04/05/2008, 10:14 AM

GFR Wrote:bo198214 Wrote:Do you mean ... fainting?GFR Wrote:I hope that everybody would be patient enough to read it, without ... fainting,At least I do it!

No, no, no! That meant: I (am patient enough to) read it without fainting!!!

Quote: By the way, did you mean "at least" or "at last"?

And probably I was even the first one that read it!

Quote:In fact, in my opinion, this would mean that zeration does not exist at all, as a binary operation.First of all a binary operation is just an operation that takes two inputs and delivers one output. Whether the output depends on both operands is another question. So in this sense the mother-law-zeration is a binary operation.

Quote:And this would be very sad, at least for me.

But why? The hyper operations become simpler and simpler with decreasing degree and an operation that only depends on the right operand shows an extreme simplicity as one would expect from zeration.

But mathematics is mathematics, I also find it sad that I can not divide by zero, but its just the way it is. Thats a higher joy.

Quote:Only to that, for any "a"?We have always to be a bit cautious of the base. As long as we work with inverse operations we have to ascertain that they indeed exist (which is the base requirement in quasigroups). Thatswhy I assume in all my derivations that the base is in the bijectivity domain. That means arbitrary x for x[1]y, that means x>0 for x[2]y, and that means x>1 for x[n]y, n>2. Let R(n) denote the corresponding base range, then I think the functions f(x)=x[n]y : R(n) -> R(n) are always bijective. And the functions g(x)=b[n]x : R(0) -> R(n) , b in R(n), are also bijective. The exception is however n=0, where f is not bijective.

Quote:Quote:And now under the assumption that y[s]0=1:Which, of course, should read y = x/ [s]2. You see? In the best families!

y [s] 2 = y [s] (1+1+0) = y[s-1](y[s-1](y[s]0)) = x

y = x [s]/ 2

Haha, yes, though I dont know even whether my family belongs to the best

Quote:But, ... ... Henryk! This is the Slash Algebra.Its quasigroups!

Quote:The point is that the limit of the sequence y <= (y[0](y[1]\ x)) /[1]2 = y[0](x-y) -2 is not x /[2] 2 (if it exist at all).bo198214 Wrote:However the assumption is wrong for s=2, y[2]0=0.Yes, it is wrong, but I don't see the point.

Quote:bo198214 Wrote:Stop a moment, please! I don't believe in a generalisation of: a[n+1]1 = a. In fact, for n=0, a[1]1 = a+1 = a is wrong!Quote:Pillar 4 - The Hyper-means.The hyper means (a[n]a)/[n+1] 2 = a

follow directly from the assumption that a[n]a=a[n+1]2 which is equivalent to a[n+1]1=a.

Yes exactly, and thatswhy (a[0]a)-2=a is wrong.

Quote:As I already wrote, we have a[n]1=a for all n>1 and are not allowed to generealize to a[1]1=a, in the same way we have a[n]a=a[n+1]2 for all n>0 and are not allowed to generalize to a[0]a=a[1]2.bo198214 Wrote:Asserting that a[0]a=a+2 is a bit like asserting that a[1]1=a.Not really! At level a[s]1, we have different behaviours. In fact, as we have seen:

Do you see the similarity?

a[2]1 = a*1 = a

a[1]1 = a+1 (and tht's it!)

while:

a[2]a = a*a = a^2

a[1]a = a+a = a*2

Quote:But, Henryk, the problem is that the formula mentioned in Pillar 4 may work, and the choice of -oo as neutral element of zeration (as we defined it) is shared by other researchers in this field.-oo as well as +oo is a neutral element for your zeration as well as for the mother-law-zeration. This comes from -oo and +oo not being a number (not being in the bijectivity domain). You know you can prove everything (e.g. 1=0) by treating infinities like numbers. In so far my pillar 5 was a bit misleading as for s=0 the limit does not exist (is not a number).

However I dont know about other publishing researchers considering zeration. Why dont they post on this forum in this thread?