I'm trying to use cardinalities to prove that $$\forall a,b {\in} \mathbb{Q} \:\:\exists x {\in}\mathbb{R}{\setminus} \mathbb{Q}\:\:a \lt x \lt b.$$ Is the following proof of correct?
Let $ I = \mathbb{R} \setminus \mathbb{Q} $
Assume $ \exists a,b \in \mathbb{Q}\left(\left(b \gt a\right) \land \forall x \in I\left(x \gt b\lor x \lt a\right)\right)$
Therefore, $ \left[a, b\right] \subseteq \mathbb{Q} \Rightarrow \left|\left[a, b\right]\right| \le \aleph_0$. However, $ \forall r,m \in \mathbb{R}\left(\left(r \lt m\right) \Rightarrow \left(\left|\left[r,m\right]\right| = \aleph\right)\right) $, and that is a contradiction.
