Talk:Binary operation: Difference between revisions
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Binary operation definition in Israeil textbooks is not the same as this article. |
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I think you are onto something, Axel. Binary operation on a set requires closure for the result, and the elements chosen must also be from the set, so (S x S ->S) makes more sense. |
I think you are onto something, Axel. Binary operation on a set requires closure for the result, and the elements chosen must also be from the set, so (S x S ->S) makes more sense. |
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WMORRIS |
WMORRIS |
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My textbook doesn't agree with the definition used on this article. I guess it is rather a convention or terminology problem than a real issue. |
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It defines a "binary operation: f:AxA -> B". Where closure isn't required. The definition of a [[Group|Group (Mathematics)]] includes the requirement for closure of course, again, conflicting with the Group article. This is the convention in Israel, I guess.. -- [[User:Rotem Dan|Rotem Dan]] 14:20 13 Jul 2003 (UTC) |
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Revision as of 14:20, 13 July 2003
There's also binary function. Merge them? -- JanHidders
I'm not sure they really mean the same thing. A binary operation is usually an algebraic operation, and is often denoted more like a*b than f(a,b). Probably the article ought to explain this. Also, if I had written the binary operation article from scratch I would have only allowed it to cover functions of the form f : S x S -> S, rather than the general f : S x T -> U. I didn't like to change the original too much, but perhaps it should be changed. In any case it would be a good idea to cross-link binary function and binary operation.
Zundark, 2001-08-08
I agree, binary operations are S x S -> S. This article simply describes functions with two arguments. I think it should be changed, and the popular infix notation a*b for *(a,b) should be mentioned. --AxelBoldt
Oh, you guys don't consider the vector scalar product (V * V -> R) or scaling of vectors (R * V -> V) or matrices ( R * M -> M ), etc to be binary operations? --Buz Cory
Perhaps we should ask "what would Eric Weisstein" have done?" :-) But he doesn't seem to be sure either. There is
and there is
which doesn't explicitly require the input domains to be the same. I know that in my own field (computer science) the term is used for any operator that needs two arguments. Perhaps it should be someting like this:
- begin with S x S -> S definition
- something about the notation
- a remark that sometimes also S x T -> U is possible, with Buz' examples and ref. to binary function
-- JanHidders
Maybe we should distinguish between a binary operation on a set (S x S->S) and a binary operation as such (S x T->U)? I don't know. --AxelBoldt
I think you are onto something, Axel. Binary operation on a set requires closure for the result, and the elements chosen must also be from the set, so (S x S ->S) makes more sense. WMORRIS
My textbook doesn't agree with the definition used on this article. I guess it is rather a convention or terminology problem than a real issue. It defines a "binary operation: f:AxA -> B". Where closure isn't required. The definition of a Group (Mathematics) includes the requirement for closure of course, again, conflicting with the Group article. This is the convention in Israel, I guess.. -- Rotem Dan 14:20 13 Jul 2003 (UTC)