 4) o (2 + 7) = 6 o 9 = 10, etc., but: 2 o (4 + 7) = 2 o 11 = 12 and: (2 o 4) + (2 o 7) = 5 + 8 = 13, !!!!!!!! It is also useful to notice that, at the 2 / 3 ranks (the TIMES-POW level), we have: a ^ (b * c) >< (a ^ b) * (b ^ c) and: a * (b ^ c) >< (a * b) ^ (a * c) both wrong equalities, because the ‘level’ is not distributive. [In this case, for levels including ranks s > 2, the distributivity does not hold, even if the operations are defined, or described, by lower rank hyperoperations]. Second misunderstanding. Having said so, we may see that the PLUS-ZER distributivity holds, even if this would not be absolutely necessary to consider zeration as consistent, because (anyway) it doesn’hold at higher ranks. However, there is a problem about considering -oo (minus infinite) as the ‘unit element’ of zeration (like it also happens in the ‘tropical’ Max-Plus algebra), just because -oo is NaN (Not a Number) and it does not belong to the set of the ‘Reals’. We neeed to define the actual oo, different from the oo of the classical ‘Limit Theory’ of Mathematical Analysis. By the way, nobody ever said that there is zeration 'inside' Max-Plus algebra, but it is easy to see that the 'maximation' Max-Plus operation is a fundamental component of the RR-zeration definition. Thank you for your interest and your kind attention. Best regards. tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 10/02/2014, 09:27 PM (10/02/2014, 02:44 PM)GFR Wrote: Hi, Tommy 1729! (10/01/2014, 08:40 AM)tommy1729 Wrote: 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. tommy1729 Dear Tommy1729, Hi again! Concerning the supposed inconsistency of zeration, due to the non-validity of the distributive property at the ZER-PLUS levels, because it is definable via higher rank hyperoperations, I should like to start by clarifying that there seems to be a peculiar double misunderstanding in your above-mentioned example. In fact, I suppose by mis-attention, you have mentioned a correct relationship, together with a wrong one, such as: a * (b + c) = a*b + a*c correct! where the stronger operation ‘distributes’ over the weaker, as well as a wrong ‘equality’ such as: a [0] (b + c) = a [0] b + a [0]c, impossible, (first misunderstanding) because [0] is weaker than [+]. Let us see how this works in the framework of the ‘Field’ of ‘Real Numbers’ and, therefore, at the 0 - 3 ranks of the HOH (Hyper-Operations Hierarchical family). Actually, let us start at the 1 / 2 ranks (the PLUS-TIME level), where we have: a * (b + c) = a*b + a*c, correct, because TIMES (stronger) distributes over PLUS, but: a + b*c >< (a + b) * (a +c) the equality would be wrong, because PLUS is weaker. [It would also do it, but in the idempotent Boolean Algebra, not in the ‘Real Field’]. Considering, now, the 0 / 1 ranks (the ZER-PLUS level), we have (sorry for the brackets), with [0]=o: a + (b o c) = (a + b) o (a + c) correct, PLUS ‘dominates’ ZER, but: a o (b + c) >< (a o b) + (a o c) the equality would be wrong, because ZER is ‘dominated’. Example - let’s put: a = 2, b = 4, c = 7, we get: 2 + (4 o 7) = 2 + 8 = 10; and: (2 + 4) o (2 + 7) = 6 o 9 = 10, etc., but: 2 o (4 + 7) = 2 o 11 = 12 and: (2 o 4) + (2 o 7) = 5 + 8 = 13, !!!!!!!! It is also useful to notice that, at the 2 / 3 ranks (the TIMES-POW level), we have: a ^ (b * c) >< (a ^ b) * (b ^ c) and: a * (b ^ c) >< (a * b) ^ (a * c) both wrong equalities, because the ‘level’ is not distributive. [In this case, for levels including ranks s > 2, the distributivity does not hold, even if the operations are defined, or described, by lower rank hyperoperations]. Second misunderstanding. Having said so, we may see that the PLUS-ZER distributivity holds, even if this would not be absolutely necessary to consider zeration as consistent, because (anyway) it doesn’hold at higher ranks. However, there is a problem about considering -oo (minus infinite) as the ‘unit element’ of zeration (like it also happens in the ‘tropical’ Max-Plus algebra), just because -oo is NaN (Not a Number) and it does not belong to the set of the ‘Reals’. We neeed to define the actual oo, different from the oo of the classical ‘Limit Theory’ of Mathematical Analysis. By the way, nobody ever said that there is zeration 'inside' Max-Plus algebra, but it is easy to see that the 'maximation' Max-Plus operation is a fundamental component of the RR-zeration definition. Thank you for your interest and your kind attention. Best regards. Hi. Thanks for your swift reply. There are indeed some misunderstandings , but I think most are on your side. Since there are no lower operations than zeration defined , I could not write a distributive property where zeration was - what you call - STRONGER. Im unaware of a nice consistant definition of zeration such that a + ( b [0] c ) = ( a + b ) [0] ( a + c ). So I tried to reverse everything : define zeration in