Talk:Wheel theory

References and definitions

Any references, definitions?

Charles Matthews 08:37, 4 May 2004 (UTC)[reply]

The link appears to give a reference to a forthcoming article by the contributor. While it appears to be perfectly legit, it seems like too much a case of original research at the moment...maybe wait a couple years until after publication, or once the definition (which isn't given here) has been used by a few others. Revolver 09:19, 15 Oct 2004 (UTC)

I've looked at an online paper; this area does seem to be of possible interest, though as you say this formulation is yet young. Charles Matthews 10:47, 15 Oct 2004 (UTC)

I've been able to find several other names associated with this. It seems to go back to the mid 1990s at least. I don't see any problem, but given it's pretty new, a list of references would be good. Revolver 09:40, 19 Oct 2004 (UTC)

Does anyone speak sweedish?

Quite a few Sweedes, I understand - but they're mostly in Sweeden, the country next to Noorway.

I started this article once. It is true that I am the author of the abovementioned paper, published in Mathematical Structures in Computer Science 14(1):143–184, Cambridge University Press, 2004. I have now learned that it is against the rules to write wiki articles about ones own inventions. I did it because people contacted me, saying that it was stated in Wikipedia that you cannot divide by zero, and they frequently thought that my results had disproved this claim. I had to explain to them that in the ordinary sense of division, you cannot divide by zero. I edited the article about division by zero, explaining roughly the same thing. I inserted a link there to this article, and I wrote it. Now, different people have expanded on the topic. A few minutes ago, I added a couple of sentences in the end. As long as noone objects, I will let the article remain here.

Jesper Carlström

No objections here. The reason so-called vanity articles are not allowed on Wikipedia is to ensure that the article does not contain original research, that the content is verifiable and notable, and that the article is written from a neutral point of view. Since you're citing a doctoral thesis and don't include weasel words, these requirements are definitely met.

— Kwi | Talk 19:57, 11 October 2006 (UTC)[reply]

Constants "0" and "1"

Are the constants "0" and "1" in wheels additive and multiplicative identities, respectively, as they are in fields? If so, shouldn't this be mentioned as an axiom? mike4ty4 00:12, 17 May 2007 (UTC)[reply]

Its in there

Addition and multiplication are commutative and associative, with 0 and 1 as units respectively

--Salix alba (talk) 00:29, 17 May 2007 (UTC)[reply]

"Unit" means something has a multiplicative inverse, but this does not specify what the multiplicative identity or additive identity is. "Unity" refers to the multiplicative identity, although I think "unit" can also be used to refer to identity, however it is confusing nonetheless. This is at least how it is in ring theory. Even then, in the article ring (mathematics) it is mentioned as a distinct axiom in defining a ring. Is the situation different for a wheel? Regardless, it is still ambiguous and definitions need to be crisp. mike4ty4 07:24, 17 May 2007 (UTC)[reply]

I think it was me who put it in there, so I've fixed it now. --Ørjan 00:49, 18 May 2007 (UTC)[reply]

Thanks! mike4ty4 08:31, 1 June 2007 (UTC)[reply]

0 = 1

Is the algebra still a wheel in the case where zero equals one? This seems rather degenerate. 76.190.157.141 (talk) 00:32, 26 February 2009 (UTC)[reply]

Well the axioms are variety axioms, so none of them require anything to be distinct, and a degenerate wheel is thus still a wheel.

A different question is whether 0=1 means everything else is also equal. Hm. We get

x = 1x = 0x
x + y = 0x + 0y = 0xy = xy (first derived identity in article)
/0 + x = 0/0 + x = 0/0 = /0
/x = /(0x) = /0/x = /0 + /x = /0
x = //x = //0 = 0

So 0=1 makes everything else equal. --Ørjan (talk) 01:31, 26 February 2009 (UTC)[reply]

x/0

Consider the wheel made by extending the real numbers into a wheel. Are 1/0 and 2/0 distinct? Or are they the same? I can't seem to derive either result from the rules. Only that /0 + /0 = 0/0.

Is it that /0 + finite = /0 ? How do you prove that?

Or are these things not provable, but rather just one possible wheel that you can make out of the real numbers? --AndreRD (talk) 21:51, 1 December 2014 (UTC)[reply]

Let a be any value (e.g. any nonzero original real number) such that 0/a = 0, then

a/0 = //a /0 = /0 //a = /(0/a) = /0

Also, if 0z = 0 then

/0 + z = 1/0 + z + 0 0 = (1 + 0z)/0 = /0

--Ørjan (talk) 07:09, 3 December 2014 (UTC)[reply]

History and applications

It would be nice if this article had sections describing the history and applications.—Anita5192 (talk) 01:24, 17 November 2018 (UTC)[reply]

Has this concept ever appeared outside the articles that introduced it?

Also, in what way (if at all) does it clarify the question of division by 0. 73.89.25.252 (talk) 23:22, 15 June 2020 (UTC)[reply]

@73.89.25.252: I added two more sources by the same guy, but, to be honest, I suspect this article does not meet Wikipedia's notability standards. I will revisit it later, to at least achieve understanding of the mathematics for myself ;) Notrium (talk) 23:43, 15 June 2020 (UTC)[reply]

Thanks. The abstract for the second article says that the 1/0 can be interpreted as "ERR", which does not sound like a formalism that accomplishes anything. Abstract for first article says that it solves (with 1/0 defined as 0) some problem of equational presentation "without hidden functions". I don't know what hidden functions means here, but this sounds potentially more interesting. 73.89.25.252 (talk) 23:54, 15 June 2020 (UTC)[reply]

/ used as binary operator

The article gives "0/0 + x = 0/0" as an example, but / is a unary operator, so this is incorrect, so I assume this is a typo somehow. 46.223.43.217 (talk) 20:34, 24 October 2020 (UTC)[reply]

I think what happening is that multiplication is represented as juxtaposition "x y". So using "*" for multiplication "x/y" is really "x*(/y)" i.e. the multiplication of x with the inverse of y. You can think of normal division in this way, the multiplication of x with the reciprocal of y. --Salix alba (talk): 21:26, 24 October 2020 (UTC)[reply]