4 ms·
You're right. It really would've been nice for the author to offer a direct proof. However, while I'm a bit rusty, I think it basically has to be true. For the
by hyperpape 8y ago
You're right. It really would've been nice for the author to offer a direct proof.
However, while I'm a bit rusty, I think it basically has to be true. For there to be an inconsistency, 1/0 = 0 must either
a) imply the negation of some previous theorem of arithmetic or
b) imply that 1/0 = x, for x != 0.
I think a) can only be true by way of b), since no existing theorem of arithmetic involves the expression y/0 for any y. I confess, I don't know the right way to prove that b cannot be a consequence, but it doesn't look like one.
Edit: I think it's just trivial to take a model of the existing axioms, then add n/0 = 0 to the division relation.
- j2kun 8y agoYou have to add extra "except when"s to every theorem involving division. E.g.: (a+b)/c = a/c + b/c (in particular, for C != 1, as the author claimed it would work not just for 0 for for any real number) Now to say this is true you have to say "unless c = 0", whereas before that was automatic from the definition of division.
- hyperpape 8y agoThe same theorem forall c != 0, (a+b)/c = a/c + b/c is true before and after. It may be helpful to observe that while using the same symbol '/' in both cases is confusing, mathematical theorems are not about symbols, but about particular mathematical objects. The old theorem is about the division relation defined for real numbers != 0, the new one is a relation defined for all real numbers, that just happens to coincide.
- j2kun 8y agoThat is exactly my point. Now you have to disambiguate a poorly-chosen symbol. The theorem doesn't need the stipulation that c != 0 since you have already excluded c from the domain.