# A divisibility property

• Jun 30th 2010, 11:23 AM
Demandeur
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?
• Jun 30th 2010, 11:44 AM
melese
Quote:

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