Prove or disprove: the average of any two odd integers is always odd

Question

Prove or disprove the following statement: the average of any two odd integers is always odd.

The proof that this statement is false is given below. Please, verify whether the one is valid or not.

$\mathrm{(I)}$ Formal restatement of the problem: $$\forall m, n \in \mathbb{Z}, m\ and\ n\ are\ odd\ \rightarrow \frac{m + n}{2}\ is\ odd.$$

$\mathrm{(II)}$ Proof that this statement is true for some odd integers. More precisely, when two odd integers are equal.

$$\frac{(2k + 1) + (2k + 1)}{2};$$ $$\frac{4k + 2}{2};$$ $$\frac{2(2k + 1)}{2};$$ $$2k + 1.$$

$2k + 1$ is odd by definition, so we proved that the statement $\mathrm{(I)}$ is true for some odd integers.

$\mathrm{(III)}$ Now let us consider the case when two odd integers are not equal. Let $2k + 1$ represent the first odd integer and $2r + 1$ represent the second one.

We want to show that their average is equal to yet another odd integer which we will denote as $2s + 1$.

$$\frac{(2k + 1) + (2r + 1)}{2} = 2s + 1;$$ $$\frac{2k + 2r + 2}{2} = 2s + 1;$$ $$\frac{2(k + r + 1)}{2} = 2s + 1;$$ $$k + r + 1 = 2s + 1;$$ $$k + r = 2s;$$ $$s = \frac{k + r}{2}.$$

By definition of an odd number, $s$ must belong to $\mathbb{Z}$, but that's not always the case since $\mathbb{Z}$ is not closed under division. For example, if $k = 1$ and $r = 2$, then $s = 1.5$ and $s \notin \mathbb{Z}$.

The statement $\mathrm{(I)}$ is false since it is not true for all odd numbers.

Q.E.D.

Answer 1

Let $4k+1$ and $4t+3$, $k,t \in Z$ be two odd integers of the form.

$\text{average}=\frac{(4k+1)+(4t+3)}{2}=2(k+t+1)$ is an even number.

The assumption is not true for all pairs of odd numbers.

Answer 2

To disprove the statement, it is enough to produce a single counterexample, rather than a family of counterexamples (as you do in your post - this is overkill and does not add anything to the proof.) For a quick counterexample, the average of $1$ and $3$ is $\frac{1+3}{2} = 2$ which is even.