Suppose a 2-adic metric is defined. Showing that if $d(x,y)$ has a midpoint, then $x=y$ Let $\mathbb{Z}$ be the integers. Recall 2-adic metric $$ d(x,y) =  \begin{cases} 
      0 & x=y \\
      \frac{1}{2^{n}} & x \ne y\ \text{and}\ 2^{n} \text{is the largest power of 2 that divides}\ x-y. 
   \end{cases}
$$
Show that if $x-y$ has a mid point then $x=y$. 
Here is my attempt at the proof. In general, I want to show that if p=2, then $p^{n}$ must divide the distance from $x$ to the midpoint and the distance from $y$ to the midpoint as well as $d(x,y)$, but this is impossible. 
Claim:
If $\exists$ a point $m$ s.t. $$d(x,y) = d(x,m) + d(m,y) \implies d(x,y) = 0 \implies x=y$$
Proof: 
Suppose $x \ne y$ and $\exists\ m$ s.t.  $d(x,y) = d(x,m) + d(m,y)$
$\implies d(x,y) = \frac{1}{2^{n}}$ 
$$\implies 2^{n}|(x-m)\ \text{and}\ 2^{n}|(m-y)\ \text{and}\ 2^{n}|(x-y)$$ but this is impossible since $2^{n}$ is the largest power of $2$ that divides $x-y$ this is impossible unless it is the case that $$x=y.$$
End proof.
This is my intuition, but I am not fully understanding the details of my argument. 
 A: A preliminary comment: a mathematical proof is a piece of expository prose, consisting of (paragraphs of) sentences. You’ll be much clearer if you use more words and fewer symbols for the ‘connective tissue’ of your argument — things like if ... then, therefore, etc.
Starting out by supposing (to get a contradiction) that there are points $x,y$, and $m$ such that $x\ne y$, $d(x,y)=d(x,m)+d(m,y)$, and $d(x,m)=d(m,y)$ is fine. However, you have to be more careful about just what this means. I’d say something like this:

Let $\frac1{2^n}=d(x,m)=d(m,y)$. This means that $2^n\mid x-m$, $2^n\mid m-y$, $2^{n+1}\nmid x-m$, and $2^{n+1}\nmid m-y$. Let $x-m=2^nr$ and $m-y=2^ns$, where $r$ and $s$ are odd integers.

You don’t really explain at all what the resulting contradiction is. Here’s one way to write it up:

Then $(x-y)=(x-m)+(m-y)=2^nr+2^ns=2^n(r+s)$. But $r$ and $s$ are both odd, so $r+s$ is even, and $2^{n+1}\mid x-y$. It follows that $$d(x,y)\le\frac1{2^{n+1}}<\frac1{2^n}+\frac1{2^n}=d(x,m)+d(m,y)\;,$$ contradicting the hypothesis that $d(x,y)=d(x,m)+d(m,y)$. Thus, no such distinct points $x$ and $y$ exist.

