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?


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