Equivalence Relation on Natural Numbers such that Quotient is Power of Two/Equivalence Class of Prime

Theorem
Let $\alpha$ denote the relation defined on the natural numbers $\N$ by:
 * $\forall x, y \in \N: x \mathrel \alpha y \iff \exists n \in \Z: x = 2^n y$

We have that $\alpha$ is an equivalence relation.

Let $\eqclass p \alpha$ be the $\alpha$-equivalence class of a prime number $p$.

Then $\eqclass p \alpha$ contains no other prime number other than $p$.

Proof
That $\alpha$ is an equivalence relation is proved in Equivalence Relation on Power Set induced by Intersection with Subset.

Let $p$ be a prime number whose $\alpha$-equivalence class is $\eqclass p \alpha$.

$\eqclass p \alpha$ contains a v $q$ such that $q \ne p$.

Then:
 * $p = 2^n q$

for some $n \in \Z$

If $n = 0$ then $p = q$ which contradicts $q \ne p$.

Thus $n \ne 0$.

Let $n > 0$.

Then:
 * $p = r q$

for $r \in \Z$ such that $r = 2^n$.

Thus $p$ is a composite number with divisors $q$ and $r$ (and possibly more).

This contradicts the supposition that $p$ is prime.

Let $n < 0$.

Then:
 * $p = \dfrac q {2^m}$

where $m = -n$ and so $m > 0$.

Thus:
 * $q = r p$

for $r \in \Z$ such that $r = 2^m$. Thus $q$ is a composite number with divisors $p$ and $r$ (and possibly more).

This contradicts the supposition that $q$ is prime.

Thus all cases of $n$ lead to a contradiction.

It follows that no $\alpha$-equivalence class can contain more than one prime number.