Characterization of normed space Let $\mathbb R^n$ be vector space over $\mathbb R$. Then we know that all norms over $\mathbb R^n$ are homeomorphism. Is it true for $\mathbb Q^2$ over $\mathbb  Q$ For instance are the Euclidean norm and $p$-adic norm  two non-homeomorphism norms over $\mathbb Q^2$?
 A: You seem a little bit confused about your definitions, so I'll try and make some things a bit clearer.  
Definition 1 If $K$ is a field, an absolute value on $K$ is a map $|.|\colon K\to \mathbb R_{\ge 0}$ such that the following axioms apply:


*

*$|x|=0$ if and only if $x=0$.

*For all $x,y\in K$, $|xy|=|x||y|$

*For all $x,y\in K$, $|x+y|\le |x|+|y|$.  


We say two absolute values $|.|,|.|'$ on a field $K$ are equivalent if there exists some $\alpha$ such that $|x|'=|x|^\alpha$ for all $x\in K$.  
In the case that $K=\mathbb Q$, a theorem of Ostrowski classifies all the absolute values:
Theorem 1 The absolute values on $\mathbb Q$, up to equivalence, are:


*

*The usual absolute value (the modulus)

*The trivial absolute value given by $|x|=1$ for all $x\ne0$

*The $p$-adic absolute value for some $p$.  


Now we can start talking on norms on vector spaces over fields like $\mathbb Q$.  
Definition 2 Let $K$ be a field, and let $|.|$ be an absolute value on $K$.  If $V$ is a vector space over $K$, then a norm on $V$ (with respect to $K$) is a map $\|.\|\colon V\to \mathbb R_{\ge0}$ satisfying the following axioms:


*

*$\|x\|=0$ if and only if $x=0$

*If $\lambda\in K$ and $x\in V$, then $\|\lambda x\|=|\lambda|\|x\|$, where $|\lambda|$ means our absolute value applied to $\lambda$.  

*If $x,y\in V$, then $\|x+y\|\le \|x\|+\|y\|$.  


We can define equivalence of norms in the usual way.  Then a generalization of the result you mentioned over $\mathbb R$ is:
Theorem 2 Let $K$ be a field, and let $|.|$ be an absolute value on $K$ such that $K$ is complete with respect to the metric given by $d(x,y)=|x-y|$.  Then any two norms defined on $K^n$ with respect to $|.|$ are equivalent.  
Now $\mathbb Q$ is never complete with respect to any of its non-trivial absolute values.  So this result does not immediately apply.  However, if $|.|$ is an absolute value on $\mathbb Q$, we can complete it to get the smallest possible extension of $(\mathbb Q,|.|)$ that is complete.  
If we start with the usual absolute value, the completion we get is $\mathbb R$.  If we start with the $p$-adic absolute value, we get the $p$-adic numbers $\mathbb Q_p$.  All norms on $\mathbb Q_p^n$ are equivalent with respect to the $p$-adic norm on $\mathbb Q_p$.  So we can pull this norm back to $\mathbb Q^n$ and get a classification of some of the norms on $\mathbb Q^2$.  
Another way we get a partial classification is if we identify the vector space $\mathbb Q^2$ with a quadratic field extension $K$ of $\mathbb Q$ and require that the norm on $\mathbb Q^2$ give us an absolute value with respect to $K$.  In other words, we are looking for absolute values on $\mathbb K$ that extend the absolute value on $\mathbb Q$.  Then we can again classify all norms:


*

*If we start with the usual absolute value on $\mathbb Q$, then the absolute values on $K$ extending this absolute value arise from embedding $K$ into the real or complex numbers, and taking moduli as usual.  

*If the original absolute value on $\mathbb Q$ was the $p$-adic absolute value $|.|_p$, the absolute values on $K$ extending $|.|_p$ are the $\mathfrak p$-adic absolute values defined as follows:

*Let $x\in K$.  Then we have a unique factorization of the fractional ideal $x\mathcal O_K$ into prime ideals $\mathfrak q_1^{a_1},\dots,\mathfrak q_n^{a_n}$.  If $\mathfrak p$ is a prime ideal of norm $p$, then $|x|_{\mathfrak p}$ is the reciprocal of the coefficient of $\mathfrak p$ in this factorization.  
Update (requested by OP): We show that two absolute values $|.|,|.|'$ on a field $K$ are equivalent (as defined above) if and only if they give rise to the same topology on $K$ via the metric $d(x,y)=|x-y|$.  
Theorem 3 Let $K$ be a field, and let $|.|_1,|.|_2$ be two non-trivial absolute values on $K$.  Then the following are equivalent:
i) $|.|_1,|.|_2$ give rise to the same topology on $K$.
ii) For all $x\in K$, $|x|_1<1$ if and only if $|x|_2<1$.
iii) There exists $\alpha>0$ such that for all $x\in K$, $|x|_2=|x|_1^c$.
Proof: (i) $\Rightarrow$ (ii): Suppose $|.|_1,|.|_2$ give rise to the same topology on $K$.
\begin{align}
|x|_1<1 &\Leftrightarrow x^n\to 0 \textrm{ with respect to } |.|_1\\
&\Leftrightarrow x^n\to 0 \textrm{ with respect to } |.|_2\\
&\Leftrightarrow |x|_2<1
\end{align}
(ii) $\Rightarrow$ (iii): Since $|.|_1$ is non-trivial, we may choose some $a\in K^*$ with $|a|_1<1$.  Let $x\in K^*$, and let $m,n\in\mathbb Z$, with $n>0$.  
Now
\begin{align}
\frac{\log(|x|_1)}{\log(|a|_1)}>\frac{m}{n} &\Leftrightarrow \left|\frac{x^n}{a^m}\right|_1<1\\
&\Leftrightarrow \left|\frac{x^n}{a^m}\right|_2<1\\
&\Leftrightarrow \frac{\log(|x|_1)}{\log(|a|_2)}>\frac{m}{n}
\end{align}
But now $\frac{m}{n}\in\mathbb Q$ is arbitrary.  Therefore, we must have
$$
\frac{\log(|x|_1)}{\log(|a|_1)}=\frac{\log(|x|_2)}{\log(|a|_2)}
$$
So $\log(|x|_2)=c\log(|x|_1)$ for some $c$ independent of $x$.
(iii) $\Rightarrow$ (i): Clear.  $\Box$
