I'm trying to proof that a number can't be both odd and even at the same time. The definitions are the follows:
- A number is called even if there's $m\in\mathbb{Z}$ that satisfies $k = 2m$.
- A number is called odd if there's $m\in \mathbb{Z}$ that satisfies $k = 2m - 1$.
I wrote a proof by contradiction that uses the field axioms and I wonder if I haven't missed anything. The proof goes as follows: Lets assume by contradiction that this state is invalid, which means let $k\in\mathbb{Z}$ be a number that is both odd and even. That means: $2m = k = 2k - 1$ and by the transitivtiy it means $2m = 2m - 1$. In that case, I can add the negative number of $2m$ to both side of the equation and get:
$2m + (-2m) = 2m - 1 + (-2m)$
$(2m + (-2m)) = (2m + (-2m)) - 1$
$0 = -1$
And since we already proofed $0\ne1$ we get a contradiction. $Q.E.D.$
I wonder if my proof is correct? or do I miss anything?
That is not correct because you are assuming the same $m$ for your odd and even integers.
You may approach the problem with $$ 2m=2n-1$$ and see where you can take it.