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Quick group theory proof with some minor inquiries

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    See attachment, problem #19a.


    2. Relevant equations



    3. The attempt at a solution

    a) Let j ∈ S X S be arbitrary. Then j is an ordered pair of the form (a,b) for some a,b ∈ S. Now let c = a + b + ab ∈ S. Then clearly a*b = c. Now let d ∈ S and assume a*b = d. But then it follows that c = a + b + ab = d.

    ^My only concern with this proof is that it doesn't explicitly demonstrate that * is closed on S. That is, that a*b is always not -1. I figured this may not have to be shown or just has to do with the properties of addition on the set S, but thought I'd inquire if this is a valid qualm?
     

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    Last edited: Sep 26, 2011
  2. jcsd
  3. Sep 26, 2011 #2
    Hello,

    Indeed, you need to show that a*b is never -1. You can show this using reductio ad absurdum.
     
  4. Sep 27, 2011 #3
    since a, b are in S , you know [itex]a\neq -1 \;, b\neq -1[/itex] , what you have to prove
    is an implication

    [tex][(a\neq -1)\wedge(b\neq -1)\Rightarrow(a+b+ab\neq -1)] [/tex]

    use contrapositve

    [tex][(a+b+ab=-1)\Rightarrow (a=-1)\vee(b=-1)] [/tex]

    prove this ....
     
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