I thought I answered this. Two zeros and the property of zero as an additive identity ends up breaking closure as defined by Apostol.
if there existed two 0s 01=x, 02=0 then 01+02=01. then define now 02=x, 01=0, 02+01=02. then Apostol seems to take these 2 equations and aruge that by commutativity 01+02=02+01 and therefore there is a unique 0? but I don't see how this proves the hypothesis by contradiction. haven't you proven only that x=x by tautology with 01+02=01, 03+02=03, 01=03? maybe I'm missing something in the proof.
I guess a universe with 2 0s would be kind of interesting like polarized to positive and negative 0
so in other words the sum 01+02 isn't in the original set? it's kind of nice to have a unique zero since you can equate orthogonal dot products. I've already done most of Stewart but Apostol is more difficult. there's an interesting proof in volume 1 using number theory sums to prove the integral of x^2 instead of epsilon-delta and so you can see how constant multiples can be taken outside the integral
I heard also Hubbard is more advanced than Apostol and maybe baby Rudin?
No, that's not what ThePerfectHacker is saying. (For one thing "01+ 02" does not exist because there are NOT two additive identities.) He's saying that if 01 is an additive identity then a+ 01= a for all a. In particular, taking a= 02 gives 02+ 01= 02. And if 02 is an identity then 02+ a= a for all a. In particular, taking a= 01 give 02+ 01= 01.
Since both 01+ 02 is equal to both 01 and 02, we must have 01= 02.
it's kind of nice to have a unique zero since you can equate orthogonal dot products. I've already done most of Stewart but Apostol is more difficult. there's an interesting proof in volume 1 using number theory sums to prove the integral of x^2 instead of epsilon-delta and so you can see how constant multiples can be taken outside the integral
I heard also Hubbard is more advanced than Apostol and maybe baby Rudin?
thanks! so by finding that 02=01, 02 and 01 are the same additive identity and is sufficient to prove that there is a unique 0 since you can use the same argument to prove that 02=03=04=0n
there is a group proof, Let (G, +) be a group and let 0 and 0' in G both denote additive identities, so for any g in G,
0 + g = g = g + 0 and 0' + g = g = g + 0'
It follows from the above that (0') = (0') + 0 = 0' + (0) = (0) and so 0'=0