Talk:Complex Numbers cannot be Totally Ordered
- 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)
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".
- 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.
- 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)
- Yes, this is what I was taking for granted as a result: Properties of Ordered Ring. Still needs completing and restructuring. --prime mover (talk) 10:35, 11 December 2016 (EST)