Simple property of a valuation on a field

Let $F$ be a field and $v:F\rightarrow G\cup\{\infty\}$ be a valuation on $F$ so $G$ is a totally ordered abelian group with $\infty$ having the properties $\infty+\infty=g+\infty=\infty+g=\infty$ and $\infty\leq\infty$ for all $g\in G$, $\infty \notin G$.

Show if $v(a)\neq v(b)$ then $v(a+b)=\min(v(a),v(b)$).

I must be missing some trick here on how to show this is true.

My approach has been to use that $G$ is a totally ordered group so I start by assuming $v(a)\neq v(b)$ and letting my first case be $v(a)<v(b)$ as trichotomy holds in $G$. By $v$ being a valuation on $F$, we then have that $v(a+b)\geq \min(v(a),v(b))=v(a)$. From here what I've tried doing doesn't seem to help much.

I've said lets assuming $v(a+b)>\min(v(a),v(b))=v(a)$ and we also have $$v(a+b)> v(a) \Rightarrow v(a+b)+v(a^{-1}) > 0 \Rightarrow v(1+ba^{-1}) > 0$$ and $$v(a)<v(b) \Rightarrow 0< v(b)+v(a^{-1}) \Rightarrow 0 < v(ba^{-1})$$ also $$v(1+ba^{-1})\geq\min(v(1),v(ba^{-1}))=\min(0,v(ba^{-1}))=0.$$

How does one solve this?

Solution 1:

Let $a,b\in F$ be such that $v(a)<v(b)$. As any totally ordered group is torsion-free, we have $v(\zeta)=0$ for all roots of unity $\zeta\in F$, and in particular $v(-1)=0$. Then $v(-b)=v(b)$, and so $$v(a)=v(a+b-b)\geq\min(v(a+b),v(b))\geq\min(v(a),v(b))=v(a).$$ It follows that $\min(v(a+b),v(b))=v(a)$ and hence that $v(a+b)=v(a)$. By symmetry, if $v(b)<v(a)$ we find that $v(a+b)=v(b)$, which shows that $v(a+b)=\min(v(a),v(b))$ if $v(a)\neq v(b)$.