> [!NOTE] Lemma > Let $(R,\partial)$ be a [[Euclidean Domain|Euclidean domain]] and $a,b\in R\setminus \{ 0 \}$. If $b$ is not a [[Unit in a Ring|unit]] then $\partial(a)<\partial(ab)$. **Remark**: recall that by definition $\partial(a)\leq \partial(ab)$ so this statement is equivalent to its contrapositive which reads if $\partial(ab)=\partial(a)$ then $b$ is a unit and so $ab$ and $a$ are [[Associates in Integral Domain are Necessarily Unit Multiples of Each Other|associates]]. ###### Proof We prove the contrapositive. Sufficient to show that $ab\mid a$. By (EF2), we may write $a=abq+r$ with $r=0$ or $\partial(r)<\partial(ab)$. If $r = 0$ then $a=abq$ so $ab\mid a$ as required. Suppose $r\neq 0$. Since $a\mid a$ and $a\mid ab$ we must have $a\mid r$ so by (EF1), $\partial(a)\leq \partial(r)$. But by (EF2), $\partial(r)<\partial(ab)$ which is a contradiction.