Why is my logical statement wrong? "Express the following sentence symbollically, using only quantifiers for real numbers, logical connectives, the order relation < and the symbol Q having the meaning 'x is rational'" 
I have to translate the sentence "There is a rational number between any two unequal real numbers". I worked a bit on it and eventually deduced the following:
$$(\forall x,y\in \mathbb{R})[x> y](\exists q\in \mathbb{Q})[q>y \wedge q< x]$$
In light of some comments a correct version of my incorrect statement should be:
$$(\forall x,y\in \mathbb{R})[x≠ y \Rightarrow (\exists q\in \mathbb{Q})[q>y \wedge x> q]\vee[y>q \; \wedge \;q>x]]$$
Can you help me understand why my answer is wrong?
 A: I think that it should actually be $\forall{x,y}(x\neq{y}\implies{\exists{q}(Q(q)\wedge(\neg(x<q)\wedge{x\neq{q}}\wedge{y<q})\vee(\neg{(y<q)}\wedge{y\neq{q}}\wedge{x<q}})))$.
This answer is written assuming that your variables range accross $\mathbb{R}$, since you technically don't have $\epsilon$ symbol in your language and $x\neq{y}$ is shorthand for $\neg(x=y)$
A: In your first formula, following @Danul G's advise, I assume $\mathbb R$ as the domain of the variables, and I'll rewrite it with the predicate $Q(x)$ (for $x$ is rational) to stay with the original formulation of the problem :

$\forall x \forall y ( x > y ) \exists q (Q(q) \land [q>y \land q<x])$

there are basically two mistakes :
(i) one substantial: as pointed in the above comments, in this way you are assuming that $x > y$, but your problem says $x \ne y$; so you must correct the formula accordingly.
(ii) the other one is connected to the way you have written the formula; juxtaposition of the two sub-formulas is not formally correct: at most you can read it as a conjunction.
But in this way you are saying that :

$\forall x \forall y ( x > y )$ and $\exists q (Q(q) \land [q>y \land q<x])$

i.e.that for two real numbers whatever, the first is greater than the second (false ! : take $0$ as $x$ and $1$ as $y$) and ... and the resulting statement is false, being the conjunction of two formulas, one of which is false.
A: Try the next. Using $\sim$ for denying: translate Trichomoty Law, use $\forall x,y\in\mathbb{R}\colon \phi(x,y)$ and $\phi(x,y)\equiv\sim(x=y)\rightarrow[[x<y]\vee[x>y]]$. Then we also have $\forall x\in\mathbb{R}\forall{q}\in Q\colon \phi(x,q)$ and as the antecedent is fullfilled we have $[x<q]\vee[x>q]$. The same holds for $\phi(q,y)$. Now the existence of q. Take p.e. $x<q$ and $q<y$, fix x and y; then you will have two sets of rationals for which the cut is not empty.  
A: $$\forall x,y\in\mathbb{R} x\not=y \implies \exists q\in\mathbb{Q} (y-q)(q-x) > 0 $$
