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Left and right Identities

  1. Jan 16, 2012 #1
    If an operation has two left identities, show that it has no right identity.
    [itex]_{}[/itex]
    pf/
    Let e[itex]_{1}[/itex] and e[itex]_{2}[/itex] be left identities such that e[itex]_{1}[/itex]≠e[itex]_{2}[/itex]. Assume there exist a right identity and call it r.

    Then we have that
    e[itex]_{1}[/itex]x=x
    e[itex]_{2}[/itex]x=x and
    xr=x.


    From here I want to try and show that there can not be a right identity but I don't see where to go.
     
  2. jcsd
  3. Jan 16, 2012 #2

    gb7nash

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    Homework Helper

    This is fine so far.

    Try evaluating e1r. What two pieces of information can you conclude? Similarly...
     
  4. Jan 16, 2012 #3
    wouldnt I get
    e[itex]_{1}[/itex]r=e[itex]_{1}[/itex]=r

    and

    e[itex]_{2}[/itex]r=e[itex]_{2}[/itex]=r

    So we get e[itex]_{2}[/itex] and e[itex]_{1}[/itex] are equal contradicting that they were distinct.



    Is that right?
     
  5. Jan 16, 2012 #4

    gb7nash

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    Correct.
     
  6. Jan 16, 2012 #5
    thanks, I was trying to figure it out with the x's and I couldnt come to any contradiction
     
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