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Algebraic structure : Group

  1. Oct 10, 2011 #1
    Please I have a problem with natural log for group set as follow:


    a*b=eln(a)*ln(b)

    1- Show that the group law * is associative and commutative
    2- Show that the group law * accept an element e (Identity element)

    Thank you !
     
    Last edited: Oct 10, 2011
  2. jcsd
  3. Oct 10, 2011 #2

    lavinia

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    what are you stuck on?
     
  4. Oct 10, 2011 #3
    Sorry, I mean to achieve to result that a*b = b*a, and a*e=a.
    But I'm stuck on.
    Thanks
     
  5. Oct 10, 2011 #4
    You answered lavinia's question...by not answering it. As a start, use the definition you wrote in your first post to write out what a*b, b*a, and a*e are.
     
  6. Oct 12, 2011 #5

    Deveno

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    is elna*lnb the same as, or different than elnb*lna?

    show us your work so far
     
  7. Oct 12, 2011 #6
    I tested with Maple e^ln(a)*ln(b), then the result is a^ln(b).
    so I tested with 2 different numbers, example:
    3^ln(2) = 2^ln(3) ==> Gives the same result.
    So I found that the e^ln(a)*ln(b) is commutative.
     
  8. Oct 12, 2011 #7

    micromass

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    No, you have tested it with two arbitrary numbers. You have to test it for all numbers!!!!
    Just checking it with two numbers does not suffice at all!!

    You must show that a*b=b*a. Write out the definition of * and show us what it means.
     
  9. Oct 13, 2011 #8

    Deveno

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    get a piece of paper, and use your brain. Maple won't solve this one for you.
     
  10. Oct 13, 2011 #9
    Last edited by a moderator: Apr 26, 2017
  11. Oct 14, 2011 #10
    Thanks for all
     
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