| I can give you a simple proof by contradiction. 1. Let F be a field containing an element x =/= 0. 2. Suppose we have defined division by zero in F such that, for all x in F, there exists an element y = x/0 (i.e. F adheres to the field axiom of multiplicative closure). Note that at this point it does not matter how we have defined division by 0, we will just generously continue and assume you've done it in a way that maintains the other field axioms. 3. Since y = x/0, it follows that the product of y and 0 is equal to x, because division is the inverse of multiplication. By the field axioms, division does not exist if there is no multiplicative inverse with which to multiply. 4. But by the field axioms this implies that x = 0, which contradicts our initial assumption. Likewise, since we can repeat this procedure with any element x in F, this demonstrates that there exists no nonzero element x in F, and in fact F = {0}. The failure in the article's refutation is that this proof is designed to permit you to assume you have suitably defined division by zero, then proceed to demonstrate without any loss of generality that you could not possibly have unless 1) F is not a field, or 2) F contains only 0. The fundamental algebraic property you sacrifice by defining division by zero is uniqueness, and uniqueness is a hard requirement in fields with nonzero elements. |
Even if you could come up with another formalization that does cause a problem, e.g. `∀ x ∈ dom(1/t) . x(1/x) = 1` (and I would say that this is the only formalization that causes an issue, and it requires the use of a language with a dom operator, something that is absolutely not required for theories of fields), it won't matter because the question is not whether one could come up with a formalization that leads to contradiction, but whether there are reasonable formalizations of fields where this does not happen, and there are (in fact, most of them satisfy this, as they do not rely on a dom operator).
In addition, it is not true that "by the field axioms, division does not exist if there is no multiplicative inverse with which to multiply." It's just that the field axioms do not define what the meaning of division is in that case. Defining it, however, does not lead to contradiction with the axioms, at least not a contradiction you've point out. In fact, most common languages of mathematics cannot even explicitly express the statement "x is not in the domain of f." All they can do is define f(x) for values of x in the domain, and not define f(x) for values outside it. The "exist" in your statement does not refer to ordinary mathematical existence (usually formally expressed with the existential quantifier) but to an informal notion of definedness (discussed by Feferman) that has no formal counterpart in most formal systems, because it is very rarely needed; it is certainly not needed to state the field axioms.