There is an odd number between any two even numbers 
Write the complete mathematical logical expression of the following: there is an odd number between any two even numbers.

My attempt:
$E(x): x$ is even.
$O(x): x$ is odd.
$$∀m∀n∃k[(E(m) ∧ E(n) ∧ O(k)) ∧ (m < k < n)]$$
This looks like in English I'm saying: For all m and for all n there is a k where m is even, n is even, k is odd, and m is less than k is less than n, meaning that there is some odd number between any two even numbers.
Or, do I need to specify that n could be less than m as well?
EDIT: Attempt 2:
$$∀m∀n∃k[(E(m) ∧ E(n) ∧ O(k)) ∧ ((m < k < n) v (n < k < m))]$$
 A: 

there is an odd number between any two even numbers.


First, let's rearrange the sentence into a more logical ordering: $$\text{Given any two even numbers, there is some odd number between them.}$$ So: $$\forall (m,n){\in}\mathbb E^2,\;\exists k{\in}\mathbb O\,\Big(\text{$k$ is between $m$ and $n$}\Big)$$ But wait, this formalisation admits couples like $(2,2),(16,16),$ etc.! Assuming that the author actually means “two distinct even numbers” (in which case the given statement is true), we have to refine it (adopting the usual precedence convention to minimise my number of parentheses): $$∀m∀n \;\bigg(Em ∧ En ∧ m\ne n \to ∃k\: \Big(Ok ∧ (m < k < n \;\lor\;  n < k < m)\Big)\bigg).$$ Notice that I did not insist that the quantifiers all be lumped together in front; it's easier to make sense of the translation when they are placed at their natural locations. Finally, if the prenex form is preferred: $$∀m∀n∃k\; \bigg(Em ∧ En ∧ m\ne n \to Ok ∧ \big(m < k < n \;\lor\;  n < k < m\big)\bigg).$$ Do note that although it seems that I have simply pulled the quantifiers to the front, the actual algorithm for converting a logical formula to prenex form is a little less straightforward than this.
Alternatively, the previous formula can be rewritten without the conditional: $$∀m∀n∃k\; \bigg(¬Em ∨ ¬En ∨ m=n ∨ \Big( Ok ∧ \big(m < k < n \;\lor\;  n < k < m\big)\Big)\bigg).$$
When formalising a statement (analogous to writing an essay in drafts), you don't have to do it in a single step, and the preliminary steps need not be fully correct.

Attempt 2: $$∀m∀n∃k[(E(m) ∧ E(n) ∧ O(k)) ∧ ((m < k < n) \;\lor\;   (n < k < m))]$$

This sentence is false: for $(m,n)=(1,2),$ there is no $k$ for which $$E(m) ∧ E(n) ∧ O(k) ∧ (m < k < n \;\lor\;  n < k < m)$$ is true.
A: Let $E(x)$: $m$ is even
$O(x)$: $m$ is odd
where $x\in\mathbb N$.
Then, the statement is:
$\forall m,n\in\mathbb N, \exists k\in(m,n)$ s.t. $(E(m)\land E(n))\implies O(k)$
