hyper 0

[MphLee]

Ok, at the beginning I was convinced that the grouping operation on fingers was just the arithmetic mean.

To start you would put a fingers in one hand and b fingers in the other. Next you would figure out how many groups of a fingers you had total. Then you would perform the sum (# of groups) + (# in each group ) which is the same as (# of groups) + a.

In that interpretation we have that given a sequence \( a_i\in\mathbb C \) for \( i\in I \), we can see this sequence as \( |I| \) group of fingers where every group \( i\in I \) has \( a_i \) fingers in it. So we define an "hyper 0" operator \( \underset{i\in I}{\rm O} \)

\( \underset{i\in I}{\rm O}a_i=\frac{\sum_{i\in I}a_i}{|I|}+|I| \)

if \( \forall i,j\in I: a_i=a_j \) then  \( \underset{i=1}{\overset{n}{\rm O}}a=a+n \)

for \( |I|=2 \) (the 2-ary version) we get

\( a_1{\rm O}a_2=\frac{a_1+a_2}{2}+2 \)   and   \( a{\rm O}a=a+2 \)

But then an example of computation proposed is

3[0]2
You have 5 fingers total.That is 1 and 2/3s groups of 3. So the answer is 5/3+3 = 14/3

So the groups are meant to be weighted and the operation is clearly not commutative anymore. In fact the operation proposed is the following. Let \( a_i\in\mathbb C \) for \( I=\{1,2,3,..., n\} \). Define \( \lambda_i:=a_i/a_1 \)

\( \underset{i=1}{\overset{n}{\rm O}}a_i:=a_1+\sum_{i=1}^n\lambda_i \)

if \( 1\leq\forall i,j\leq n: a_i=a_j \) then  \( \underset{i=1}{\overset{n}{\rm O}}a=a+\sum_{i=1}^n1=a+n \)

for \( n=2 \),  \( a_1=a\neq 0 \) and \( a_2=b \)

\( a{\rm O}b=a+(1+\frac{b}{a}) \)   and   \( a{\rm O}a=a+2 \)

It is clear that the solutions work in some way for preaddition. It is not clear to me how these two solutions can meet the requirment of fundamentality

3. This operation should be something that is fundamental

since they require summation and ratios to be defined.