Example $x$, $y$ and $z$ values for $x\uparrow^\alpha y=z$ where $\alpha\in \Bbb R-\Bbb N$ $\uparrow^n$ and $G(n,\cdot,\cdot)$ are notations for hyperoperation.
http://en.m.wikipedia.org/wiki/Hyperoperation

$n$ is the hyperoperations rank.

Can example $x$, $y$ and $z$ values be provided for either the following formula?
The formula can be notated:
$$z = x\uparrow^{-3}y$$
$$\text{or}$$
$$z = G(-1, x, y)$$

Or, alternatively, for this formula?
The formula can be notated:
$$z = x\uparrow^{-0.5}y$$
$$\text{or}$$
$$z = G(1.5, x, y)$$

I prepose a possible name for $\uparrow^{-0.5}$ to be addiplication, also resulting in the words addiply, addiplicative and addiplicativly.
 A: Start with the definition
$$
  G(n,a,b) =  
   \begin{cases}
    b + 1 & \text{if } n = 0 \\
    a &\text{if } n = 1, b = 0 \\
    0 &\text{if } n = 2, b = 0 \\
    1 &\text{if } n \ge 3, b = 0 \\
    G(n-1,a,G(n,a,b-1)) & \text{otherwise}
   \end{cases}
$$
for non-negative integers $n$, $a$, and $b$.
If we can extend this definition to negative integers $n$, then presumably
we still want it to satisfy the recurrence
$G(n,a,b) = G(n-1,a,G(n,a,b-1))$ when $a\geq 0$ and $b > 0$.
In particular, for any $a\geq 0$ and $b > 0$,
$$G(0,a,b) = G(-1,a,G(0,a,b-1)) = G(-1,a,b)$$
since $G(0,a,b-1) = b.$
That is,
$$G(-1,a,b) = G(0,a,b) = b+1.$$
We can take this even further: for an integer $k \geq 0$, suppose
$G(-k,a,b) = b+1$ for all $b \geq k$. 
Then $G(-k,a,b-1) = b$ for all $b \geq k+1,$ and
$$G(-(k+1),a,b) = G(-(k+1),a,G(-k,a,b-1)) = G(-k,a,b) = b+1.$$
By induction, $G(n,a,b) = b+1$ for all $n\leq 0$ and all $b\geq -n.$
If you assume from the beginning that $G(0,a,b) = b+1$ for any integer $b$,
then you end up with $G(n,a,b) = b+1$ for any integer $b$
and any integer $n\leq 0.$
I do not have any good idea yet how to deal with non-integer values of $n$.
