What is the correct negation of the Statement "For every rational number $x$, $x \lt x + 1$ " They statement is $:-$

For every rational number $x$,  $x \lt x + 1$

At first glance my answer was $:-$

There exists a rational number $x$ such that $x \geq x + 1$

But then i saw this

$p : \sqrt{11}$ is rational
~$p$ : $\sqrt{11}$ is not rational
same as ~$p$ : $\sqrt{11}$ is irrational

I just wonder why not,
For every irrational number $x$,  $x \lt x + 1$
is a correct negation of the first statement ?
Sorry for this silly question i can't seem to find a answer in my textbook.
 A: This is a statement about rational numbers. Whatever properties irrational numbers have is irrelevant to the truth value of this statement. This statement is only talking about a property rational numbers have. Therefore, the negation of this statement is the existence of a rational number without this property because that would be a contradiction of the statement that all rational numbers have this property. Irrational numbers have nothing to do with the negation.
Now, to make this more clear, let's use your example: Clearly, the following statement is true:

For every rational number $x$, $x<x+1$

Now, by your logic, the negation of this is the following:

For every irrational number $x$, $x<x+1$

However, this statement is also clearly true. Therefore, by this logic, the statement and its negation are both true which can't be possible, so this is the wrong way to find the negation.
A: If you know the rules of predicate logic, you can prove that your initial answer is correct, as follows.
Suppose $\neg \forall x: [x \in Q \implies x\lt x+1]$
Changing the quantifier and removing the resulting double negation, we obtain:
$\exists x:\neg [x \in Q \implies x\lt x+1]$
Applying  the definition of $\implies$ and $\ge$ and removing the resulting double negation, we obtain:
$\exists x: [x\in Q \land x\ge x+1]$
