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 Zeration
02/14/2008, 06:38 PM (This post was last modified: 02/20/2008 10:54 PM by GFR.)
Post: #1
 GFR Member Posts: 174 Joined: Aug 2007
Zeration
Dear Henryk and dear Participants,

As promised (and also on behalf of KAR), I should like to introduce “zeration”, a subject considered as controversial by standard and non-standard mathematicians and that, despite the various debates (I must say: around ... the world), still remains fresh and new. Some of those who, based on correct existing mathematical standards, said “It’s just ... wrong”, are now coming back to their statement, thinking that perhaps, despite the new language and the lack of precision in expressing these new concepts, there might be something behind those ideas, still needing a careful attention.

An article on zeration was submitted sometime ago to Wikipedia and it was rejected, mainly because “Wikipedia is not the place for information on new scientific terminology and controversies and zeration is not present in the classical encyclopedias”. Moreover, due to the fact that the original research on it was covered by scientific literature in Russian (Kostantin Rubcov, 1989), the article was presented in two versions, in Russian and in English. This fact was considered as “spamming” by the distinguished Wikipedia Revisers.

Zeration, as part of the hyper-operation hierarchy was described in a paper cited in the bibliography of mentioned in the Stephen Wolfram’s book “A New Kind of Science”, under the title
Ackermann’s Function and New Arithmetical Operation”, K. A. Rubtsov, G. F. Romerio.
(see: http://www.wolframscience.com/reference/...raphy.html ).
The same subject was also submitted twice to the International Congress of Mathematicians, during ICM-1994, Zurich, and ICM-2006, Madrid, where a poster on zeration and tetration was jointly presented by me and Konstantin Rubcov (Rubtsov). A thread on the subject was submitted by KAR and GFR to the Wolfram Research Institute NKS Forum, since Jan 6th, 2006, with a detailed report under the title: "Hyper-operations. Progress Report, Zeration".

Recently, zeration has also been discussed in the Republic of Korea, during a seminar organized by the National Institute of Mathematical Sciences (NIMS), on “Tetration and Zeration”.
(see: http://www.nims.re.kr/news/news_seminar_...600&seq=99

At the very simple common sense level, zeration is an attempt for finding an operation filling the gap in the following operations’ sequence:

a ^ a = a [3] a = a [4] 2 = a # 2 exponentiation <-> tetration
a . a = a [2] a = a [3] 2 = a ^ 2 multiplication <-> exponentiation
a + a = a [1] a = a [2] 2 = a . 2 addition <-> multiplication

which should be logically completed by a new operation that we may call zeration (indicating it by the infixed operation sign “°”) and that should at least have the following “strange” property:

a ° a = a [0] a = a [1] 2 = a + 2 zeration <-> addition.

The choice of the word “zeration” is justified by the fact that its rank is immediately lower than the rank of addition (normally considered at rank s=1) and by the common acceptance that iterative exponentiations produce the tetration operation, normally supposed to be at rank 4. The theoretical way for justifying such new operation is provided by the Ackermann Function. In fact, the definition of Ackermann Function (AF) can be summarised as follows:

A(0, n) = n + 1
A(s, 0) = A(s-1, 1)
A(s, n) = A(s-1, A(s, n-1))

By analysing the AF, it is known that we can find the following pattern:

in row s=0, by definition: A(0, n) = n + 1
in row s=1: A(1, n) = 2 + (n + 3) – 3 = n + 2;
in row s=2: A(2, n) = 2 . (n + 3) – 3 = 2n + 3;
in row s=3: A(3, n) = 2 ^ (n + 3) – 3 = 2n+3 – 3;
in row s=4: A(4, n) = 2 # (n + 3) – 3 = n+32 – 3;

With the provisional exception of row s=0, we could re-define Ackermann’s Function as follows:

A(s, n) = 2 [s] (n + 3) – 3, or:
2 [s] n = A(s, n-3) + 3

For s=0 we have:
A(0, n) = 2 [0] (n + 3) – 3 = n + 1 (zeration)
which gives: 2 ° (n+3) = n + 4
therefore: 2 ° n = n + 1 , (for: n ≥ 3)
to which we may add: 2 ° 2 = 2 + 2
and: n ° n = n + 2

We can start using these expressions in order to find out the first properties of the “zeration” operation, which can be described as follows:

a ° b = a + 1 , if a > b
a ° b = b + 1 , if a < b
a ° b = a + 2 = b + 2 , if a = b

