Does every non-archimedean absolute value on field take value in $\mathbb{Q}$ Let $K$ be a field, a non-archimedean absolute value is defined to be a map $K\to \mathbb{R}$ satisfying $|x|=0\Rightarrow x=0$, $|x|\cdot|y|=|xy|$ and $|x+y|\leq\max(|x|,|y|)$. 
Is there an example that the valuation of a non-archimedean field not take value in $a^\mathbb{Q}$ for any fixed 
real number $a$?
 A: Think to $x^{\sqrt{2}}$ as a function $(0,\infty)\to \Bbb{R}$, no problem to look at the ring $\Bbb{R}[x,x^{\sqrt2}]$ and its fraction field $K$ with the natural valuation $$v(\sum_{n,m}c_{n,m}x^{n+m\sqrt2})=\inf \{ n+m\sqrt2, c_{n,m}\ne 0\}, \qquad |f/g|=2^{v(f)-v(g)}$$
A: Yes (to the question in your post, not in the title). Here is an example: Choose your favourite prime $p$, your favourite positive number $r\notin p^\mathbb{Q}$, and define the following absolute value on the polynomial ring $\mathbb{Q}[t]$:
$$|\sum_{i=0}^n a_i t^i| := \max (|a_i|_p r^i)$$
where $|\cdot|_p$ is the usual $p$-adic absolute value on $\mathbb{Q}$. Extend it multiplicatively to the field of fractions $K:=\mathbb{Q}(t)$. We have
$$|K| = \{0\} \cup p^\mathbb{Z}\cdot r^\mathbb{Z}.$$
In particular both $p$ and $r$ are in $|K^*|$, but if there were an $a\in \mathbb{R}$ with $p \in a^\mathbb{Q} \ni r$, it would follow that $r \in p^\mathbb{Q}$, contradiction.
Note that of course there are other non-archimedean (and non-trivial) absolute values on that field such that the value group $|K^*|$ is contained in $a^\mathbb{Q}$. A follow-up question would be: Are there any fields $K$ such that every non-trivial nonarchimedean value on $K$ has a value group that is not contained in some $a^\mathbb{Q}$?
A: Hint: Any power of a non-archimedian absolute value is again an absolute value.  Power by a real number.  See if you can arrange irrational values that way.
A: Is there an example of a non-Archimedean valuation that does not take values in $a^\mathbb{Q}$ for any fixed real number $a$?
The answer is no. In fact, if $|\cdot|$ is the trivial valuation, then it takes values in $a^\mathbb{Q}$ for $a=0$ and $a=1$.
If  $|\cdot|$ is not the trivial valuation, then there exists $x\in K$ such that $|x|\not\in\{0,1\}$. Thus, $|x|^\mathbb{Z}\subset|K|$. 
But maybe the question that Qixiao had in mind was: Is there an example of a non-Archimedean valuation over a field $K$ such that $|K|\subset a^\mathbb{Q}$ for no real number $a$? I do not know the answer to this second question.
