Choice of a square root in $\mathbb{Q}_7$ While studying a calculation in Koblitz's P-adic Numbers, P-adic Analysis, and Zeta-Functions, he remarks the following: 

...we were sloppy when we wrote $4/3 = (1 + 7/9)^{1/2}$. In both $\mathbb{R}$ and $\mathbb{Q}_7$ the number $16/9$ has two square roots $\pm 4/3$. In $\mathbb{R}$, the series for $(1 + 7/9)^{1/2}$
  converges to $4/3$, i.e., the positive value is favored. But in $\mathbb{Q}_7$ the square root congruent to $1 \text{ mod } 7$, i.e., $-4/3 = 1 - 7/3$, is favored. 

So, my question is why is, in $\mathbb{Q}_7$, the root $-4/3$ "favoured"?
 A: Consider the polynomial $1+X$ in the ring $A=\mathbb Q[[X]]$ of power series over $\mathbb Q$. The degree 2 equation $T^2 = 1+X$ has no solution over $\mathbb Q[X]$ but it 
has two solutions over the domain $A$. Mainly, $T = \pm\left( 1+\frac{X}{2}-\frac{X^2}{8}+\frac{X^3}{16}-\frac{5 X^4}{128}+\frac{7 X^5}{256}-\frac{21 X^6}{1024}+\frac{33 X^7}{2048}-\frac{429 X^8}{32768}+\dots\right)$
The "positive" or "favoured" one is 
$$
B_{1/2}(X)= 1+\frac{X}{2}-\frac{X^2}{8}+\frac{X^3}{16}-\frac{5 X^4}{128}+\frac{7 X^5}{256}-\frac{21 X^6}{1024}+\frac{33 X^7}{2048}-\frac{429 X^8}{32768}+\dots
$$
This is the Taylor expansion of $\sqrt{x+1}$ around $0$ in the classical-euclidean sense, seen purely as a formal power series.
It is easy to check that $B_{1/2}\in \mathbb Z[1/2][[X]] \subset \mathbb Z_p[[X]$ for every odd prime $p$. In particular, for every odd $p$ one can consider $B_{1/2}$ as a continuous function
$$
B_{1/2} : p\mathbb Z_p \rightarrow 1+ p\mathbb Z_p 
$$
and also
$$
B_{1/2} : (-1,1)\subset \mathbb R\rightarrow [0,\infty) \subset\mathbb R.
$$
Koblitz points out that $7/9$ is in both $(-1,1)$ and $7\mathbb Z_7$. Hence 
$B_{1/2}(7/9) = 4/3$ as function of $(-1,1)\subset \mathbb R$ while $B_{1/2}(7/9)=-4/3$ as function of $7\mathbb Z_7$.
