Let Q(x,y) denote "x + y = 0". What are the truth values of the
quantifications ∃y∀xQ(x,y) and ∀x∃yQ(x,y), where the domain for all
variables consists of all real numbers?
So from this problem, I take it to mean that ∃y∀xQ(x,y) translates to "There is a real number y such that for every real number x, x + y = 0." In other words, if you were to pick y = π, the statement says that every real number x + π = 0. So 1 + π should equal to 0. Obviously, this is false so the statement is false.
∀x∃yQ(x,y) means "For every real number x there exists a real number y such that x + y = 0." That means if you pick a single real number x, there is a single y such that x + y = 0. This is true.
Am I understanding this correctly?
And to further check that I am translating this correctly, here is another problem.
Let Q(x,y,z) be the statement "x + y = z". What are the truth values
of the statements ∀x∀y∃zQ(x,y,z) and ∃z∀x∀yQ(x,y,z), where the domain
of all variables consists of all real numbers?
∀x∀y∃zQ(x,y,z) means that "For all real numbers x and all real numbers y, there exists a real number z such that x + y = z." That means whichever x and y you pick, you will get a unique number z, which is true.
∃z∀x∀yQ(x,y,z) means that "There exists a real number z such that for all real numbers x and all real numbers y it is true that x + y = z". This means if you were to pick z = 50 for example, the statement asserts that no matter which x and y you pick, added together you should get 50. So according to the statement, 1 + 2 should equal 50? Which is false.
Best Answer
Your understanding of $\exists$ is not quite right, at least not in your reasoning. If you pick an element, say $x$, and the rest of the statement is false you cannot conclude that the whole statement is false. You may not have picked the right $x$. If, on the other hand, the rest of the statement is true the whole statement is true, because you found a $x$. (Also: $\exists$ makes no statement about uniqueness).
Consider $\exists y \forall x\colon (x - x = y)$. In your reasoning you could pick $y = 2$ and then $x - x = 2$ is false. But the statement is not false because there still exists an $y$ with the asked property (namely $y = 0$).
To argue that $\exists y \forall x Q(x,y)$ is false, where $Q(x,y)$ denotes $x+y=0$ you should prove that the negation of the statement is true, which is $\forall y \exists x \neg Q(x,y)$. So let $y \in \mathbb R$ be arbitrary. You have to consider two cases: If $y = 0$ choose $x=1$, otherwise $x = 0$. In both cases $x+y\neq0$. (Note that the order of the quantifiers is important in the reasoning: You could choose $x$ depending on $x$ because it's $\forall y \exists x$, you could not do that with $\exists x \forall y$.)