Let $n$ be a integer not equal to zero, $p$ a prime and $$|n|_p = p^{-\operatorname{ord}_p(n)}; |0|_p=0$$
Proposition: $|xy|_p=|x|_p|y|_p$
We want to show: $\operatorname{ord}_p(ab)=\operatorname{ord}_{p}(a)+\operatorname{ord}_p(b)$ So if we suppose there is a largest integral $v$ such that $p^{v}|(ab)$, then we can split up v such that $p^{v_1}|a,p^{v_2}|b$. If we suppose that there exist $v_1,v_2$ with them both being the largest integers such that $p^{v_1}|a, p^{v_2}|b$, with $v_1+v_2= : v $ then $p^v|ab$. (I don't believe this to be a formal proof but rather my best own "idea" I can come up with...)
The script also mentions that the padic valuations' sums are bounded: $|x+y|_p\le \sup\{|x|_p,|y|_p\}$, how to show this?
Edit:
According to Robert Israel and Bill Dubuques answers, the first proposition is proven, if I understood correctly:
Given $ord(a)= v$ and $ord(b) = u$ which is the same as $a:= p^{v}m, b:= p^{u}n$ with $gcd(m,p)=gcd(n,p)=1$, because $ab=p^{u+v}mn$ with $gcd(mn,p)=1$ it follows that $ord(ab)=ord(a)+ord(b)$ and so $|xy|_{p}=p^{-ord(xy)}=p^{-(ord(x)+ord(y))}=|x|_{p}|y|_{p}$
Now to show: $|x+y|_{p}\le sup\{|x|_{p},|y|_{p} \}$
and the second can be tackled like this: set $a:=p^{v}m. b:=p^{u}n$, then checking two cases $1:(u=v)$ and $2:(v
and in case of $(v
It seems that I did something wrong in both cases...