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, $c=\overline{\frac{1}{q}},\;c,q\notin R$.

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

  2. $a + (-a) = c$

  3. $a (c) = c $

  4. 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.

  1. 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.

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    $\begingroup$ Then what meaning does $c=0$ have? Since they don't work the same with respect to division, they cannot be the same thing. But having $c=0$ implies they are the same. $\endgroup$ – Joe Johnson 126 Nov 29 '14 at 18:44
  • $\begingroup$ @JoeJohnson126 Yes, $c$ is supposed to work just like $0$, except with division. $\endgroup$ – Jonathan Cender Nov 29 '14 at 18:45
  • $\begingroup$ @JoeJohnson126 $c$ is a replacement for $0$. They are both designed to be numbers of nothing. $c$ just extends the number system a little further than $0$ did. $\endgroup$ – Jonathan Cender Nov 29 '14 at 18:47
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    $\begingroup$ You should explicitly describe the set of numbers, and then the addition and multiplication operations on this set. Is the set $\mathbb R$? Or $\mathbb R \cup \{c\}$? Or $(\mathbb R \setminus \{0\}) \cup \{c\}$? Where does this $q$ thing come from? $\endgroup$ – Chris Culter Nov 30 '14 at 5:40
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    $\begingroup$ From 5a, we have $q+cq=q$. Does it follow that $cq=0$? (Or, I suppose, $cq=c$?) $\endgroup$ – Chris Culter Dec 2 '14 at 6:46

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:


and we would hope that the function $1\overline{\div}x$ would reverse both of these inequalities, but instead we get


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.

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  • $\begingroup$ The new operation does work well on the right. $1/(c \cdot c) = (1/c)/c = q^2$. Note that $c\cdot c=c$, but in the denominator $c(c)=q^2$ since $1/q$ becomes "present" per rule 4a. $1/c$ doesn't "just" turn into $q$. "$a/b=aq$ if $b=0$" is too simple. As a divisor, $c$ becomes $q$ only after becoming present as $1/q$. Since $2(c)$ is in the denominator it is equal to $2/q$ so both sides equal $(1/2)q$. I hope $1/-c=-(1/c)$ makes sense now. Not sure what your intent is in noting that $1/(1/c)\neq c$. This is true. Following normal rules for complex fractions it equals $1/q$. $\endgroup$ – Jonathan Cender Dec 7 '14 at 23:26
  • $\begingroup$ The parentheses in $1/(c \cdot c)$ mean that we compute $c \cdot c$ first, call it $x$, and then compute $1/x$. If $c \cdot c = c$, then the result is $1/c$. This is just a matter of syntax; it doesn't have anything to do with the meaning of $c$ or the meaning of $/$. $\endgroup$ – Chris Culter Dec 8 '14 at 1:09
  • $\begingroup$ Well, I want to make it clear somehow that when a zero is a divisor, or, in this case, a member of a divisor, rule 4a applies. Maybe rewriting this without parens helps. Parens in your example specify $\frac{1}{c\cdot c}$. This to me says multiply the zeros as divisors, not as dividends. $\endgroup$ – Jonathan Cender Dec 8 '14 at 2:32
  • $\begingroup$ I think I see where you're coming from, but by definition, a binary operation like $\cdot$ can't be context-sensitive. It's impossible for $c\cdot c$ to act differently from $c$ in one context, while we have $c\cdot c=c$ in a different context. I think you're proposing to split the operation of multiplication into two distinct operations: one operation for multiplying numbers "as divisors" and another operation for multiplying numbers "as dividends". That would get really complicated! If you want to pursue that route, though, you should probably use different symbols for the two operations. $\endgroup$ – Chris Culter Dec 8 '14 at 5:50
  • $\begingroup$ 1)Then do you see rule 4a as valid? Esp. step $\div(\overline{\frac{1}{q}}\times \frac{q}{1})$ 2)Wondering why you refer to $q$ as a variable. It's intended to be a number like the imaginary $i$. Like the amount of digits depend on the base, $q$ does depend on the selected universe, e.g., $\mathbb{R}$ versus $\mathbb{Z}$ vs $\mathbb{Q}$, but it's not a variable. 3)Re order - these are not usual infinitesimals. Strange, but there is good reason to think that, altho not worked out in detail, the correct order is $c\leq q^{-1}<q^{-2}$. So their inverses do reverse $q\geq q>q^2$. $\endgroup$ – Jonathan Cender Dec 13 '14 at 2:25

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