> The problem this and the other replies miss is that the standard definition of division is multiplication by the inverse.
Try to state this definition formally. The statement: ∀ x,y . x/y = xy⁻¹ is not a theorem of fields or a definition of division. However, ∀ x, y . y ≠ 0 ⇒ x/y = xy⁻¹ is, but is completely unaffected by defining division at 0. Those who think they see a problem rely on informal and imprecise definitions. Could you formally state a theorem that is affected? That would help you get around issues that are merely artifacts of imprecision.
But let's entertain you, and state that what we really mean by the informal and vague statement, "division is the inverse of multiplication," could be stated formally as:
∀ x ∈ dom(1/t). x(1/x) = 1
You are absolutely correct that this equational theorem is broken by extending the domain of division. However, there is absolutely no way to say that the formalization of this theorem isn't actually
∀ x ≠ 0 . x(1/x) = 1
because the two are equivalent. You cannot then claim that something is necessarily broken if you choose to pick a formalization that is indeed broken, while an equivalent formalization exists, that is not broken (not to mention that the formalization that is broken requires a strictly richer language). All that means is that your formalization in this case is brittle, not that laws are broken.
Try to state this definition formally. The statement: ∀ x,y . x/y = xy⁻¹ is not a theorem of fields or a definition of division. However, ∀ x, y . y ≠ 0 ⇒ x/y = xy⁻¹ is, but is completely unaffected by defining division at 0. Those who think they see a problem rely on informal and imprecise definitions. Could you formally state a theorem that is affected? That would help you get around issues that are merely artifacts of imprecision.
But let's entertain you, and state that what we really mean by the informal and vague statement, "division is the inverse of multiplication," could be stated formally as:
You are absolutely correct that this equational theorem is broken by extending the domain of division. However, there is absolutely no way to say that the formalization of this theorem isn't actually because the two are equivalent. You cannot then claim that something is necessarily broken if you choose to pick a formalization that is indeed broken, while an equivalent formalization exists, that is not broken (not to mention that the formalization that is broken requires a strictly richer language). All that means is that your formalization in this case is brittle, not that laws are broken.