What is $x^\bot$? Is $\zeta(\bot)=\bot$ for Riemann's zeta function $\zeta$ and wheel theory's $\bot$? Background:
A wheel is an algebraic structure $(W,0,1,+,\cdot, /)$ where:

*

*$W$ is a set,

*$0,1\in W,$

*$+$ and $\cdot$ are binary operations,

*$/$ is a unary operation,

and

*

*$+,\cdot$ are associative, commutative, and with identities $0$ and $1$, respectively,

*$//x=x$,

*$/(xy)=/x/y$,

*$xz+yz=(x+y)z+0z$,

*$(x+yz)=x/y+z+0y$,

*$0\cdot 0=0$,

*$(x+0y)z=xz+0y$,

*$/(x+0y)=/x+0y$,

*$0/0+x=0/0$.

We denote $0/0$ by $\bot$.
The final axiom can be written as
$$\bot+x=\bot.$$
See
What are the mathematical properties of ⊥ in wheel theory?
Therefore,
$$\sum_{i=1}^\infty a_i=\bot\tag{$\Sigma$}$$
is $\bot$ whenever $a_i=\bot$ for at least one $i\in\Bbb N$.
The Question:

Can we go any further than $(\Sigma)$? What is $x^\bot$? In particular, is $$\zeta(\bot)=\bot$$ for Riemann's zeta function?

Thoughts:
Due to the argument in the question linked to, my intuition is that, yes, we can go further; for instance,
$$\prod_{i=1}^\infty a_i=\bot\tag{$\Pi$}$$
whenever $a_i=\bot$ for at least one $i\in\Bbb N$, where $\Pi$ is defined in the obvious manner; but
$$
\zeta(\bot)
:=\sum_{n=1}^\infty 1\cdot/(n^\bot)
$$
requires some notion of what $n^\bot$ means. I guess it should be
$$x^\bot=\bot.\tag{$\bot$}$$
But breaking this down:
$$
x^\bot =x^{0/0}
=x^{0\cdot /0}
=(x^0)^{/0}
=1^{/0},$$
which has me stumped. Should we define
$$1^{/0}:=\bot?\tag{1}$$
Further Context:
This is just for fun. I don't think anything deep is going on here.
I have no formal training in wheel theory.

Please help :)
 A: Here's some thoughts:
In order to define $x^\bot$, we need to define exponentiation on wheels. This isn't something we can do analogously to rings, because for a general ring $R$ there isn't some well-defined exponentiation $R\times R \rightarrow R$. Even if $R=\mathbb Q$ we're in trouble. I guess the wheel-ized version of $\mathbb R$ is $$
\mathbb R \cup\{/0,0/0\}
$$
analogous to how the wheel of fractions over $\mathbb Z$ is $\mathbb Q \cup\{/0,0/0\}$. Then we'd have the usual addition and multiplication and $/x = x^{-1}$ for $x\ne 0$, then for the extra stuff: $x + /0 = x\cdot/0 = /0$ for $x\notin\{/0,0/0\}$ and $x + 0/0 = x\cdot 0/0 = 0/0$ and $0\cdot /0= 0/0$.
To define exponentiation on $\mathbb{R}$, we usually start with the function $$
\exp(x) =\sum_{n=0}^\infty\frac{x^n}{n!}
$$
The standard proof that $\exp(x+y) = \exp(x)\exp(y)$ works just as well here. We have, of course, $$
\exp(/0) = /0 \text{ and }\exp(0/0) = 0/0
$$
Thus $\exp$ is a bijection from $\mathbb{R}\cup\{/0,0/0\}$ to $\mathbb{R}^+\cup\{/0,0/0\}$, so we have $\log : \mathbb{R}^+\cup\{/0,0/0\} \rightarrow \mathbb{R}\cup\{/0,0/0\}$ is well defined. Thus we can define exponentiation $a^b$ with $a>0$ or $a\in\{/0,0/0\}$ by $$
a^b = \exp((\log a)b)
$$
from which we can see right away $a^{0/0}=0/0$ for any $a$, so as you suspect $\zeta(\bot) = \bot$.
