04/23/2008, 09:17 AM

Dear Friends!

I reiterate how glad I am for having the possibility to exchange views with all of you at this "fronteerland" of Mathematics. Nevertheless, I am more and more feeling the need of putting some order in my brain, concerning the various laws that we choose for describing the hyperoperation hierarchy. Perhaps it is only a personal problem, for which I badly need your kind assistance.

Before submitting these few lines to the Ockham's (or Occam's) Razor, which can sometimes be also a boomerang, I should like to share with you my concern, starting from the basic right/wrong stipulations. In fact, in general, I don't think it is possible to say that an assessment such as x = x + 1 is wrong, if we don't perfectly define the context and the limits of the debate.

As a matter of fact, and with all my apologies for writing such basic "Banalitaeten", we could have the following different situations:

a) - Expression x = x+1 is an equality, i.e. a statement stipulating that the two members of the algebraic equality have identical value, despite the values assumed by the letter symbols. In this case, obviously, x = x+1 is indeed absolutely wrong (come on ... for any x's ... ??!!??). But this (... sorry, Henryk, nothing ... personal), I presume, is also the case of an expression such as x = e^x, where y = e^x and y = x have not even one real common point (intersection) for any real x. Therefore, also in this second example, the expression would be wrong, if it is an equality. An example of a well known correct (right) equality is for instance (x+a)^2 = x^2 + 2ax + a^2.

b) - Expression x = x+1 is an equation, i.e. a statement that can be transformed into an equality for some (at least one) real or complex values assumed by variables, defined as the "unknown variables". If x = x+1 is an equation, then it is easy to verify that it is satisfied by both "non standard" solutions x = {-oo, +oo}. In fact, y = x and y = x+1 can be represented in a cartesian plot by two parallel lines, "crossing" themselves in one of the improper points of the plane, with the mentioned infinite coordinates. In this situation, we cannot say that x = x+1 is wrong, but that, since it implies that it must be x-1 = x+1, it can be true only for infinite values of x. The case of x = a^x is even more evident, because we know that this equation, with a as parameter, is studied for defining the infinite towers and has a double solution x = - W(-ln a)ln a, with W representing the two real branches of the Lambert function. However, in both the mentioned cases, it is not appropriate to say if an equation is right or wrong, but if it has or not any solution. An old good example of equation is x^2 +2ax + c = 0. The problem, in the case of x = x+1, is that, while we can easily accept that x-1 = x = x+1 (with x: plus/minus infinite), we should forbid to move the infinity from one to the other member of the equality, otherwise we shall get: x-x = 1 (!!!???!!!) which I have some problems to accept. The situation would be different if we consider x = a+x, with x as a parameter and a as the unknown variable, to be determined. The solution of the equation would then be a = 0. Traffic problems, I suppose. ... !

c) - Expression x = x+1 is a programming instruction (e.g. in Basic or in Fortran). In this case it is neither right nor wrong, but it has just to be executed (take x, add 1, put it in x). I goes without saying, but it goes much better by saying it! It is not a synthetical a priory judgement (V. I. Kant) but a simple a posteriori decision (Befehl ist Befehl...!). Probably, it is not even mathematics. But, this is life! With my pocket calcolator, I should better write it as: x+1 "store" x. Much better.

d) - Expression x = x+1 is a recursive statement. This would mean that x = 1+x = 1+(1+x) = 1+(1+(1+x)) = 1+(1+(1+(1+x))) = ... and so on. Actually, by trying not to create more than necessary confusion, we might think that it could mean that 1+x = x = 1*(+oo) = +oo. Or, better, we could say that: IF x = x+1, THEN x = +oo. A very famous equivalent situation is found in expression x = e^x, intended as a recoursive statement. In fact it would mean that: x = e^x = e^(e^x) = e^(e^(e^x)) = ... and so on. In other words: IF x = e^x, THEN x = e#(+oo). This result is similar to what described in situation "b", with the only difference that here the negative infinity has no apparent meaning.

Well, in our discussions, we should try to avoid crossing frontiers, without showing the ... passport. I mean that we are not authorized to start a speach under situation "d" and then stop if we find a strange equation, which is not ... an equality. Something like: "Ops, sorry! I found x = x+1 and I stop because it is clearly wrong". Probably you should (I mean: ... stop), but please, let's discuss a little bit before. We are walking in a wild territory!! Think of a honest man who, put in front of an instruction like x = x+1, would say: "I don't do this kind of things!"

Concerning the laws and orders to be applied in the study of the hyperoperations (I am convinced to have launched this name, associated to the G. hierarchy, together with KAR, but I don't insist) I should like to observe that we don't only have the mother law (ML), but also something else. I shall indicate by "s" the hyperoperation rank and by "r" the number of iterations, whenever appropriate.

