Can anyone gives me an example of $V$ such that there exists at least one pair of elements $ a,c(\neq 0 ,1) \in D$ such that $V(a) > V(ac)$ $D$ is an Integral Domain.

$V$ only satisfies the first Euclidean property, i.e. for all $\,a,b\in D\,$ if $\,b\neq 0\,$ then there are $\,q,r\in D\,$ such that $\, a = qb +r\,$ with either $r=0$ or $\, V(r) &lt; V(b),\,$ where $V$ maps $D$ into  (well-ordered)  $\,\Bbb N$.
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Can anyone gives me an example of $V$ such that there exists at least one  pair of elements $ a,c(\neq 0 ,1) \in D$ such that $V(a) &gt; V(ac)$?

I have been really trying hard to construct one. But I can't.
Let $D=\Bbb Q[X]$ and
$$ V(f)=\begin{cases}
2\deg f&amp;\forall x\gg 0\colon f(x)&gt;0 \\
2\deg f+1&amp;\forall x\gg 0\colon f(x)&lt;0 \\
\text{whatever}&amp;f=0\end{cases}$$

For $a,b\ne0$, we can find by polynomial division $q,r$ with $a=qb+r$ and either $r=0$ or $\deg r&lt;\deg b$, in which case also have $V(r)&lt;V(b)$.
Thus the desired conditins hold.
Now let $a=c=-1$ and note that $V(a)=1$, $V(ac)=0$.
Let $D=\Bbb Q[X]$ and
$$ V(f)=\begin{cases}
2\deg f&amp;\forall x\gg 0\colon f(x)&gt;0 \\
2\deg f+1&amp;\forall x\gg 0\colon f(x)&lt;0 \\
\text{whatever}&amp;f=0\end{cases}$$

For $a,b\ne0$, we can find by polynomial division $q,r$ with $a=qb+r$ and either $r=0$ or $\deg r&lt;\deg b$, in which case also have $V(r)&lt;V(b)$.
Thus the desired conditins hold.
Now let $a=c=-1$ and note that $V(a)=1$, $V(ac)=0$.