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Binary operation

  1. Oct 7, 2006 #1

    radou

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    In the definition of a group on mathworld, http://mathworld.wolfram.com/Group.html, it is stated that the group operation is a binary operation, and it is stated that elements of a group must satisfy the four properties, including closure. Now, the definition of a binary operation, http://mathworld.wolfram.com/BinaryOperation.html, implies closure, so, isn't it unnecessary to talk about the property closure in the definition of a group?
     
    Last edited: Oct 7, 2006
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  3. Oct 7, 2006 #2

    AKG

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    What do you mean "isn't it sufficient to talk about the property of closure in the definition of a group?" It is not necessary to talk about the property of closure in the defintion of a group. But "not necessary" is not the same thing as "sufficient". Anyways, although it is not necessary, in theory, to talk about the property of closure, you are often just given a set S with a function * with domain SxS, and you have to verify that * is indeed an operation, that is, that closure does indeed hold.
     
  4. Oct 7, 2006 #3

    radou

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    Yes, I corrected that, I meant 'not necessary'. OK, I get it, it's more general, since we don't always know if * is a binary operation.
     
  5. Oct 7, 2006 #4

    matt grime

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    There is often redundancy in how axioms. The statements are usually made so as to appear as clean as possible. One can, and often does, define a group without invoking the strict definition of binary operation, and just takes it to mean some operation that takes two objects and gives a third without necessarily saying where the third lies.
     
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