Proving n is a multiple of 3 if and only if 2n is a multiple of 3 I want to prove that for all natural numbers $n$, $n$ n is a multiple of 3 if and only if $2n$ is a multiple of 3.
I started by writing:
$$x\equiv n=3k$$
$$y\equiv 2n=3k'$$
(where $k$ and $k'$ is any integer)
Since we want to prove $x\iff y$, it can be done by proving $x\implies y$ and $y\implies x$ individually.
How do you prove this formally?

Clearly if $2n = 3k$ then $k$ must be even, so $k = 2K$ and $2n = 3k = 6K$ and so $n = 3K$ as desired.
Conversely, if $n = 3k$, ... can you complete this showing $2n = 3K$ for some $K$?
