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Stuck, finding inverse in element in ring Z

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data

    I need to find the inverse to element in 7 in ring [tex]Z_{13}[/tex]

    2. Relevant equations

    7^-1 in [tex]Z_{13}[/tex]

    3. The attempt at a solution

    Needs to find X so that, 7x=1 in [tex]Z_{13}[/tex] => 7x=1+k*13

    And then my notes was messed up :(
     
  2. jcsd
  3. Sep 11, 2010 #2

    Dick

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    The general method is the extended Euclidean algorithm. But Z_13 is small enough it's probably easier to just guess the answer. Try that first.
     
  4. Sep 12, 2010 #3
    This was a note from the class, I've forgotten what k means in this one. May I please ask you what you think it means?
     
  5. Sep 12, 2010 #4

    Dick

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    It's some integer that you want to find, just like x. If you can find integers x and k such that 7x=1+k*13 then if you reduce both side mod 13, you'll see x is 7^(-1). Like I said, see if you can find values by guessing.
     
  6. Sep 12, 2010 #5

    HallsofIvy

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    You want to find x and k such that 7x= 13k+ 1. That is the same as 7x- 13k= 1.
    7 divides into 13 once with remainder 6. That says 13= (1)(7)+ 6 or (1)(13)+ (-1)(7)= 6

    6 divides into 7 once with remainder 1. That says 7= (1)(6)+ 1 or (1)(7)+ (-1)(6)= 1.
    Replacing "6" in that from the previous equation, (1)(7)+ (-1)((1)(13)+ (-1)(7))= (2)(7)- 13(1)= 1.

    But, as Dick said, 13 is small enough that it's probably simpler to just look at 7(1), 7(2), 7(3), etc. You do know what 7 times 2 is, don't you?
     
  7. Sep 12, 2010 #6
    So z13= 0 to 13 in the ring. First I didn't get what a ring was, then I saw an example of the clock that z12 is 12=0 and 11+1=12=0 and 12+1=13-1=0. Something in that style. Now I got it.

    inverse elements to 7 in ring z13 = 2

    Thanks guy.
     
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