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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)

$|nx|=1\cdot|nx|=|-1||nx|=|-nx|\le|-n|\cdot|x|$

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$n$ 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.

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

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