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 $K$?
Your proof should take the following shape (assuming that you want to prove this directly from the definition of divisibility).
Suppose $n$ is a multiple of 3, say $k$ is such that $n = 3k$. [Now somehow construct a clever $k'$; this is very easy.]. Therefore $2n = 3k'$, which shows that $2n$ is a multiple of 3.
Conversely, suppose that $2n$ is a multiple of 3, say $k'$ is such that $2n = 3k'$. [Now somehow construct a clever $k$; slightly harder than the previous case; either use that $k'$ is even because $2n$ is; if that's something you don't want to use, $3n = 3n$ instead.] Therefore $n = 3k$, which shows that $n$ is a multiple of 3.
let n be a multiple of 3. Then n=3k. so 2n=3(2k).
Now let 3 divide 2k. By euclid's lemma 3 divides 2 or k. So 3 divides k.
