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Math Help - Proof of a congruence

  1. #1
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    Proof of a congruence

    Prove (w/o induction) that B^n and 0 are congruent mod B.
    Hint from professor: let n = m + 1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post
    Prove (w/o induction) that B^n and 0 are congruent mod B.
    Hint from professor: let n = m + 1
    Hint isn't really necessary as long as n \ge 1, which i think you are assuming

    note that B^n \equiv 0 ~(\bmod ~B) \Longleftrightarrow B \mid (B^n - 0)

    this is obviously the case.

    So begin by saying B^n = B^{m + 1} = B \cdot B^m

    where can we go with this?
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  3. #3
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    I guess I should've post this in my post...but I actually got that far.

    Since B^n = B^(m+1) then B|(BB^m) - 0 or B|BB^m.

    But, Im not sure what to do from here.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post
    I guess I should've post this in my post...but I actually got that far.

    Since B^n = B^(m+1) then B|(BB^m) - 0 or B|BB^m.

    But, Im not sure what to do from here.
    since B^n = B \cdot B^m we have that B | B^n \Longleftrightarrow B|(B^n - 0) which means?
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  5. #5
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    You did say without induction this time, didn't you?

    B is congruent to 0, mod n, if and only B is a multiple of n: say B= na where a is some integer.

    If B is congrunt to 0 mod n, then B= na. B^m= (na)^B= n^ma^m= n(n^{m-1}a^n) is sufficient.

    And, of course, 0= 0n.
    Last edited by Jhevon; March 19th 2009 at 07:37 PM. Reason: fixed LaTeX
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