How to prove: $|nx|\le |n|\cdot|x|$, for $x\in K$ and $n \in \mathbb Z$ ?
The absolute value here is a nonnegative function from a field $K$ to $\mathbb R$ and in the definition there's a point;
$|xy|=|x|\cdot |y|,\quad \forall x,y\in K$ ?
What is not working for my case ? Is $n$ not necessarily contained in $K$ ? Are $1$ and $-1$ always in $K$ ?
Can I prove it inductively using $|-1|=|1|=1$ (this is already known)
$|nx|=1\cdot|nx|=|-1||nx|=|-nx|\le|-n|\cdot|x|$
$n$ is not necessarily contained in $K$. But $n\in\mathbb{N}$ acts on $K$ in the usual way $n\cdot x=x+x+\dots+x$ (sum $x$ $n$-times)
I guess you also have triangle inequality for your norm. So, for $n\geq 0$, you have $$|n\cdot x|=|x+\dots +x|\leq |x|+\dots +|x|=n|x|$$
Do the same thing for $n$ negative.