a) Determine the truth value of the statement $∃y∀x¬V(x,y)$ where $V(x,y) = x+y=2xy$
b) Determine the truth value of the statement $∃yV(1,y)$ where $V(x,y) = x+y=2xy$
My process for figuring out a) is to translate it into English as There exists a value of y where all x values return false in V(x,y). I interpret this as there is a value of y that fits for an instance where x can be any number in $V(x,y)$ to make it false which is true from my understanding (say $x=1$; there are values of y that make $1+y=2y$ false.
For b), I again translate it into English: There exists a value of y that returns true in V(1,y). To me, this means I should try to solve for a value of y, and if possible, then $V(x,y)$ is true since a given y-value fits.
Am I on the right track? My tentative answers are True and True for a) and b) respectively.
I am particularly new to math logic and I must admit math is not my area of strength. The textbook I am using is less than helpful.
b) is certainly true because $V(1,1)$.
For a), you need to find a $y$ such that $\neg V(x,y)$ for all $x$. That is, does there exist $y$ such that $x+y\neq 2xy$ for all $x$? Suppose we fix $y$. Then $$ x+y=2xy\iff y=x(2y-1)\iff x=\frac{y}{2y-1}. $$ So for any $y$, there exists $x$ such that $V(x,y)$ ASSUMING THAT $\frac{y}{2y-1}$ is in your set for all $y$. So if all $\frac{y}{2y-1}$ exist in your set for whatever $y$, then a) is true. Otherwise, a) is false.