Proof: If $x$ is odd, then $x+2$ is odd.

Question

I'm fairly new to writing proofs so any advice can help.

I'm asked to prove the following statement: "If $x$ is odd, then $x+2$ is odd". Here is my proof:

We will prove this by contraposition: if $x+2$ is not odd, then $x$ is not odd.

Let there be an integer $k$ such that $x+2 = 2k$.

Thus, \begin{align} x & = 2k-2 \\ & = 2(k-1) \end{align}

Then $x = 2(k-1)$ is an even number.

Since the contrapositive is true, the statement "If $x$ is odd, then $x+2$ is odd" is true by logical equivalency.

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The problem is: I don't know if my proof is enough or how to properly tackle them. Any advice?

Answer 1

This seems fine as long as you know that "not odd" is the same as even for integers. Also, for your opening sentence in the proof, I might say "If $x+2$ is even then we can write $x+2=2k$ for some integer $k$."

You can also just prove this directly if you know that odd integers are of the form $2k+1$. That is, if $x=2k+1$, then $x+2=2(k+1)+1$.

Answer 2

An integer $m$ is even if and only $2 \mid m$; otherwise, $m$ is odd.

If $x + 2$ is even, then

$2 \mid x + 2; \tag 1$

clearly,

$2 \mid 2; \tag 2$

thus

$2 \mid (x + 2) - 2 = x; \tag 3$

that is, $x$ is even, contrary to hypothesis; thus, $x + 2$ cannot be even, hence it is odd.

By the way, our OP's proof, being essentially the same as mine, is fine.