Let $G$ be an ordered group; then $G$ is Archimedean if and only if the following condition holds: $$\text{if} \space a, b \in G \space \text{with} \space a>0, \space \text{ there exists a natural number} \space n \space \text{such that} \space na>b $$ (Matsumura page 76, Theorem 10.6)
There were two things that confused me...
1)
In the parenthesis, it says if $y \geq 0$, "this is clear by assumption". However the assumption says $a, b \in G \space \text{with} \space a>0 \implies \text{ there exists a natural number} \space n \space \text{such that} \space na>b$...it does not assume that $b \geq 0$. So why does the proof differentiate between $y \geq 0$ and $y < 0$? Are they not both true by the assumption?
2)
I wasn't quite sure how $(n+n')x \leq 10^r(y+y') < (n+n'+2)x$ $\implies$ $\phi(y+y') - (n+n')10^{-r} < 2.10^{-r}$.
Thanks in advance
If $y>0$, the hypothesis gives us a natural number $n$ such that $nx>y$; since the natural numbers are well-ordered, this automatically gives us a least $m\in\Bbb N$ such that $mx>y$. Setting $n=m-1$ then gives us the largest integer $n$ such that $nx\le y$. If $y=0$, then $nx\ge 0$ for all $n\in\Bbb N$, and $n=-1$ is evidently the largest integer such that $nx<y$. If $y<0$, then $nx>y$ for all $n\in\Bbb N$, and there’s nothing that immediately tells us that there even is an integer $n$ such that $nx\le y$; that’s what the argument in parentheses does.
Suppose that $(n+n')x\le 10^r(y+y')<(n+n'+2)x$. This inequality says that the number obtained by taking the first $r$ decimal places of $\varphi(y+y')$ is at least $n+n'$ and less than $n+n'+2$, which in turn says that $(n+n')\cdot 10^{-r}\le\varphi(y+y')<(n+n'+2)\cdot 10^{-r}$ and hence that $\varphi(y+y')-(n+n')\cdot 10^{-r}<2\cdot 10^{-r}$. (Note that the dot in $2.10^{-r}$ and $4.10^{-r}$ indicates multiplication; the notation on this page is inconsistent, since the dot in $(n+n')\cdot 10^{-r}$ is centred.)