What is the negation of this statement?
Let the sequences $\{x_{n}\}$ and $\{y_{n}\}$ be given. There exist a positive rational $a$ and a positive integer $N$ such that $x_{n} - y_{n} \geq a$ for all positive integer $n$ with $n \geq N$.
My answer is,
Let the sequences $\{x_{n}\}$ and $\{y_{n}\}$ be given. For every positive rational $a$ and every positive integer $N$, there is a positive integer $n$ with $n \geq N$ such that $x_{n} - y_{n} < a$.
Is this correct?
What problem ? Your translation is fine.
Assuming that the statement of the problem regards the convergence of a couple of "given" sequences, we have that the sentence :
There exist a positive rational $a$ and a positive integer $N$ such that for all positive integer $n$ with ...
has the "form :
$\exists a \exists N \forall n \varphi$,
where $\varphi$ is : $x_n−y_n ≥ a$
Negating it we get :
$\lnot \exists a \exists N \forall n \varphi$, i.e. $\forall a \forall N \exists n \lnot \varphi$
and $\lnot \varphi$ is $\lnot (x_n−y_n ≥ a)$, i.e. $x_n−y_n < a$, which fits exactly with your translation.
Your negated statement does not start out correctly. It should be "there are sequences ...", not "let the seqeunces ... be given".
Write the statement using quantifiers, then negate it. For notational brevity, set $\mathcal{S}$ be the set of all sequences.
I take the phrase " let the sequences be given ..." to mean "given any sequences ...". The original statement is $$ \forall (x_n), (y_n) \in \mathcal{S} \, \exists a \in \mathbb{Q} \cap (0, \infty) \, \exists N \in \mathbb{Z}^+ \, \forall n \in \mathbb{Z} \; n \ge N \Longrightarrow x_n - y_n \ge a $$ Negating this results in $$ \exists (x_n), (y_n) \in \mathcal{S} \, \forall a \in \mathbb{Q} \cap (0, \infty) \, \forall N \in \mathbb{Z}^+ \, \exists n \in \mathbb{Z} \, \ni \, n \ge N \wedge x_n - y_n < a $$ Now translate this into words.
