Proof. We first see that $1\in R$ since $|1| = 1$ . Let then $a,\,b$ be any two elements of $R$ . The non-archimedean triangle inequality shows that $|a-b| \leqq \max\{|a|,\,|b|\} \leqq 1$ , i.e. that the difference $a-b$ belongs to $R$ . Using the multiplication rule 4 of inequalities we obtain $$|ab| = |a|\cdot|b| \leqq 1\cdot 1 = 1$$ which shows that also the product $ab$ is element of $R$ . Thus, $R$ is a subring of the field $K$ , and so an integral domain. Let now $c$ be an arbitrary element of $K$ not belonging to $R$ . This implies that $1 < |c|$ , whence $|c^{-1}| = |c|^{-1} < 1$ (see the inverse rule 5). Consequently, the inverse $c^{-1}$ belongs to $R$ , and we conclude that $R$ is a valuation domain. The presentations $a = \frac{a}{1}$ and $c = \frac{1}{c^{-1}}$ make evident that $K$ is the field of fractions of $R$ .