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Prove: ab|ac, then b|c

  1. Nov 10, 2013 #1
    I'm really having trouble with this proof. at first I thought, oh easy the a's cancel then I realized I am proving that property so that was no help at all. Here is my work so far:

    --snip--
    Let a, b, and c be integers with a≠0.
    If ab|ac, we know from the definition of "divides" that there is an integer k, such that ac=ab⋅k

    Then (a - c⋅k) = 0 so k(a - b⋅k) = 0

    Since we know that a≠0, then b|c.
    --snip--

    I'm pretty sure that I'm off the rails after "such that ac=ab⋅k" any one care to help me? please?
     
  2. jcsd
  3. Nov 10, 2013 #2

    mathman

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    Gold Member

    Where did that come from????

    ac=ab.k => c=b.k
     
  4. Nov 10, 2013 #3
    I got a hint from the professor and it really threw me for a loop

    --snip--
    Then (�� - ��⋅�) = 0 so �(� - �⋅�)
    --snip--

    so, I was trying to make that form work.
     
  5. Nov 11, 2013 #4

    mathman

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    ac-kab=0 therefore a(c-kb)=0. Since a≠0, c-kb=0 or b|c.
     
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