GML - The Grand-mother Law, considered by all people approaching this matter for the first time, as a first definition of the hierarchy, after applying the "priority to the right" traffic rules to the hyperops operators. It sounds like:

a[s]<r>a = a[s+1](r+1).

ML - The Mother Law, discovered (or ... assumed) by most of thew the Researchers, during a deeper hyperops stydy. This can be written as follows:

a[s+1](x+1) = a[s](a[s+1]x).

DL - The Daughter Law (this is a ... new one !! Haha! Well, not really!). In fact, the recursive application of the hyperops operator gives, as a consequence of situation "b":

a[s]x = x = a[s+1]oo.

Its left-inverse operations can give:

x = x/[s]x = x/[s+1]oo,

which means:

[s]-self-hyperroot (x) = [s+1]-oo-hyperroot (x), e.g.:

self-rt (x) = oo-srt (x) = x^(1/x), and, at another rank:

x/x = oo-rt (x) = 1;

x-x = x/oo = 0 (ops, sorry, see "d" ... special case).

On the other hand (what do you mean by: "which one|!?!"), if both GML and ML are together valid (Andydude, please forgive me!), we should have (by putting r = x):

a[s]<r>a = a[s](a[s+1]r), implemented by:

(s=1, r=0) a = a+(a*0) = a

(s=1, r=1) a+a = a+(a*1) = 2a

(s=1, r=2) a+(a+a) = a+(a*2) = 3a.

On the third (... ???) hand, we have:

(s=0, r=0) a[0]<0>a = a°(a+0) = a°a ... how strange ... !!

(s=0, r=1) a°a = a°(a+1) = a+2 (particularly for GML)

(s=0, r=2) a°(a°a) = a°(a+2) = a+3.

What is pointing out is that, for s=0 and r=0, we have:

a°a = a+2, for GML

a°a = a+1 for ML

a°a = a[0]<0>a, for GML and ML together valid.

The provisional conclusion is, in my opinion, that the zero level iteration of the zeration operator is singular. I mean that a[0]<0> is not a neutral operator.

So:

a[s]<0>a = a , for any s>0 is right, but:

a[0]<0>a = a is wrong (it is a wrong equality).

I am very tired and I go drinking a "makkiato". I should write it correctly as "macchiato", but I fear that Gottfried will pronouce it as Mc Shadow, who, together with Mc Roney, are Irish people and neither have anything to do with Italy or with coffee (particularly the second one).

For to-day "ich habe fertig" (Trapp-@-Tony).

GFR

I reiterate how glad I am for having the possibility to exchange views with all of you at this "fronteerland" of Mathematics. Nevertheless, I am more and more feeling the need of putting some order in my brain, concerning the various laws that we choose for describing the hyperoperation hierarchy. Perhaps it is only a personal problem, for which I badly need your kind assistance.

Before submitting these few lines to the Ockham's (or Occam's) Razor, which can sometimes be also a boomerang, I should like to share with you my concern, starting from the basic right/wrong stipulations. In fact, in general, I don't think it is possible to say that an assessment such as x = x + 1 is wrong, if we don't perfectly define the context and the limits of the debate.

As a matter of fact, and with all my apologies for writing such basic "Banalitaeten", we could have the following different situations:

a) - Expression x = x+1 is an equality, i.e. a statement stipulating that the two members of the algebraic equality have identical value, despite the values assumed by the letter symbols. In this case, obviously, x = x+1 is indeed absolutely wrong (come on ... for any x's ... ??!!??). But this (... sorry, Henryk, nothing ... personal), I presume, is also the case of an expression such as x = e^x, where y = e^x and y = x have not even one real common point (intersection) for any real x. Therefore, also in this second example, the expression would be wrong, if it is an equality. An example of a well known correct (right) equality is for instance (x+a)^2 = x^2 + 2ax + a^2.

b) - Expression x = x+1 is an equation, i.e. a statement that can be transformed into an equality for some (at least one) real or complex values assumed by variables, defined as the "unknown variables". If x = x+1 is an equation, then it is easy to verify that it is satisfied by both "non standard" solutions x = {-oo, +oo}. In fact, y = x and y = x+1 can be represented in a cartesian plot by two parallel lines, "crossing" themselves in one of the improper points of the plane, with the mentioned infinite coordinates. In this situation, we cannot say that x = x+1 is wrong, but that, since it implies that it must be x-1 = x+1, it can be true only for infinite values of x. The case of x = a^x is even more evident, because we know that this equation, with a as parameter, is studied for defining the infinite towers and has a double solution x = - W(-ln a)ln a, with W representing the two real branches of the Lambert function. However, in both the mentioned cases, it is not appropriate to say if an equation is right or wrong, but if it has or not any solution. An old good example of equation is x^2 +2ax + c = 0. The problem, in the case of x = x+1, is that, while we can easily accept that x-1 = x = x+1 (with x: plus/minus infinite), we should forbid to move the infinity from one to the other member of the equality, otherwise we shall get: x-x = 1 (!!!???!!!) which I have some problems to accept. The situation would be different if we consider x = a+x, with x as a parameter and a as the unknown variable, to be determined. The solution of the equation would then be a = 0. Traffic problems, I suppose. ... !

