Do these arithmetic rules work? They extend the number system by a zero not based on the empty set that is a divisor with unique quotients. These rules are part of an attempt to define an additive identity in terms of division in basic standard arithmetic. The difficulties with defining division by $0$ are well known. In order to circumvent the difficulties, this additive identity has not been based on the empty set. The $0$ defined in terms of subtraction has been replaced with a zero back engineered to make division feasible. The new zero is in two parts: one part indicates the absence of some numbers, the other indicates the absent numbers. I do want to focus on the division rules themselves as much as possible, especially any similarities to existing mathematics. Or to any fatal flaws in the rules. Go here for further background such as a geometric interpretation, or go to the link on my profile.
Symbol definition: Absence bar - a bar indicating that what is below it is absent. Used in constructing a number zero not based on the empty set. Example: $\overline{\frac{1}{q}}$ indicates $\frac{1}{q}$ is absent.
$\mathbb{R}^{\pm}\cup \{\overline{\frac{1}{q}}\};+; \times;<;\overline{\div}$ where $\overline{\div}$ indicates that division by the alternate zero is defined.
For brevity in what follows, $c=\overline{\frac{1}{q}},\;c,q\notin R$.


*

*$a + c = a,\; a \in R$.

*$a + (-a) = c$ 

*$a (c) = c $

*Division rules for making the absent additive identity $ c $ ''present''.
4a. $ 1 / c = c^{-1} = 1\div \overline{\frac{1}{q}} = (1 \times \frac{q}{1}) \div (\overline{\frac{1}{q}} \times \frac{q}{1}) = 1q \div 1 = q $
where $q\notin R $ and $\frac{1}{q} \;$ becomes present. Note that $c \neq \frac{1}{q}\;$ unless $c$ is a divisor.  
4b. $a / c = (a)(1q) = ((a)(1))q = aq$ 
where $aq \notin R$.
4c. $q(\frac{1}{q}) =  1 $ so that $aq(\frac{1}{q}) = aq^0 = a$
4d. $c/c = c((1)q) = ((c)(1))q = cq  \;\;$     (by rule 3) 
so $c/c \neq 1 ;\ cq \notin R$ 
4e. $c / q = cq^{-1}$ 
4f. $c / a = c \;\;$ (by rule 3) 
4g. $aq / c =  a(q|2)$
The notation $q|2$ indicates division by zero a second time instead 
of exponents.


*Rules for ''$q$'' numbers. Note that $q$ is like the imaginary $i$ in that it always appears with a Real number. So, $7i$ and likewise $7q$.


5a. $ aq + cq = (a+c)q = aq $
5b. $ bq + a = a + bq $ and is in lowest terms.
5c. $ aq + bq = (a+b)q,\;a,b \in R$
5d. $ a \times bq = (a\times b)q $
5e. $ aq \times bq = (a\times b)(q \times q) = (ab)q^2 $
5f. $ q^a \times q^b = q^{a+b}$
5g. $ q^0 = 1 $
5h. $ aq^n \times bq^{-p} = (ab)q^{n-p},\; n,p\in \mathbb{N}  $
Rules given above are for ''q '' numbers orthogonal to R . Further rules for other ''q '' numbers are not covered here. They are similar to rules for polynomials.
 A: Based on the edits and comments, it sounds like you're working with the field of rational functions in one variable over $\mathbb R$, which we can call $\mathbb R(q)$. You used the Latin letter $c$ to denote the additive identity, but I'll use the more common Arabic numeral $0$ to denote the same thing. Along with the usual operations on this field, you define a new operation $\overline{\div}:\mathbb R(q)\times\mathbb R(q)\to\mathbb R(q)$ by
$$a\overline{\div}b=\begin{cases}
a/b&\text{if}\ b\neq0,\\
aq&\text{if}\ b=0.
\end{cases}$$
This operation has nice associtivity and distributivity properties on the left. For all $a,b,c\in\mathbb R(q)$, we have:
$$(ab)\overline{\div}c=a(b\overline{\div} c)$$
$$(a\pm b)\overline{\div}c=a\overline{\div}c\pm b\overline{\div}c$$
In particular,
$0\overline{\div}a=0,$
which is nice.
On the other hand, the new operation doesn't combine so well on the right:
$$1\overline{\div}(1\overline{\div}0)\neq0$$
$$1\overline{\div}(0\times0)\neq(1\overline{\div}0)\overline{\div}0$$
$$1\overline{\div}(0\times2)\neq(1\overline{\div}0)\overline{\div}2$$
$$1\overline{\div}(-0)\neq-(1\overline{\div}0)$$
So that's unfortunate.
We can also make $\mathbb R(q)$ into an ordered field in such a way that $q$ is infinite and $q^{-1}$ is infinitesimal, but $\overline{\div}$ doesn't play all that well with the order. We have:
$$0<q^{-2}<q^{-1}$$
and we would hope that the function $1\overline{\div}x$ would reverse both of these inequalities, but instead we get
$$q<q^2>q.$$
So on one hand, $\overline{\div}$ can be given a simple definition which enjoys some nice properties, even when $0$ appears on the right. On ther other hand, it doesn't extend other properties one expects from a "division" operation, and that's probably going to make it unwieldy in practice.
