Understanding a statement about equivalent norms ($||\cdot ||_2 \sim||\cdot||_1 $) I am trying to understand the following statement from an analysis book:

Two norms are equivalent ($||\cdot ||_2 \sim||\cdot||_1 $) if they
  induce equivalent metrics.

At first I thought this was straight forward but then the book has the following proposition that leads me to believe that I have a misunderstanding:

Let $||\cdot||_1$ and $||\cdot||_2$ be two norms on a field F. Then
  $||\cdot ||_2 \sim||\cdot||_1$ iff there exists a positive real number
  $\alpha$ such that $||x||_2=||x||_1^\alpha , \ \forall x\in F$.

How can it be that $||x||_2=||x||_1^\alpha$ if x is any other value other that 1 if $\alpha\ne$1? Maybe this is related to my misunderstanding... can norms and metrics be thought of as the same thing? Thanks!
 A: They aren't equal only equivalent, which metrics are if they induce the same topology. This means that they have the same open sets, and since a field is also a one-dimensional vector space, this just means they need to have the same unit balls.
But then $$\lVert \cdot \rVert_1\le 1 \iff \lVert \cdot\rVert_2\le 1$$
so choose any $y$ such that $0\ne \lVert y\rVert_1>1$--if this is not possible the proposition is trivial since both give the discrete topology where every non-zero vector has norm $1$--and consider $x\ne 0$ so that $\lVert x\rVert_1=\lVert y\rVert_1^\alpha$, for some $\alpha\in\Bbb R$. Let $m_i/n_i\downarrow \alpha$ be a sequence of rational numbers with $n_i$ positive. Then
$${\lVert x^{n_i}\rVert\over\lVert y^{m_i}\rVert_1}<1\implies {\lVert x^{n_i}\rVert\over \lVert y^{m_i}\rVert_2}<1$$
so that $\lVert x\rVert_2\le \lVert y\rVert_2^{m_i/n_i}$ so that the same holds in the limit giving $\lVert x\rVert_2\le\lVert y\rVert_2^\alpha$. Similarly using a sequence $m_i'/n_i'\uparrow\alpha$ we get the reverse inequality $\lVert x\rVert_2\ge \lVert y\rVert_2^\alpha$.
Then logging both of 
$$\begin{cases}\lVert x\rVert_1=\lVert y\rVert_1^\alpha \\ \lVert x\rVert_2=\lVert y\rVert_2^\alpha\end{cases}$$
we get
$${\log \lVert x\rVert_1\over\log \lVert x\rVert_2}={\log \lVert y\rVert_1\over\log \lVert y\rVert_2}=s$$
and using that $y$ is arbitrary and $>0$, we get that $s>0$ and that the equality holds on all of the space.
