Talk:Order of Squares in Totally Ordered Ring without Proper Zero Divisors

When I first found this theorem, its title claimed it worked for totally ordered rings, its stated premises called only for ordered rings, and its proof relied on a (partially) ordered field. I moved that to a page with an appropriate title, and I came up with this version. I failed to prove a version for totally ordered rings, and I doubt there is one, but are welcome to try to prove/disprove it. --Dfeuer (talk) 04:28, 14 January 2013 (UTC)

What's wrong with the punctuation, exactly? --Dfeuer (talk) 10:53, 14 January 2013 (UTC)


 * Looking at it again, it seems only to fail to precede lines of equations with a colon. On some other pages you wrote mainly yesterday, there are some sentences not properly ending with a dot. --Lord_Farin (talk) 11:02, 14 January 2013 (UTC)

While you're looking at it, I found the structure of this proof rather inelegant as I wrote it. I pondered possible lemmas about ordered rings without zero divisors, but didn't come up with anything good just then. --Dfeuer (talk) 10:55, 14 January 2013 (UTC)


 * 't Would appear that a proof can be crafted using $R$ is totally ordered, hence $x \not\le y \iff x > y$, after which it only remains to remark that $x^2 = y^2 \implies x = y$ because $x,y$ are positive. Such seems to drop the "without zero divisors condition". --Lord_Farin (talk) 11:02, 14 January 2013 (UTC)
 * Hm, not sure what I use in that last step... --Lord_Farin (talk) 11:04, 14 January 2013 (UTC)
 * In general, it may be good to first put in a rename request and an invitation for other people to shed their light on the matter before moving things. This reduces the amount of maintenance work. --Lord_Farin (talk) 11:03, 14 January 2013 (UTC)
 * OK. I also don't know what you use in that last step. As far as I can tell, if a ring has zero divisors, $(0 < a) \land (0 < b) \not\implies  (0 < a \circ b)$, because $a \circ b$ could be $0$, and that makes a mess of things. --Dfeuer (talk) 11:08, 14 January 2013 (UTC)


 * No zero divisors is essential for my step. $x^2 - y^2 = 0 \iff (x+y)(x-y) = 0$. Since $x,y> 0, x+y> 0$, hence $x-y=0$. --Lord_Farin (talk) 11:20, 14 January 2013 (UTC)


 * Also: Difference of Squares requires a commutative ring. --Dfeuer (talk) 11:21, 14 January 2013 (UTC)

By the way, if $R$ has a non-null trivial subring, then $0_R \le y^2 \implies 0_R \le y$ is clearly false. --abcxyz (talk) 18:05, 14 January 2013 (UTC)

My revert
I wanted to get the proof back to what I think was a correct state as quickly as possible. Either of us can fix your version up. I don't think it will take much. --Dfeuer (talk) 17:35, 14 January 2013 (UTC)


 * If $0 \le y < x$, then $x > 0$. What's wrong? --abcxyz (talk) 17:37, 14 January 2013 (UTC)


 * Nothing. --Lord_Farin (talk) 17:37, 14 January 2013 (UTC)


 * You're absolutely right. I must have been hallucinating again. Dammit! If Lord_Farin didn't reinstate your version, and you didn't, then I'll do it right now. --Dfeuer (talk) 17:46, 14 January 2013 (UTC)


 * Dfeuer, if you would be as kind as to give other people the opportunity to comment on changes. Now that you've been promoted to admin status, you bear a greater responsibility for the well-faring of this web site. Such includes, when it comes to actual content (thus, mathematics), not jumping the gun and instead placing comments and perhaps Template:Questionable calls. In addition to that, you are also to be held more personally responsible for your edits and newly posted proofs - in particular in the house style department. As asked before, please increase your efforts to adhere to it. I won't be chasing your tail to help you out in that regard forever - (sometimes) I have responsibilities in the real world, and I like to contribute by posting up more results, not by cleaning up all the time. --Lord_Farin (talk) 17:48, 14 January 2013 (UTC)


 * I apologize. --Dfeuer (talk) 18:17, 14 January 2013 (UTC)

Statement regarding the usual number fields
The original version of this page (by Prime.mover) eschewed the $x^2$ notation for the ring product $x \circ x$. Perhaps it's conventionally used only for certain things? I really don't know myself, so I just left that as it was. The statement about the usual fields $\R,\Q$, etc. then put it in a more familiar notation. --Dfeuer (talk) 18:17, 14 January 2013 (UTC)


 * Hmm, somehow I thought I saw it somewhere else. I can't find it now, so I guess I'll change it back. --abcxyz (talk) 18:20, 14 January 2013 (UTC)

Presentation
Using a LaTeX object as a section title is not optimal as it's not obvious it's a section title. --prime mover (talk) 17:41, 14 January 2013 (UTC)


 * Agree. When I first saw the page, I was actually quite confused. --abcxyz (talk) 17:48, 14 January 2013 (UTC)