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Math Help - Modular Arithmetic question

  1. #1
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    Modular Arithmetic question

    Hey, Im not quite too sure about this problem what the value of [7]^(-1) is. I would believe [3] just because of addition and multiplication tables. But how do you determine what operation to use for the inverse? Thats where I get lost.

    "Find all values of [x] in the set Z_m, 0<=x<m, that satisfy the equation."

    [7]^(-1) - [2] = [x] in the set Z_10

    x=1..? Help appreciated!
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  2. #2
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    Re: Modular Arithmetic question

    Hey Jtmartz.

    There are some formulas to calculate inverses but the general way is to look at the multiplication group table for the congruence classes since [ab] = [a][b] (Mod n) which means if you have [a] and need to get [b] then [ab] = 0 (mod n).

    If the group operation is [a] + [b] instead [ab] then use that but the idea is the same.
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  3. #3
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    Re: Modular Arithmetic question

    Yeah that's the problem I'm just not getting - I've written everything about the question above, I don't know what operation the inverse of [7]^(-1) is, it doesnt say multiplication or addition - my thought was that both the inverse of [7] in addition and multiplication is 3 - would I just assume thats the value theyre getting at? I still feel like it's vague, in case the operation is subtraction or division, but it doesnt specify :S - another thought, since it asks find all values of x that satisfy the equation, do I consider the inverse of [7] in the set Z_10 for all operations?

    (that is,{+ [7]^(-1)=3}, {- [7]^(-1)=7}, {* [7]^(-1)=3}, {/ [7]^(-1)=7})
    Last edited by Jtmartz; October 29th 2012 at 09:16 PM. Reason: mistake
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  4. #4
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    Re: Modular Arithmetic question

    What is the modulus for the congruence class?
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  5. #5
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    Re: Modular Arithmetic question

    mod 10 - sorry i thought I included that
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  6. #6
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    Re: Modular Arithmetic question

    Yes sorry, you did (with the Z_10) include that.

    It's kind of hard to know what this group refers to (multiplicative or additive) but if its multiplicative then [7]*[7]^(-1) = [1] in Z_10. Now [7] is any integer where x = 7 (mod 10) and in your table you should have [7] in both the headers for the rows and columns. If its multiplicative then you have anything with remainder 7 times some other number with some remainder giving remainder of 1.

    As an example I'll go through 7 times something: we get 0,7,14,21,28,35,42,49,56,63 and the one with 1 remaining mod 10 is 21 which means under multiplication [7]^(-1) = [3]. Under addition, our identity is not [1] but [0] and the inverse is more straight-forward (inverse of [1] is [6] and so on).

    I'm afraid you'll have to figure out what the group is and whether its multiplicative or additive.
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