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Math Help - A divisibility property

  1. #1
    Newbie Demandeur's Avatar
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    A divisibility property

    Prove that if d\mid{a} and a \ne 0, then |d|\le|a|.

    If d\mid{a} then a = qd, for some q\in\mathbb{Z}, and d = \frac{a}{q}, therefore |d|\le|a|. Is this argument correct?
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  2. #2
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    Quote Originally Posted by Demandeur View Post
    Prove that if d\mid{a} and a \ne 0, then |d|\le|a|.

    If d\mid{a} then a = qd, for some q\in\mathbb{Z}, and d = \frac{a}{q}, therefore |d|\le|a|. Is this argument correct?
    I think it is correct, but your transition from d=\frac{a}{q} to |d|\le|a| is not so clear to me.

    Here is another solution: Starting like you did, if d\mid{a} then a = qd. Now, |a|=|dq|=|d|\cdot|q| and since |q|\geq1, you have |a|=|d|\cdot|q|\geq|d|.
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