proposition with implication EDIT: I first designated $x$, $y$ as irrational numbers. I mean rational.
I have this, In the question it says: For every $x$, $y$ being rational,there exists $z$ being rational so that: $x<z$ or $z<y$  Now, I have this: $\forall(x,y) \in\Bbb Q^2, \exists z\in\Bbb Q/(x<z)∨(z<y)$ Does this the signify the same as $\forall(x,y)\in\Bbb Q^2 \Rightarrow\exists z\in\Bbb Q/(x<z)\lor(z<y)$
 A: The second statement you have written down is not a well-formed statement in first-order logic. 
One way to see this is to look at any logic textbook where they inductively define what statements are admissible, and see that it's impossible to create something that looks like what you've written down.
Another way, though, is to think about the interpretation of $\rightarrow$. We say classically that $A\rightarrow B$ is true iff $B$ is true or $A$ is false and both $A$ and $B$ are statements. So for your statement, let's check whether the left hand side of the implication is true. 
But the left hand side is just the fragment $\forall (x,y) \in \mathbb{Q}^2$, which is not a statement by itself. 
EDIT: I apologize, there are conflicting conventions on this issue, relating to whether the variant of first order logic you're using has sorts. In some presentations $(\forall (x,y)\in \mathbb{Q}^2)A(x,y)$ is shorthand for $\forall x \forall y ((x,y)\in \mathbb{Q}^2 \wedge A(x,y))$.
A: Your first version says "For all rational ordered pairs (x, y), there exists a rational number z such that $x\lt z \lor z\lt y$."
Here is another way of stating what you seem to want to express, in the form of an implication: 
$$\forall x, \forall y \Big((x\in \mathbb Q \land y \in \mathbb Q)\to \exists z\big(z\in \mathbb Q \land ((x\lt z)\lor (z\lt y)\big)\Big)\tag 1$$
$$\forall (x, y) \Big(\big (x, y) \in \mathbb Q^2)\to \exists z\big(z\in \mathbb Q \land ((x\lt z)\lor (z\lt y))\big)\Big)\tag 2$$
You've done most of the work, but in, example, $(1)$ above, we have (in loglish):  For all $x, y$, (if $x$ and $y$ are rational, then there exists a rational number $z$ such that $(x\lt z \lor z\lt y))$.  

But you pretty much said the same thing, in your first proposal. My versions show the explicit implication operator; but your expression is another way to say the same thing.  In math, you'll often see statements written in your fashion.  It a more strict logic class, we use set membership as a predicate.
