How to negate an implication in English? How to negate this proposition:
"If $xy$ is irrational then either $x$ is irrational or $y$ is irrational. "
Because the negation of $p\Rightarrow q$ is $p \wedge \text{not } q$. If I translate this sentence into English, it would be "$xy$ is irrational and $x$ and $y$ are rational." It is so strange. 
 A: $P$ in this case is "$xy$ is irrational" and $Q$ is "either $x$ is irrational or $y$ is irrational". You wrote that $\lnot(P \implies Q)$ is $(P \wedge \lnot Q)$. So, colloquially, I would suggest that $\lnot Q$ is "neither $x$ nor $y$ are irrational". This colloquially gives us:
$P \wedge \lnot Q$: "$xy$ is irrational and neither $x$ nor $y$ are irrational."
A: This is a bit of a weird question, because the statement you are trying to negate is true!
But you are right. The negation of this statement is: 
There exists x and y which are rational such that xy is irrational.
A: "If xy is irrational then either x is irrational or y is irrational."
Negates to:
"It is not the case that if xy is irrational, then either x is irrational or y is irrational."
We have to assume equivalences or definitions to make the sort of transformations you did.  We simply don't know if those equivalences hold... we only know that we can negate such a statement. 
A: It's not strange at all: in ordinary language, the negation of the expression

"If $xy$ is irrational then either $x$ is irrational or $y$ is irrational." 

$(xy \notin \mathbb{Q}\rightarrow (x \notin \mathbb{Q} \vee y \notin \mathbb{Q}))$


is stated as:

"Is not the case that if $x y$ is irrational then either $x$ is irrational or $y$ is irrational. "

$\neg ((x y \notin \mathbb{Q}) \rightarrow x \notin \mathbb{Q} \vee y \notin \mathbb{Q})$


which is logically equivalent to:

" $x y$ is irrational and is not the case that either $x$ is irrational or $y$ is irrational. "

$(xy \notin \mathbb{Q}) \wedge \neg(x \notin \mathbb{Q} \vee y \notin \mathbb{Q})$


that implies:

" $x y$ is irrational and $x$ is not irrational and $y$ is not irrational.

$(xy \notin \mathbb{Q}) \wedge (\neg x \notin \mathbb{Q} \wedge \neg y \notin \mathbb{Q})$"


You can see that from the above statement we can actually find two contradictions, and it is therefore an invalid one. In a formal proof, you could also use this reasoning to prove that your original statement is valid.
Also note that the above statement, in classic logic, also implies that:

" $xy$ is irrational and $x$ is rational and $y$ is rational.

$x y\notin \mathbb{Q} \wedge (x \in \mathbb{Q} \wedge y \in \mathbb{Q})$"


A: The negated statement sounds strange because the original statement implicitly uses quantifiers (that is, they are omitted, but they are still there):

For all $\bf x$ and $\bf y$, if $xy$ is irrational, then either $x$ is
irrational or $y$ is irrational.

The negated version, however, omits them. The correct version of the negation must include the correct quantifiers:

There exist $x$ and $y$ such that both of the following statements are true:

*

*$xy$ is irrational,

*$x$ and $y$ are rational.


