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    Okay i'm having a bit of trouble starting this problem..

    a, b, c are integers 0, ±1, ±2.... with a|b (a divides b) and a|c. Show that for any m, n (also integers), a|mb + nc ?


    I'm having trouble working out the angle of attack on this beast. Other than trial and error (which proves nothing really), i can't see a way to prove it. Possibly, trying to contradict it?

    I know that from the properties of divisability if a|b and a|c, then b|c.

    Any help is appreciated. Cheers.
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    (Original post by ForeverOptimistic)

    I know that from the properties of divisibility if a|b and a|c, then b|c.
    This is false.

    Counterexample: 2|6 and 2|4 but 6 does not divide 4.
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    (Original post by Get me off the £\?%!^@ computer)
    This is false.

    Counterexample: 2|6 and 2|4 but 6 does not divide 4.
    You're right, i was getting carried away with myself. Any help then? I'm completely at a loss.

    Cheers.
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    (Original post by ForeverOptimistic)
    You're right, i was getting carried away with myself. Any help then? I'm completely at a loss.

    Cheers.
    b=ka for some integer k.

    similar for c.

    then mb+nc = mka+.... = a(.......)
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    (Original post by Get me off the £\?%!^@ computer)
    b=ka for some integer k.

    similar for c.

    then mb+nc = mka+.... = a(.......)

    So, c = pa for some integer p.

    then mb + nc = mka + npa = a (mk + np)

    And, m, k, n and p are all integers. Two integers added together make another, and if (mk + np) is an integer and two interers multiplied make another, then a(mk + np) must also be an integer.

    So we've proved mb + nc is an integer, right?

    To prove 'a| mb + nc' , could we say a = b/k and then (b/k) | mk(b/k) + np(b/k) = b | mb + npb. Hmm..not sure where to go there. I'm getting taken in circles, not sure how to finish the proof.
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    If mb+nc =integer*a then by definition a divides mb+nc.
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    (Original post by Get me off the £\?%!^@ computer)
    If mb+nc =integer*a then by definition a divides mb+nc.
    Cheers dude, i really appreciate it!
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    (Original post by ForeverOptimistic)
    Cheers dude, i really appreciate it!
    No problem. It keeps me out of trouble.
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    b=ka c=la,so mb=mka,nc=cla,both of them can be divided by a,so mb+nc can be divided by a
 
 
 
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