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Divisibility proof

  1. Sep 18, 2006 #1
    I'm having trouble with this question, I need to prove or disprove this statement: If c divides (a+b), but c does not divide a, then c does not divide b.

    what i have so far is ck = (a+b) where k is some integer. Next I have a=ck-b and b=ck-a. I tried doing things like a = ck-(ck-a) but that got me nowhere. Any ideas? Thanks in advance.
  2. jcsd
  3. Sep 19, 2006 #2


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    If c divides b, try writing b in terms of c, then go back to a=ck-b
  4. Sep 19, 2006 #3


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    Try proof by contradiction: Suppose c does divide b. Then b= cn for some n so c= a+ cn. What does a equal? Why is that a contradiction?
  5. Sep 19, 2006 #4
    Thanks for the help. I think theres a problem though, shouldn't C = (a+b)n for some integer n? Instead of c= a + cn, it should be c divides a + cn.

    I tried the following c = (a +cn)k where k is another integer, so
    a = (c/k) - cn but I can't figure out where to go from there. Thanks
  6. Sep 19, 2006 #5


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    Orikon, there's a small typo in Halls' post.

    c divides b => nc = b
    c divides a+b => kc = a+b = a+nc {not the other way round}

    So what can you say about a?
  7. Sep 20, 2006 #6
    kc = a + nc
    a = kc - nc
    a = c(k-n)

    which means c must also be divisible by a when c is divisible by b. That's the contrapositive of the original statement, so it must be true, right? :biggrin: Thank you so much!
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