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Math Help - Help with a proof - divisor of a linear combination

  1. #1
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    Help with a proof - divisor of a linear combination

    Hi guys.

    The teacher in our class made some claim and didn't prove it for some reason, and I really want to see the proof for that.

    let n,m,d\in \mathbb{Z}.
    if d|n and d|m then d|(am+bn) for every a,b\in\mathbb{Z}.

    thanks in advanced.
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  2. #2
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    Re: Help with a proof - divisor of a linear combination

    d|n means n = dx for some x \in \mathbb{Z}. d|m means m = dy for some y \in \mathbb{Z}. So, am+bn = a(dy) + b(dx) = d(ay+bx). Since ay+bx \in \mathbb{Z}, am+bn is divisible by d.
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