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Counterexample to uniqueness of identity element?

  1. Jun 25, 2011 #1
    (Hopefully, this question falls under analysis. I was unable to match it well with any of the forums.)
    The proof that the identity element of a binary operation, f: X x X [itex]\rightarrow[/itex] X, is unique is simple and quite convincing: for any e and e' belonging to X, e=f(e,e')=f(e',e)=e'.
    However, if we take f(m,n)=max(m,n), it appears that any m will have multiple identity elements--the elements of the set of numbers n less than m.
    There must be something that I'm missing here. Any help would be appreciated. Thanks!
     
  2. jcsd
  3. Jun 26, 2011 #2

    HallsofIvy

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    You are missing the fact that individual elements of the set do not "have" their own identities. In order that "e" be the identity, we must have f(e, a)= f(a, e) for all a in the set. f(a,b)= max(a, b), over some ordered set, has an identity if and only if the set has an largest member.
     
  4. Jun 26, 2011 #3
    Thank you. :smile:
     
  5. Jun 26, 2011 #4
    Wouldn't the "max identity" be the smallest element? For example in the natural numbers 0, 1, 2, ..., we have max(0, n) = n for all n, so 0 is the identity of the max operation.
     
  6. Jun 26, 2011 #5
    I was thinking the same thing. Maybe that's what HallsofIvy meant.
     
  7. Jun 26, 2011 #6

    gb7nash

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    Correct. Assuming the smallest element is e, then max(e,a) = max(a,e) = a for all a in the set. If you're dealing with min, then you do need a largest element in order to have an identity that works for all numbers.
     
  8. Jun 27, 2011 #7

    HallsofIvy

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    Yes, of course. I don't know why I said "maximum".
     
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