# Help $|nx|\le |n|\cdot|x|$, for $x\in K$ and $n \in \mathbb Z$?

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.
• I must have $|n|\cdot |x|$, so $n$ should not be pulled out – counterfeit May 14 at 17:08
• I was doing it for $n$ positive. For $n$ negative your action is $n\cdot x=-x-x\dots -x$ ($-n$ times) – Julian Mejia May 14 at 17:45