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How does this proof look?

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Theorem:

    Let A, B, C, M, and N be integers.

    If A divides both B and C, A divides NB+MC.

    2. Relevant equations



    3. The attempt at a solution

    Proof:

    Since we have defined A to divide both B and C, there exists an M in the integers such that B = AM, and there exists an N in the integers such that C = AN.

    So, it follows that:

    A = B/M and A = C/N

    B/M + C/N = A + A

    B/M + C/N = 2A

    B + MC/N = 2AM

    NB + MC = 2AMN

    It is shown that 2AMN = NB + MC, and since A divides 2AMN, A divides NB+MC.

    Therefore, A divides NB + MC.
     
  2. jcsd
  3. Feb 8, 2012 #2

    Mark44

    Staff: Mentor

    I think that this is not as direct as it could be.

    You are given that A|B and A|C, which means that B = rA and C = sA for integers r and s.
    Then NB + MC = N*rA + M*sA = A(Nr + Ms).
     
  4. Feb 8, 2012 #3
    I see... for some reason I felt compelled not to introduce any new integers, but I like what you did.
     
  5. Feb 8, 2012 #4

    Mark44

    Staff: Mentor

    You should avoid doing stuff like this, with a whole separate equation that does nothing more than show that A + A = 2A.

    Instead, you could do this:
    B/M + C/N = A + A = 2A
     
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