Why is my logical statement wrong? "Express the following sentence symbollically, using only quantifiers for real numbers, logical connectives, the order relation &lt; 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:

$$(\forall x,y\in \mathbb{R})[x&gt; y](\exists q\in \mathbb{Q})[q&gt;y \wedge q&lt; x]$$

In light of some comments a correct version of my incorrect statement should be:
$$(\forall x,y\in \mathbb{R})[x≠ y \Rightarrow (\exists q\in \mathbb{Q})[q&gt;y \wedge x&gt; q]\vee[y&gt;q \; \wedge \;q&gt;x]]$$

Can you help me understand why my answer is wrong?
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 :


  $\forall x \forall y ( x &gt; y ) \exists q (Q(q) \land [q&gt;y \land q&lt;x])$


there are basically two mistakes :

(i) one substantial: as pointed in the above comments, in this way you are assuming that $x &gt; y$, 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 :


  $\forall x \forall y ( x &gt; y )$ and $\exists q (Q(q) \land [q&gt;y \land q&lt;x])$


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.
$$\forall x,y\in\mathbb{R} x\not=y \implies \exists q\in\mathbb{Q} (y-q)(q-x) &gt; 0 $$