Zeration can therefore be defined as a new binary arithmetical operation belonging to the Grzegorczyk hierarchy and coinciding, in some particular cases, with the “successor” unary operation. It is not associative, but it is commutative and, therefore, it has a unique inverse operation, called “deltation”. In fact, zeration follows the general hyperoperation rules:

a [s] a = a [s+1] 2 , s = 0, 1, 2, 3, 4, ....

and, in particular:
a [0] x = a ° x = x ° a = x [1] 1 = x + 1 for x > a
x [0] x = x ° x= x [1] 2 = x + 2 for any x
and, also, for a = 2:
2 ° 2 = 2 + 2 = 2 * 2 = 2 ^ 2 = 2 # 2 = .... = 4

Zeration follows its own algebraic rules and generates, with its inverse, a new set of numbers, called the “delta numbers”, which can be put in bijection with the logarithms of negative numbers. Functions built with the help of zeration are discontinuous and singular. For these reasons, zeration can be used for defining Boolean operators and fundamental discontinuous functions, such as the Dirac function and the Heaviside function. Zeration and Delta numbers are usable for the construction of approximated algorithms concerning tetration (e.g. in y = b[4]a, with b > 1 and a < -2)

Thanking you very much in advance for your possible interest and expected ... patience , please find attached the detailed progress report on zeration, presented as an attachment to the NKS Forum thread of 2006-01-06.

The authors encourage the readers of the attachment to freely use the terminology and symbols that they have proposed. They will welcome any suggestion and comment and, in case of partial or total text citations, they would appreciate the reference to: “C. (or K.) A. Rubtsov – G. F. Romerio; Hyper-operations. Progress Report. Zeration. The Wolfram Research Institute NKS Forum, Jan 6th, 2006

GFR

Attached File(s)
02/14/2008, 08:06 PM (This post was last modified: 02/14/2008 08:08 PM by bo198214.)
Post: #2
 bo198214 Administrator Posts: 1,365 Joined: Aug 2007
RE: Zeration
GFR Wrote:a ° b = a + 1 , if a > b
a ° b = b + 1 , if a < b
a ° b = a + 2 = b + 2 , if a = b
This operation has no inverse, you can not even cancel it:
aob = cob as long as b>a,c

Also the above rules surely dont follow completely from
aoa=a+2

To be consistent with the other hyperoperations we would surely demand that
aoa=a+2
ao(aoa)=a+3
ao(ao(aoa))=a+4

or generally ao(a+n)=a+n+1 in analogy to $a^{^n a} = {^{n+1}}a$ and $a\times a^n = a^{n+1}$ and $a+a\times n = a\times (n+1)$

ao(a+0)=a+1 as opposed to aoa=a+2
02/14/2008, 08:10 PM (This post was last modified: 02/14/2008 10:22 PM by Ivars.)
Post: #3
 Ivars Long Time Fellow Posts: 366 Joined: Oct 2007
RE: Zeration
GFR

This is a great attempt to generalized "operations" space that in my opinion is complementary to structure (number, function etc.) spaces. Only after that their interactions will become understandable, and, may be, number/function/structure spaces will need to be redefined.

Have You considered "+" zeration and "-" zeration?

The idea comes as follows:

If we imagine, there might be fractional operations (halfation, etc), also negative (not saying explicitly yet what a negative operation rank might mean)- then they would tend to "converge" to zeration from both positive (1/n) and negative (-1/n) sides as they get more fractional with n-> infinity.

Ivars
02/14/2008, 10:30 PM (This post was last modified: 02/14/2008 10:42 PM by GFR.)
Post: #4
 GFR Member Posts: 174 Joined: Aug 2007
RE: Zeration
bo198214 Wrote:This operation has no inverse, you can not even cancel it:
aob = cob as long as b>a,c
Indeed, you cannot cancel, under ... any situatiom.
aob = cob = b+1, if b>a,c
aob = cob --> a = c, if b < a,c
Moreover, let us see these examples:
3#2 = 27, 2#3 = 16; non-commutable;
3^2 = 9, 2^3 = 8; non-commutable;
3*2 = 6, 2*3 = 6 ; commutable;
3+2 = 5, 2+3 = 5 ; commutable;
3o2 = 4, 2o3 = 4 ; commutable.

slog(/3)27 = 2 ; srt(\2)27 = 3; slog(/2)16 = 3 ; srt(\3)16 = 2;
log(/3)9 = 2 ; rt(\2)9 = 3 ; log(/2)8 = 3 ; rt(\3)8 = 2;
6/3 = 2 ; 6/2 = 3;
- but ... attention, now ...
4ò3 = 2 ; 4ò2 = 3 (with "ò": inverse operator of "o"), and indeed also:
4ò2 = 2 (because 2o2 = 4 and 2o3 = 3o2 = 4).

