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Thread: A divisibility property

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

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

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

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

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