Let $ A =\{n\in\mathbb{Z}: 2 | n\}$ and $B=\{n\in\mathbb{Z}: 4 | n\}$. Prove that $n\in (A - B)$ if and only if $n=2k$ for some odd integer k.

Question

Let $A = \{n\in\mathbb{Z}: 2\, |\, n\}$ and B={$n\in\mathbb{Z}: 4 \, | \, n$}. Prove that $n\in(A - B)$ if and only if $n=2k$ for some odd integer $k$.

I'm not sure how to prove this 'correctly'. Any help would be greatly appreciated, thanks!

Answer 1

Say $n\in A-B$. Clearly, $A-B=\{...,-6,-2,2,6,...\}$ contains multiples of $2$ that are not multiples of $4$, that is, integers of the form $4m+2$ for some $m\in\mathbb Z$. Note that $n=4m+2=2(2m+1)$ and $2m+1$ is odd $\ \forall m\in\mathbb Z$.

Now, let $n=2k$, where $k$ is an odd integer. We can write $k=2a+1, a\in\mathbb Z\implies n=2(2a+1)=4a+2\implies2|n, 4\nmid n\implies n\in A-B$

Answer 2

Suppose $n\in A - B$. Then $2$ divdes $n$ so $n = 2k$ for some integer $k$.

We proceed to prove that $k$ is odd by contradiction.

Suppose $k$ were even. Then $k = 2l$ for some integer $l$. Hence, $n = 4l$ and so $n$ is divisible by $4$. However, by assumption $n\not\in B$ and hence $n$ cannot be divisible by $4$. Therefore $k$ must be odd.

Putting this together, $n = 2k$ for an odd integer $k$, as required.