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)


 * It be noted however, that to a certain extent (in casu the placement of apt maintenance templates) new results are more important than their first form. If you find yourself noting that something doesn't adhere to house style and incapable of resolving that, it'd help a great deal if you'd indicate that yourself. From my point of view, that's an acceptable intermediate step towards mastering said house style. I'm not trying to burn you down here, just putting in some critique that hopefully gives direction to your journey of discovery and skill acquisition. --Lord_Farin (talk) 18:33, 14 January 2013 (UTC)


 * Noting the change to the result you just made while I was typing this, it is generally considered that equation labels should be on the left of the equation. Maybe the indentation of equation lines can be dropped if the line has a label as well; such prevents awkwardly large indentation. --Lord_Farin (talk) 18:33, 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)

Precedes vs. strictly precedes
Do we need to note explicitly that $<$ is the reflexive reduction of $\le$, or use some terminology Lord_Farin was trying to work out a bit ago to express their relationship in a friendlier fashion? --Dfeuer (talk) 18:34, 14 January 2013 (UTC)

Subtraction
The original version of this page used $a+(-b)$ where I've used $a-b$. Since we're dealing with a ring, so that $+$ is commutative, this strikes me as unnecessary and confusing, but I figured I should check to make sure there wasn't a good reason for it. --Dfeuer (talk) 19:32, 14 January 2013 (UTC)


 * Yes there is a good reason. --prime mover (talk) 21:26, 14 January 2013 (UTC)


 * ... namely? --Lord_Farin (talk) 21:27, 14 January 2013 (UTC)


 * We are talking about a general ring. There is always a danger, when using conventional mathematical symbols to denote the operations on such a ring, to consider them as "being" those mathematical operators. Therefore it is a good idea to divorce the thinking as far as possible from the operations on numbers that readers may associate them with.


 * To that end, on a general ring, the "subtraction" symbol is not defined as a primitive: what you have got is the $+$ operation and there is the symbol $-a$ meaning the inverse of the element $a$ under the $+$ operation. While it is generally understood that $a-b$ is a non-primitive abbreviation for $a+(-b)$, I would counsel against using it.


 * But it has nothing to do with me. Do what you have to do. My views, as pointed out and explained in full elsewhere on this website, are "absurd". --prime mover (talk) 21:41, 14 January 2013 (UTC)


 * I think that general notion is valid, but I see it as also important for students to learn just where certain notation does become valid. Addition becomes valid once you have a semigroup. Subtraction becomes valid once you have an abelian group. Multiplication becomes valid once you have a ringoid. Division doesn't become valid until you have a full field. --Dfeuer (talk) 21:48, 14 January 2013 (UTC)


 * Incidentally, do you have much experience in teaching this subject, and if so at what level? --prime mover (talk) 22:21, 14 January 2013 (UTC)


 * I still believe it is not helpful to start using $a - b$ while it is still being established what properties $a + b$ and $a+(-b)$ have. Note that $a-b$ is not commutative or associative and therefore the usual proofs that apply to commutative and associative operations cannot be used - and as the object being discussed here is not an integral domain yet, it is prudent to be as explicit as possible. If you don't like it, then you will overrule me as always by using argumentum ad passiones, so what I'll do is wait till you're bored with this site then come back and change it back again. --prime mover (talk) 22:09, 14 January 2013 (UTC)

This one's a tough call. I don't feel strongly and there are good arguments for both sides. While personally I'd go with Dfeuer, I think it's best for the site to go with PM. --Lord_Farin (talk) 22:19, 14 January 2013 (UTC)