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I'm reading Wikipedia about operations and binary operations . Intuitively I always thought that a binary operation is a operation that takes two arguments. But Wikipedia defines a binary operation as something where the domain has the form codomain times codomain. In this sense, subtraction with input natural numbers, is not a binary operation. (As Malice points out, I mean the operation $$-:\mathbb{N}\times\mathbb{N}\to\mathbb{Z}$$). This sound strange to me. Because if it is not a binary operation, what would you call it then? An operation with two arguments? It sounds more intuitive to me, to call it a binary operation, and subtraction with input integers, a closed binary operation.

I'm wondering if this is just a strange definition from Wikipedia, or that it is common to define a binary operation in this way.

Willemien
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Kasper
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  • It is common.${}$ – Git Gud Jan 28 '14 at 01:14
  • Sorry, misread the question at first. No, this is not just a weird Wikipedia thing, as far as I know. The usual meaning of a "binary operation on $X$" does seem to be a closed binary function. The function $\in:X\times\mathcal{P}(X)\to 2$ is not what most people would call a binary operation, for example. – Malice Vidrine Jan 28 '14 at 01:18
  • subtraction doesn't map ${\Bbb{N}}\times{\Bbb{N}}\to{\Bbb{N}}$ cuz for example, where does it go $(1,2)$? – janmarqz Jan 28 '14 at 01:18
  • @janmarqz: I think the OP is wondering why $-:\mathbb{N}\times\mathbb{N}\to\mathbb{Z}$ wouldn't be a "binary operation" (though I think their wording on this could be clearer). – Malice Vidrine Jan 28 '14 at 01:27
  • @MaliceVidrine yes, that is what I'm asking – Kasper Jan 28 '14 at 14:38
  • you are misreading the article http://en.wikipedia.org/wiki/Binary_operation a bit later it is more explained in detail it gives the example of devision (you cannot devide by zero) so devision is not a complete function – Willemien Feb 03 '14 at 16:06
  • very related: http://math.stackexchange.com/questions/796374/binary-operation-english-terminology ;) – MattAllegro Jun 04 '16 at 08:06

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In my experience, the definition of a binary operation as $f\colon S\times S\to S$ is standard. Certainly you would want $f\colon S\times S\to T$. Mapping back to $S$ though is very useful because you want to be able to repeatedly apply the map (say, to form a group). I guess the thing is that we just aren't usually interested in giving a name to $f\colon S\times S\to T$. It might be in some sense better to call that a binary relation and then talk about closed ones, but that gives the longer name to the thing we want to talk about more often.

Jessica B
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