I'm learning about equivalence relations and equivalence classes and wanted to ask about an exercise. The exercise is as follows;
Let $R$ be a relation given by $$xR y \iff \exists n \in \Bbb Z: \ \ x = 2^ny. \ \ \ \ \ (x, y \in \Bbb Z)$$ Show that no $R$-class contains more than one prime.
I presume this has something to do with the fundamental theorem of arithmetic (i.e. uniqueness of prime factorisation).
Any $x$ that we choose has to be expressable as $x = 2^ny$ which means that, for instance, if $x = 10$ then the $R$-class of $10$ is
$$R_{10} = \lbrace y : 10 = 2^ny\rbrace.$$
The only way we can write $10$ as a product of prime factors is $2 \times 5$, so the only prime in $R_{10}$ is $5$.
Can we split this into cases, the case where $x$ is prime, so the only prime in $R_x$ is $x$ and the case where $x$ is composite, so $x$ has prime factors $2^n$ and another prime $y$?