terms of higher operators , instead of lower , and define distributivity in the REVERSE way. It was therefore done intentionally ! NOT to try to ridicule or dismiss zeration but rather as an attempt to " fix it ". I understand it looks like a silly mistake but it was not. Have a bit more confidence in me Im not sure what you consider as the MUST HAVE PROPERTIES OF ZERATION. That matters alot if we want to talk about the same thing. It seems you too want distributivity ? It also appears you have already picked a particular interpretation of zeration ? So what definition of zeration do you use such that a + ( b [0] c ) = ( a + b ) [0] ( a + c ) for all reals ? That is the key question. Otherwise we might be talking about different things due to lack of consensus and agreed definitions. --- Secondly and I quote : By the way, nobody ever said that there is zeration 'inside' Max-Plus algebra, but it is easy to see that the 'maximation' Max-Plus operation is a fundamental component of the RR-zeration definition (end quote) Then if zeration is not within Max-Plus , why does Max-Plus matters to zeration or vice versa ? That is not clear to me. Also I do not know what "RR-zeration definition" is. The Max-Plus algebra is nice and I also like the amoebe. I discussed variants of those 2 with Timothy Golden , mainly in the direction of (algebraicly closed) higher dimensional numbers and uncountable sets. Amoebe set theory and Amoebe ring theory if you like. That in the far past, although Im still considering them today. " Amoebe numerical methods " is what I am currently considering. ( of course those terms are made up by me , Im unaware of official namings for these concepts or prior deep investigations ) My friend Mick had some ideas to apply Amoebe for physics. But that is background and Im going a bit offtopic. regards tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 10/02/2014, 10:02 PM (This post was last modified: 10/02/2014, 10:07 PM by MphLee.) $a[0]b:=max(a,b)+1$ $if a \neq b$ $a[0]b:=max(a,b)+2$ $if a = b$ This is the Rubtsov's definition of Zeration. It satisfies the distributivity. Addition distributes over the addition (easy to see). As I said before Quote:Using a classic definition over the naturals (starting from the addition) leaves an opening for a non-trivial solution for zeration below the addition because if we try to define it starting from the higher Hos. (the addition) we have to use the cutoff subtraction ${-}^{*}$ (not defined if $a\lt n$ ). $a[0]n=(a+((a{-}^{*}n)+1))$ This means that for the recursive definition of the hyperoperation sequence on the naturals we have an infinite ammount of functions that "models" Zeration (some are commutative!) and each of those contains the successor function: in other worlds, those models should behave as the successor function over a restricted subset of $\mathbb{N} \times \mathbb{N}$). the recursive definition of the hyperoperations over the naturals make us able to have alot of different solution for Zeration, the most interesting solutions are probably the ones definable via Max operator. example: Rubtsov's Zeration $a[0]_{RR}b:=max(a,b)+1$ $if a \neq b$ $a[0]_{RR}b:=max(a,b)+2$ $if a = b$ Trappman's Zeration $a[0]_{HT}b:=max(a+2,b+1)$ Both are possible models of Zeration inside the Rubtsov-Romerio Hyperoperations Sequence. Inside the classic sequence we could use $a[0]_{Classic}b:=max(a,b)+1$ Those three solutions are related witht he max operator. We could call them "tropical solutions for Zeration" maybe. MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 10/02/2014, 10:58 PM It feels a bit strange ... A so called new concept " zeration " being almost equivalent to max[a,b]. Max[a,b] does not seem so intresting as a function. Max[12,100] = Max[13,100] = Max[91,100] Nothing special. zeration also still seems undefined or inconsistant for complex arguments. (15 + 13 i) [0] (- 41 +91 i) = ?? Consider Trapmann's solution : Max(a+2,b+1) This fails to be commutative ! Also Max does not have AN INTRESTING DERIVATIVE. Also iterations of max are usually not intresting. So , no complex numbers , no intresting calculus , no intresting dynamics. Also there is not much algebra or geometry about Max. Max+ algebra being a big exception. It seems tetration is more intresting than zeration. Zeration seems to loose properties rather then to gain ? A possible solution to that is finding intresting properties afterall for max or finding another zeration. Or ... Finding addional must-have properties. 