Is $\mathbb{Q} \times \mathbb{R}$ countable or uncountable? Problem:

Is $\mathbb{Q} \times \mathbb{R}$ countable or uncountable?

Attempt:
It's known that $\mathbb{Q}$ is countable, and $ \# \mathbb{R} = \#(0,1)$ is uncountable. I then use the fact that:
$$\mathbb{Q} \times \mathbb{R}=\bigcup_{x \in (0,1)}(\mathbb{Q} \times \{x\})>\mathbb{N\times N}$$
which shows that $\mathbb{Q} \times \mathbb{R}$ is uncountable.
Am i on the right track?
 A: Note that
$$\mathbb{Q}\times\mathbb{R} \supseteq \{0\}\times\mathbb{R} \cong \mathbb{R}$$
Since $\mathbb{R}$ is uncountable, so must $\mathbb{Q}\times\mathbb{R}$
A: You were almost on the right track to prove the statement easily (you can proceed by your method, but it requires some more work to complete it).
Observe that $$\mathbb Q\times \mathbb R=\bigcup_{q\in\mathbb Q}(\{q\}\times\mathbb R)$$ Now suppose there exists a bijection $\phi$ between $\mathbb Q\times\mathbb R$ and $\mathbb N$.
Pick up one of the sets in the union, say $\{0\}\times\mathbb R=:A$. Then $A\subset \mathbb Q\times\mathbb R$ and is clearly in bijection with $\mathbb R$ say by $\psi$). Then $A$ is in bijection with $\phi(A)\subseteq \mathbb N$. Compose this with $\psi$ (or $\psi^{-1}$ as per your definition), and you get a bijection between a subset (proper or improper doesn't matter) of $\mathbb N$ with $\mathbb R$, which is impossible.
Thus there exists no bijection between $\mathbb Q\times\mathbb R$ and $\mathbb N$ and hence the former set is uncountable.
A: Depending on your definition of (un)countable, this could be simplified.
Definition: A set $S$ is said to be uncountable if there exists a surjective map $f:S\longrightarrow\Bbb{R}$ (or any other uncountable set you desire).

To show that $\Bbb{Q}\times\Bbb{R}$ is uncountable, consider $f:\Bbb{Q}\times\Bbb{R} \longrightarrow \Bbb{R}$ where $(a,\alpha)\mapsto \alpha.$ It is clear that $f$ is well defined. To show it is surjective, let $\beta \in \Bbb{R}.$ Then any element $(b,\beta)\in \Bbb{Q}\times\Bbb{R}$ has the property that $f(b,\beta)=\beta.$ Thus, $f$ is surjective. Since $f: \Bbb{Q}\times\Bbb{R} \longrightarrow \Bbb{R}$ is surjective, then $\Bbb{Q}\times\Bbb{R}$ must bet uncountable.
A: Cartesian product of countable and uncountable is uncountable.
Proof: If $A$ is countable and $B$ isn't, then
$$A \times B \supseteq \{a\} \times B$$ QED

Btw, Cartesian Product of Countable Sets is Countable.
