Q:

showing $d(a)<d(ab)$ for $b\neq 0$ and $b$ not unit in Euclidean ring R.

Let $R$ be an Euclidean ring and $a, b \in R$. If $b\neq 0$ and $b$ is not unit in R, show that $d(a)<d(ab)$.

Here is outline of the proof on my book.
Let $A = (a) = \{ x a|x \in R\}$. Then $d(a) \le d(xa)$. If for some $b$, $d(ab) = d(a)$, then the value of $d(ab)$ is minimal in $A$ and $a = abr$ for some $r\in R$. Since $R$ is an integral domain $br = 1$ which shows that $b$ is unit in $R$ contradticting $d(a) = d(ab)$. Hence the proof.
Why does $br =1$ imply $b$ is unit?

A:

The term "unit" is overloaded in ring theory. It can refer either to the multiplicative identity (or neutral element) usually denoted by $1,\,$ or it can refer more generally to any invertible element. If an identity element exists it is unique, so it is often referred to as the unit, which helps to distinguish between the two usages. Generally one easily infers from the context which meaning is intended. See also  this Wikipedia page.

