Are there axioms for dividing by zero that don't destroy math?

Question

I concluded that - in a certain sense - all fractions are made up. That is, when we say $$\frac{a}{b}=x$$ We are merely declaring the number $x$ to satisfy the equation: $$b\cdot x=a$$ Of course there is much more to fractions than this. But it naturally came next - why not define division by zero to be exist? So write that $$\omega=\ "\frac{1}{0}" \quad\text{meaning simply that}\quad \omega\cdot 0 =1$$ We declare $\omega$ to have this property $\textit{and only this property}$. If we were to give $\omega$ the same properties every other number has then math is quickly "destroyed" - meaning every number is equal to every other number.

The next goal is to give $\omega$ as many properties of other integers as possible without destroying algebra.

For an example failed attempt, let's give $\omega$ $\textbf{distributivity}$ and an $\textbf{additive inverse}$. We have $$1=\omega\cdot 0=\omega\cdot(1-1)=\omega-\omega=0$$ and it happened.

It seems safe to give $\omega$ the following commutative properties $$\omega \cdot a=a\cdot\omega$$ $$\omega + a = a + \omega = \omega$$

Is there a foolproof way to check if giving $\omega$ some particular property is "safe"? (i.e. leaves us with a consistent number theory)

Answer 1

Even without an additive inverse of $\omega,$ it is not safe to give it mutiplicative distribution over addition: $$ 1 = \omega \cdot 0 = \omega \cdot (0+0) = \omega \cdot 0 + \omega \cdot 0 = 1 + 1 = 2. $$

Suppose you allow associativity of multiplication involving $\omega.$ Then $$ 1 = \omega \cdot 0 = \omega \cdot (0 \cdot 0) = (\omega \cdot 0) \cdot 0 = 1 \cdot 0 = 0. $$

If $1 + \omega = \omega$ then we must give up either the additive inverse of $\omega$ or associativity of addition with $\omega$, or else $$ 0 = \omega + (-\omega) = (1 + \omega) + (-\omega) = 1 + (\omega + (-\omega)) = 1 + 0 = 1. $$

What about the equation $x \cdot 0 = 2$? Is $x = 2\cdot \omega$ the solution of this equation? We cannot use associativity to prove this, so do we just take it as a definition that $(a\cdot\omega) \cdot 0 = a$ for all $a$?

What about $(x + x)\cdot 0 = 2$? Can we still rewrite the left-hand side as $x\cdot 0 + x\cdot 0$? Even when it might turn out that $x=\omega$?

What about $x \cdot (0 + 0) = 2?$ We already ran into trouble with this one; do we therefore have to make special rules restricting the operations we can perform on arbitrary polynomials, just in case we decide to put one equal to some positive number when ordinary arithmetic would reduce it to the zero polynomial?

So it's not just "properties of $\omega$" we have to give up; we have trouble even with arithmetic we used to do with ordinary numbers. I realize all of the above doesn't answer the main question (whether there's a foolproof way to tell which properties are "safe" for $\omega$), but I think that question may not even come close to covering all of the difficulties caused by adjoining the definition $\omega \cdot 0 = 1$ to ordinary arithmetic.

Answer 2

Regarding a foolproof method: Produce a model of the set of sentences (properties) you wish to hold. A theory $T$ is consistent if and only if there exists a model of $T$. (See, for example, the second form of completeness mentioned in the answer to Question about the proof of consistency iff satisfiability of a theory)

I interpret your question as being about a signature $(+, \cdot, 0, 1, \omega)$. Informally, what "produce a model" means in this context is to find a set $X$ and two binary operations $+$,$\cdot$, and a mapping of the constant symbols $0$,$1$,$\omega$ into $X$ that satisfy whatever list of sentences you have in mind. As mentioned in the comments, it's not hard to produce examples if you don't require distributivity.

For example, consider the set $\{a,b\}$ with $+$ exhaustively given by $a + a = a$, $a + b = b$, $b + a = b$, $b + b = a$ and $\cdot$ exhaustively given by $a \cdot a = b$, $a \cdot b = a$, $b\cdot a = a$, and $b \cdot b = b$. Interpreting $0$ as $a$, $1$ as $b$, and $\omega$ as $a$, we see $\omega \cdot 0 = 1$ in this model and I believe we can hand-check all commutative ring axioms except for distributivity. (This is a disguised version of XOR as $+$ and NOT-XOR as $\cdot$.) If I haven't messed up, this implies the commutative ring axioms without distributivity together with a constant $\omega$ satisfying $\omega \cdot 0 = 1$ won't produce an inconsistency.