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Does an operation have to be closed?

  1. Jan 21, 2015 #1
    According to both Wikipedia and Wolfram MathWorld, a binary operation must be closed. Wiki does leave room for a so-called external binary operation, i.e. a function from (K X S) to S, but not from (S X S) to K. (This would make the operators in physics actually "external operators", no? Do we edit all the physics books?) Wikipedia, for example, says in its site on binary operations that the inner product is hence not a binary operation. Yet elsewhere (on the site "inner product") Wiki defines the inner product as an "algebraic operation...". If the latter article is only fudging because the author didn't find the proper word, what word is appropriate for an inner product? My guess would be a map (since a map, being a function, can have domain ≠ codomain). Or does one just leave it as "relation"?
     
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  3. Jan 21, 2015 #2

    Mark44

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    "Operator" is a term that can have different meanings in different contexts. I'm not familiar with the physics usage, but an "operator" in mathematics can map a function to another function in one context, or more generally, can map a vector space to itself.
    "Operation" is such a general term that it can mean just about anything.
    Either function or map would be fine. An inner product maps a pair of vectors from some space to whatever the relevant field of scalars happens to be.
     
  4. Jan 21, 2015 #3

    Stephen Tashi

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    Both Wikipedia and Wolfram require that a "binary operation on a set" be closed. One can talk about a binary operation without saying "on a set".

    A function whose domain is a sets of ordered pairs of elements from a set D and whose codomain C is a different set C can be considered a binary operation. For example the function that computes inner product of two vectors is a binary operation. It's true that such binary operations are usually named by some special terminology that doesn't incorporate the phrase "binary operation".
     
  5. Jan 21, 2015 #4
    thanks for the replies, Mark44 and Stephen Tashi. First, Stephen, you apparently meant "cannot" instead of "can" in your sentence
    Yes, I recognize that I was a bit sloppy.
    When you said
    you then contradict Wikipedia, and I suppose, by extracting from the two examples
    also Mark44.
    But in sum, it appears that both you and Mark44 are saying that the usage of the word is, well, vague, not as precise as Wiki would have me believe.
     
  6. Jan 21, 2015 #5

    Stephen Tashi

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    No, I mean: One can talk about a binary operation without saying "on a set". i.e. It is possible to call something a binary operation without saying it is a binary operation on a set.
     
  7. Jan 21, 2015 #6
    I presume you mean that the "on a set" would be implicit.
     
  8. Jan 21, 2015 #7

    Stephen Tashi

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    No. I mean that there can be "binary operations" that are not "binary operations on a set".
     
  9. Jan 21, 2015 #8
    Could you give me an example?
     
  10. Jan 21, 2015 #9

    Fredrik

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    To me, a binary operation is a map from ##S\times S## into ##S##, where ##S## is some set. I think it would be pretty strange to allow a different codomain, but if someone wants to do that, I can't say that it's wrong. It's just terminology. It's not like every math book agrees about the meaning of every term.

    As others have said, "map" and "function" are both fine. Since you asked specifically about inner products, you may also want to look up the terms "bilinear form" and "sesquilinear form".
     
    Last edited: Jan 21, 2015
  11. Jan 21, 2015 #10
    Thanks, Fredrik. Indeed, "bilinear form" and "sesquilinear form" are just the terms I was looking for. :D
     
  12. Jan 21, 2015 #11

    Stephen Tashi

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    As I said, the function that takes the inner product of two vectors.
     
  13. Jan 22, 2015 #12
    Thanks, Stephen Tashi. First, it seems from the rest of the discussion that one may avoid calling the function that takes the inner product of two vectors a binary operation, by using the terms "bilinear form" and "sesquilinear form". But be that as it may, your example is a function (from the set of ordered pairs of vectors to the desired field), and a function is defined as a set. No?
     
  14. Jan 22, 2015 #13

    HallsofIvy

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    I am not sure where you are going with this. Yes, you can define a function as a set. In particular, you can define a "binary" function, from [itex]X\times Y[/itex] to Z as a subset of the set of all ordered triples, {(x, y, z)} where x and y can be any members of X and Y, respectively, and z is the resulting value of the function. But anything can be defined in terms of sets. This is very different from say "on a set" which implies the same set for all variables.

    And while you can "avoid calling the function that takes the inner product of two vectors a binary operation" that does not mean it is NOT one! Another binary operation that is not "on a set" is the operation that maps (x, y), where x is any integer and y is any non-zero integer, to the rational number [itex]\frac{x}{y}[/itex]. That cannot be called a "bilinear form" or any kind of "linear, sequilinear form" of any kind of "linear form" because it is not linear. A binary operation "on a set" is specifically one that maps an ordered pair, (x, y), with each a member of the same set, to a member of that same set.
     
    Last edited: Jan 22, 2015
  15. Jan 22, 2015 #14

    Fredrik

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    Most of us will never have a reason to go outside the branch of mathematics called ZFC set theory, and as long as we stay inside it, pretty much every single "object" that we encounter is a set. This includes functions, relations, integers, real numbers, elements of sets, ordered pairs, etc. The only exceptions I can think of are terms for things in the language that we use to talk about sets. For example, the term "series" is sometimes defined so that a series is a notation that represents either a sequence or the limit of that sequence, depending on context.
     
  16. Jan 22, 2015 #15

    Stephen Tashi

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    Yes, all functions have a set that is their domain. Are you hinting that all operations defined by a function of two variables are "binary operations on a set"? No, that isn't true. Unfortunately, we can't always deduce mathematical definitions by manipulating the English language.
     
  17. Jan 22, 2015 #16
    Many thanks, HallsofIvy, Fredrik and Stephen Tashi.
    First, HallsofIvy, points taken: I did not know that "on a set" means that domain=codomain X codomain, so I stand corrected. (I like to be corrected.)
    Fredrik, your suggestion sounds to me like model theory's idea of a set, whereby the sets are entities in the model (interpretation, semantics) and the theory (syntax) is not (at least with respect to that model/theory combination).
    That question was the one that started this thread. You are right that natural language is not usually mathematically rigorous, although it can be, so I asked for a rigorous definition of binary operation. A couple of posts emphasized that the term is used rather loosely, and the more rigorous posts implied that your example is not a binary operation. So while not all functions of two variables are binary operations, all binary operations are functions of two variables. Adding to that the point that was made that "on a set" for a function of two variables meant that the domain = codomain X codomain (which corrected my impression that "on a set" meant that the domain was a set), the conclusion is that if you use the word "operation" in the same (strict) way in both instances of that question -- that is, if you are defining an operation (in the sense of a function SXS to S) in terms of a function, then that function must be on a set in order that your definition meets the requirements to be a binary operation. However, if you were using "operation" in the question in two ways-- the first "operation" in the loose fashion, and the second one in the strict fashion -- then no, not all operations defined by a function of two variables are binary operations (on a set, by definition of binary operation).
     
  18. Jan 22, 2015 #17

    Stephen Tashi

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    For the sake of completeness (as the saying goes), the web says that people use the phrase "external binary operation" , referring to an operation more a general than a "binary operation on a set" but less general than a "binary operation".
     
  19. Jan 22, 2015 #18
    Thanks, Stephen Tashi. That would be one of those terms that remains implicit in most people's use of the term, supposedly understood from the context. But implicit distinctions sometimes get forgotten, leading to confusion -- or at least to my confusion.
     
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