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Take the two sets

$A=\{p \in \mathbb{Q}| p^{2}\leq 2\}$ and $B=\{p \in \mathbb{Q}| p^{2}\geq 2\}$. Show that $A \cup B =\mathbb{Q}$.

For some reason I am stuck on how to show this, I feel like each time I try something I end up using what I have to prove which is that the union of A and B equals the rational line.

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    Given $p \in \Bbb Q$, we have $p^2 \in \Bbb Q$. As $\Bbb Q$ is totally ordered, we have $p^2 \le 2$ or $p^2 \ge 2$. – 2017-01-21

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