# Is it true that an equivalent 'absolute value' is an absolute value?

I've a very basic question on absolute values on fields. If $K$ is a valued field with absolute value $|- |:K\to \mathbb R_{\geq0}$ then is the map $|-|':K\to \mathbb R_{\geq0}$ defined by $|x|'=|x|^r$ for some $r> 0$ also an absolute value? How do I show that the triangle inequality still holds for $|-|'$? Maybe this is not possible?

Many thanks.

The triangle inequality holds if $r\leq 1$, but may not hold if $r>1$. To see that it may not hold if $r>1$, you may use $\mathbb{Q}$ with the standard absolute value as an example. If $r<1$, then note that for $x$ and $y$ positive with $y\leq x$ we have that for some $c$ with $x\leq c\leq x+y$ $$(x+y)^r-x^r=ryc^{r-1}\leq ryx^{r-1}\leq ry^r$$ by the mean value theorem, so $$(x+y)^r\leq x^r+ry^r\leq x^r+y^r\mathrm{.}$$
Thus $$|a+b|^r\leq (|a|+|b|)^r\leq |a|^r+|b|^r$$