Write the negation of the statement "xy is even implies that x or y is even." I figured that implications statements can be written as:
P implies Q meaning that if P then Q
And I thought the negation of if-then statements can be written in the format: p and not q
Hence I thought that the negation would be:
"xy is even and x or y is not even"
But I am not sure about the answer and would appreciate it if anyone could help.
 A: Presumably this is quantified over $x$ and $y$, and $x$ and $y$ are integers. So the statement would read
"For all $x \in \mathbb{Z}$, for all $y \in \mathbb{Z}$, if $xy$ is even, then $x$ is even or $y$ is even."
You can write this as sentence in first order logic like:
$$\forall (x \in \mathbb{Z})\forall (y \in \mathbb{Z})[Even (xy) \implies Even(x) \vee Even(y) ]$$
You can specify what $Even(n)$ means more precisely, but that is not the relevant issue right now. Then, to negate this, just put a negation sign in front:
$$\neg \forall (x \in \mathbb{Z})\forall (y \in \mathbb{Z})[Even (xy) \implies Even(x) \vee Even(y) ]$$
So, you could rightfully declare that the negation is...
"It is not the case that for all $x \in \mathbb{Z}$, for all $y \in \mathbb{Z}$, if $xy$ is even, then $x$ is even or $y$ is even."
Not very helpful, right? We can put it in a different form by following the rules for moving negation signs through quantifiers
$$\neg \forall (x \in \mathbb{Z})\forall (y \in \mathbb{Z})[Even (xy) \implies Even(x) \vee Even(y) ]$$
$$ \exists(x \in \mathbb{Z})\neg \forall (y \in \mathbb{Z})[Even (xy) \implies Even(x) \vee Even(y) ]$$
$$\exists (x \in \mathbb{Z})\exists (y \in \mathbb{Z})\neg [Even (xy) \implies Even(x) \vee Even(y) ]$$
"There exists $x \in \mathbb{Z}$ and a $y \in \mathbb{Z}$ such that it is not the case that if $xy$ is even, then $x$ or $y$ is even."
But we can do better and negate the conditional "if-then" if we want:
$$\exists (x \in \mathbb{Z})\exists (y \in \mathbb{Z})[Even (xy) \wedge \neg [Even(x) \vee Even(y)] ]$$
"There exists $x \in \mathbb{Z}$ and a $y \in \mathbb{Z}$ such that $xy$ is even, and it is not the case that $x$ or $y$ is even."
$$\exists (x \in \mathbb{Z})\exists (y \in \mathbb{Z})[Even (xy) \wedge  [\neg Even(x) \wedge \neg Even(y)] ]$$
"There exists $x \in \mathbb{Z}$ and a $y \in \mathbb{Z}$ such that $xy$ is even, and $x$ is not even and $y$ is not even."
A: The negation is $xy$ is even and $x$ and $y$ are odd.
That's because, as you noted, $\lnot (p\implies q)$ is $ p\land \lnot q$.
A: The negation would simply be "xy is even does not imply that x or y is even." The best way to go about writing negations is to add "not" before your q.
The way you wrote it doesn't really have a p or a q. There is no conclusion, but rather two variables: "xy is even", and "x or y is not even".
Hope this helps!