In conclusion, we could say that:
4ò2 = {2,3}. Strange, but not unique, in fact:
rt(\2)4 = sqrt 4 = {-2,+2}.

It happens, in the best families!

bo198214 Wrote:To be consistent with the other hyperoperations we would surely demand that
aoa=a+2
ao(aoa)=a+3
ao(ao(aoa))=a+4

or generally ao(a+n)=a+n+1 in analogy to $a^{^n a} = {^{n+1}}a$ and $a\times a^n = a^{n+1}$ and $a+a\times n = a\times (n+1)$

ao(a+0)=a+1 as opposed to aoa=a+2

Well, as a matter of fact, for n=0, we have:
ao(a+0) = aoa = a+2 , but:
ao(a+n) = (a+n)+1 , for n>0, which gives:

ao(a+1) = (a+1)+1 = a+2 (again, but as ... successor of a+1))
ao(a+2) = (a+2)+1 = a+3
ao(a+3) = (a+3)+1 = a+4
..............................

And, of course:
ao(ao(ao(aoa))) = a+5

The definition of zeration has a discontinuous and multi-valued character, which we should be prepared to accept in the execution of such operation. In fact:

ao(a+0) = ao(a+1) = a+2

The result of zeration is given by the successor of the greatest of the operands, if they are different, or by the second successor of any of them, if they are equal.

Così è, .... se vi pare (Pirandello, ... I presume).

GFR
02/14/2008, 10:41 PM
Post: #5
 GFR Member Posts: 174 Joined: Aug 2007
RE: Zeration
Ivars Wrote:Have You considered "+" zeration and "-" zeration?

Not really. But there is some work carried on for "mapping" the numbers obtained by its inverse operation. They ... risk ... to be multivalued objects, similar, in some cases, to the logarithms of negative real numbers. There might perhaps also exist hyperoperations with negative ranks, giving, for instance:

a@a = ao2
aoa = a+2
a+a = a*2
......

But, this requires a ... more active intervention of KAR in this discussion.

GFR
02/14/2008, 10:58 PM
Post: #6
 bo198214 Administrator Posts: 1,365 Joined: Aug 2007
RE: Zeration
But see Gianfranco, the postulate
aob = max{a,b}+1 for a!=b
aob = a+2=b+2 for a=b
aoa=a+2
nor does it follow from the extended set of rules
ao(a+x)=a+x+1, x>0
I am not completely sure whether it follows from the last rules, but even if it would, the restriction x>0 is arbitrary, imho.

Finding a unique continuous extension is already as we have seen from tetration a nearly hopeless task. But *without* continuity as you propose for zeration, I think you can construct arbitrary nasty solutions to the functional equation ao(a+x)=a+x+1.
02/18/2008, 09:05 AM
Post: #7
 KAR Junior Fellow Posts: 2 Joined: Aug 2007
RE: Zeration
bo198214 Wrote:But see Gianfranco, the postulate
aob = max{a,b}+1 for a!=b
aob = a+2=b+2 for a=b
aoa=a+2

It not a postulate!
Definition Zeration follows from similar:
a+a=a*2
a+a+a=a+3
a+a+a+a=a+4
......
a+...+a=a*n (n>1)

a*a=a^2
a*a*a=a^3
...................
If calculations to start at the left:
aoa=a+2
aoaoa=a+3=(aoa)oa=(a+2)oa, a+2>a
aoaoaoa=a+4=((aoa)oa)oa=(a+3)oa, a+3>a
......
ao...oa=a+n, a+(n-1)>a if n>1.
From definition Zeration for "a" it is possible to write for b>a:
boa=b+1 and boa=a+2=b+2 if a=b.
This base definition

The rule of evaluation Zeration is a corollary from extended base definition, but not a postulate!