1) commutative and distributive solution : max(a,b) + kroneckerdelta(a,b) + 1 Notice this is not (even) differentiable near a = b. Can we do better ? Im sure there are many nice algorithms that use max. ( even without max+ algebra ) Maybe one of those could help us out. --- Assume we agree on zeration(a,b) = max(a,b) + kroneckerdelta(a,b) + 1. What would then be the next question ?? " -1-ation " ? abel functions / superfunctions of zeration ? numerical methods based on zeration ? ( not equal to those from max+ algebra ofcourse ... remember we want to talk about NEW things right ? ) regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 10/02/2014, 11:11 PM (This post was last modified: 10/02/2014, 11:15 PM by tommy1729.) But what bothers me most is that zeration , unlike addition and multiplication does not have an inverse !!! a - b , a / b a [0]^-1 b ??? Take the equations : a + b = c a = c - b a * b = c a = c/b max(a,b) + 1 = c a = max^-1(c - 1,b) ?? fail. Let C = c - 1. max(a,b) = C so a = C or b = C. (a-C)(b-C) = 0. => a [0]^-1 b => kroneckerdelta(a,C) kroneckerdelta(b,C) = 0 ?? now since C = a or b , that delta product is not very surprising and hardly a " computation ". this is the best we can do ?? --- Saying zeration is a new concept and writing max ... it seems ... you know. regards tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 10/03/2014, 09:20 AM (This post was last modified: 10/03/2014, 04:44 PM by MphLee.) You are right, Trappmann solution isn't commutative, but following the original Zeration Thread we can see that Trapmann did propose that solution in order to outline the non-uniqueness of the Rubtsov solution: there is a non-commutative solution too! By the way he later proposed an extension to the complex too, but i don't find this really interesting. The real questions should be: A-Why A1-Why, between the infinite solutions for Zeration, we should chose the non-trivial ones? A2-Is it an ideological and arbitrary choice? A3-Is the request of commutativity a sufficently strong push? A4-Is the existence of solutions definable via max operator a good reason for chosing a non-trivial solution? Comment: the superfunction problem seems to suggest that the solution should be trivial (the successor)...but the quest for a commutative solution seems enough interesting and the tropical solutions give something interesting imho. Note also that, even if the idempotency of max implies that if you iterate it nothing happens, the Rubtsov's and the Trapmann's solutions are subfunctions of the addition even if over a restricted domain (because they behave as the successor on that restricted domain). If we use tropical solutions we have an usefull tool: the Litinov-Maslov dequantization can be used to reach, imho, the zerative inverses through a limit process (im still thinking about this). ------------------------- B-How (Uniqueness) B1-Once we have chosen to go for a non-trivial solution to zeration, the Trappmann solution remind us that there is not uniqueness... how we can chose additional must-have properties in order to reach uniqueness? B2-Do we really need associativity, commutativity and/or distributivity (higher ranks don't have them)? Comment:the distributivity, as GFR said, doesn't hold for the higher ranks thus the requirement of distributivity (and associativity and commutativity) of hyperoperations lower then addition seems like something we want but probably isn't needed at all. Anyways is interesting to see that the Rubtsov's solution make us able to define a semifiels where the semifield addition is the RR-Zeration and the semifield multiplication is the common addition. ------------------------- If we want the solution of Zeration to satisfies some additional properties we have to work on the definition of the family: like adding "axioms" or using new recursion formulas. But if we do it we are talking about a totally different Hyperoperations family (or an exension):is like if we are "changing theory" (ex. from semirings to rings) This bring us to the last problem: C-How (Hyperoperations sequences) C1-How we should modify the definition of Hyperoperations? C2-Once we have a great ammount of different Hyperoperations sequences definitions (i call them theories) wich one is the "real" one? Comment:from a formalistic point of view there is not a real one. From a more platonist point of view those families are all real, we should just investigate their relationships. These are hard questions...but maybe only for who doesn't like trivial solutions . MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 10/03/2014, 09:32 PM Thanks MphLee. How was zeration for complex numbers defined ?? Question : a [-1] b = ?? a [0] ( b [-1] c ) = (a [0] b) [-1] (a [0] c) downation ? regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 10/03/2014, 09:41 PM As for the exp^[a]( ln^[a](x) + ln^[a](y) ) case , I think I have something intresting ; numerical methods independant of a. Need more research. Looks promising. Ok ok , I say a bit more : solving for x when given a and b : exp^[a]( 2 ln^[a] (x) ) = b seems to have a method. regards tommy1729 « Next Oldest | Next Newest »

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