1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.

    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


    User Avatar
    Science Advisor
    Homework Helper

    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:

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook