A divisibility property

**Demandeur** · June 30th 2010, 11:23 AM

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?

**melese** · June 30th 2010, 11:44 AM

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 $|a|=|d|\cdot|q|\geq|d|$ .

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