How to prove the strong triangle inequality? I'm trying to prove that p-adic space is a metric space(also a ultrametric space), but I find it difficult to prove the triangle inequality. So if one can prove the strong triangle inequality, then the triangle inequality will be easily solved. Does anyone know the proof of the Strong Triangle Inequality or any helpful hint?
 A: The case in which $a=0$ or $b=0$ is easy, so assume that $a\ne 0\ne b$. Let $a=\pm p^mr$ and $b=\pm p^ns$, where $m$ and $n$ are integers, and $r$ and $s$ are rationals whose numerators and denominators in lowest terms are not divisible by $p$. (This is the representation used by Rotman; I just collapsed all of the factors not involving $p$ into single rationals $r$ and $s$.)
By definition $\|a\|_p=p^{-m}$ and $\|b\|_p=p^{-n}$. Without loss of generality assume that $m\le n$. Then $p^m\le p^n$, so $p^{-m}\ge p^{-n}$, and $\max\{\|a\|_p,\|b\|_p\}=p^{-m}$. Thus, we must show that $\|a+b\|_p\le p^{-m}$.
Now $a+b=\pm p^mr\pm p^ns=\pm p^m\left(r\pm p^{n-m}s\right)$, where $n-m\ge 0$. Now let’s write $r$ and $s$ as fractions in lowest terms, say as $r=\frac{i}j$ and $s=\frac{k}\ell$, where $p$ does not divide $i,j,k$, or $\ell$. Then 
$$r\pm p^{n-m}s=\frac{i}j\pm\frac{p^{n-m}k}\ell=\frac{i\ell\pm p^{n-m}kj}{j\ell}\;.\tag{1}$$
The denominator of this last fraction clearly has no factors of $p$. The numerator has no factors of $p$ if $n-m>0$ (why?). Thus, if $n-m>0$, $\pm p^m(r\pm p^{n-m}s)$ expresses $a+b$ as a product of a power of $p$ and a rational whose numerator and denominator in lowest terms have no factors of $p$, and by definition $\|a+b\|_p=p^{-m}$, which is fine.
If $n-m=0$, the numerator in the last fraction of $(1)$ might have factors of $p$, but if so, then $a+b=\pm p^{m'}t$ for some $m'>m$, where $t$ in lowest terms has no factors of $p$ in either numerator or denominator, and $\|a+b\|_p=p^{-m'}<p^{-m}$, which is also fine.
