If $n$ is an integer and $3n+2$ is even, then $n$ is even.

Question

I am still new to proofs and just starting to get some practice in. Here is what the homework problem asks.

Prove that if $n$ is an integer and $3n+2$ is even, then $n$ is even.

Proof attempt by contradiction: Suppose $3n+2$ is even and $n$ is odd. If $n$ is odd, then $n=2k+1$ for some $k \in \mathbb{Z}$. Then, $3n+2=3(2k+1)+2=2(3k)+5$. Since $k \in \mathbb{Z}$, then $3k \in \mathbb{Z}$. Let $3k=m \Rightarrow 3n+2=2m+5$, which is odd. Thus, contradicting our assumption that $3n+2$ is even. $\blacksquare$

I feel as if I am missing some steps and/or grazing over something important. Any feedback, tips/suggestions, or words of wisdom would be greatly appreciated. Thank you in advance.

Answer 1

Yes, your proof is OK.

Say $n$ is odd, so $3n$ is odd, so $3n+2$ is odd, a contradiction.

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Or you can do like this: since $2\mid 3n+2$ we have $2\mid (3n+2)-2 = 3n$. But $\gcd(3,2)=1$ so $2\mid n$.

Answer 2

Yes it is a correct proof by contradiction, more simply we can conclude from here

$$2(3k)+5=2(3k)+4+1=2(3k+2)+1$$

but it is really a detail.

Answer 3

Your proof by contradiction is fine. This is a direct proof: if $3n+2$ is even then there is $k\in\mathbb{Z}$ such that $3n+2=2k$. Hence, since $n$ is an integer, it follows that $$n=2k-2-2n=2(\underbrace{k−1−n}_{\in \mathbb{Z} })$$ which means that $n$ is even.

Answer 4

If $3n+2$ is even, then $3n+2\equiv0 \mod 2\Rightarrow 3n\equiv0 \mod 2$. Where $n\in\mathbb{Z}_{2}=\{0,1\} $.

If $n\equiv1 \mod 2$ then $3(1)\equiv0 \mod 2$ this is a contradiction, since $1\not \equiv0 \mod 2$.

Therefore $n\equiv 0 \mod 2$ then $n$ is even.