# Tetration Forum

Full Version: Poweroids
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OK, so I'm done with finals, so I thought I would get back into the quasigroup thing (as Henryk started here). It took me a while to figure out how to deal with exponentiation as a quasigroup, specifically because

$(\mathbb{C}, \uparrow, \log, \sqrt{?})$ does not form a quasigroup.
Proof.
$b^a = b^{a + \frac{2 i \pi}{\ln(b)}}$, so $\log_b(b^a)$ is not uniquely defined as a. Since $\log_b(b^a)$ must be uniquely defined for it to be a quasigroup, it follows that $\mathbb{C}$ does not form a quasigroup under exponentiation.$\qed$

$(\mathbb{R}, \uparrow, \log, \sqrt{?})$ does not form a quasigroup.
Proof.
$(-1) \in \mathbb{R}$ and $1/2 \in \mathbb{R}$, but $(-1)^{1/2} \not\in \mathbb{R}$, so $\mathbb{R}$ does not form a quasigroup under exponentiation.$\qed$

So I started thinking about why it didn't form a quasigroup, and what could be done about it. Then I realized that it was because the two domains were being forcefully equated. Conceptually, the base and the exponent may come from domains that do not even intersect! Consider, for example bases in the interval $(0, 1)$ and positive integer exponents. This does not form a quasigroup, but you can see that $(\uparrow)$ is at least closed over these domains. So if separating the domains is required to explain how exponentiation relates to quasigroups, then this function-with-2-domains is no longer a binary operation! However, I have heard people refer to this kind of binary function as a scalar multiplication and an external binary operation, so we can call these things external quasigroups.

A related structure to quasigroups is that of a loop, but in no way, shape or form does exponentiation form a loop, since there does not exist an identity element such that $1^y = y$ for all y, granted, there does exist a function $I(y) = \sqrt[y]{y}$ such that $I(y)^y = y$, but this follows directly from the quasigroup structure. However, exponentiation does have a right-identity element that acts as $x^1 = x$ for all x. So this gives me some justification to make the following definition.

A poweroid $(X, Y, \uparrow, 1)$ is an external quasigroup with right-identity.

In other words, let $X, Y$ be subsets of $Z$. If $(X, Y, *, \backslash, /, 1)$ is a poweroid, then the operations $(*, \backslash, /)$ satisfy:


\begin{align}
&(*)\ :\ X \times Y \to X &\text{external binary operation} \\
&(\backslash)\ :\ X \times X \to Y &\text{log, left-inverse operation} \\
&(/)\ :\ X \times Y \to X &\text{root, right-inverse operation} \\
&\exists{1 \in Y}, \forall{x \in X}, (x \backslash x) = 1 &\text{right-identity element}
\end{align}

For starters, poweroids do exist, namely the trivial poweroid $(\{u\},\ \{1\},\ \uparrow,\ 1)$ where
$u^1 = \sqrt[1]{u} = u$ and $\log_u(u) = 1$.

Some of the things that I have found recently are that
• $(\mathbb{R}^{+}-\{1\},\ \mathbb{R}^{+},\ \uparrow,\ 1)$ forms a poweroid.
• $((0, 1),\ \mathbb{R}^{+},\ \uparrow,\ 1)$ forms a poweroid.
• $((1, \infty),\ \mathbb{R}^{+},\ \uparrow,\ 1)$ forms a poweroid.
• For any complex number u such that $0 \ne |u| \ne 1$, let $Y$ be the complex unit circle, and let $X = u^Y$, then $(X,\ Y,\ \uparrow,\ 1)$ forms a poweroid.

I'm starting to think that this looks a lot like iteration (semi)groups, but I don't have a clear idea of how they're related at the moment.

Andrew Robbins
The more I think about it, I think it is possible to show that:
• $((1, \infty), \mathbb{R}^{+}, x^{x^{y-1}}, 1)$ is a poweroid if the inverse functions are understood a bit more.
• $((1, \infty), \mathbb{R}^{+}, {\uparrow}{\uparrow}, 1)$ is a poweroid, regardless of the extension. The only requirement I can think of is monotonicity.
• $((1, \infty), \mathbb{R}^{+}, {\uparrow}{\uparrow}{\uparrow}, 1)$ is a poweroid, etc.

I think there are also poweroids with these operations (lower tetration, standard tetration, etc.) over complex domains, such as the one for exponentiation with the unit circle... but I'm not sure how exactly it would work for an arbitrary tetration extension.
(12/18/2009, 08:13 AM)andydude Wrote: [ -> ]
• $(\mathbb{R}^{+}-\{1\},\ \mathbb{R}^{+},\ \uparrow,\ 1)$ forms a poweroid.
...
Andrew Robbins

SCRATCH THAT! it is false.

$\{1/2,\ 2\} \subset \mathbb{R}^{+}-\{1\}$, but $\log_2(1/2) = -1$ which is not a member of the positive reals. However, with the slight modification
• $(\mathbb{R}^{+}-\{1\},\ \mathbb{R}-\{0\},\ \uparrow,\ 1)$ forms a poweroid.
I believe the statement is true.

The reason why 0 and 1 must be excluded is so that $\sqrt[y]{x}$ is uniquely defined. If we include 1, 0 in the base domain, exponent domain respectively, then roots are not uniquely defined, at which point it stops being a poweroid.