Where the domain is irrational numbers, could ∀x∃y(x+y∈ℚ) be the disproof of ∀x∀y(x+y∉ℚ)? The title really says it all. I am aware that the the negation provided isn't accurate (i.e. the latter statement in negated form would be $\exists x \exists y(x+y\in\Bbb Q))$ and it would suffice to simply provide a counterexample. But is it any less correct to show that for any x there exist a y that disproves the statement? This is implies Universal Generalization - if it's true for every object, it's true for all object: P($c$) for any arbitrary $c \to
∴∀x\text P(x)$
I provided the answer in the title for class and was not awarded points because, in summary, I negated incorrectly and I didn't provide a counterexample. I'll take the hit, I did provide the wrong negated statement, but I extrapolated on the negated statement and proved that by universal generalization we can go from "there exist an $x$" to "for any $x$".
Thanks in advance for your thoughts!
 A: 
Where the domain is irrational numbers, could ∀x∃y(x+y=ℚ) be the disproof of ∀x∀y(x+y≠ℚ)?

No.  I assume that by 'disproof to a statement' you mean a counterexample to the statement, but just providing some statement $X$ by itself is not a counterexample to a different statement $Y$.
This is because even if the two statements are contrary (i.e. cannot both be true), there is nothing that says that that statement $X$ can possibly be true, and hence that it is possible for $Y$ to be false, and it's the latter that you need to show.
Consider this: Suppose I need to provide a disproof for $\forall x \forall y \ x = y$, and suppose I provide the statement $\forall x \forall y \ x \neq y$. Well, it is true that these two statements are contrary, but the problem is that $\forall x \forall y \ x \neq y$ cannot be made true.
Now, of course you say that $\forall x \exists y(x+y=\mathbb{Q})$ is not self-contradictory, and can be made true. OK, you're right ... but you still haven't shown that. And to show that, you need to come up with a concrete model. But note that that model will also be a counterexample to claim $\forall x \forall y(x+y \neq \mathbb{Q})$.  So .. you might as well immediately provide a counterexample.
One more note: you said that you negated incorrectly ... and yes, if you negate a statement and you can show that you can make that negation true, then you have thereby also shown a counterexample to the original statement.  However, you do not need to negate a statement to find a counterexample. As long as your statement is contrary to the original statement, and you show that your statement can be made true, then you have thereby also shown a counterexample to the original statement.
So: points were correctly taken off if the task was to show a counterexample (which you did not).  But if the instructor thought that you had to negate the original statement and then find a model for that negation in order to find a counterexample, then the instructor was wrong about that. Of course, the instructor is in their full right to demand that you first negate the statement as part of the exercise/instructions. So if the instructions were to negate the statement, and then to use that to find a counterexample, then points should be taken off your answer as well.
A: Use Parenthesis to make it easy to negate : $ ( \forall x ( \forall y ( x+y \ne ℚ ) ) ) [[ EXP 1 ]] $
(A) What is the Correct Negation :
We want to negate it : $ ( \lnot ( \forall x ( \forall y ( x+y \ne ℚ ) ) ) ) $
$ ( \lnot ( \forall x ( \forall y ( x+y \ne ℚ ) ) ) ) \Leftrightarrow  ( \exists x ( \lnot ( \forall y ( x+y \ne ℚ ) ) ) ) $
$ ... \Leftrightarrow ( \exists x ( \exists y ( \lnot ( x+y \ne ℚ ) ) ) ) $
$ ... \Leftrightarrow ( \exists x ( \exists y ( x+y = ℚ ) ) ) [[ EXP 2 ]] $
Hence , this is the Correct Negation.
Here , assuming the "$=$" is "$\in$" , we can see that this will satisfy the condition :
$ ( x = 1+\pi ) $ , $ ( y = 1-\pi ) $
$ ( x+y = 2 \in Q ) $
(B) What is wrong with the given wrong negation :
This is not the Correct Negation : $ ( \forall x ( \exists y ( x+y = ℚ ) ) ) [[ EXP 3 ]] $
It says that each & every $x$ must have this negative Property , whereas , it is enough to have even a single $x$ where this negative Property is true.
In the current context , it is indeed true that each & every $x$ has at least one $y$ such that this negative Property is true [[ simply take $ ( y = -x+z ) $  where $z$ is rational ]] but that may not be the case in other contexts , hence we can not try this to get the Correct Negation.
(C) Can we check that with the logically equivalent chain ?
Yes , we can negate the given wrong negation $ ( \lnot ( \forall x ( \exists y ( x+y = ℚ ) ) ) ) $ to show that we do not get the original.
$ ( \lnot ( \forall x ( \exists y ( x+y = ℚ ) ) ) ) \Leftrightarrow ( \exists x ( \lnot ( \exists y ( x+y = ℚ ) ) ) ) $
$ ... \Leftrightarrow ( \exists x ( \forall y ( \lnot ( x+y = ℚ ) ) ) ) $
$ ... \Leftrightarrow ( \exists x ( \forall y ( x+y \ne ℚ ) ) ) [[ EXP 4 ]] $
This is not Equivalent to the original $ ( \forall x ( \forall y ( x+y \ne ℚ ) ) ) $
Hence that was not the Correct Negation.
(D) Given that EXP 1 is not true , while EXP 2 is true & very close to EXP3 which is also true & EXP 4 is not true , is there a way to salvage the Situation to complete the Exercise ?
In this context , where we add two irrational numbers , to check whether we get irrational number or rational number , we can use Extra "Domain" knowledge to proceed.
We can try this :
$ EXP 2 \Leftrightarrow \lnot EXP 1 $ ( no Extra "Domain" knowledge ! )
Prove EXP3 ( using Extra "Domain" knowledge ! )
$ EXP 3 \implies EXP 2 $ ( Changing $\forall$ to $\exists$ )
Hence, $ EXP 3 \implies \lnot EXP 1 $
We can not have $ EXP 3 \Leftrightarrow \lnot EXP 1 $
Here , OP did not Prove EXP3 , hence , Instructor is right to Deduct Marks.
