Help with Proof Involving Sets

Question

Below is my solution to a proof; questions follow.

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

Observe that $A-B=\{n \in \Bbb Z:4|n-2\}$.

First, we show that if $n\in A-B$ then $n=2k$ for some odd integer $k$. Assume $n\neq 2k$ for some odd integer $k$. Then, it must be that $n=2k+1$ for some odd integer $k$. Let $k=2x+1$ for some $x\in \Bbb Z$. Then, $n=2k+1=2(2x+1)+1=4x+3$. Therefore, $4\not|n-2$ and $n\not\in A-B$.

Next, we show that if $n=2k$ for some odd integer $k$, then $n\in A-B$. Assume $n=2k$ for some odd integer $k$. Let $k=2y+1$ for some $y\in\Bbb Z$. Then, $n=2k=2(2y+1)=4y+2$. Thus, $4|n-2$ and $n\in A-B$.

$$\tag*{$\blacksquare$}$$

I have several questions:

1. Is this proof correct?
2. I arrived at the fact that $A-B=\{n \in \Bbb Z:4|n-2\}$ through experimentation. Is there a more formulaic way to handle operations between sets defined using the divisibility of their elements? Do I have to do more to prove what $A-B$ equals?
3. Do I have to do more to prove that $4\not|n-2$, or is that obvious from the fact that $n$ may be written as $4x+3$?
4. Any general criticisms/suggestions regarding my proof writing technique.

Thanks

Answer 1

This is not correct.

For one, you need to justify your assertion that $A-B=\{n \in \Bbb Z:4|n-2\}$. This is approximately the same difficulty as the original problem, so you can't just assume it to be true or take it as obvious, at least at this level. It requires proving both $$A-B\subseteq \{n \in \Bbb Z:4|n-2\}$$ and $$\{n \in \Bbb Z:4|n-2\} \subseteq A-B.$$

Furthermore, the proof of the first implication is not correct:

Assume $n\neq 2k$ for some odd integer $k$. Then, it must be that $n=2k+1$ for some odd integer $k$.

This does not follow. It may be that $n = 2k$ for an even integer $k$. All you know is that it is not true that there is an odd integer $k$ so that $n = 2k$.

My advice would be to not use the assertion $A-B=\{n \in \Bbb Z:4|n-2\}$ and instead prove what you want to prove directly. Assume that $n \in \mathbb{Z}$ with $n \in A-B$, and prove that $n = 2k$ for an odd $k$. Conversely, assume that $n = 2k$ for some odd $k$ and prove that $n \in A-B$.

Answer 2

$n\in A-B$ if and only if $n \in A$ and $n\not \in B$.

So $n \in A-B$ if and only if $2|n$ and $4\not\mid n$.

Since $2|n$ then there is a $k$ so that $n = 2k$.

If $k$ is even we get a contradiction: $k = 2m$ for some $m$ and so $n=2k=4m$ and $4|n$. So $k$ must be odd.

Alternatively:

If $n$ is an integer it will have a remainder when divided by $4$. That remainder will by either $0,1,2$ or $3$.

If the remainder is $0$ then $4|n$ and $n \in B$ so $n \not \in A-B$.

If the remainder is $1$ or $3$ then $2\not \mid n$ so $n \not \in A$ sso $n \not \in A-B$.

If the remainder is $2$ then $2\mid n$ but $4\not \mid n$. So $n \in A$ and $n \not \in B$ so $n \in A-B$.

So $n \in A-B$ if and only if $n = 4k +2 = 2(2k+1)$ for some integer $k$.