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Abstract Algebra any help is appreciated!

  1. Feb 12, 2008 #1
    1. On the set of real numbers, R the following operation is defined:
    *RxR implies (arrow) R, (x,y) implies (arrow) x*y=2(x+y)-xy-2
    Find the neutral element of this operation.




    3. since we know x*e=x, e*x=x, so i attempted:
    using e as y, because it would just mean y.

    so
    2(x+e)-xe-2= 2x+2e-xe-2 = 2x+e(2-x)-2 = 2

    so wouldnt my e be 2?

    i think i am right, but am not too sure, any input would be appreciated!
     
  2. jcsd
  3. Feb 12, 2008 #2

    Dick

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    Try that again. You should solve e*y=y, or 2*(e+y)-ey-2=y. Solve that for e.
     
  4. Feb 12, 2008 #3
    i know where i made my mistake the first time, so e=1...im thinking.
    this is how i got to my second conclusion:
    2(e+y)-ey-2=y
    2e+2y-ey-2=y
    2e-ey+2y-2=y
    e(2-y)=y-2y+2
    e(2-y)=(2-y)
    e=(2-y)/(2-y)
    e=1

    so my neutral element would be 1.
    thanks for the help!
     
  5. Feb 12, 2008 #4

    HallsofIvy

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    Since this is not symbolic logic, that [itex]\rightarrow[/itex] is not "implies". It simply indicates that the function if "from" R x R "to" R and changes (x, y) into 2(x+ y)- xy- 2.
     
  6. Feb 12, 2008 #5

    Hurkyl

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    Two comments

    This step is sketchy -- this step is only permissible if 2-y is known to be (globally) nonzero. Some methods for getting around this problem by handling 2-y=0 as a separate case, making an explicit assumption that 2-y is not zero, managing to translate the problem appropriately into univariate polynomials, or my favorite: factoring rather than dividing.


    That's not what you've shown! You've shown that if there is a neutral element, it has to be 1. You have not actually shown there exists a neutral element. Fortunately, it's easy to prove that 1 is, in fact, a neutral element.
     
    Last edited: Feb 12, 2008
  7. Feb 12, 2008 #6
    thank you!
     
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