Let $K$ be a field with a valuation $\mathrm{val}$. Show that if $a,b \in K$ with $\mathrm{val}(a) \ne \mathrm{val}(b)$ then $\mathrm{val}(a+b) = \min(\mathrm{val}(a),\mathrm{val}(b))$.

Suppose $a,b \in K$ are such that $\mathrm{val}(a) \ne \mathrm{val}(b)$. Without loss of generality, one may assume that $\mathrm{val}(a) > \mathrm{val}(b)$. For purposes of contradiction, assume that $\mathrm{val}(a+b) > \min(\mathrm{val}(a),\mathrm{val}(b))=\mathrm{val}(b)$, so $\mathrm{val}(a+b)=\mathrm{val}(b)$. Now, $\mathrm{val}(b)=\mathrm{val}(a+b-a)> \min(\mathrm{val}(a+b),\mathrm{val}(-a))=\min(\mathrm{val}(a+b),\mathrm{val}(a))$. There are two possibilities. First, suppose the minimum is $\mathrm{val}(a)$. This is a contradiction, for one then has $\mathrm{val}(b)>\mathrm{val}(a)$, but the reverse was true. Thus, one must have $\mathrm{val}(b) \geq \mathrm{val}(a+b)$. However, this contradicts the assumption that $\mathrm{val}(a+b) > \min(\mathrm{val}(a),\mathrm{val}(b))=\mathrm{val}(b)$. Thus, it must be the case that $\mathrm{val}(a+b) = \min(\mathrm{val}(a),\mathrm{val}(b))$, as desired.

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