02/20/2008, 11:30 PM

bo198214 Wrote:..........Hi, Henryk!

the bracketing must be to the right. So

ao(ao(aoa))=a+4

ao(aoa)=a+3

aoa=a+2

a = a+1 ???

As a matter of fact, your last line is not a criticism to the poor new zeration "baby", but it is a general "problem" of all the elements of the hierarchy, including a very old and respectable operation. Let me try to show it as follows, starting from the behaviour of the "right unit element u" of each hyperop.

a[4]u = a#u = a ---> u = 1, as we have just correctly shown,

a[3]u = a^u = a ---> u = 1, as we know from ... centuries;

a[2]u = a*u = a ---> u = 1, as we knew ... even before;

a[1]u = a+u = a ---> u = 0, bingo !, Deviation from the "rule"! However, nobody complained, until now.

a[0]u = a°u = a ---> u = -oo re-bingo! But, here, we must discuss a little bit.

As a first observation, we may see that, in case of a general hyperoperation, i.e.:

a[s]u = a ---> u = 1 for s > 1; u = 0 for s = 1; and u = -oo for s = 0.

The suggestion to remain cool is coming from the fact that the "rule" u = 1 is violated also by the old very respectable addition. So, we see that, taking this into consideration, we are not allowed to write

neither a ° 1 = a , nor a + 1 = a.

But, this should not be a reason for doubting of zeration.

Concerning the "left unit element v" (and showing it in the inverse ranking order), we have something even more peculiar:

v[0]a = v°a = a ---> v = -oo, and .. we shall discuss;

v[1]a = v+a = a ---> v = 0, OK, as above;

v[2]a = v*a = a ---> v = 1, idem, as above;

v[3]a = v^a = a ---> v = [a/]rt a = selfrt a.

Well, we know, selrt a it's not an "element", because it is changing with a. For rank 4 (tetra), we should have something similar to:

v[4]a = v#a = a ---> v = [a/]srt a = sselfrt a.

A deep analysis of this left/right behaviour is strongly needed.

bo198214 Wrote:Can you btw show me, how you derived the commutativity of zeration?Via the analysis of homomorphisms (see KAR). For example, for the calculations of the square superroot, the squqreroot and the half of a variable z we may use the following iterative formulas:

x = ssqrt z ≈ sqrt(p * [p/] rt z) , with x -> p

x = sqrt z ≈ ( p + z/p) / 2 , with x -> p

x = z / 2 ≈ (p ° (z – p)) – 2 , with x -> p, and p integer.

By a change of variables (x-p ↔ p), we can assume that the following relation is also verified:

x = z / 2 ≈ ((z – p) ° p) – 2

and that, therefore, zeration must be commutative. But, here, KAR might probably add something else.

In conclusion, we came again to:

a ° b = a + 1 if a > b

a ° b = b + 1 if a < b

a ° b = a + 2 = b + 2 if a = b.

Gianfranco