discrete math Quantifiers? In the following exercises, V(x,y) means "x + y = 2xy", where x and y are integers. Determine
the truth value of the statement: ∀x∃y¬V(x,y)
what this says is for every x, there exists an y such that negation of V? what is the negation v?
and also for this question, U(x,y) means "2x + 3y = xy", where x and y are integers. Determine
the truth value of the statement: ∃x∀yU(x,y)
so there exists an x such that for every y, 2x+3y=xy?
does that mean there is one x that for any value y, 2x+3y=xy? or?
 A: Negation of V means that $V(x,y)$ fails to hold, i.e. $x+y\neq 2xy$.  To prove that $\forall x\exists y \neg V(x,y)$ is true, it is your task that for any $x$ you must find a $y$ where $x+y\neq 2xy$.  
Here's how I would do it.  If $x\neq 0$, then we take $y=0$ and $x+y=x\neq 0$, but $2xy=0$.  If however $x=0$ then we take $y=1$ and $x+y=1$ but $2xy=0$.
Also, your interpretation of the $U(x,y)$ question is correct.  
A: The negation of V is $x + y \neq 2xy$. $\forall x \exists y¬V(x,y)$ means that whatever number $x$ you are given, you can find another (non necessarily distinct) number $y$ for which $x + y \neq 2xy$. Do you think this is true or false? This is true, think of a simple way of proving it!
For $∃x∀yU(x,y)$, mean that there is one $x$ (at least one, I should say) such that $U(x,y)$ is true, for all integers $y$! So you are correct.
This statement seems less reasonable. If, say you think it is false, a good way of showing it is to prove its negation $¬(∃x∀yU(x,y))=\forall x \exists y ¬U(x,y)$ is true.
