prove that $R$ is an equivalence relation $$\forall a,b \in \mathbb{Q} \quad aRb \Leftrightarrow \quad \exists k \in \mathbb{Z}: \quad b=2^ka$$
1) Reflexivity:
$\forall a \in \mathbb{Q}\quad aRa \Leftrightarrow \quad \exists k \in \mathbb{Z}: \quad a=2^ka $ 
choosing $k=0 \quad  \Rightarrow a=2^0a=a \Rightarrow aRa \Rightarrow R \text{ is reflexive}$
2) Symmetry
3) Transitivity:
$\forall a,b,c \in \mathbb{Q}:$
$$ aRb \Leftrightarrow \quad \exists k \in \mathbb{Z}: \quad b=2^ka $$ 
$$ bRc \Leftrightarrow \quad \exists h \in \mathbb{Z}: \quad c=2^hb $$ 
Then
$$ aRc \Leftrightarrow \quad \exists p \in \mathbb{Z}: \quad c=2^pa $$ 
so $aRb,bRc \Rightarrow c=2^hb=2^h2^ka=2^{h+k}a$
choosing $p=k+h \Rightarrow c=2^pa \Rightarrow aRc \Rightarrow \text{ R is transitive}$ 
Can anyone confirm that 1) and 3) are correct?
I tried to prove 2) but is ended like transitive proof and I think that is entirely wrong, I have no idea how to succeed, can anyone provide some hints/proof/solution?. thanks in advance

Your proofs for parts (1) and (3) are correct.  For symmetry, suppose $aRb$, so that
\begin{equation}
b = 2^ka
\end{equation}
for some $k\in\mathbf{Z}$.  Can you think of an integer $l$ so that
\begin{equation}
a = 2^lb?
\end{equation}
(Hint: Remember that negative integers are integers too!)  Once you have such an integer, you can conclude that $bRa$, meaning that $R$ is symmetric.
