How do I prove the following statements or their negations in the context where $x$ and $y$ are rational numbers in the closed interval $[-\sqrt{2}, \sqrt{2}]$?
Statement 1: $\forall x \exists y\; x < y$ ; Negation of Statement 1: $\exists x\forall y\; x \geq y$
Statement 2: $\exists y \forall x\; x < y$ ; Negation of Statement 2: $\forall y \exists x\; x \geq y$
Statement 3: $\forall y \exists x\; x < y$ ; Negation of Statement 3: $\exists y\forall x\; x \geq y$
Statement 4: $\exists x \forall y\; x < y$ ; Negation of Statement 4: $\forall x \exists y\; x \geq y$
Proof of the Negation of Statement 1 for real numbers: The negation of Statement 1 says that “There is some $x$ such that for all $y$, $x$ is greater than or equal to $y$.” We can choose an x in the interval in which x would be greater than y to satisfy this statement. If we choose that $x = \sqrt{2}$, then $y$ can be any number in the given interval and the statement will always be true since $\sqrt{2}$ is the upper bound of this interval.
I believe that all the negations are true for the set of real numbers, but I am unsure on how to edit my proofs using real numbers to prove or disprove the above statements if they must be rational numbers on this interval.
Statement 1 is true. Let $x\in [-\sqrt{2},\sqrt{2}]$ be rational. We know that $x\neq \sqrt{2}$. Then there is another rational number in $(x,\sqrt{2})$. Call this number $y$. Then $x<y$.
Statement 2 is false. Its negation is true; just let $x=y$.
Statement 3 is true. Its proof is similar to the proof of 1.
Statement 4 is false. Its proof is similar to the proof of 2.