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Math Help - Divides proof

  1. #1
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    Divides proof

    I'm struggling with the following question:

    let a, b, c, k, l be integers
    Prove that if: a|b and a|c then a|(kb + lc)

    I'm not looking for the answer outright just a little help.

    I realise that:
    b = ka
    c = la

    (kb + lc) = m x a

    The approach I've been taking so far is to go: b + c = ka + kl and then try to rearrange (ka + kl) to (kb + lc) since I know that
    if a|b and a|c then a|(b + c)

    Am I on the right track? If so then I know it's my algebra skills which are the problem.
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  2. #2
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    Re: Divides proof

    k and l are arbitrary integers. It doesn't make sense to let b = ka and c = la. Plus it confuses readers like me....

    You can use different integers, e.g. b = qa and c = ra, in which kb + lc = k(qa) + l(ra) = a(kq + lr), clearly, a divides this.

    An alternate solution is to use mods, which works just as well (and it doesn't need random variables!).
    Last edited by richard1234; August 19th 2012 at 10:12 PM.
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  3. #3
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    Re: Divides proof

    Thanks for that Richard, makes much more sense.

    We have been studying mods lately so out of interest, how would you approach this using mods?
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  4. #4
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    Re: Divides proof

    Basically, b \equiv 0 (\mod a) and c \equiv 0 (\mod a), so for any integers k and l,

    bk \equiv 0 (\mod a)
    cl \equiv 0 (\mod a)

    Therefore the sum bk + cl is also congruent to 0 (mod a).
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