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Modulo Arithmetic ( Classes )

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve in [tex]K=\mathbb{Z}/_{5}\mathbb{Z}\cdot\mathbb{Z}/_{5}\mathbb{Z}[/tex] the following system of equations:

    [tex]\begin{cases}
    2^{-}x-3^{-}y=1^{-} & 1^{-}x+2^{-}y=2^{-}\end{cases}[/tex]



    Not that the ^- means a number with a bar over it. ( I don't know how to input it in the latex software that I use.

    3. The attempt at a solution

    I solve this system of equations by using the second equation to get x=2-2y and i substituted it for x in the first equation and i got a solution of S={(-6^-,4^-)}

    Is that solution correct??
     
    Last edited: May 2, 2012
  2. jcsd
  3. May 2, 2012 #2

    Mark44

    Staff: Mentor

    I get (4, 4) as the only solution. Since -6 ##\equiv -1## (mod 5) ## \equiv ## 4 (mod 5), I think we arrived at the same thing.
     
  4. May 3, 2012 #3

    I like Serena

    User Avatar
    Homework Helper

    As Mark already said, your solution is correct. :smile:


    For your information, \bar or \overline will put a bar on top of a symbol.
    Like this: ##\bar{12}## or ##\overline{12}##.


    Btw, your set should be ##\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}##.

    ##5\mathbb{Z}## is the set with the number 5 multiplied with any whole number (the 5-folds).

    ##\mathbb{Z}/5\mathbb{Z}## is the so called quotient set with all 5-folds "divided away".

    ##A \times B## is the so called cartesian product of sets A and B that consists of ordered pairs.
     
  5. May 3, 2012 #4
    Thanks for the extra information. Too bad my teacher doesn't tell us any of this stuff.
     
  6. May 3, 2012 #5

    I like Serena

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    Homework Helper

    Ah well, at least we can give some added value then, here on PF. :wink:
     
  7. May 3, 2012 #6
    YEP:redface:
     
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