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Right inverse, left inverse, binary operations

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data
    If * is a binary operation on a set B, and the domain of definition is B^2, if * is associative and the neutral element is p. If r and l are elements of b we can say that r is a left inverse of l under * iff r * l = p, and l is a right inverse of r iff l * r = p. Show that if an element of B has a left and right inverse, then they are equal.


    3. The attempt at a solution
    Does the neutral element have anything to do with finding the answer, also what does associativity have to do with finding the answer? All I can think of is since there is only one neutral element in *, r = p and l = p, but I don't think that is the answer.
     
    Last edited: Feb 10, 2014
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  3. Feb 10, 2014 #2

    Dick

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    Sure, you have to assume there is a neutral element p, so that for any x, x*p=p*x=x. Now suppose r is a right inverse of x (so x*r=p) and l is a left inverse of x (so l*x=p). Associativity tells you (l*x)*r=l*(x*r), yes? So?
     
  4. Feb 10, 2014 #3
    Does it matter than the question say l * r = p and r * l = p, where does the x come from? would this mean (l * r) * l = l * (r * l)?
     
    Last edited: Feb 10, 2014
  5. Feb 10, 2014 #4

    Dick

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    The letters were confusing and I changed them. Your writing (l * r) * l = l * (r * l) shows why. The definitions you've got are using 'l' to be a symbol for both the right and left inverse. They are two different things. Those definitions mean something independent of the particular symbols used. Suppose I tell you that b is a right inverse of c. Let's keep the symbol 'p' to mean the neutral element, though I probably would have used the symbol '1' instead. It's more suggestive. What does that mean?
     
    Last edited: Feb 10, 2014
  6. Feb 10, 2014 #5
    Then why in the question is l * r = p, and r * l = p defined? btw r and l do not mean right and left inverse, they are arbitrary variables, the question says, "l is left inverse of r under * iff l * r = p, and l is a right inverse of r under * iff r * l = p". It is kind of confusing.
     
  7. Feb 10, 2014 #6

    D H

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    First off, look at some numbers you know, such as the rationals. With * being the multiplication operator you learned long ago and p being 1, what is the left inverse of 2 ? The right inverse? The answer is obviously 1/2 in both cases: 1/2*2 = 2*1/2 = 1. The goal of this problem is to show that if an element of B has both left and right inverses and if * is associative then there is only one inverse (i.e., the left and right inverses are one and the same).

    Note that this result does not necessarily mean that * is commutative. What it does mean is that an element of B commutes with its multiplicative inverse.

    Dick took the correct approach by giving that element of B that has both a left and right inverse a new symbol, x.
     
  8. Feb 10, 2014 #7

    Dick

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    Yes, they are arbitrary variables, and yes, it's kind of confusing. The reason it's confusing is because they made such a poor choice of arbitrary symbols. Try to unscramble it. You didn't answer my question. I'll repeat it. Suppose I tell you that b is a right inverse of c. What does that mean? Bypass the whole 'l' and 'r' thing.
     
    Last edited: Feb 10, 2014
  9. Feb 10, 2014 #8
    c * b = p, where p is the neutral element
     
  10. Feb 10, 2014 #9

    Dick

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    Bingo! Now I'll tell you that d is a left inverse of c. Same question. And then can you show b=d?
     
    Last edited: Feb 10, 2014
  11. Feb 10, 2014 #10
    (d * c) * b = d * (c * b)
    d * c = c * b = p
    p * b = d * p
    so this implies d = b, right?
     
  12. Feb 10, 2014 #11

    Dick

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    Right. Couldn't be clearer. I'm almost wondering if they didn't choose the symbol choice in original definitions to try to test you and throw you off track.
     
  13. Feb 10, 2014 #12
    Thanks for the help!
     
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