1. ## 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?

2. Originally Posted by Demandeur
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 $|a|=|d|\cdot|q|\geq|d|$.