Archimedean absolute value on reals equivalent to usual one 
On $\mathbb R$, every archimedean absolute value $|\cdot|$, such that $(\mathbb R, |\cdot|)$ is complete, is equivalent to the usual absolute value defined for $x\in \mathbb R$ by:
  $$\mathbb |x|_{\mathbb R}=
 \begin{cases} 
      x & \text{if } x\ge 0 \\
      -x & \text{if } x\le 0
   \end{cases}$$

Now, I am trying to show that $\forall x, |x|=\mathbb |x|^a_{\mathbb R}$, for some $a>0$. Using the same steps as in the proof of the Ostrowski Theorem for $\mathbb Q$ in the archimedean case, one knows that there exists an $a>0$ such that $|r|=\mathbb |r|^a_{\mathbb R}, \forall r\in \mathbb Q$. Now I want to show that this "$a$" does work for all reals, not just the rationals. So, for an arbitrary $x\in \mathbb R$, I was thinking to start with a sequence of rationals $(r_i)$, such that $r_i\rightarrow x$ with respect to "usual" absolute value. Now, since $|r|=\mathbb |r|^a_{\mathbb R}$, it means that $(r_i)$ is Cauchy w.r.t. $|\cdot|$, hence converges to some $t\in \mathbb R$, because of completeness asumption. Now I want to show that $t=x$, but I am not able to do that.
Note: I use the definition of absolute value from Neukirch, chapter 2.
 A: It seems that we really have to make use of more extra structure on $\mathbb{R}$, namely, the usual order $<$.
So let $t \in \mathbb{R}$, wlog $t \ge 0$, and $r_i$ a sequence of rationals that converge to $t$ with respect to $|\cdot|$. By replacing $r_i$ with $2t-r_i$ if necessary, we can make sure that 
(*) for all indices $i\in \Bbb{N}$, we have $r_i \le t$ for odd $i$ and $r_i\ge t$ for even $i$; in particular, $t$ lies between $r_i$ and $r_{i+1}$.
I think you already got the following, and it does not need (*): $r_i$ is a Cauchy sequence w.r.t. $|\cdot|$. Then the sequence of values $|r_i| = |r_i|_\Bbb{R}^a$ is a Cauchy sequence in $\mathbb{R}$ w.r.t. the usual value (! -- this is how the metric is defined), hence converges to some $x_0 \in \mathbb{R}_{\ge 0}$. So $|t| = x_0$ by continuity of the value. On the other hand, if $|r_i| = |r_i|_{\mathbb{R}}^a \to x_0$, then by continuity of $a$-exponentiation, $|r_i|_{\mathbb{R}} \to x_0^{1/a}$. In other words, with respect to the usual value, $r_i$ converges to $x := x_0^{1/a}$. So $|t| = x^a = |x|_\mathbb{R}^a$.
So what is left to show is $x=t$. But by (*), we have $t\in \bigcap_{n\in \Bbb{N}} [r_{2n-1}, r_{2n}]$, and because the sequence is Cauchy w.r.t the usual value, that intersection is actually a singleton, hence $\{t\}$. But one also sees (going to monotonous subsequences in the odd resp. even indices) that $x\in \bigcap_{n\in \Bbb{N}} [r_{2n-1}, r_{2n}]$. Hence $x=t$ and $|x|=|x|_\mathbb{R}^a$.
