Talk:Complex Numbers cannot be Totally Ordered

I think I added many details instead of drastically changing the framework of the proof... --kc_kennylau (talk) 06:09, 11 December 2016 (EST)

Difficult to tell. The first proof is more readable and can be followed easily. The structure and direction of the new one are unclear. --prime mover (talk) 06:20, 11 December 2016 (EST)
The problem is that the first statement of this proof is not yet demonstrated. In fact it is a combination of the antisymmetry of ordering and compatibility with addition. --kc_kennylau (talk) 06:22, 11 December 2016 (EST)
I'm inclined to agree with Kennylau on this one. If we keep the original version, we have to make it clear exactly what the contradiction is, and what is implied by the assumption thst there is an ordering. --GFauxPas (talk) 06:52, 11 December 2016 (EST)
Okay, I'll see what I can do to keep us all happy -- leave it with me, and we can take this forward when I've had a play with it, yeah? --prime mover (talk) 06:53, 11 December 2016 (EST)

I will say this though: pasting this up as a Featured Proof was a bit premature if it was of such poor quality in the first place. The accepted form is to offer it up as a POTW candidate, and give other contributors a chance to discuss it. I would not have made it POTW myself, but since the suggestion was made 16 microseconds before putting it live, the chance to discuss it was limited. --prime mover (talk) 06:57, 11 December 2016 (EST)

Sorry about that. I didn't realize that there was a probem. It was a PotW Candidate for 4 years, though. It's been a long time since the POTW was changed, so I picked one from the existing candidates that has been a candidate for a long time, without realizing that the proof could have been improved. --GFauxPas (talk) 07:09, 11 December 2016 (EST)
Was it? Silly me, I thought it had just been made a candidate. I apologise, I should have been wearing a head for a change.
How about this? Instead of saying "$0 \prec z \lor 0 \prec -z$, but not both", say "$0 \prec z \lor z \prec 0$, but not both".
Then, the contradiction will be "from $0 \prec -1$ we have $1 \prec 0$, by compatibility with addition." With that amendation, I'd say both proofs become the same. --GFauxPas (talk) 07:47, 11 December 2016 (EST)
The first proof shows that $0 \prec 1$ and $0 \prec -1$, contradicting the initial statement (a one-step process), and no more need be said. The contradiction has been demonstrated and the proof is complete.
The second proof takes two steps: once to prove $0 \preceq -1$ and the second to prove that $1 = 0$. Completely different. --prime mover (talk) 08:12, 11 December 2016 (EST)
I have added some more detail to $(1)$, so as to emphasise the specific result and the specific statement that leads to its own contradiction. --prime mover (talk) 08:21, 11 December 2016 (EST)
You will find that they are the same once you try to justify hypothesis $(1)$. --kc_kennylau (talk) 08:24, 11 December 2016 (EST)
No. We have specified that $\preceq$ is an ordering. Definition of an ordering: $a \preceq b \land b \preceq a \implies a = b$. A strict ordering is defined as $a \prec b \iff a \preceq b \land a \ne b$. Thus $(1)$. We do not need to wade through all that on this page. If you feel the need, link to whatever existing proofs already exist in order theory to justify the statement of $(1)$. --prime mover (talk) 08:31, 11 December 2016 (EST)
I linked to a result I made. This, I feel, keeps pm's proof the same but also alleviates kc's concerns. I'll leave it up to you guys to decide if kc's proof is worth keeping, because as far as I am aware he only made his proof to address the equivalence between what is now called condition $(1)$ and condition $(1')$. --GFauxPas (talk) 09:36, 11 December 2016 (EST)