# Reference Request: Had this line of solution about division by zero been investigated before?

If we add a single transfinite cardinal number $$\aleph$$ to the non negative reals, weaken the arithmetical operators to be ternary relations instead of functions, and define subtraction and division as converse relations of addition and multiplication; define division by zero as self subtraction. Then in some sense the problem of division by zero would find a solution! The results are:

$$x/0 = \aleph$$, for $$x \neq 0$$

$$x/\aleph=0$$, for $$x\neq \aleph$$

$$0/0$$ range over all numbers.

$$\aleph/\aleph$$ range over all numbers.

The formal workup involves:

Define: $$[(x \ \tilde o \ y) \to z] \iff \tilde o(x,y,z)$$

Where $$\tilde o$$ stand for any arithmetical operator. (the arrow $$\to"$$ is not to be confused with implication, it only means "is a result of", so $$(x+y) \to z"$$ means: $$z \text{ is a result of summation of } x \ and \ y$$).

If $$x',y'$$ are particular numbers and if there is a unique $$z'$$ such that $$(x' \ \tilde o \ y') \to z'$$, then and only then this can written in functional manner as: $$x' \ \tilde o \ y' = z'$$

If there is no unique $$z'$$, then we write that using inequalities, like for example:

$$k \leq x' \ \tilde o \ y' \leq l$$

Of course $$x,y,z,x',y',z',k,l$$ above are meant to be numbers and not sets of numbers. However with the inequalities the expression $$x \ \tilde o \ y$$ can be seen to correspond to the set of all numbers within the range of those inequalities.

Define: $$(x - y) \to z \iff (z+y) \to x$$, for $$y \leq x$$

Define: $$(x \div y) \to z \iff (z \times y) \to x$$

Define: $$(x \times 0) \to y \iff (x-x) \to y$$

Now we take the following to be rules:

$$\forall x [x + \aleph = \aleph]$$

$$\forall x [x \times \aleph = \aleph]$$, for $$x \neq 0$$

If we add the rest of Cantor's transfinite cardinal numbers, then we'd get:

$$x/0 \geq min \ \aleph_{\alpha} \ (x \leq \aleph_{\alpha})$$; for $$x \neq 0$$

So we'll have many $$x/0$$ $$(x \neq 0)$$ each can be thought to correspond to a set of a numbers! However, all of those would have ONE inverse, that is $$0$$.

Other results are:

$$x/\aleph_{\alpha}=0 \iff x < \aleph_{\alpha}$$

$$x/\aleph_{\alpha} = x \iff x > \aleph_{\alpha}$$

$$\aleph_{\alpha}/\aleph_{\alpha} \leq \aleph_{\alpha}$$

This would entail that $$0/0$$ corresponds to the largest indeterminate set, that is the set of all numbers, while each $$\aleph_{\alpha}/0$$ would correspond to a proper subset of that, and the least is the $$\alpha$$ value the bigger is that subset; i.e., the more indeterminate it is!

We can easily extend this method to have negatively signed rationals as well.

Question: had this method been investigated before?