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Hrm Congruence Proofs. Don't remember the rules

  1. May 1, 2012 #1
    Hrm... Congruence Proofs. Don't remember the "rules"

    1. The problem statement, all variables and given/known data

    Take equals sign as congruence or equals based on context, please. Itex does not work in Opera.

    Prove that for all n, 8^n = 1 (mod 7)

    2. Relevant equations



    3. The attempt at a solution

    This will be a proof by induction.

    Consider the case where n = 1.

    Since 8^1 = 8, and 8 - 1 = 7, which is divisible by 7, we see that 8 = 1 (mod 7).

    Now, assume that this theorem holds for some k in the natural numbers.

    Then,

    8^k = [1]

    (8)8^k = [8][1]
    8^(k+1) = [1][1] (since we know that 8 = 1(mod 7))
    8^(k+1) = [1]

    Thus, 8^k+1 = 1 (mod 7)

    Therefore, the theorem holds for all n.


    Been a while... sorry if it's really terrible.
     
  2. jcsd
  3. May 1, 2012 #2

    Mark44

    Staff: Mentor

    Re: Hrm... Congruence Proofs. Don't remember the "rules"

    Your logic is sound, but the work could be cleaned up in places. For example, you switch from 1 (mod 7) near the beginning to the [1] equivalence class. It would be better if you were consistent.

    Also, what you wrote as 8^k+1 needs to be written as 8^(k + 1). This seems to have been a momentary lapse, as you wrote it correctly in the preceding work.

    You might already know this, but 8^k+1 would be interpreted to mean 8k + 1, which isn't what you want.
     
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