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Integer mod proof

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that either [tex][a]_{n}[/tex][tex]\cap[/tex][tex]_{n}[/tex]=empty set or [tex][a]_{n}[/tex]=[tex]_{n}[/tex].


    2. Relevant equations



    3. The attempt at a solution
    I want to assume there is an element x in [tex][a]_{n}[/tex][tex]\cap[/tex][tex]_{n}[/tex] and show this implies [tex][a]_{n}[/tex]=[tex]_{n}[/tex].
    This tells me x is in [tex][a]_{n}[/tex] and [tex]_{n}[/tex].
    That's where I get stuck.
     
    Last edited: Sep 23, 2010
  2. jcsd
  3. Sep 23, 2010 #2
    sorry, I meant to show this implies [a]=.
     
  4. Sep 23, 2010 #3

    Office_Shredder

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    If x=a mod n then n|x-a. x=b mod n means n|x-b.

    If x|x-a and n|x-b.... what can you say about a-b?
     
  5. Sep 23, 2010 #4
    x-a=x-b
    x-x=a-b
    0=a-b
    b=a
     
  6. Sep 23, 2010 #5

    Office_Shredder

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    x-a is probably not equal to x-b
     
  7. Sep 23, 2010 #6
    Ok then I'm not really sure where to go with a-b then.
     
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