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proof of theorem on equivalent valuations

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It is easy to see that $| \cdot |$ and $| \cdot |^c$ are equivalent valuations for any constant $c > 0$ --- it follows from the fact that $0 \le x^c < 1$ if and only if $0 < x \le 1$.

Assume that the valuations $| \cdot |_1$ and $| \cdot |_2$ are equivalent.  
Let $b$ be an element of $K$ such that $0 < |b|_1 < 1$.  Because the valuations are assumed to be equivalent, it is also the case that $0 < |b|_2 < 1$.  Hence, there must exist positive constants $c_1$ and $c_2$ such that $|b|_1^{c_1} = {1 \over 2}$ and $|b|_2^{c_2} = {1 \over 2}$.  

We will show that show that $| x |_1^{c_1} = | x |_2^{c_2}$ for all $a \in K$ by contradiction.

Let $a$ be any element of $k$ such that $0 < |a|_1 < 1$.  Assume that $| a |_1^{c_1} \neq | a |_2^{c_2}$.  Then either $| a |_1^{c_1} < | a |_2^{c_2}$ or $| a |_1^{c_1} > | a |_2^{c_2}$.  We may assume that $| a |_1^{c_1} < | a |_2^{c_2}$ without loss of generality.

Since $| a |_2^{c_2} / | a |_1^{c_1} > 1$, there exists an integer $m > 0$ such that $(| a |_2^{c_2} / | a |_1^{c_1})^m > 2$.  Let $n$ be the least integer such that $2^n |a|_2^{m c_2} > 1$.  Then we have
 $$2^n |a|_1^{m c_1} < 2^{n-1} |a|_2^{m c_2} < 1 < 2^n |a|_2^{m c_2}.$$
Since $2 = |b^{-1}|_1^{c_1} = |b^{-1}|_2^{c_2}$, this implies that
 $$\left| {a^m \over b^n} \right|_1^{c_1} < 1 < \left| {a^m \over b^n} \right|_2^{c_2},$$
but then
 $$\left| {a^m \over b^n} \right|_1 < 1$$
and
 $$\left| {a^m \over b^n} \right|_2 > 1,$$
which is impossible because the two valuations are assumed to be equivalent.
\rightline{Q.E.D}
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