Prove this is an equivalence relation. Define a relation of $\mathbb{Q} -$ {$0$} as follows:
$x$ ~ $y$ $\Leftrightarrow$ $\dfrac {x} {y} = 2^k $ for some $k  \in \mathbb{Z}$
Prove this is an equivalence relation.
ATTEMPT:
Reflexive: For any $x\in\mathbb Z$, $\dfrac {x} {x}=1$ and $2^0 = 1$.
Symmetric: For any $x,y\in\mathbb Z$, if $\dfrac {x} {y}= 2^k$ for some $k\in\mathbb Z$, then $\dfrac {y} {x}= 2^k$ where $k\in\mathbb Z$.
Transitive: For any $x,y,z\in\mathbb Z$, if $\dfrac {x} {y}= 2^k$ and $\dfrac {x} {z}= 2^l$ for some $k,l\in Z$, then $\dfrac {x} {z}= 2^\frac{k} {l} -  ^y$ where $\frac{k} {l}-y\in\mathbb Z$.

As we're working with the rationals of form $\frac{x}{y}$ here, we'll need the restriction that $x,y \neq 0$, else we'd get $\frac{x}{0}$, etc. I will work with the assumption that $x,y \neq 0$ in all problems.
Reflexivity:
$x \sim x \Rightarrow \frac{x}x = 2^k$. Obviously, $\frac{x}x = 1 \forall x \in \mathbb{Z}$, so let $k = 0 \in \mathbb{Z}$.
Symmetry:
WTS if $x \sim y$ then $y \sim x$, i.e. $\frac{x}{y} = 2^k$ implies $\frac{y}{x} = 2^l$ for some $k,l \in \mathbb{Z}$. Note that $\frac{y}{x} = \frac{1}{x/y}$. So take $\frac{x}{y} = 2^k$. Then $\frac{1}{x/y} = \frac{1}{2^k}$. Note that $\frac{1}{2^k} = 2^{-k}$. Let $-k = l$, where $l \in \mathbb{Z}$ because of additive inverses. Then we have $\frac{y}{x} = 2^l$.
In this discussion, it is important to note that our fractions $\frac{1}{2^k}, \frac{1}{x/y}$, etc. are never equal to $0$ (because of nonzero numerators, and we have $\frac{x}{y}$ defined as nonzero, and $2^k \neq 0 \forall k$.), so they are in $\mathbb{Q} \setminus \{0\}$.
Transitivity:
Note that your approach is confused: either you made a typo, or you were trying to show that if $x \sim y$ and $x \sim z$, then you can draw some conclusion about $x\sim z$.
WTS: For any $x,y,z \neq 0 \in \mathbb{Z}$, if $x \sim y$ and $y \sim z$, then $x \sim z$. So WTS if $\frac{x}{y} = 2^k$ and $\frac{y}{z} = 2^l$, then $\frac{x}{z} = 2^m$ for some $k,l,m \in \mathbb{Z}$.
Take $\frac{y}{z} = 2^l$. Then $y = 2^l \cdot z$. Now take $\frac{x}{y} = 2^k$. By substitution, we get $\frac{x}{2^l \cdot z} = 2^k$. Now note $\frac{x}{2^l \cdot z} = \frac{x}{z} \cdot \frac{1}{2^l}$. So $\frac{x}{z} \cdot \frac{1}{2^l} = 2^k$. So $\frac{x}{z} = 2^l \cdot 2^k = 2^{l+k}$. 
I.e. if $\frac{x}{y} = 2^k$ and $\frac{y}{z} = 2^l$, then $\frac{x}{z} = 2^{k+l}$. Done.