c) - Expression x = x+1 is a programming instruction (e.g. in Basic or in Fortran). In this case it is neither right nor wrong, but it has just to be executed (take x, add 1, put it in x). I goes without saying, but it goes much better by saying it! It is not a synthetical a priory judgement (V. I. Kant) but a simple a posteriori decision (Befehl ist Befehl...!). Probably, it is not even mathematics. But, this is life! With my pocket calcolator, I should better write it as: x+1 "store" x. Much better.

d) - Expression x = x+1 is a recursive statement. This would mean that x = 1+x = 1+(1+x) = 1+(1+(1+x)) = 1+(1+(1+(1+x))) = ... and so on. Actually, by trying not to create more than necessary confusion, we might think that it could mean that 1+x = x = 1*(+oo) = +oo. Or, better, we could say that: IF x = x+1, THEN x = +oo. A very famous equivalent situation is found in expression x = e^x, intended as a recoursive statement. In fact it would mean that: x = e^x = e^(e^x) = e^(e^(e^x)) = ... and so on. In other words: IF x = e^x, THEN x = e#(+oo). This result is similar to what described in situation "b", with the only difference that here the negative infinity has no apparent meaning.

Well, in our discussions, we should try to avoid crossing frontiers, without showing the ... passport. I mean that we are not authorized to start a speach under situation "d" and then stop if we find a strange equation, which is not ... an equality. Something like: "Ops, sorry! I found x = x+1 and I stop because it is clearly wrong". Probably you should (I mean: ... stop), but please, let's discuss a little bit before. We are walking in a wild territory!! Think of a honest man who, put in front of an instruction like x = x+1, would say: "I don't do this kind of things!"

Concerning the laws and orders to be applied in the study of the hyperoperations (I am convinced to have launched this name, associated to the G. hierarchy, together with KAR, but I don't insist) I should like to observe that we don't only have the mother law (ML), but also something else. I shall indicate by "s" the hyperoperation rank and by "r" the number of iterations, whenever appropriate.

GML - The Grand-mother Law, considered by all people approaching this matter for the first time, as a first definition of the hierarchy, after applying the "priority to the right" traffic rules to the hyperops operators. It sounds like:

a[s]<r>a = a[s+1](r+1).

ML - The Mother Law, discovered (or ... assumed) by most of thew the Researchers, during a deeper hyperops stydy. This can be written as follows:

a[s+1](x+1) = a[s](a[s+1]x).

DL - The Daughter Law (this is a ... new one !! Haha! Well, not really!). In fact, the recursive application of the hyperops operator gives, as a consequence of situation "b":

a[s]x = x = a[s+1]oo.

Its left-inverse operations can give:

x = x/[s]x = x/[s+1]oo,

which means:

[s]-self-hyperroot (x) = [s+1]-oo-hyperroot (x), e.g.:

self-rt (x) = oo-srt (x) = x^(1/x), and, at another rank:

x/x = oo-rt (x) = 1;

x-x = x/oo = 0 (ops, sorry, see "d" ... special case).

On the other hand (what do you mean by: "which one|!?!"), if both GML and ML are together valid (Andydude, please forgive me!), we should have (by putting r = x):

a[s]<r>a = a[s](a[s+1]r), implemented by:

(s=1, r=0) a = a+(a*0) = a

(s=1, r=1) a+a = a+(a*1) = 2a

(s=1, r=2) a+(a+a) = a+(a*2) = 3a.

On the third (... ???) hand, we have:

(s=0, r=0) a[0]<0>a = a°(a+0) = a°a ... how strange ... !!

(s=0, r=1) a°a = a°(a+1) = a+2 (particularly for GML)

(s=0, r=2) a°(a°a) = a°(a+2) = a+3.

What is pointing out is that, for s=0 and r=0, we have:

a°a = a+2, for GML

a°a = a+1 for ML

a°a = a[0]<0>a, for GML and ML together valid.

The provisional conclusion is, in my opinion, that the zero level iteration of the zeration operator is singular. I mean that a[0]<0> is not a neutral operator.

So:

a[s]<0>a = a , for any s>0 is right, but:

a[0]<0>a = a is wrong (it is a wrong equality).

I am very tired and I go drinking a "makkiato". I should write it correctly as "macchiato", but I fear that Gottfried will pronouce it as Mc Shadow, who, together with Mc Roney, are Irish people and neither have anything to do with Italy or with coffee (particularly the second one).

For to-day "ich habe fertig" (Trapp-@-Tony).

GFR