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Proof: integers divisibility property

  1. Nov 14, 2007 #1
    Someone please help me with this qiestion:

    Prove that for all integers a, b, and c, if a divides b but not c then a does not
    divide b + c, but the converse is false.

    Last edited: Nov 14, 2007
  2. jcsd
  3. Nov 14, 2007 #2


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    Looks like a homework problem to me! First write out the definitions: If "a divides b" then b= an for some integer n. I would use "proof by contradiction". Suppose you know that a divides b+c, that is, that b+c= am for some integer m and that b= an for some integer n. Can you use that to contradict "a does not divide c"?

    The "converse" of that statement is, of course, "if a does not divide b+ c, then a divides b but not c". It should be fairly simple to find a counter example for that, or the more general "converse", "if a does not divide b+ c, then it must divide one but not the other".
  4. Nov 14, 2007 #3


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    I imagine the question does refer to the more general converse. The specific converse can be disproved simply by symmetry arguments, without using a shred of number theory.
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