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Math Help - Finite Math Help!!!!?

  1. #1
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    Question Finite Math Help!!!!?


    Not sure it this is the appropriate area to post this, but I am having a really hard time figuring out modular arithmetic. Every time I think i have it I realize that I missed something. I need to know how modular arithmetic relates to time. I know that it is clock math, but I don't know how to find the right amount of numbers that should go around the clock. EX: 10mod(6), I know the answer is four, but that's only because I divided 10 by 6 and 4 is the remainder. I also don't know how to add (positive/negative) or multiply (positive/negative) in mod(7). I need someone to break this down for me like I'm a 2 yr old, because I'm obviously missing something. Please help. ANY relevant assistance in this matter would be greatly appreciated. Thank you in advance.
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  2. #2
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    Re: Finite Math Help!!!!?

    The appropriate forums for problems about modular arithmetic are Discrete Math and Number Theory.

    Quote Originally Posted by DanaT View Post
    EX: 10mod(6), I know the answer is four, but that's only because I divided 10 by 6 and 4 is the remainder.
    That's the right way to do it.

    Quote Originally Posted by DanaT View Post
    I also don't know how to add (positive/negative) or multiply (positive/negative) in mod(7).
    You demonstrated that you can find remainders. To add numbers mod 7, add them normally and take the remainder when the sum is divided by 7. Can you show an example you are having difficulties with?
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  3. #3
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    Re: Finite Math Help!!!!?

    There are two different, but equivalent, ways to think of "m modulo n". One is to think of the "numbers" as 0 to n-1. If any arithmetic operation gives a number not among those, divide by n and take the remainder. The other is to think of the "numbers" as sets of integers, each set consisting of numbers that have} the same remainder when divided by n.

    For example, we can think of "modulo 7" dealing only with 0, 1, 2, 3, 4, 5, and 6. 5+ 4= 9 which is larger than 7. 9= 7+ 2 so 5+ 4= 2 (mod 7). 2*6 = 12= 7+ 5 so 2*6= 5 (mod 7).

    Or think of the "numbers" as being the sets
    0= {0, 7, 14, 21, ...}
    1= {1, 8, 15, 22, ...}
    2= {2, 9, 16, 23, ...}
    3= {3, 10, 17, 24, ...}
    4= {4, 11, 18, 25, ...}
    5= {5, 12, 19, 26, ...}
    6= {6, 13, 20, 27, ...}
    Since every number larger than 6 is already in one of those sets, we stop there. Now we see that 5+ 4= 9 and 9 is in the set represented by 2: 5+ 4= 2 again. Also 2*6= 12 and 12 is in the set "represented" by 5 so 2*6= 5 (mod 7).
    Thanks from topsquark
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