Originally Posted by

**emakarov** I agree that extending equality from Z x Z* to Z x Z using the same formula, i.e., (a, b) = (c, d) iff ad = bc, does not work because this relation is not transitive: (a, 0) = (0, 0) = (c, d) for any c and d, but (a, 0) ≠ (c, d) if a ≠ 0 and d ≠ 0. My idea had nothing to do with rational numbers per se. I was saying that if we add the axiom $\displaystyle \forall x.\,x\ne 0\to x^{-1}=0$ to field axioms, we will get a consistent theory. So, given any field, including the field of rational numbers with the *standard carrier and definition of equality*, we can in addition define x / 0 to be 0 for x ≠ 0 and not get a contradiction. In particular, one cannot derive ∀a,b,c,d (a / b + c / d = (ad + bc) / (bd)) from field axioms even with this additional axiom; one needs the proviso b ≠ 0 ⋀ d ≠ 0.

I did not know about wheels, but what I am saying seems to be much less interesting than wheel theory. Wikipedia says that in wheels 0 * x is not necessarily 0 and x - x is not necessarily 0, so this is a substantially different mathematical object from a field.