KAR
02/19/2008, 12:24 PM (This post was last modified: 02/19/2008 12:50 PM by bo198214.)
Post: #8
 bo198214 Administrator Posts: 1,365 Joined: Aug 2007
RE: Zeration
Hey Konstantin, I am pleased to see you writing your first article here.
I really honor this, knowing that you nearly had no previous knowledge of the english language.

KAR Wrote:
bo198214 Wrote:But see Gianfranco, the postulate
aob = max{a,b}+1 for a!=b
aob = a+2=b+2 for a=b
aoa=a+2

It not a postulate!

Ok, that was perhaps a wrong usage of the word "postulate". I meant that this evaluation rule does not follow from the initial rule, but is rather an interesting suggestion for an operation that satisfies the initial rule.

Quote:aoa=a+2
aoaoa=a+3=(aoa)oa=(a+2)oa, a+2>a
aoaoaoa=a+4=((aoa)oa)oa=(a+3)oa, a+3>a
......
ao...oa=a+n, a+(n-1)>a if n>1.

But to be compatible with the definition of tetration (and also Ackermann function)
a^a = a[4]2
a^(a^a) = a[4]3
a^(a^(a^a)) = a[4]4
(as a note to Gianfranco: I really like the ASCII notation a [n] b, which was I think introduced by you)

the bracketing must be to the right. So
ao(ao(aoa))=a+4
ao(aoa)=a+3
aoa=a+2
a = a+1 ???

One notices that the last equation, though being the logical progression of the previous lines, is not possible; while for all higher operation we have:

a+a=a*2
a=a*1

a*a=a^2
a=a^1

a^a=a[4]2
a=a[4]1

Quote:From definition Zeration for "a" it is possible to write for b>a:
boa=b+1 and boa=a+2=b+2 if a=b.
This base definition

The rule of evaluation Zeration is a corollary from extended base definition, but not a postulate!

No, thats not a strict consequence. The following is a strict consequence:
ao(ao(ao......))) = a+n for integer n>1 (and a being real or complex)
then we conclude
ao(a+n)=a+n+1

If we assume a and b to be an integer, then we can write for b>a+1
aob=ao(a+(b-a))=a+(b-a)+1=b+1.

So what to do then about real number arguments? For tetration we have a similar situation:
(1) a^(a[4]n)=a[4](n+1)
(2) a[4]1=a

and we just demanded that the above rule (1) also shall be valid for real (or even complex) n, and that the function f(x)=b[4]x shall be analytic (or at least continuous). For Zeration we probably would look for an operation (on the reals) that satisfies
(1') ao(a+x)=a+x+1 for each real x>1
(2') aoa=a+2

So we dont need any continuity or analyticity to see that
aob=b+1 for b>a+1 (a,b real). If we would now demand that f(x)=aox is analytic, then it would follow that f(x) must be x+1 for all complex x (because an analytic function is already determined by being defined on a set which is dense at one point, and here it is even defined for all x>a+1). So this would contradict f(x)=x+2 for a=x. The next try would be to keep f(x) at least continuous or infinitely differentiable, but I see no canonic definition that one would chose for f(x)=aox on the yet undefined interval (a,a+1].

Can you btw show me, how you derived the commutativity of zeration?
02/19/2008, 01:20 PM
Post: #9
 bo198214 Administrator Posts: 1,365 Joined: Aug 2007
RE: Zeration
So my continuous counter proposal for zeration would be the following operation:

$a\circ b = \begin{cases}b+1 & : b>a+1\\a+2&: b\le a+1\end{cases}$

This is continuous in both arguments and satisfies
$a\circ a=a+2$
$a\circ (a\circ a)=a+3$
$a\circ (a\circ (a\circ a))=a+4$
etc. for all real $a$
02/19/2008, 09:41 PM
Post: #10
 andydude Long Time Fellow Posts: 470 Joined: Aug 2007
RE: Zeration
In my view, it is not a matter of proof, but a matter of choice. If we choose to hold $a [N-1] b = a [N] (\text{hy}N\text{log}_a(b) + 1)$ to be true for all N, then the natural consequence is that $a [0] b = b + 1$ for all a (because $a [1] (\text{sub}_a(b) + 1) = a + ((b - a) + 1) = b + 1$). If we choose to hold $a [N-1] a = a [N] 2$ true for all N, then we get $a [0] b = b + 2$ for a == b. Both cannot be true, so we must make a choice.

Andrew Robbins
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