Let $R$ be a discrete valuation ring on a fraction field $K$ of $R$. If $x,y\in K$ such that $v(x)< v(y)$, prove that $v(x+y)=\min(v(x),v(y))$.
By definition of discrete valuation, we have that $v(x+y)\geq \min(v(x),v(y))$.
So, we must show the other direction, but I'm not sure how to do this.
I tried something like $v(x)<v(y)$ implies $v(x)-v(y)<0$ implies $-v(xy)<0$, but don't know how that helps.
We can assume that $v(x), v(y)<\infty$, since if, say $v(y)=\infty$, then $y=0$, so $v(x+y)=v(x)=\min(v(x), v(y))$. Now consider $v(x)=v((x+y)-y)\ge \min(v(x+y), v(y))\ge \min(v(x), v(y))=v(x)$, so if $v(x+y)\ge v(y)$, then $v(x)\ge v(y)\ge v(x)$, so $v(y)=v(x)$ a contradiction. Thus we have $v(x)\ge v(x+y)\ge v(x)$, so $v(x)=v(x+y)$, as desired.