Determine whether it is finite, countably infinite, or uncountably infinite. Justify
$$\Big\{\Big(\frac{m}{2}, \frac{n}{3}\Big) \in \mathbb{R}^2 \mid m,n \in \mathbb{Z}\Big\}$$
The set is countably infinite, I'm struggling to write a justified answer.
$(m/2, n/3)$ are rational numbers, so does that mean $\mathbb{Q}^2$ can be a subset of $\mathbb{R}^2?$
EDIT: after reviewing hints in the given answer my idea is:
HINT:
A set $X$ is countably infinite if you can construct a bijection between the set of of natural numbers $\mathbb{N}$ and $X$. A bijection is a function $\phi$ which is both injective and surjective.
So in essence you are looking for a function
$$ \phi:\mathbb{Z}\times\mathbb{Z}\to \mathbb{Q}\times \mathbb{Q}, \quad \phi(m,n)=\Big(\frac{m}{2}, \frac{n}{3}\Big), \quad m,n\in\mathbb{Z} $$
So one way could be to try to prove that the $\phi$ above in a bijection. Since we know that $\mathbb{Z}\times \mathbb{Z}$ is countable (the set of fractions) so there already exists a bijection $\psi:\mathbb{N}\to\mathbb{Z}\times\mathbb{Z}$. But for completeness sake you could also prove this.
Another way to look at it could be to consider the two sets $$ \Big\{\frac{m}{2}\mid m\in\mathbb{Z} \Big\}\quad \Big\{\frac{n}{3}\mid n\in\mathbb{Z} \Big\} $$ and determine whether these are countably infinite and try to reason about the countability of the Cartesian product of two countable/not countable sets.
Hope this helped a bit.